Mensuration

Given,

Base of triangle = 6 cm

Height of triangle = 4 cm

We know that,

Area of triangle = $\frac{1}{2}$ × base × height

Substituting the values we get,

Area of triangle = $\frac{1}{2}$ × 6 × 4

= 6 × 2

= 12 cm².

Hence, area of triangle = 12 cm².

Consider a = 3 cm, b = 4 cm and c = 5 cm

We know that

Semi perimeter (s) = $\frac{(a + b + c)}{2}$

Substituting the values we get,

s = $\frac{(3 + 4 + 5)}{2} = \frac{12}{2}$ = 6 cm.

Area of triangle = $\sqrt{s(s - a)(s - b)(s - c)}$

Substituting values we get,

$$\begin{aligned}A = \sqrt{6(6 - 3)(6 - 4)(6 - 5)} \\[1em] = \sqrt{6 \times 3 \times 2 \times 1} \\[1em] = \sqrt{36} \\[1em] = 6 \text{ cm}^2.\end{aligned}$$

Hence, area of triangle = 6 cm².

Consider a = 29 cm, b = 20 cm and c = 21 cm

We know that,

Semi perimeter (s) = $\frac{(a + b + c)}{2}$

Substituting the values we get,

s = $\frac{(29 + 20 + 21)}{2} = \frac{70}{2}$ = 35 cm.

Area of triangle = $\sqrt{s(s - a)(s - b)(s - c)}$

Substituting values we get,

$$\begin{aligned}A = \sqrt{35(35 - 29)(35 - 20)(35 - 21)} \\[1em] = \sqrt{35 \times 6 \times 15 \times 14} \\[1em] = \sqrt{44100} \\[1em] = 210 \text{ cm}^2.\end{aligned}$$

Hence, area of triangle = 210 cm².

Consider a = 12 cm, b = 9.6 cm and c = 7.2 cm

We know that,

Semi perimeter (s) = $\frac{(a + b + c)}{2}$

Substituting the values we get,

s = $\frac{(12 + 9.6 + 7.2)}{2} = \frac{28.8}{2}$ = 14.4 cm.

Area of triangle = $\sqrt{s(s - a)(s - b)(s - c)}$

Substituting values we get,

$$\begin{aligned}A = \sqrt{14.4(14.4 - 12)(14.4 - 9.6)(14.4 - 7.2)} \\[1em] = \sqrt{14.4 \times 2.4 \times 4.8 \times 7.2} \\[1em] = \sqrt{1194.39} \\[1em] = 34.56 \space \text{cm}^2.\end{aligned}$$

Hence, area of triangle = 34.56 cm².

Consider 34 cm, 20 cm and 42 cm as the sides of triangle.

a = 34 cm, b = 20 cm and c = 42 cm

We know that,

Semi perimeter (s) = $\frac{(a + b + c)}{2}$

Substituting the values we get,

s = $\frac{(34 + 20 + 42)}{2} = \frac{96}{2}$ = 48 cm.

Area of triangle = $\sqrt{s(s - a)(s - b)(s - c)}$

Substituting values we get,

$$\begin{aligned}A = \sqrt{48(48 - 34)(48 - 20)(48 - 42)} \\[1em] = \sqrt{48 \times 14 \times 28 \times 6} \\[1em] = \sqrt{112896} \\[1em] = 336 \text{ cm}^2.\end{aligned}$$

Here the shortest side of the triangle is 20 cm. Let height = h cm be the corresponding altitude.

We know that,

Area of triangle = $\frac{1}{2}$ × base × height

Substituting the values we get,

⇒ 336 = $\frac{1}{2}$ × 20 × h

⇒ h = $\frac{336 \times 2}{20}$

⇒ h = $\frac{336}{10}$

⇒ h = 33.6 cm.

Hence, area of triangle = 336 cm² and length of the altitude corresponding to the shortest side = 33.6 cm.

Consider, a = 975 m, b = 1050 m and c = 1125 m.

We know that,

Semi perimeter (s) = $\frac{(a + b + c)}{2}$

Substituting the values we get,

$$\begin{aligned}s = \frac{(975 + 1050 + 1125)}{2}\\[1em] = \frac{3150}{2}\\[1em] = 1575 \text{ m}.\end{aligned}$$

By formula,

Area of triangle (A) = $\sqrt{s(s - a)(s - b)(s - c)}$

Substituting values we get :

$$\begin{aligned}A = \sqrt{1575(1575 - 975)(1575 - 1050)(1575 - 1125)}\\[1em] = \sqrt{1575 \times 600 \times 525 \times 450}\\[1em] = \sqrt{(525 \times 3) \times (150 \times 2 \times 2) \times (525) \times (150 \times 3)}\\[1em] = \sqrt{525^2 \times 150^2 \times 2^2 \times 3^2}\\[1em] = 525 \times 150 \times 2 \times 3\\[1em] = 472500 \text{ m}^2 \\[1em] = \frac{472500}{10000} \\[1em] = 47.25 \text{ hectares}.\end{aligned}$$

We know that,

Selling price of 1 hectare field = ₹ 10 lakh.

∴ Selling price of 47.25 hectare field = ₹ 10,00,000 × 47.25 = ₹ 4,72,50,000.

Hence, selling price of the triangular field = ₹ 4,72,50,000.

It is given that,

ABC is a right angled triangle.

From figure,

BC = 12 cm and AC = 13 cm

Using the Pythagoras theorem,

AC² = AB² + BC²

Substituting the values we get,

⇒ 13² = AB² + 12²

⇒ AB² = 13² – 12²

⇒ AB² = 169 – 144 = 25

⇒ AB = $\sqrt{25}$ = 5 cm.

We know that,

Area of triangle ABC = $\frac{1}{2}$ × base × height.

Substituting the values we get,

A = $\frac{1}{2} \times 12 \times 5$ = 30 cm².

Perimeter of triangle ABC (P) = AB + BC + CA

Substituting the values we get,

= 5 + 12 + 13

= 30 cm.

Hence, area of triangle = 30 cm² and perimeter = 30 cm.

Given,

Side of equilateral triangle = 8 m.

We know that,

Area of equilateral triangle = $\frac{\sqrt{3}}{4}$(side)²

Substituting the values we get,

$$\begin{aligned}A = \frac{\sqrt{3}}{4} \times (8)^2 \\[1em] = \frac{\sqrt{3}}{4} \times 64 \\[1em] = 16\sqrt{3} \\[1em] = 16 \times 1.732 \\[1em] = 27.71\end{aligned}$$

Hence, area of equilateral triangle = 27.71 m².

We know that,

Area of equilateral triangle = $\frac{\sqrt{3}}{4}$(side)²

Substituting the values,

$$\begin{aligned}\Rightarrow 81\sqrt{3} = \frac{\sqrt{3}}{4}(side)^2 \\[1em] \Rightarrow (side)^2 = \frac{81\sqrt{3} \times 4}{\sqrt{3}} \\[1em] \Rightarrow (side)^2 = 81 \times 4 \\[1em] \Rightarrow (side)^2 = 324 \\[1em] \Rightarrow \text{side} = \sqrt{324} = 18 \text{ cm}.\end{aligned}$$

So the perimeter of equilateral triangle = 3 × side

= 3 × 18 = 54 cm.

Hence, perimeter of equilateral triangle = 54 cm.

We know that,

Perimeter of an equilateral triangle = 3 × side.

Substituting the values,

⇒ 36 = 3 × side

⇒ side = $\frac{36}{3}$ = 12 cm.

Area of equilateral triangle = $\frac{\sqrt{3}}{4}(side)^2$

Substituting the values we get,

$$\begin{aligned}A = \frac{\sqrt{3}}{4} \times (12)^2 \\[1em] = \frac{\sqrt{3}}{4} \times 144 \\[1em] = 36\sqrt{3} \\[1em] = 36 \times 1.732 \\[1em] = 62.4 \text{ cm}^2\end{aligned}$$

From figure,

In triangle ABD,

Using Pythagoras Theorem,

AB² = AD² + BD² .......(1)

The perpendicular from a vertex of an equilateral triangle to the opposite side, bisects it.

So, BD = $\frac{12}{2}$ = 6 cm.

Substituting the values in (1) we get,

⇒ 12² = AD² + 6²

⇒ 144 = AD² + 36

⇒ AD² = 144 – 36 = 108

⇒ AD = $\sqrt{108}$ = 10.4 cm.

Hence, area of triangle = 62.4 cm² and height = 10.4 cm.

Let a, b and c be the sides of the triangle.

Given,

Ratio of the sides are 3 : 4 : 5.

Let a = 3x cm, b = 4x cm and c = 5x cm.

Given,

⇒ Perimeter = 48 cm

⇒ a + b + c = 48

⇒ 3x + 4x + 5x = 48

⇒ 12x = 48

⇒ x = $\frac{48}{12}$ = 4.

Substituting value of x,

a = 3x = 3 × 4 = 12 cm,

b = 4x = 4 × 4 = 16 cm,

c = 5x = 5 × 4 = 20 cm.

We know that,

Semi perimeter (s) = $\frac{(a + b + c)}{2}$.

$$\begin{aligned}s = \frac{12 + 16 + 20}{2} \\[1em] = \frac{48}{2} \\[1em] = 24 \text{ cm}.\end{aligned}$$

Area of triangle = $\sqrt{s(s - a)(s - b)(s - c)}$

Substituting values we get,

$$\begin{aligned}A = \sqrt{24(24 - 12)(24 - 16)(24 - 20)} \\[1em] = \sqrt{24 \times 12 \times 8 \times 4} \\[1em] = \sqrt{9216} \\[1em] = 96 \text{ cm}^2.\end{aligned}$$

Hence, area of triangle = 96 cm².

Given,

Sides of a triangle are in the ratio = 3 : 5 : 7

Perimeter = 300 m

Let a = 3x cm, b = 5x cm and c = 7x cm.

Given,

⇒ Perimeter = 300 m

⇒ a + b + c = 300

⇒ 3x + 5x + 7x = 300

⇒ 15x = 300

⇒ x = $\frac{300}{15}$ = 20.

Substituting value of x,

a = 3x = 3 × 20 = 60 m,

b = 5x = 5 × 20 = 100 m,

c = 7x = 7 × 20 = 140 m.

We know that,

Semi-perimeter (s) = $\frac{\text{Perimeter}}{2} = \frac{300}{2}$ = 150 m.

Area of triangle = $\sqrt{s(s - a)(s - b)(s - c)}$

Substituting values we get,

$$\begin{aligned}A = \sqrt{150(150 - 60)(150 - 100)(150 - 140)} \\[1em] = \sqrt{150 \times 90 \times 50 \times 10} \\[1em] = \sqrt{6750000} \\[1em] = 2598.07 ≈ 2598 m^2.\end{aligned}$$

Hence, area of triangle = 2598 m².

It is given that

AB = AC = 4 cm

From figure,

Using the Pythagoras theorem,

BC² = AB² + AC²

Substituting the values we get,

⇒ BC² = 4² + 4²

⇒ BC² = 16 + 16 = 32

⇒ BC = $\sqrt{32} = 4\sqrt{2}$ cm.

Let perpendicular from A to BC be h cm.

Area of △ABC = $\frac{1}{2}$ × base × height

= $\frac{1}{2}$ × AC × AB

= $\frac{1}{2} \times 4 \times 4$

= 8 cm.

From figure,

Area of △ABC = $\frac{1}{2}$ × BC × h.

$$\begin{aligned}\therefore 8 = \frac{1}{2} \times BC \times h \\[1em] \Rightarrow 8 = \frac{1}{2} \times 4\sqrt{2} \times h \\[1em] \Rightarrow 8 = 2\sqrt{2}h \\[1em] \Rightarrow h = \frac{8}{2\sqrt{2}} \\[1em] \Rightarrow h = 2\sqrt{2} = 2.83\end{aligned}$$

Hence, area of △ABC = 8 cm² and length of perpendicular from A to BC = 2.83 cm.

Consider △ABC as the isosceles triangle.

Here, AB = AC = 12 cm.

Perimeter = 30 cm

⇒ AB + AC + BC = 30

⇒ 12 + 12 + BC = 30

⇒ BC = 30 - 24 = 6 cm.

We know that,

Semi-perimeter (s) = $\frac{\text{Perimeter}}{2} = \frac{30}{2}$ = 15 cm.

Area of an isosceles triangle = $\frac{1}{4}b\sqrt{4a^2 - b^2}$, where a is length of equal sides and b is the length of other side.

Substituting values we get,

$$\begin{aligned}A = \frac{1}{4} \times 6 \times \sqrt{4(12)^2 - 6^2} \\[1em] = \frac{1}{4} \times 6 \times \sqrt{4 \times 144 - 36} \\[1em] = \frac{1}{4} \times 6 \times \sqrt{540} \\[1em] = \frac{1}{4} \times 6 \times 23.24 \\[1em] = \frac{139.44}{4} \\[1em] = 34.86 \text{ cm}^2.\end{aligned}$$

Hence, area of isosceles triangle = 34.86 cm².

Given,

base = 6 cm and perimeter = 16 cm

Consider △ABC as an isosceles triangle in which,

Let, AB = AC = x cm.

So, BC = 6 cm.

We know that,

Perimeter of △ABC = AB + BC + AC

Substituting the values we get,

⇒ 16 = x + 6 + x

⇒ 16 = 2x + 6

⇒ 16 – 6 = 2x

⇒ 10 = 2x

⇒ x = $\frac{10}{2}$ = 5 cm.

Area of an isosceles triangle = $\frac{1}{4}b\sqrt{4a^2 - b^2}$, where a is length of equal sides and b is the length of other side.

Substituting values we get,

$$\begin{aligned}A = \frac{1}{4} \times 6 \times \sqrt{4(5)^2 - 6^2} \\[1em] = \frac{1}{4} \times 6 \times \sqrt{4 \times 25 - 36} \\[1em] = \frac{1}{4} \times 6 \times \sqrt{100 - 36} \\[1em] = \frac{1}{4} \times 6 \times \sqrt{64} \\[1em] = \frac{1}{4} \times 6 \times 8 \\[1em] = 12 \text{ cm}^2.\end{aligned}$$

Hence, area of isosceles triangle = 12 cm².

Consider △ABC as a right angled triangle.

AB = 5x cm and BC = (3x – 1) cm

We know that,

Area of △ABC = $\frac{1}{2}$ × base × height = $\frac{1}{2}$ × BC × AB

Substituting the values we get,

⇒ 60 = $\frac{1}{2}$ × (3x – 1) × 5x

⇒ 120 = 5x(3x – 1)

⇒ 120 = 15x² – 5x

⇒ 15x² – 5x – 120 = 0

⇒ 5(3x² – x – 24) = 0

⇒ 3x² – x – 24 = 0

⇒ 3x² – 9x + 8x – 24 = 0

⇒ 3x(x – 3) + 8(x - 3) = 0

⇒ (3x + 8)(x - 3) = 0

⇒ 3x + 8 = 0 or x - 3 = 0

⇒ 3x = -8 or x = 3

⇒ x = $-\frac{8}{3}$ or x = 3

Since, x cannot be negative as length of a side cannot be negative. So, x = 3.

AB = 5 × 3 = 15 cm

BC = (3 × 3 – 1) = 9 – 1 = 8 cm

In right angled △ABC,

Using Pythagoras theorem,

⇒ AC² = AB² + BC²

Substituting the values we get,

⇒ AC² = 15² + 8²

⇒ AC² = 225 + 64 = 289

⇒ AC² = 17²

So, AC = 17 cm.

Hence, the hypotenuse of the right angled triangle is 17 cm.

Given,

AB = (2x + 1) cm

BC = (x + 1) cm

We know that,

Area of △ABC = $\frac{1}{2}$ × base × height

= $\frac{1}{2}$ × BC × AB

Substituting the values we get,

⇒ 60 = $\frac{1}{2}$ × (x + 1) × (2x + 1)

⇒ 60 × 2 = (2x + 1)(x + 1)

⇒ 120 = 2x² + 3x + 1

⇒ 2x² + 3x + 1 – 120 = 0

⇒ 2x² + 3x – 119 = 0

⇒ 2x² + 17x – 14x – 119 = 0

⇒ x(2x + 17) – 7(2x + 17) = 0

⇒ (x – 7)(2x + 17) = 0

⇒ x – 7 = 0 or 2x + 17 = 0

⇒ x = 7 or 2x = -17

⇒ x = 7 or x = $-\frac{17}{2}$

Since, x cannot be negative as length of a side cannot be negative. So, x = 7.

⇒ AB = (2x + 1) = 2 × 7 + 1 = 15 cm

⇒ BC = (x + 1) = 7 + 1 = 8 cm.

In right angled △ABC,

Using Pythagoras Theorem,

⇒ AC² = AB² + BC²

Substituting the values we get,

⇒ AC² = 15² + 8²

⇒ AC² = 225 + 64

⇒ AC² = 289

⇒ AC² = 17²

⇒ AC = 17 cm

Perimeter of △ABC = AB + BC + AC = 15 + 8 + 17 = 40 cm.

Hence, perimeter of △ABC = 40 cm.

Let △ABC be the right angle triangle.

We know that,

Perimeter of a right-angled triangle = 60 cm

Hypotenuse = 25 cm

So, the sum of other two sides of triangle = 60 – 25 = 35 cm

Let base (BC) = x cm

So, AB = (35 - x) cm

Using the Pythagoras theorem,

⇒ AC² = AB² + BC²

⇒ 25² = (35 - x)² + x²

⇒ 625 = 1225 + x² - 70x + x²

⇒ 2x² - 70x + 600 = 0

Dividing by 2 on both sides,

⇒ x² - 35x + 300 = 0

⇒ x² - 15x - 20x + 300 = 0

⇒ x(x – 15) - 20(x - 15) = 0

⇒ (x - 15)(x - 20) = 0

⇒ x - 15 = 0 or x - 20 = 0

⇒ x = 15 or x = 20.

If x = 15, then 35 - x = 35 - 15 = 20 cm.

If x = 20, then 35 - x = 35 - 20 = 15 cm.

So, length of other two sides apart from hypotenuse are 15 cm and 20 cm.

Area = $\frac{1}{2}$ × base × height

Substituting the values we get,

A = $\frac{1}{2}$ × 15 × 20 = 150 cm².

Hence, area of triangle = 150 cm².

Let the length of equal sides be x cm.

Length of base = $\frac{2}{3}(x + x) = \frac{2}{3} \times 2x = \frac{4x}{3}$ cm.

Given,

Perimeter = 40 cm

$$\begin{aligned}\therefore x + x + \frac{4x}{3} = 40 \\[1em] \Rightarrow 2x + \frac{4x}{3} = 40 \\[1em] \Rightarrow \frac{6x + 4x}{3} = 40 \\[1em] \Rightarrow 10x = 120 \\[1em] \Rightarrow x = 12 \text{ cm}.\end{aligned}$$

Length of Base = $\frac{4x}{3}$ = $\frac{4}{3} \times 12$ = 16 cm

Hence, length of equal sides = 12 cm and length of base = 16 cm.

Let the length of equal sides be a cm and length of base be b cm.

Area of isosceles △ABC = $\frac{1}{4}b\sqrt{4a^2 - b^2}$, where a is length of equal sides and b is the length of other side.

Substituting values in above equation we get,

$$\begin{aligned}\Rightarrow 60 = \frac{1}{4}b\sqrt{4(13)^2 - b^2} \\[1em] \Rightarrow 240 = b\sqrt{4 \times 169 - b^2} \\[1em] \Rightarrow 240 = b\sqrt{676 - b^2}\end{aligned}$$

Squaring both sides we get,

$$\begin{aligned}\Rightarrow 57600 = b^2(676 - b^2) \\[1em] \Rightarrow 676b^2 - b^4 - 57600 = 0 \\[1em] \Rightarrow b^4 - 676b^2 + 57600 = 0 \\[1em] \Rightarrow b^4 - 576b^2 - 100b^2 + 57600 = 0 \\[1em] \Rightarrow b^2(b^2 - 576) - 100(b^2 - 576) = 0 \\[1em] \Rightarrow (b^2 - 100)(b^2 - 576) = 0 \\[1em] \Rightarrow (b^2 - 100) = 0 \text{ or } (b^2 - 576) = 0 \\[1em] \Rightarrow b^2 = 100 \text{ or } b^2 = 576 \\[1em] \Rightarrow b = 10 \text{ or } b = 24.\end{aligned}$$

Hence, the length of base = 10 cm or 24 cm.

Given,

Cost of cultivating the field at the rate of ₹25 per 100 m² = ₹ 60000

In ₹25, area of field cultivated = 100 m²

∴ In ₹60000, area of field cultivated = $\frac{100 \times 60000}{25}$ = 240000 m².

∴ Area of field = 240000 m².

Let base of field = b meters and height = 3b meters.

Area of triangle = $\frac{1}{2} \times$ base × height

Substituting values we get,

$$\begin{aligned}\Rightarrow 240000 = \frac{1}{2} \times b \times 3b \\[1em] \Rightarrow 240000 = \frac{3b^2}{2} \\[1em] \Rightarrow b^2 = \frac{240000 \times 2}{3} \\[1em] \Rightarrow b^2 = 160000 \\[1em] \Rightarrow b = \sqrt{160000} = 400 \text{ m} \\[1em] \Rightarrow \text{height } = 3b = 3 \times 400 = 1200 \text{ m}.\end{aligned}$$

Hence, base = 400 m and height = 1200 m.

It is given that,

ABC is a triangular park with sides 120 m, 80 m and 50 m.

Here, the perimeter of △ABC = 120 + 80 + 50 = 250 m

Portion at which a gate is built = 3 m

Remaining perimeter = 250 – 3 = 247 m.

So, the length of the fence required around the park = 247 m.

Rate of fencing = ₹ 200 per metre

Total cost of fencing = 200 × 247 = ₹ 49,400.

We know that,

Semi perimeter (s) = $\frac{\text{Perimeter}}{2} = \frac{250}{2}$ = 125 m.

By formula,

Area of triangle (A) = $\sqrt{s(s - a)(s - b)(s - c)}$

Substituting values we get,

$$\begin{aligned}A = \sqrt{125(125 - 120)(125 - 80)(125 - 50)}\\[1em] = \sqrt{125 \times 5 \times 45 \times 75}\\[1em] = \sqrt{(25 \times 5) \times 5 \times (5 \times 3 \times 3) \times (25 \times 3)} \\[1em] = \sqrt{25^2 \times 5 \times 5^2 \times 3^2 \times 3}\\[1em] = 25 \times 5 \times 3 \times \sqrt{5 \times 3}\\[1em] = 375 \sqrt{15} \text{ m}^2.\end{aligned}$$

Hence, area needed for plantation = $375 \sqrt{15}$ m² and cost of fencing = ₹49,400.

Semi-perimeter (s) = $\frac{a + b + c}{2} = \frac{20 + 50 + 50}{2} = \frac{120}{2}$ = 60 cm.

Total 10 triangular pieces of cloth are required. So, 5 of each colour.

Area of triangle = $\sqrt{s(s - a)(s - b)(s - c)}$

Substituting values we get,

$$\begin{aligned}A = \sqrt{60(60 - 20)(60 - 50)(60 - 50)} \\[1em] = \sqrt{60 \times 40 \times 10 \times 10} \\[1em] = \sqrt{240000} \\[1em] = 200\sqrt{6} \text{ cm}^2.\end{aligned}$$

There are 5 triangular pieces. So area of 5 pieces = $5 \times 200\sqrt{6} = 1000\sqrt{6}$ cm².

Hence, $1000\sqrt{6}$ cm² of each colour cloth is required.

Given,

ABC is an equilateral triangle of side = 10 cm

We know that,

Area of equilateral triangle ABC = $\frac{\sqrt{3}}{4}\text{ (side)}^2$

Substituting the values we get,

$$\begin{aligned}A = \frac{\sqrt{3}}{4} \times (10)^2 \\[1em] = \frac{\sqrt{3}}{4} \times 100 \\[1em] = 25\sqrt{3} \\[1em] = 43.3 \text{ cm}^2.\end{aligned}$$

In right angled triangle BCD,

⇒ BC² = BD² + CD²

⇒ 10² = BD² + 6²

⇒ BD² = 100 - 36

⇒ BD² = 64

⇒ BD = 8 cm.

We know that,

Area of right angled triangle = $\frac{1}{2}$ × base × height.

Area of △BCD = $\frac{1}{2} \times CD \times BD = \frac{1}{2} \times 6 \times 8$ = 24 cm²

From figure,

Area of shaded portion = Area of triangle ABC - Area of triangle BCD

Substituting the values we get,

Area of shaded portion = 43.3 - 24 = 19.3 cm².

Hence, area of shaded region = 19.3 cm².

From figure,

In right angle triangle ADE,

Using pythagoras theorem,

⇒ DE² = AD² + AE²

⇒ DE² = 3² + 3²

⇒ DE² = 9 + 9

⇒ DE² = 18

⇒ DE = $\sqrt{18} = 3\sqrt{2}$ cm

Since, DEFG is a rectangle.

∴ GF = DE = $3\sqrt{2}$ cm.

In △DBG and △ECF,

DB = EC = 4 cm

DG = EF (Opposite sides of rectangle are equal)

∠DGB = ∠EFC = 90° (∵ DEFG is a rectangle)

Hence, by RHS axiom △DBG ≅ △ECF.

So, BG = FC (By C.P.C.T.)

Let BG = FC = x.

In right angle triangle ABC,

⇒ BC² = AB² + AC²

⇒ BC² = 7² + 7²

⇒ BC² = 49 + 49

⇒ BC² = 98

⇒ BC = $\sqrt{98} = 7\sqrt{2}$ cm

From figure,

BG + GF + FC = BC

⇒ BG + GF + FC = $7\sqrt{2}$

⇒ x + $3\sqrt{2}$ + x = $7\sqrt{2}$

⇒ 2x = $7\sqrt{2} - 3\sqrt{2}$

⇒ 2x = $4\sqrt{2}$

⇒ x = $2\sqrt{2}$ cm.

In right angle triangle DBG,

⇒ DB² = BG² + DG²

⇒ 4² = $(2\sqrt{2})$² + DG²

⇒ 16 = 8 + DG²

⇒ DG² = 16 - 8 = 8

⇒ DG = $\sqrt{8} = 2\sqrt{2}$ cm.

Area of right angle triangle DBG = $\frac{1}{2}$ x BG x DG

= $\frac{1}{2}$ x $2\sqrt{2}$ x $2\sqrt{2}$

= $\frac{1}{2}$ x 8 = 4 cm².

Since, △DBG ≅ △ECF.

∴ Areas of both triangle are equal.

Area of right angle triangle ADE = $\frac{1}{2}$ x AD x AE

= $\frac{1}{2}$ x 3 x 3

= $\frac{9}{2}$ = 4.5 cm².

Area of shaded region = Area of (△ADE + △DBG + △ECF)

= 4.5 + 4 + 4 = 12.5 cm².

Hence, area of shaded region = 12.5 cm².

Consider ABCD as a quadrilateral in which BD = 20 cm, AY = 15 cm and CX = 9 cm.

Area of quadrilateral ABCD = Area of triangle ABD + Area of triangle BCD

Area of triangle = $\frac{1}{2}$ × base × height

∴ Area of quadrilateral ABCD = $\frac{1}{2}$ × BD × AY + $\frac{1}{2}$ × BD × CX

Substituting the values we get,

Area of quadrilateral ABCD = $\frac{1}{2}$ x BD x (AY + CX)

= $\frac{1}{2}$ x 20 x (15 + 9)

= 10 x 24

= 240 cm²

Hence, area of quadrilateral = 240 cm².

Consider ABCD as a quadrilateral in which the diagonals AC and BD intersect each other at M at right angles.

From figure,

AC = 18 cm and BD = 12 cm

When diagonals of a quadrilateral intersect at right angles,

Area of quadrilateral = $\frac{1}{2}$ x d₁ x d₂, where d₁ and d₂ are diagonals.

Substituting the values we get,

Area of quadrilateral ABCD = $\frac{1}{2}$ x 12 x 18

= 6 x 18

= 108 cm²

Hence, area of quadrilateral = 108 cm².

From figure,

ABCD is a quadrilateral field.

In triangle BAD,

∠A = 90°

Using the Pythagoras Theorem

⇒ BD² = AB² + AD²

Substituting the values we get,

⇒ BD² = 40² + 9²

⇒ BD² = 1600 + 81 = 1681

⇒ BD = $\sqrt{1681}$ = 41 m

We know that,

Area of quadrilateral ABCD = Area of △BAD + Area of △BDC

Calculating area of △BDC,

In △BDC,

Let a = BD = 41 m, b = BC = 28 m and c = CD = 15 m.

Semi-perimeter (s) = $\frac{a + b + c}{2} = \frac{41 + 28 + 15}{2} = \frac{84}{2}$ = 42 m.

By Heron's formula,

Area of triangle = $\sqrt{s(s - a)(s - b)(s - c)}$

Substituting values we get,

$$\begin{aligned}A = \sqrt{42(42 - 41)(42 - 28)(42 - 15)} \\[1em] = \sqrt{42 \times 1 \times 14 \times 27} \\[1em] = \sqrt{15876} \\[1em] = 126 \text{ m}^2.\end{aligned}$$

Calculating area of △BAD,

$$\begin{aligned}\text{Area of △BAD } = \frac{1}{2} \times \text{ base × height} \\[1em] = \frac{1}{2} \times BA \times AD \\[1em] = \frac{1}{2} \times 40 \times 9 \\[1em] = 180 \text{ m}^2\end{aligned}$$

Area of quadrilateral ABCD = Area of △BAD + Area of △BDC

= 180 + 126

= 306 m².

Hence, area of quadrilateral ABCD = 306 m².

In right-angled △ABC,

Using Pythagoras theorem,

⇒ AB² = AC² + BC²

Substituting the values we get,

⇒ 13² = 12² + BC²

⇒ BC² = 13² – 12²

⇒ BC² = 169 – 144 = 25

⇒ BC = $\sqrt{25}$ = 5 cm.

Calculating area of △BCA,

$$\begin{aligned}\text{Area of △BCA } = \frac{1}{2} \times \text{ base × height} \\[1em] = \frac{1}{2} \times AC \times BC \\[1em] = \frac{1}{2} \times 12 \times 5 \\[1em] = 30 \text{ cm}^2.\end{aligned}$$

Calculating area of △ACD,

$$\begin{aligned}\text{Area of △ACD } = \frac{\sqrt{3}}{4} \times \text{ (side)}^2 \\[1em] = \frac{\sqrt{3}}{4} \times (12)^2 \\[1em] = \frac{\sqrt{3}}{4} \times 144 \\[1em] = 36\sqrt{3} \\[1em] = 62.35 \text{ cm}^2 \\[1em]\end{aligned}$$

From figure,

Area of quadrilateral ABCD = Area of △BCA + Area of △ACD

= 30 cm² + 62.35 cm²

= 92.35 cm².

Hence, area of quadrilateral ABCD = 92.35 cm².

In △ABC,

Using Pythagoras theorem,

AC² = AB² + BC²

Substituting the values we get,

⇒ AC² = 6² + 8²

⇒ AC² = 36 + 64 = 100

⇒ AC² = 10²

⇒ AC = 10 cm

Calculating area of △ADC,

In △ADC,

Let a = AD = 13 cm, b = DC = 13 cm and c = AC = 10 cm.

Semi-perimeter (s) = $\frac{a + b + c}{2} = \frac{13 + 13 + 10}{2} = \frac{36}{2}$ = 18 cm.

By Heron's formula,

Area of triangle = $\sqrt{s(s - a)(s - b)(s - c)}$

Substituting values we get,

$$\begin{aligned}\text{Area of △ADC} = \sqrt{18(18 - 13)(18 - 13)(18 - 10)} \\[1em] = \sqrt{18 \times 5 \times 5 \times 8} \\[1em] = \sqrt{3600} \\[1em] = 60 \text{ cm}^2.\end{aligned}$$

Calculating area of △ABC,

$$\begin{aligned}\text{Area of △ABC } = \frac{1}{2} \times \text{ base × height} \\[1em] = \frac{1}{2} \times AB \times BC \\[1em] = \frac{1}{2} \times 6 \times 8 \\[1em] = 24 \text{ cm}^2\end{aligned}$$

From figure,

Area of quadrilateral ABCD = Area of △ADC + Area of △ABC

= 60 + 24

= 84 cm².

Hence, area of quadrilateral ABCD = 84 cm².

We know that,

Perimeter of rectangle = 2 × (l + b) = 96 cm

Substituting the values we get,

⇒ 2(l + 18) = 96

⇒ (l + 18) = 48

⇒ l = 48 - 18 = 30 cm.

Area of rectangular cardboard = l × b

Substituting the values we get,

Area = 30 × 18 = 540 cm².

Hence, length = 30 cm and area of rectanglular cardboard = 540 cm².

Let breadth = x meters

So, length = (x + 5) meters

We know that,

Area of rectangular hall = length × breadth

Substituting the values we get,

⇒ 594 = x(x + 5)

⇒ 594 = x² + 5x

⇒ x² + 5x – 594 = 0

⇒ x² + 27x – 22x – 594 = 0

⇒ x(x + 27) – 22(x + 27) = 0

⇒ (x – 22)(x + 27) = 0

⇒ x – 22 = 0 or x + 27 = 0

⇒ x = 22 or x = -27.

Since, length of side cannot be negative so, x ≠ -27.

∴ Breadth = x = 22 m and Length = (x + 5) = 22 + 5 = 27 m.

Perimeter = 2(l + b)

Substituting the values we get,

Perimeter = 2(27 + 22) = 2 × 49 = 98 m.

Hence, perimeter of hall = 98 m.

We know that,

Area of rectangle = length × breadth

Area of square = side × side.

From figure,

Area of shaded portion = Area of rectangle ABCD + Area of rectangle PQRS – Area of square LMNO

Substituting values we get,

Area of shaded portion = AB × AD + PR × RS - LM × MN

= 50 × 5 + 35 × 5 - 5 × 5

= 250 + 175 - 25

= 400 m².

Hence, area of shaded region = 400 m².

We know that,

Area of rectangle = length × breadth

Area of square = side × side.

From figure,

Area of shaded portion = Area of large rectangle - 5 × Area of a small square.

Substituting values we get,

Area of shaded portion = (8 × 6) - (5 × 2 × 2)

= 48 - 20

= 28 cm².

Hence, area of shaded region = 28 cm².

Consider ABCD as a plot.

Length of plot = 20 m and breadth of plot = 14 m.

Let PQRS be the plot to be covered with grass.

From figure,

PQ = 20 - (2 × 2)

= 20 - 4

= 16 m

QR = 14 - (2 × 2)

= 14 - 4

= 10 m

Area of rectangular plot PQRS = length × breadth

Substituting the values we get,

Area = 16 × 10 = 160 m².

Hence, area to be laid with grass = 160 m².

(i) Let PQRS be the lawn.

From figure,

QR = BC - BQ - RC = 30 - 2 - 2 = 26 m.

SR = CD - DS = 12 - 2 = 10 m.

Length of PQRS = QR = 26 m and,

Breadth of PQRS = SR = 10 m.

Hence, length and breadth of lawn are 26 m and 10 m respectively.

(ii) From figure,

Area of flower beds = Area of rectangle ABCD - Area of rectangle PQRS

= (AD × DC) - (QR × SR)

= (30 × 12) - (26 × 10)

= 360 - 260

= 100 m².

Hence, area of flower-beds = 100 m².

Consider ABCD as a rectangular field having, length = 50 m and breadth = 38 m.

Let x meters be the width of foot path.

We know that,

Area of rectangular = l × b

From figure,

Area of path = Area of rectangle ABCD - Area of rectangle PQRS

Substituting the values we get,

Area of path = (AB × BC) - (PQ × QR) ........(1)

From figure,

PQ = AB - x - x = (50 - 2x) m,

QR = BC - x - x = (BC - 2x) m.

Substituting the values in equation 1 we get,

⇒ 492 = (50 × 38) - (50 - 2x) (38 - 2x)

⇒ 492 = 1900 - [50(38 - 2x) - 2x(38 - 2x)]

⇒ 492 = 1900 - (1900 - 100x - 76x + 4x²)

⇒ 492 = 1900 - 1900 + 100x + 76x - 4x²

⇒ 492 = 176x - 4x²

⇒ 492 = 4(44x - x²)

⇒ 123 = 44x - x²

⇒ x² - 44x + 123 = 0

⇒ x² - 41x - 3x + 123 = 0

⇒ x(x - 41) - 3(x - 41) = 0

⇒ (x - 3)(x - 41) = 0

⇒ x - 3 = 0 or x - 41 = 0

⇒ x = 3 m or x = 41 m.

Since, width of path cannot be greater than breadth of field,

So, x ≠ 41 m.

Hence, width of the footpath is 3 m.

Given,

Length = 100 m.

Let breadth = x meters.

By formula,

Perimeter of rectangle = 2(l + b)

Substituting the values we get,

Perimeter of rectangular garden = 2(100 + x) = (200 + 2x) m.

Given,

Cost of enclosing fence = ₹ 150 per meter.

∴ Cost of enclosing fence all round the rectangular garden = ₹150(200 + 2x) = ₹(30,000 + 300x).

Given, total cost of fencing = ₹ 54,000

∴ 30,000 + 300x = 54,000

⇒ 300x = 54,000 – 30,000

⇒ 300x = 24,000

⇒ x = $\frac{24,000}{300}$

⇒ x = 80 m.

∴ Breadth of garden = 80 m.

So, the area of rectangular garden = length × breadth

= 100 × 80

= 8000 m².

Hence, the area of rectangular garden = 8000 m².

Let ABCD be the rectangular floor and PQRS be the carpet.

Area of floor = l × b = 15 × 8 = 120 m² = 120 × (100 cm)² = 1200000 cm²

Area of a tile = 50 cm × 25 cm = 1250 cm²

No. of required tiles = $\frac{\text{Area of rect. floor}}{\text{Area of a tile}}$

Substituting the values we get,

No. of required tiles = $\frac{1200000}{1250}$ = 960.

From figure,

Length of carpet (PQ) = 15 – 1 – 1

= 15 – 2

= 13 m

Breadth of carpet (QR) = 8 – 1 – 1

= 8 – 2

= 6 m

Area of carpet = l × b

= 13 × 6

= 78 m².

Area of floor which is uncovered by carpet = Area of floor – Area of carpet

= 120 – 78

= 42 m²

Fraction of floor uncovered = $\frac{\text{Area of floor uncovered}}{\text{Area of floor}}$

= $\frac{42}{120} = \frac{7}{20}$.

Hence, number of tiles required to cover the floor = 960 tiles and $\frac{7}{20}$ is the fraction of floor uncovered.

Given,

Length of rectangular room = x meters

Width of rectangular room = $\frac{3}{5}x$ meters.

Perimeter = y meters.

We know that,

Perimeter = 2(l + b)

Substituting the values we get,

$$\begin{aligned}\Rightarrow y = 2\Big[x + \frac{3}{5}x\Big] \\[1em] \Rightarrow y = 2\Big[\frac{5x + 3x}{5}\Big] \\[1em] \Rightarrow 5y = 2 \times 8x \\[1em] \Rightarrow 5y = 16x \text{ ........ (1)}\end{aligned}$$

The above equation is the required relation between x and y.

Given, perimeter = y = 32 m.

Now substituting the value of y in equation (1)

⇒ 16x = 5 × 32

⇒ x = $\frac{160}{16}$ = 10 m,

⇒ Breadth = $\frac{3}{5} \times x = \frac{3}{5} \times 10$ = 6 m.

Floor area of the room = l × b

= 10 × 6

= 60 m².

Hence, 16x = 5y is the equation connecting x and y and the floor area of room = 60 m².

Let ABCD be a rectangular garden.

Length = 10 m and Breadth = 16 m.

Area of garden ABCD = l × b

= 10 × 16 = 160 m²

Given, width of the walk = x meters.

From figure,

Length of rectangular garden PQRS = 10 + x + x = (10 + 2x) m

Breadth of rectangular garden PQRS = 16 + x + x = (16 + 2x) m

From figure,

⇒ Area of walk = Area of rectangle PQRS - Area of rectangle ABCD

⇒ 120 = (10 + 2x)(16 + 2x) - 160

⇒ 120 = 160 + 20x + 32x + 4x² - 160

⇒ 120 = 4x² + 52x

⇒ 4x² + 52x - 120 = 0

⇒ 4(x² + 13x - 30) = 0

⇒ x² + 13x - 30 = 0

The above equation is the equation in x.

Solving further,

⇒ x² + 15x - 2x - 30 = 0

⇒ x(x + 15) - 2(x + 15) = 0

⇒ (x - 2)(x + 15) = 0

⇒ x = 2 or x = -15.

Since, length cannot be negative.

∴ x = 2.

Hence, equation is x² + 13x - 30 = 0 and x = 2 metres.

It is given that

Length of rectangular room = 6 m

Breadth of rectangular room = 4.8 m

Height of rectangular room = 3.5 m

By formula,

Inner surface area of four walls = 2(l + b) × h

= 2(6 + 4.8) × 3.5

= 2 × 10.8 × 3.5

= 21.6 × 3.5

= 75.6 m².

Hence, inner surface area of four walls = 75.6 m².

Let ABCD be the rectangular plot.

Given,

Length of rectangular plot = 41 metres,

Breadth of rectangular plot = 22.5 metres.

Height of boundary wall = 2 metres.

Boundary wall is built at a distance of 1.5 m. It means wall is built on base PQRS.

From figure,

Length of plot PQRS = 41 + 1.5 + 1.5 = 44 m.

Breadth of plot PQRS = 22.5 + 1.5 + 1.5 = 25.5 m.

By formula,

Inner surface area of the boundary wall = 2(l + b) × h

= 2 (44 + 25.5) × 2

= 2 × 69.5 × 2

= 278 m².

Hence, inner surface area of boundary wall = 278 m².

From figure,

Area of rectangle ABFG = l × b

= BF × AB

= (BC + CF) × AB

= (4 + 1.5) × 2

= 5.5 × 2

= 11 m².

Area of rectangle CDEF = l × b

= CD × DE

= 4 × 1.5

= 6 m².

Total Area = Area of rectangle ABFG + Area of rectangle CDEF

= 11 + 6 = 17 m².

From figure,

AG = BF, GF = AB and FE = CD (As opposite sides of rectangle are equal.)

Perimeter of figure = AB + BC + CD + DE + EF + FG + AG

= 2 + 4 + 4 + 1.5 + 4 + 2 + 5.5

= 23 m.

Hence, perimeter = 23 m and area = 17 m².

The points are labelled on the figure as shown below:

Area of rectangle ABIJ = l × b

= AB × BI

= AB × (BC + CI)

= 3 × (5 + 2)

= 3 × 7 = 21 m².

Area of rectangle EFGH = l × b

= EF × FG

= 2 × 7 = 14 m².

Area of rectangle CDHI = l × b

= CD × DH

= 8 × 2 = 16 m².

Total area = Area of rectangle ABIJ + Area of rectangle EFGH + Area of rectangle CDHI

= 21 + 14 + 16

= 51 m².

Perimeter of figure = AB + BC + CD + DE + EF + FG + GH + HI + IJ + JA

= 3 + 5 + 8 + 5 + 2 + 7 + 2 + 8 + 3 + 7

= 50 m.

Hence, perimeter = 50 m and area = 51 m².

The points are labelled on the figure as shown below:

Area of rectangle BCDE = l × b

= ED × CD

= 5 × 2

= 10 cm².

Area of rectangle FGHI = l × b

= FG × GH

= 3 × 2

= 6 cm².

Area of rectangle JKLM = l × b

= JK × KL

= 5 × 2

= 10 cm².

Area of rectangle ABMN = l × b

= AB × AN

(From figure AB = AC - BC = 7 - 5 = 2 cm.)

= 2 × 12

= 24 cm².

From figure,

Total area = Area of rectangle BCDE + Area of rectangle FGHI + Area of rectangle JKLM + Area of rectangle ABMN

= 10 + 6 + 10 + 24

= 50 cm².

From figure,

The vertical distance between C and L will be equal to vertical distance between A and N,

So ignoring the vertical sides in right side and replacing it will CL.

Perimeter = AC + DE + FG + HI + JK + LN + NA + CL

= 7 + 5 + 3 + 3 + 5 + 7 + 12 + 12

= 54 cm.

Hence, area = 50 cm² and perimeter = 54 cm.

Area of rectangle = l × b = 12 × 9

= 108 cm².

Given,

Area of triangle = one-third the area of rectangle.

Substituting the values we get,

Area of triangle = $\frac{1}{3}$ × 108 = 36 cm².

Consider h cm as the height of triangle.

By formula,

Area of triangle = $\frac{1}{2}$ × base × height

Substituting the values we get,

⇒ 36 = $\frac{1}{2}$ × 9 × h

⇒ 36 × 2 = 9 × h

⇒ h = $\frac{72}{9}$

⇒ h = 8 cm.

Hence, height of triangle is 8 cm.

Let ABCD be the square plot having area 484 m².

Let length of each side of the plot be x meters.

We know that,

Area of square = side × side

Substituting the values we get,

⇒ 484 = (x)²

⇒ x = $\sqrt{484}$ = 22 m.

Since, each angle = 90° in a square.

In right angle triangle ABC,

Using Pythagoras Theorem,

⇒ AC² = AB² + BC²

⇒ AC² = 22² + 22²

⇒ AC² = 484 + 484 = 968

⇒ AC = $\sqrt{968} = 22\sqrt{2}$

⇒ AC = 22 × 1.414 = 31.11 m.

Hence, length of side = 22 m and length of diagonal = 31.11 m.

Let ABCD be a square with side x metres.

Perimeter of square = 4 × side

Substituting the values we get,

⇒ 56 = 4x

⇒ x = $\frac{56}{4}$ = 14 m.

Since, each angle = 90° in a square.

In right angle triangle ABC

Using Pythagoras theorem,

⇒ AC² = AB² + BC²

⇒ AC² = 14² + 14²

⇒ AC² = 196 + 196 = 392

⇒ AC = $\sqrt{392}$

⇒ AC = $14\sqrt{2}$ = 14 × 1.414 = 19.80 m.

Area of square = (side)²

= 14² = 196 m².

Hence, the area of square = 196 m² and length of diagonal = 19.80 m.

Given,

Area of equilateral triangle = $36\sqrt{3}$ cm²

Consider x cm as the side of equilateral triangle

We know that,

Area of an equilateral triangle = $\frac{\sqrt{3}}{4}(\text{ side})^2$

Substituting the values we get,

$$\begin{aligned}\Rightarrow \frac{\sqrt{3}}{4}(\text{ side})^2 = 36\sqrt{3} \\[1em] \Rightarrow \text{side}^2 = \frac{36 \sqrt{3} \times 4}{\sqrt{3}} \\[1em] \Rightarrow \text{side}^2 = 144 \\[1em] \Rightarrow \text{side} = \sqrt{144} = 12 \text{ cm}.\end{aligned}$$

By formula,

Perimeter of equilateral triangle = 3 × side = 3 × 12 = 36 cm.

(i) As the same wire is now bent to form a square.

∴ Perimeter of equilateral triangle = Perimeter of square

36 = 4 × side

Side = $\frac{36}{4}$ = 9 cm.

Area of square = side × side = 9 × 9 = 81 cm².

Hence, area enclosed by wire in form of square = 81 cm².

(ii) As the same wire is now bent to form a rectangle.

∴ Perimeter of triangle = Perimeter of rectangle ........(1)

According to the condition given for rectangle,

Length is 2 cm more than its width

Let width of rectangle = x cm

∴ Length of rectangle = (x + 2) cm

Perimeter of rectangle = 2(l + b)

Substituting the values in equation 1 we get,

⇒ 36 = 2[(x + 2) + x]

⇒ 36 = 2[2x + 2]

⇒ 4x + 4 = 36

⇒ 4x = 32

⇒ x = 8 cm.

∴ Length of rectangle = x + 2 = 8 + 2 = 10 cm and Breadth of rectangle = x = 8 cm.

By formula,

Area of rectangle = length × breadth

= 10 × 8

= 80 cm².

Hence, area enclosed by wire in form of square = 80 cm².

Let ABCD be a parallelogram with side AB = 15 cm and side BC = 10 cm.

Distance between longer side DM = 8 cm

Consider DN as the distance between the shorter side

Area of parallelogram ABCD = base × height

= AB × DM = 15 × 8 = 120 cm².

Considering base BC and height DN

Area of parallelogram = BC × DN

⇒ 120 = 10 × DN

⇒ DN = $\frac{120}{10}$ = 12 cm.

Hence, the area of parallelogram = 120 cm² and the distance between shorter side = 12 cm.

In triangle ABC,

Let,

BC = a = 10 cm, AC = b = 16 cm and AB = c = 12 cm.

We know that,

Semi-perimeter (s) = $\frac{a + b + c}{2}$

= $\frac{10 + 16 + 12}{2} = \frac{38}{2}$ = 19 cm.

By Heron's formula,

$$\begin{aligned}\text{Area of triangle } = \sqrt{s(s - a)(s - b)(s - c)} \\[1em] = \sqrt{19(19 - 10)(19 - 16)(19 - 12)} \\[1em] = \sqrt{19 \times 9 \times 3 \times 7} \\[1em] = \sqrt{3591} \\[1em] = 59.9 \text{ cm}^2.\end{aligned}$$

We know that,

Diagonal of a parallelogram divides it into two triangles of equal area.

∴ Area of triangle ABC = Area of triangle ADC

∴ Area of parallelogram = 2 × Area of triangle ABC.

= 2 × 59.9

= 119.8 cm².

Let DM be the distance between the shorter sides of the parallelogram.

By formula,

Area of parallelogram = base × height = BC × DM

Substituting the values we get,

⇒ 119.8 = 10 × DM

⇒ DM = $\frac{119.8}{10}$

⇒ DM = 11.98 cm.

Hence, the distance between shorter sides = 11.98 cm and area of parallelogram = 119.8 cm².

Let ABCD be a parallelogram with AC and BD the diagonals intersecting at O.

From figure,

AB = 12 cm and DM = 6 cm.

By formula,

Area of parallelogram ABCD = base × height = AB × DM

= 12 × 6

= 72 cm².

Since, diagonals of parallelogram intersect each other so O is the mid-point of BD.

∴ AO is the median of the △ABD.

Since, median divides the triangle into two triangles of equal area,

∴ Area of △AOD = $\frac{1}{2}$ × Area of △ABD ......(1)

Since, diagonal of a parallelogram divides it into two triangles of equal area.

∴ Area of △ABD = $\frac{1}{2}$ × Area of || gm ABCD.

Substituting above value of △ABD in equation 1 we get,

Area of △AOD = $\frac{1}{2} \times \frac{1}{2}$ Area of || gm ABCD

= $\frac{1}{4} \times 72$ = 18 cm².

Hence, area of △AOD = 18 cm².

Let ABCD be a parallelogram with diagonals intersecting at O.

Since, diagonals of a parallelogram bisect each other.

∴ AO = $\frac{12}{2}$ = 6 cm and OB = $\frac{16}{2}$ = 8 cm.

In triangle AOB,

Let AB = a = 10 cm, BO = b = 8 cm and OA = c = 6 cm.

We know that,

Semi-perimeter (s) = $\frac{a + b + c}{2}$

= $\frac{10 + 8 + 6}{2} = \frac{24}{2}$ = 12 cm.

By Heron's formula,

$$\begin{aligned}\text{Area of triangle } = \sqrt{s(s - a)(s - b)(s - c)} \\[1em] = \sqrt{12(12 - 10)(12 - 8)(12 - 6)} \\[1em] = \sqrt{12 \times 2 \times 4 \times 6} \\[1em] = \sqrt{576} \\[1em] = 24 \text{ cm}^2.\end{aligned}$$

Since, diagonals of parallelogram intersect each other so O is the mid-point of BD.

∴ AO is the median of the △ABD.

Since, median divides the triangle into two triangles of equal area.

∴ Area of △AOB = $\frac{1}{2}$ × Area of △ABD ......(1)

Since, diagonal of a parallelogram divides it into two triangles of equal area.

∴ Area of △ABD = $\frac{1}{2}$ × Area of || gm ABCD.

Substituting above value of △ABD in equation 1 we get,

Area of △AOB = $\frac{1}{2} \times \frac{1}{2}$ Area of || gm ABCD

Substituting values in above equation we get,

24 = $\frac{1}{4}$ Area of || gm ABCD

⇒ Area of || gm ABCD = 24 × 4 = 96 cm².

Hence, area of || gm ABCD = 96 cm².

By formula,

Area of a parallelogram = base × height ........(1)

Substituting values of 1st || gm in above equation,

⇒ p = base × q

⇒ base = $\frac{p}{q}$ cm.

Given,

Base of 2nd || gm is r cm more than the base of 1st || gm.

∴ Base of 2nd || gm = $\Big(\frac{p}{q} + r\Big)$ cm.

Given,

Height of second parallelogram = h cm

Substituting the values of 2nd || gm in equation 1,

$$\begin{aligned}\Rightarrow p = \Big(\frac{p}{q} + r\Big) \times h \\[1em] \Rightarrow p = \Big(\frac{p + qr}{q}\Big) \times h \\[1em] \Rightarrow h = \Big(\frac{pq}{p + qr}\Big).\end{aligned}$$

Hence, h = $\Big(\frac{pq}{p + qr}\Big)$ cm.

By formula,

Area of rhombus = $\frac{1}{2}$ × d₁ × d₂, where d₁ and d₂ are diagonals.

Substituting the values we get,

Area of rhombus = $\frac{1}{2}$ × 16 × 12

= 8 × 12 = 96 cm².

Hence, area of rhombus = 96 cm².

By formula,

Area of rhombus = $\frac{1}{2}$ × d₁ × d₂, where d₁ and d₂ are diagonals.

Substituting the values we get,

⇒ 98 = $\frac{1}{2}$ × 14 × d₂

⇒ d₂ = $\frac{98 \times 2}{14}$ = 14 cm.

Hence, the length of other diagonal = 14 cm.

Let length of each side of rhombus = x cm.

Given,

Perimeter = 45 cm

⇒ x + x + x + x = 45

⇒ 4x = 45

⇒ x = $\frac{45}{4}$ cm

By formula,

Area of rhombus = base × height

Substituting the values we get,

Area of rhombus = $\frac{45}{4}$ × 8 = 90 cm².

Hence, area of rhombus = 90 cm².

From figure,

PQ is the base of rhombus PQRS and SM is the height of rhombus.

By formula,

Area of rhombus PQRS = base × height

= 3 × 2.5

= 7.5 cm².

Hence, area of rhombus PQRS = 7.5 cm².

Let ABCD be a rhombus with AC and BD as two diagonals.

Let AC = 8 cm and BD = 6 cm.

Since, diagonals of a rhombus bisect each other at right angles.

∴ AO = 4 cm and BO = 3 cm.

In right angle triangle AOB,

Using Pythagoras theorem

⇒ AB² = AO² + BO²

⇒ AB² = 4² + 3²

⇒ AB² = 16 + 9 = 25

⇒ AB = $\sqrt{25}$ = 5 cm.

Side of rhombus ABCD = 5 cm

By formula,

Perimeter of rhombus = 4 × side = 4 × 5 = 20 cm.

Hence, perimeter of rhombus = 20 cm.

(i) Let ABCD be a rhombus with AC and BD diagonals and each side = 5 cm.

Let AC = 8 cm.

Since, diagonals of rhombus bisect each other at right angles.

∴ AO = 4 cm.

In right angle triangle AOB

Using Pythagoras theorem,

⇒ AB² = AO² + BO²

⇒ 5² = 4² + BO²

⇒ 25 = 16 + BO²

⇒ BO² = 25 – 16 = 9

⇒ BO = $\sqrt{9}$ = 3 cm.

⇒ BD = 2 × BO = 2 × 3 = 6 cm.

Hence, length of other diagonal = 6 cm.

(ii) Area of rhombus = $\frac{1}{2}$ × product of diagonals

= $\frac{1}{2}$ × 8 × 6

= 4 × 6

= 24 cm².

Hence, area of rhombus = 24 cm².

(a) Construct BN perpendicular to CD.

So, BADN is a rectangle.

As opposite sides of rectangle are equal.

∴ BN = AD = 4 cm and ND = BA = 5 cm.

From figure,

CN = CD – ND = 8 - 5 = 3 cm.

In right angle triangle BCN,

Using Pythagoras theorem,

⇒ BC² = BN² + CN²

⇒ BC² = 4² + 3²

⇒ BC² = 16 + 9 = 25

⇒ BC = $\sqrt{25}$ = 5 cm.

By formula,

Area of trapezium = $\frac{1}{2}$ × sum of parallel sides × height

= $\frac{1}{2}$ × (AB + CD) × AD

= $\frac{1}{2}$ × (5 + 8) × 4

= 13 × 2 = 26 cm².

Hence, BC = 5 cm and area of trapezium = 26 cm².

(i) Construct a perpendicular from C to AD parallel to AB.

So, ABCM is a rectangle. Since, opposite sides of a rectangle are equal.

∴ AM = CB = 2 units.

From figure,

⇒ AD = AM + MD

⇒ MD = AD - AM = 8 - 2 = 6 units.

In right angle triangle MDC,

⇒ CD² = MD² + CM²

⇒ 10² = 6² + CM²

⇒ 100 = 36 + CM²

⇒ CM² = 64

⇒ CM = $\sqrt{64}$ = 8 units.

Since, ABCM is a rectangle.

∴ AB = CM = 8 units.

Hence, AB = 8 units.

(ii) Area of trapezium ABCD = $\frac{1}{2}$ × (sum of || sides) × distance between them

= $\frac{1}{2}$ × (AD + BC) × AB

= $\frac{1}{2}$ × (8 + 2) × 8

= 40 sq. units.

Hence, area of trapezium ABCD = 40 sq. units.

Consider ABCD as the cross section of canal in the shape of trapezium.

Draw a perpendicular AM from A to CD.

So, AM is the depth of canal.

Given, the area of cross-section of canal = 16.8 m².

∴ $\frac{1}{2}$ × sum of parallel sides × depth = 16.8

⇒ $\frac{1}{2}$ × (AB + DC) × AM = 16.8

⇒ $\frac{1}{2}$ × (6 + 8) × AM = 16.8

⇒ $\frac{1}{2}$ × 14 × AM = 16.8

⇒ AM = $\frac{16.8 \times 2}{14}$

⇒ AM = $\frac{33.6}{14}$ = 2.4 m

Hence, depth of canal = 2.4 meters.

Let ABCD be the trapezium in which AB || DC. Let E and F be mid-points of sides AD and BC respectively, then EF = 18 cm.

Given,

Distance between parallel sides of a trapezium is 12 cm.

∴ Height = 12 cm.

By formula,

Sum of the lengths of two parallel sides = 2 × Distance between mid-points of two non-parallel sides

⇒ AB + CD = 2 × EF = 2 × 18 = 36 cm.

Area of trapezium = $\frac{1}{2}$ × (Sum of parallel sides) × height

= $\frac{1}{2}$ × (AB + CD) × 12

= $\frac{1}{2}$ × 36 × 12

= 18 × 12

= 216 cm².

Hence, area of trapezium = 216 cm².

Given,

Area of trapezium = 540 cm²

Ratio of parallel sides = 7 : 5

Let the sides be 7x and 5x cm.

Distance between the parallel sides = height = 18 cm

By formula,

Area of trapezium = $\frac{1}{2}$ × sum of parallel sides × height

⇒ 540 = $\frac{1}{2}$ × (7x + 5x) × 18

⇒ 540 = $\frac{1}{2}$ × 12x × 18

⇒ 540 = 6x × 18

⇒ 540 = 108x

⇒ x = $\frac{540}{108}$ = 5 cm.

⇒ 7x = 7 × 5 = 35 cm and 5x = 5 × 5 = 25 cm.

Hence, the length of parallel sides are 25 cm and 35 cm.

Since, ABCD is an isosceles trapezium so, BC = AD.

Since, parallel sides of an isosceles trapezium are in the ratio 2 : 3.

∴ CD = 2a and AB = 3a.

Construct perpendicular DN from D to AB and perpendicular CM from C to AB.

Given,

Area = 60 cm²

By formula,

Area of trapezium = $\frac{1}{2}$ × sum of parallel sides × height

⇒ 60 = $\frac{1}{2}$ × (AB + DC) × DN

⇒ 60 = $\frac{1}{2}$ × (3a + 2a) × 4

⇒ 60 = 2 × 5a

⇒ 10a = 60

⇒ a = 6 cm.

⇒ AB = 3a = 3 × 6 = 18 cm and CD = 2a = 2 × 6 = 12 cm.

In △ADN and △BCM,

⇒ ∠AND = ∠CMB = 90°

⇒ DN = CM = 4 cm

⇒ AD = CB = x cm (let) (As ABCD is an isosceles trapezium).

∴ △ADN ≅ △BCM by RHS axiom.

∴ AN = MB ........(1)

Since, DNMC is a rectangle.

∴ NM = DC = 12 cm. (As opposite sides of a rectangle are equal.)

From figure,

⇒ AN + NM + MB = 18

⇒ AN + 12 + MB = 18

⇒ AN + MB = 6

⇒ 2AN = 6 (As AN = MB)

⇒ AN = $\frac{6}{2}$ = 3 cm.

⇒ MB = 3 cm.

In right angle triangle AND,

⇒ AD² = AN² + DN²

⇒ x² = 4² + 3²

⇒ x² = 16 + 9

⇒ x² = 25

⇒ x = $\sqrt{25}$ = 5 cm.

From figure,

Perimeter = AB + BC + CD + DA

= 18 + 5 + 12 + 5

= 40 cm.

Hence, perimeter of trapezium = 40 cm.

Let base = x cm

Corresponding altitude = $\frac{x}{2}$ cm

By formula,

Area of parallelogram = base × altitude

Substituting the values we get,

⇒ 98 = $x \times \frac{x}{2}$

⇒ 98 = $\frac{x^2}{2}$

⇒ x² = 98 × 2 = 196

⇒ x = $\sqrt{196}$ = 14 cm

⇒ Base = x = 14 cm

⇒ Altitude = $\frac{x}{2}$ = 7 cm.

Hence, base = 14 cm and altitude = 7 cm.

Let breadth of rectangular garden = x meters,

∴ Length = (x + 12) meters.

Area of garden = length × breadth = x(x + 12) m²

Perimeter of garden = 2(l + b)

= 2[(x + 12) + x]

= 2[2x + 12] = (4x + 24) meters.

According to question,

⇒ Area of garden = 4 × Perimeter of garden

⇒ x(x + 12) = 4 × (4x + 24)

⇒ x² + 12x = 16x + 96

⇒ x² + 12x - 16x - 96 = 0

⇒ x² - 4x - 96 = 0

⇒ x² - 12x + 8x - 96 = 0

⇒ x(x - 12) + 8(x - 12) = 0

⇒ (x + 8)(x - 12) = 0

⇒ x + 8 = 0 or x - 12 = 0

⇒ x = -8 or x = 12.

Since, breadth cannot be negative.

∴ x ≠ -8.

Breadth = x = 12 m and Length = (x + 12) = (12 + 12) = 24 m.

Hence, length and breadth of garden are 24 m and 12 m respectively.

Let ABCD be a rectangular plot of length x m and breadth y m.

By formula,

Perimeter = 2(length + breadth)

Substituting the values we get,

⇒ 68 = 2(x + y)

⇒ 34 = x + y

⇒ x = 34 - y ......... (1)

In right angle triangle ABC

⇒ AC² = AB² + BC² (By pythagoras theorem)

⇒ 26² = x² + y²

⇒ x² + y² = 676

Substituting the value of x from equation (1),

⇒ (34 – y)² + y² = 676

⇒ 1156 + y² – 68y + y² = 676

⇒ 2y² – 68y + 1156 – 676 = 0

⇒ 2y² – 68y + 480 = 0

⇒ 2(y² – 34y + 240) = 0

⇒ y² – 34y + 240 = 0

⇒ y² – 24y – 10y + 240 = 0

⇒ y(y – 24) – 10(y – 24) = 0

⇒ (y – 10)(y – 24) = 0

⇒ y – 10 = 0 or y – 24 = 0

⇒ y = 10 m or y = 24 m.

Now substituting the value of y in equation (1)

⇒ y = 10 m, x = 34 – 10 = 24 m

⇒ y = 24 m, x = 34 – 24 = 10 m

Area in both cases = xy

= 24 × 10 or 10 × 24

= 240 m².

Hence, the area of the rectangular block is 240 m².

Let length of a side of a square = x cm.

According to question,

Length of rectangle = (x + 12) cm

Breadth of rectangle = (x + 8) cm

Given,

⇒ Area of rectangle = 2 × area of square

⇒ (x + 12)(x + 8) = 2 × (x × x)

⇒ x(x + 8) + 12(x + 8) = 2x²

⇒ x² + 8x + 12x + 96 = 2x²

⇒ x² – 2x² + 8x + 12x + 96 = 0

⇒ -x² + 20x + 96 = 0

⇒ x² – 20x – 96 = 0

⇒ x² – 24x + 4x – 96 = 0

⇒ x(x - 24) + 4(x - 24) = 0

⇒ (x + 4)(x – 24) = 0

⇒ x + 4 = 0 or x - 24 = 0

⇒ x = -4 or x = 24 cm

Since, side of a square cannot be negative.

∴ x ≠ -4.

Side of square = 24 cm

Perimeter of square = 4 × side = 4 × 24

= 96 cm.

Hence, perimeter of square = 96 cm.

Perimeter of a square = 48 cm

Length of side of square = $\frac{\text{Perimeter}}{4} = \frac{48}{4}$ = 12 cm.

By formula,

Area = (side)² = 12² = 144 cm².

∴ Area of rectangle = 144 – 4 = 140 cm²

Let breadth of rectangle = x cm

∴ Length of rectangle = (x + 4) cm

Area of rectangle = l × b = x(x + 4) cm²

Substituting the values we get,

⇒ x(x + 4) = 140

⇒ x² + 4x – 140 = 0

⇒ x² + 14x – 10x – 140 = 0

⇒ x(x + 14) – 10(x + 14) = 0

⇒ (x + 14)(x – 10) = 0

⇒ x + 14 = 0 or x - 10 = 0

⇒ x = -14 or x = 10

Since, breadth cannot be negative.

∴ x ≠ -14.

Breadth = x = 10 cm and Length = x + 4 = 10 + 4 = 14 cm

Perimeter of rectangle = 2(l + b)

= 2(14 + 10)

= 2 × 24 = 48 cm.

Hence, perimeter of rectangle = 48 cm.

In △AEB,

Let AE = BE = x cm, then from right angled triangle AEB,

⇒ AB² = AE² + EB²

⇒ 10² = x² + x²

⇒ 2x² = 100

⇒ x² = 50

⇒ x = $\sqrt{50} = 5\sqrt{2}$ cm.

Area of right angled △AEB = $\frac{1}{2}$ × base × height

$$\begin{aligned}= \frac{1}{2} \times x \times x = \frac{1}{2}x^2 \\[1em] = \frac{1}{2} \times 50 \\[1em] = 25 \text{ cm}^2\end{aligned}$$

In △DGC,

Let DG = GC = y cm, then from right angled triangle DGC,

⇒ DC² = DG² + GC²

⇒ 10² = y² + y²

⇒ 2y² = 100

⇒ y² = 50

⇒ y = $\sqrt{50} = 5\sqrt{2}$ cm

Area of right angled △DCG = $\frac{1}{2}$ × base × height

$$\begin{aligned}= \frac{1}{2} \times y \times y = \frac{1}{2}y^2 \\[1em] = \frac{1}{2} \times 50 \\[1em] = 25 \text{ cm}^2\end{aligned}$$

Since, HAD and BFC are equilateral triangle with side = 8 cm.

Area of HAD = Area of BFC = $\frac{\sqrt{3}}{4}$ × (side)²

= $\frac{\sqrt{3}}{4} \times (8)^2$

= $16\sqrt{3}$ cm²

Area of rectangle ABCD = l × b = AB × CD

= 10 × 8 = 80 cm²

From figure,

Area of shaded region = Area of (△DGC + △BFC + △AEB + △HAD + rectangle ABCD)

= $25 + 16\sqrt{3} + 25 + 16\sqrt{3} + 80 = 130 + 32\sqrt{3}$ cm².

Perimeter of figure = (AE + EB + BF + FC + CG + GD + DH + HA)

= $(5\sqrt{2} + 5\sqrt{2} + 8 + 8 + 5\sqrt{2} + 5\sqrt{2} + 8 + 8)$

= $20\sqrt{2} + 32$ cm.

Hence, area of shaded region = $130 + 32\sqrt{3}$ and perimeter = $20\sqrt{2} + 32$ cm.

In right angle triangle ECF,

Using pythagoras theorem,

⇒ EF² = EC² + CF²

⇒ 5² = EC² + 3²

⇒ EC² = 5² - 3²

⇒ EC² = 25 - 9 = 16

⇒ EC = $\sqrt{16}$ = 4 cm.

Since, DEFG is an isosceles trapezium.

∴ GD = EF= 5 cm.

Since, BDEC is a rectangle,

∴ BD = EC = 4 cm and BC = DE = 6 cm.

In right angle triangle DBG,

Using pythagoras theorem,

⇒ GD² = BD² + GB²

⇒ 5² = 4² + GB²

⇒ GB² = 5² - 4²

⇒ GB² = 25 - 16 = 9

⇒ GB = $\sqrt{9}$ = 3 cm.

In trapezium,

GF = GB + BC + CF = 3 + 6 + 3 = 12 cm.

Area of trapezium DEFG = $\frac{1}{2} \times$ (sum of parallel sides) × distance between them

= $\frac{1}{2} \times (12 + 6) \times 4$

= 18 × 2

= 36 cm².

Area of equilateral triangle ABC = $\frac{\sqrt{3}}{4}\text{ (side)}^2$

= $\frac{\sqrt{3}}{4} \times (6)^2$

= $\frac{\sqrt{3}}{4} \times 36$

= 1.732 × 9

= 15.59 cm²

Area of figure = Area of trapezium DEFG + Area of equilateral triangle ABC

= 36 + 15.59 = 51.59 cm².

Hence, area of figure = 51.59 cm².

From figure,

BJ = 2 + 2 + 2 + 2 = 8 cm.

Area of rectangle ABJK = l × b

= AB × BJ = 2 × 8

= 16 cm².

From figure,

JH = KH - KI = 6 - 2 = 4 cm.

Area of trapezium FGHI = $\frac{1}{2} \times$ (sum of parallel sides) × distance between them

= $\frac{1}{2}$ × (FI + GH) × JH

= $\frac{1}{2}$ × (2 + 2) × 4

= $\frac{1}{2}$ × 4 × 4 = 8 cm².

Area of trapezium CDEF = $\frac{1}{2} \times$ (sum of parallel sides) × distance between them

= $\frac{1}{2}$ × (CF + DE) × BD

= $\frac{1}{2}$ × (2 + 2) × 4

= $\frac{1}{2}$ × 4 × 4 = 8 cm².

Total area enclosed = Area of rectangle ABJK + Area of trapezium FGHI + Area of trapezium CDEF

= 16 + 8 + 8

= 32 cm².

Hence, area enclosed by figure = 32 cm².

From figure,

Area of rectangle (I) = length × breadth

= 24 × 6

= 144 cm².

Area of rectangle (II) = length × breadth

= 24 × 6

= 144 cm².

Area of parallelogram (III) = base × height

= 8 × 6

= 48 cm².

Area of parallelogram (IV) = base × height

= 8 × 6

= 48 cm².

Area of figure M = Area of rectangle (I) + Area of rectangle (II) + Area of parallelogram (III) + Area of parallelogram (IV)

= 144 + 144 + 48 + 48

= 384 cm².

Area of cardboard = 24 × 24

= 576 cm².

Area of cardboard left = Area of cardboard - Area of figure M

= 576 - 384

= 192 cm².

Hence, area of cardboard left = 192 cm².

From figure,

The figure consist of a trapezium and a rectangle.

Area of trapezium = $\frac{1}{2} \times$ (sum of parallel sides) × distance between them

= $\frac{1}{2}$ × (0.6 + 1.5) × (1.2 + 2.4)

= $\frac{1}{2}$ × 2.1 × 3.6

= 2.1 × 1.8

= 3.78 m².

Area of rectangle = l × b

= 2.4 × 0.3 = 0.72 m².

Area of cross section = Area of trapezium + Area of rectangle

= 3.78 + 0.72 = 4.5 m².

Hence, area of cross-section = 4.5 m².

From figure,

Area of right angled △AXB = $\frac{1}{2}$ × base × height

= $\frac{1}{2}$ × BX × AX

= $\frac{1}{2}$ × 30 × 12

= 180 m².

Area of trapezium XZCB = $\frac{1}{2} \times$ (sum of parallel sides) × distance between them

= $\frac{1}{2}$ × (BX + CZ) × 15

= $\frac{1}{2}$ × (30 + 25) × 15

= $\frac{1}{2}$ × 55 × 15

= 412.5 m².

Area of right angled △CZD = $\frac{1}{2}$ × base × height

= $\frac{1}{2}$ × CZ × ZD

= $\frac{1}{2}$ × 25 × 10

= 125 m².

Area of △AED = $\frac{1}{2}$ × base × height

= $\frac{1}{2}$ × AD × EY

= $\frac{1}{2}$ × 37 × 20

= 370 m².

Area of field = Area of right angled △AXB + Area of trapezium XZCB + Area of right angled △CZD + Area of △AED

= 180 + 412.5 + 125 + 370 = 1087.5 m².

Hence, area of field = 1087.5 m².

From figure,

In right angled △DEY,

⇒ DE² = DY² + EY²

⇒ 5² = DY² + 4²

⇒ DY² = 5² - 4²

⇒ DY² = 25 - 16 = 9

⇒ DY = $\sqrt{9}$ = 3 cm.

Area of right angled △DEY = $\frac{1}{2}$ × base × height

= $\frac{1}{2}$ × EY × DY

= $\frac{1}{2}$ × 4 × 3

= 6 cm².

Area of right angle △DYC = $\frac{1}{2}$ × base × height

= $\frac{1}{2}$ × CY × DY

= $\frac{1}{2}$ × 4 × 3

= 6 cm².

From figure,

XY = DX - DY = 9 - 3 = 6 cm.

Area of trapezium ECBA = $\frac{1}{2} \times$ (sum of parallel sides) × distance between them

= $\frac{1}{2}$ × (EC + AB) × XY

= $\frac{1}{2}$ × [(EY + CY) + (AX + BX)] × XY

= $\frac{1}{2}$ × [(4 + 4) + (6 + 6)] × 6

= $\frac{1}{2}$ × 20 × 6

= 60 cm².

Area of pentagon = Area of right angled △DEY + Area of right angled △DYC + Area of trapezium ECBA

= 6 + 6 + 60

= 72 cm².

Hence, area of trapezium = 72 cm².

Let length = l metres and breadth = b metres.

Area = lb m²

Given,

If the length and the breadth of a room are increased by 1 metre, the area is increased by 21 square metres,

∴ (l + 1)(b + 1) - lb = 21

⇒ lb + l + b + 1 - lb = 21

⇒ l + b = 21 - 1

⇒ l + b = 20 ..........(1)

Given,

If the length is increased by 1 metre and breadth is decreased by 1 metre the area is decreased by 5 square metres.

∴ lb - (l + 1)(b - 1) = 5

⇒ lb - (lb - l + b - 1) = 5

⇒ lb - lb + l - b + 1 = 5

⇒ l - b = 4 ..........(2)

Adding equation (1) and (2) we get,

⇒ l + b + l - b = 20 + 4

⇒ 2l = 24

⇒ l = 12 m.

Substituting value of l in (2) we get,

⇒ 12 - b = 4

⇒ b = 12 - 4 = 8 m.

Perimeter of room = 2(l + b) = 2 × 20 = 40 m.

Hence, perimeter of room = 40 m.

Let a = 26 cm, b = 28 cm and c = 30 cm.

Semi-perimeter (s) = $\frac{a + b + c}{2} = \frac{26 + 28 + 30}{2} = \frac{84}{2}$ = 42 cm.

Area of triangle = $\sqrt{s(s - a)(s - b)(s - c)}$

$$\begin{aligned}= \sqrt{42 \times (42 - 26) \times (42 - 28) \times (42 - 30)} \\[1em] = \sqrt{42 \times 16 \times 14 \times 12} \\[1em] = \sqrt{112896} \\[1em] = 336 \text{ cm}^2.\end{aligned}$$

Since, area of parallelogram = area of triangle.

∴ Area of parallelogram = 336

⇒ base × height = 336

⇒ 28 × height = 336

⇒ height = $\frac{336}{28} = 12$ cm.

Hence, height of parallelogram = 12 cm.

Given,

Area of rectangle = 105 cm²

Length of rectangle = x cm

By formula,

Area of rectangle = length × breadth

Substituting the values we get,

105 = x × breadth

Breadth = $\frac{105}{x}$ cm.

Perimeter of rectangle = 44 cm

$$\begin{aligned}\therefore 2(l + b) = 44 \\[1em] \Rightarrow 2\Big(x + \frac{105}{x}\Big) = 44 \\[1em] \Rightarrow \Big(x + \frac{105}{x}\Big) = 22 \\[1em] \Rightarrow \frac{x^2 + 105}{x} = 22 \\[1em] \Rightarrow x^2 + 105 = 22x \\[1em] \Rightarrow x^2 - 22x + 105 = 0 \\[1em] \Rightarrow x^2 - 15x - 7x + 105 = 0 \\[1em] \Rightarrow x(x - 15) - 7(x - 15) = 0 \\[1em] \Rightarrow (x - 7)(x - 15) = 0 \\[1em] \Rightarrow x - 7 = 0 \text{ or } x - 15 = 0 \\[1em] \Rightarrow x = 7 \text{ or } x = 15.\end{aligned}$$

If x = 7 cm,

Breadth = $\frac{105}{7}$ = 15 cm

If x = 15 cm,

Breadth = $\frac{105}{15}$ = 7 cm

Hence, breadth = $\frac{105}{x}$, equation : 44 = $2\Big(x + \frac{105}{x}\Big)$ and the required dimensions of rectangle are 15 cm and 7 cm.

Let length of rectangle be x meters.

Given,

Perimeter = 180 m

∴ 2(l + b) = 180

⇒ 2(x + b) = 180

⇒ x + b = 90

⇒ b = (90 - x) m.

Area = l × b

∴ x(90 - x) = 1800

⇒ 90x - x² = 1800

⇒ x² - 90x + 1800 = 0

⇒ x² - 60x - 30x + 1800 = 0

⇒ x(x - 60) - 30(x - 60) = 0

⇒ (x - 30)(x - 60) = 0

⇒ x - 30 = 0 or x - 60 = 0

⇒ x = 30 or x = 60.

If x = 30, 90 - x = 60 and x = 60, 90 - x = 30.

Hence, breadth = (90 - x) m, equation : x(90 - x) = 1800 and length of rectangle = 60 m and breadth = 30 m.

Let radius = r cm.

By formula,

Circumference = 2πr

2πr = 44

$$\begin{aligned}\Rightarrow 2 \times \frac{22}{7} \times r = 44 \\[1em] \Rightarrow \frac{44r}{7} = 44 \\[1em] \Rightarrow r = \frac{44 \times 7}{44} \\[1em] \Rightarrow r = 7 \text{ cm}.\end{aligned}$$

Diameter = 2r = 2 × 7 = 14 cm.

Hence, length of the diameter of the circle = 14 cm.

Let radius = r cm.

By formula,

Circumference = 2πr

⇒ 2πr = 18π

⇒ 2r = 18

⇒ r = 9 cm.

Area of circle = πr²

= $\frac{22}{7} \times 9^2 = \frac{22 \times 81}{7}$

= $\frac{1782}{7} = 254\frac{4}{7}$ cm².

Hence, radius = 9 cm and area = $254\frac{4}{7}$ cm².

Perimeter of semicircular plate = (π + 2)r

$$\begin{aligned}= \Big(\frac{22}{7} + 2\Big) \times 3.85 \\[1em] = \Big(\frac{22 + 14}{7} \Big) \times 3.85 \\[1em] = \frac{36}{7} \times 3.85 \\[1em] = 36 \times 0.55 \\[1em] = 19.8 \text{ cm}.\end{aligned}$$

Hence, perimeter of a semicircular plate = 19.8 cm.

Let radius = r cm.

Area of circle = πr²

⇒ 144π = πr²

⇒ r² = 144

⇒ r = $\sqrt{144}$ = 12 cm.

Circumference of a circle = 2πr

= $2 \times \frac{22}{7} \times 12$

= $\frac{528}{7} = 75\frac{3}{7}$ cm.

Hence, radius = 12 cm and circumference = $75\frac{3}{7}$ cm.

Given,

Diameter of circle = 0.5 cm.

From figure,

Area of sheet required to cut a circle = area of a square with length of side equal to diameter.

∴ No. of discs prepared = No. of squares formed from sheet.

No. of squares that will be formed from sheet = $\frac{\text{Area of sheet}}{\text{Area of a square}}$

$$\begin{aligned}\text{No. of squares } = \frac{11 \times 2}{0.5 \times 0.5} \\[1em] = \frac{22}{0.25} \\[1em] = 88.\end{aligned}$$

Hence, no. of discs that can be prepared = 88.

Let radius = r cm.

Given,

Area of semi-circular region = 77 cm²

$$\begin{aligned}\therefore \frac{πr^2}{2} = 77 \\[1em] \Rightarrow πr^2 = 154 \\[1em] \Rightarrow \frac{22}{7} \times r^2 = 154 \\[1em] \Rightarrow r^2 = \frac{154 \times 7}{22} \\[1em] \Rightarrow r^2 = 49 \\[1em] \Rightarrow r = \sqrt{49} = 7 \text{ cm}.\end{aligned}$$

Perimeter of semi-circle = (π + 2) = πr + 2r

= $\frac{22}{7} \times 7 + 2 \times 7$

= 22 + 14

= 36 cm.

Hence, perimeter of circle = 36 cm.

(i) Let r be the radius of the circle. We know that,

Diameters of the circle divide circle into 4 equal quadrants.

Hence, area of each quadrant = $\frac{1}{4}$ πr².

Since, 2 quadrants are shaded.

∴ Area of shaded region = $2 \times \frac{1}{4}$ πr²

⇒ 308 = $\frac{1}{2} \times \frac{22}{7}$ r²

⇒ r² = $\frac{308 \times 14}{22}$

⇒ r² = 14 × 14 = 196

⇒ r = $\sqrt{196}$ = 14 cm.

Since, AC is the diameter of circle so,

⇒ AC = 2r = 28 cm.

Hence, AC = 28 cm.

(ii) Circumference of circle = 2πr

= $2 \times \frac{22}{7} \times 14$

= 88 cm.

Hence, circumference of circle = 88 cm.

Given,

AC = 16 cm, AO = $\frac{AC}{2} = \frac{16}{2}$ = 8 cm.

The diameters of the circle divide circle into 4 quadrants.

Area of each quadrant = $\frac{πr^2}{4}$

$$\begin{aligned}= \frac{3.14 \times 8^2}{4} \\[1em] = \frac{3.14 \times 64}{4} \\[1em] = 3.14 \times 16 \\[1em] = 50.24 \text{ cm}^2.\end{aligned}$$

Area of quadrant AOD + Area of quadrant BOC = 50.24 + 50.24 = 100.48 cm².

Perimeter of each quadrant = $\frac{2πr}{4} + r + r = \frac{πr}{2} + 2r$

$$\begin{aligned}= \frac{3.14 \times 8}{2} + (2 \times 8) \\[1em] = \frac{25.12}{2} + 16 \\[1em] = 12.56 + 16 \\[1em] = 28.56 \text{ cm}^2.\end{aligned}$$

Perimeter of both quadrants = 2 × 28.56 = 57.12 cm.

Hence, area of shaded region = 100.48 cm² and perimeter of shaded region = 57.12 cm.

Time in which bucket ascends = 1 minute 28 seconds = 60 + 28 = 88 seconds.

Speed of bucket = 1.1 m/sec

Distance covered by bucket while ascending = Speed × Time = 1.1 × 88 = 96.8 m.

Radius of wheel = $\frac{\text{Diameter}}{2} = \frac{77}{2}$ = 38.5 cm.

Circumference of circle = 2πr = $2 \times \frac{22}{7} \times 38.5$ = 242 cm = 2.42 m.

Let n be the no. of revolutions of wheel.

Distance covered by bucket = Distance covered by wheel

⇒ 96.8 = 2.42 × n

⇒ n = $\frac{96.8}{2.42}$ = 40.

Hence, wheel makes 40 revolutions in raising the bucket.

Radius of wheel = $\frac{\text{Diameter of wheel}}{2} = \frac{84}{2}$ = 42 cm.

Distance covered by wheel in 1 revolution = Circumference of wheel = 2πr

= $2 \times \frac{22}{7} \times 42$

= 264 cm.

Distance covered by wheel in 5 revolutions = 5 × 264 = 1320 cm.

∴ Wheel covers 1320 cm in 1 second.

$$\begin{aligned}\text{Speed } = \frac{\text{Distance}}{\text{Time}} \\[1em] = \frac{1320 \times 10^{-5} \text{ km}}{\frac{1}{60 \times 60} \text{ hr}} \\[1em] = 1320 \times 60 \times 60 \times 10^{-5} \\[1em] = 4752000 \times 10^{-5} \\[1em] = 47.52 ≈ 48\text{ km/hr}.\end{aligned}$$

Hence, speed of wheel = 48 km/hr.

(i) Let radius = r cm.

By formula,

Circumference = 2πr

2πr = 123.2

$$\begin{aligned}\Rightarrow 2 \times \frac{22}{7} \times r = 123.2 \\[1em] \Rightarrow r = \frac{123.2 \times 7}{22 \times 2} \\[1em] \Rightarrow r = \frac{862.4}{44} = 19.6 \text{ cm}\end{aligned}$$

Hence, radius = 19.6 cm.

(ii) By formula,

Area of circle = πr²

$$\begin{aligned}= \frac{22}{7} \times (19.6)^2 \\[1em] = \frac{22}{7} \times 384.16 \\[1em] = 22 \times 54.88 \\[1em] = 1207.36 ≈ 1207 \text{ cm}^2.\end{aligned}$$

Hence, area of circle = 1207 cm².

(iii) We know that,

Area of circle = πr², where r is the radius.

If radius is doubled so new radius = 2r cm.

New area of circle = π(2r)² = 4πr².

Change in area = $\frac{\text{New area}}{\text{Area of circle}} = \frac{4πr^2}{πr^2} = 4$

Hence, area becomes 4 times.

Let radius of inner circle = r cm.

From figure,

Area of shaded region = Area of outer circle - Area of inner circle

⇒ 770 = π(21)² - πr²

⇒ 770 = 441π - πr²

⇒ 770 = π(441 - r²)

⇒ 441 - r² = $\frac{770}{π}$

⇒ 441 - r² = $\frac{770}{\frac{22}{7}}$

⇒ 441 - r² = $\frac{770 \times 7}{22}$

⇒ 441 - r² = 245

⇒ r² = 441 - 245

⇒ r² = 196

⇒ r = $\sqrt{196}$ = 14 cm.

Hence, radius of inner circle = 14 cm.

Let radius of inner circle = r cm.

Circumference = 2πr

⇒ 88 = 2πr

⇒ r = $\frac{88}{2π}$

⇒ r = $\frac{44}{π}$

⇒ r = $\frac{44}{\frac{22}{7}} = \frac{44 \times 7}{22}$

⇒ r = 2 × 7 = 14 cm.

Let radius of outer circle = R cm.

From figure,

Area of shaded region = Area of outer circle - Area of inner circle

⇒ 346.5 = π(R)² - πr²

⇒ 346.5 = π(R)² - π(14)²

⇒ 346.5 = π(R² - 196)

⇒ R² - 196 = $\frac{346.5}{π}$

⇒ R² - 196 = $\frac{346.5}{\frac{22}{7}}$

⇒ R² - 196 = $\frac{346.5 \times 7}{22}$

⇒ R² - 196 = 110.25

⇒ R² = 110.25 + 196

⇒ R² = 306.25

⇒ R = $\sqrt{306.25}$

⇒ R = 17.5 cm

Hence, radius of outer circle = 17.5 cm.

Let the radius of circular plot = R meters.

Given,

Circumference of circular plot = 44 m

2πR = 44

$$\begin{aligned}\Rightarrow 2 \times \frac{22}{7} \times R = 44\\[1em] \Rightarrow \frac{44}{7} \times R = 44\\[1em] \Rightarrow R = \frac{44 \times 7}{44}\\[1em] \Rightarrow R = 7 \text{ m}.\end{aligned}$$

From figure,

Radius of outer circle = R + 3.5 = 7 + 3.5 = 10.5 m.

Area of road = Area of outer circle - Area of inner circle

= π(10.5)² - π(7)²

= 110.25π - 49π

= 61.25π

= $\frac{22}{7}$ x 61.25

= 192.50 m²

Rate of paving the road = ₹ 200 per m²

Cost of paving road = Area of road × Rate of paving the road

= 192.5 × 200

= ₹ 38,500.

Hence, cost of paving the road = ₹ 38,500.

Let radius of larger circle be R cm and smaller circle r cm.

Diameter = 2R and 2r

Given, sum of diameters of two circles is 14 cm

⇒ 2r + 2R = 14

⇒ r + R = 7 ...........(1)

Given, difference in circumferences is 8 cm

⇒ 2πR - 2πr = 8

⇒ 2π(R - r) = 8

⇒ π(R - r) = 4

⇒ $\frac{22}{7}(R - r) = 4$

⇒ $R - r = \frac{28}{22}$ .........(2)

Adding equation 1 and 2,

$$\begin{aligned}\Rightarrow r + R + R - r = 7 + \frac{28}{22} \\[1em] \Rightarrow 2R = \frac{154 + 28}{22} \\[1em] \Rightarrow R = \frac{182}{44} = \frac{91}{22}.\end{aligned}$$

Substituting value of R in Eq 2 we get,

$$\begin{aligned}\Rightarrow \frac{91}{22} - r = \frac{28}{22} \\[1em] \Rightarrow r = \frac{91}{22} - \frac{28}{22} \\[1em] \Rightarrow r = \frac{91 - 28}{22} \\[1em] \Rightarrow r = \frac{63}{22}.\end{aligned}$$

Circumference of larger circle = 2πR

= $2 \times \frac{22}{7} \times \frac{91}{22}$

= 2 x 13

= 26 cm.

Circumference of smaller circle = 2πr

= $2 \times \frac{22}{7} \times \frac{63}{22}$

= 2 x 9

= 18 cm.

Hence, circumference of larger circle = 26 cm and smaller circle = 18 cm.

Let the radius of resultant circle be r cm.

Given,

Area of resultant circle is equal to the sum of the areas of three circles with radius 2 cm, 3 cm and 6 cm.

⇒ πr² = π(2)² + π(3)² + π(6)²

⇒ πr² = π[(2)² + (3)² + (6)²]

⇒ r² = 2² + 3² + 6²

⇒ r² = 4 + 9 + 36

⇒ r² = 49

⇒ r = $\sqrt{49}$ = 7 cm.

Circumference of circle = 2πr

= $2 \times \frac{22}{7} \times 7$

= 44 cm.

Hence, circumference of circle = 44 cm.

Area of square = (side)²

Given,

Area of square = 121 cm²

∴ (side)² = 121

⇒ (side)² = (11)²

⇒ side = 11 cm.

Perimeter of square = 4 × side = 4 × 11 = 44 cm.

Circumference of the circle of same wire = Perimeter of square of same wire

Let radius of circle formed = r cm.

∴ 2πr = 44

⇒ r = $\frac{44}{2π} = \frac{44}{2 \times \frac{22}{7}} = \frac{44 \times 7}{44}$

= 7 cm.

Area of circle = πr²

= $\frac{22}{7} \times (7)^2 = 22 \times 7$

= 154 cm².

Hence, area of circle = 154 cm².

Area of equilateral triangle = $\frac{\sqrt{3}}{4}$ (side)²

Given,

Area of equilateral triangle = $121\sqrt{3}$ cm²

$$\begin{aligned}\therefore \frac{\sqrt{3}}{4}\text{ (side)}^2 = 121\sqrt{3} \\[1em] \Rightarrow \text{(side)}^2 = \frac{121\sqrt{3} \times 4}{\sqrt{3}} \\[1em] \Rightarrow \text{(side)}^2 = 484 \\[1em] \Rightarrow \text{(side)}^2 = 22^2 \\[1em] \Rightarrow \text{ side } = 22 \text{ cm}.\end{aligned}$$

Perimeter of equilateral triangle = 3 x side
= 3 x 22 = 66 cm.

Circumference of the circle of same wire = Perimeter of triangle of same wire

Let radius of circle formed = r cm.

∴ 2πr = 66

$$\begin{aligned}\Rightarrow 2 \times \frac{22}{7} \times r = 66 \\[1em] \Rightarrow r = \frac{66 \times 7}{2 \times 22} \\[1em] \Rightarrow r = \frac{21}{2} \text{ cm}^2\end{aligned}$$

Area of circle = πr²

= $\frac{22}{7} \times \frac{21}{2} \times \frac{21}{2}$

= 346.5 cm².

Hence, area of circle = 346.5 cm².

Let area of larger circle be r cm.

Radius of circle with diameter 7 cm = $\frac{\text{Diameter}}{2} = \frac{7}{2}$ = 3.5 cm.

Given,

Area of larger circle is 16 times the area of the circle with diameter 7 cm.

⇒ πr² = 16 x π x (3.5)²

⇒ r² = 196

⇒ r = $\sqrt{196}$ = 14 cm.

Circumference = 2πr = $2 \times \frac{22}{7} \times 14$ = 88 cm.

Hence, circumference of circle = 88 cm.

Let ABCD be a rectangle.

From figure,

AB = CD = 6 cm and,

AD = BC = 15 cm.

Area of rectangle = l × b = AD × AB = 15 × 6 = 90 cm².

Area of shaded portion = π(3)² + π(3)² + $\frac{π3^2}{2}$

$$\begin{aligned}= π\Big(9 + 9 + \frac{9}{2}\Big) \\[1em] = \frac{45}{2}π \\[1em] = 22.5 \times 3.14 \\[1em] = 70.65 \text{ cm}^2.\end{aligned}$$

Area of unshaded region = Area of rectangle - Area of shaded region

= 90 - 70.65 = 19.35 cm².

Hence, area of unshaded region = 19.35 cm².

Area of square = (side)² = 21² = 441 cm².

We know that,

Diagonals of a square divide it into four triangles of equal area.

Area of a triangle = $\frac{\text{Area of square}}{4} = \frac{441}{4}$ = 110.25 cm².

Area of △AOD + Area of △BOC = 110.25 + 110.25 = 220.50 cm²

Diameter of each semicircle = 21 cm

∴ Radius = $\frac{21}{2}$ = 10.5 cm.

Area of each semicircle = $\frac{πr^2}{2}$

$$\begin{aligned}= \frac{\frac{22}{7} \times 10.5 \times 10.5}{2} \\[1em] = \frac{22 \times 10.5 \times 10.5}{14} \\[1em] = \frac{2425.5}{14} \\[1em] = 173.25 \text{ cm}^2\end{aligned}$$

From figure,

Area of shaded region = Area of both semicircles + Area of △AOD + Area of △BOC

= (2 × 173.25) + 220.50

= 346.50 + 220.50

= 567 cm².

Hence, area of shaded region = 567 cm².

Area of square = (side)² = 14² = 196 cm².

From figure,

Diameter of both semicircle = side of square = 14 cm.

∴ Radius (r) = $\frac{14}{2}$ = 7 cm.

Area of both the semi-circles = $2 \times \frac{πr^2}{2}$

$$\begin{aligned}= 2 \times \frac{\frac{22}{7} \times 7^2}{2} \\[1em] = 2 \times \frac{22 \times 49}{7 \times 2} \\[1em] = 22 \times 7 \\[1em] = 154 \text{ cm}^2.\end{aligned}$$

Area of shaded region = Area of square - Area of semi-circles

= 196 - 154 = 42 cm².

From figure,

Perimeter = arc DPA + DC + arc CPB + AB

= πr + 14 + πr + 14

= 2πr + 28

= $2 \times \frac{22}{7} \times 7$ + 28

= 44 + 28 = 72 cm.

Hence, area of shaded region = 42 cm² and perimeter = 72 cm.

Let radius of each circle be r cm.

From figure,

⇒ r + r + r + r = 14

⇒ 4r = 14

⇒ r = 3.5 cm.

Since, radius is same so, each circle will have same area.

Area of each circle = πr²

= $\frac{22}{7} \times (3.5)^2$

= $\frac{22}{7} \times 3.5 \times 3.5$

= 22 × 0.5 × 3.5

= 38.5 cm²

Area of square = (side)² = 14² = 196 cm².

Area of shaded region = Area of square - Area of circles

= 196 - 4 × 38.5

= 196 - 154

= 42 cm².

Hence, area of shaded region = 42 cm².

From figure,

BD = 14 cm (Diameter of semi-circle)

AF = FE = x (let)

⇒ BD = AF + FE

⇒ 14 = x + x

⇒ 2x = 14

⇒ x = 7 cm.

Area of semi-circle BCD = $\frac{πr^2}{2}$

$$\begin{aligned}= \frac{\frac{22}{7} \times (7)^2}{2} \\[1em] = \frac{22 \times 7}{2} \\[1em] = 77 \text{ cm}^2.\end{aligned}$$

Area of quadrant ABF = Area of quadrant EDF = $\frac{πr^2}{4}$

$$\begin{aligned}= \frac{\frac{22}{7} \times (7)^2}{4} \\[1em] = \frac{154}{4} \\[1em] = 38.5 \text{ cm}^2.\end{aligned}$$

From figure, AB = ED = AF = FE = 7 cm.

Area of rectangle ABDE= AB × BD = 7 × 14 = 98 cm².

Area of shaded region = Area of rectangle ABDE + Area of semi-circle BCD - Area of quadrant ABF - Area of quadrant EDF

= 98 + 77 - 38.5 - 38.5

= 98 cm².

Hence, area of shaded region = 98 cm².

From figure,

Radius of larger semi-circle (R) = 14 cm.

Area of larger semi-circle = $\frac{πR^2}{2} = \frac{22}{7} \times (14)^2 \times \frac{1}{2}$

$= \frac{22}{7} \times 196 \times \frac{1}{2}$

= 22 × 14

= 308 cm².

Diameter of smaller semi-circle = 14 cm; radius (r) = $\frac{14}{2}$ = 7 cm.

Area of smaller semi-circle = $\frac{πr^2}{2} = \frac{22}{7} \times (7)^2 \times \frac{1}{2}$

$= \frac{22}{7} \times 49 \times \frac{1}{2}$

= 11 × 7

= 77 cm².

From figure,

Area of shaded region = Area of larger semi-circle - Area of smaller semi-circle

= 308 - 77

= 231 cm².

From figure,

Perimeter of shaded region = Circumference of larger semi-circle + Circumference of smaller circle + 14

= πR + πr + 14

= $\frac{22}{7} \times 14 + \frac{22}{7} \times 7 + 14$

= 44 + 22 + 14

= 80 cm.

Hence, area of shaded region = 231 cm² and perimeter = 80 cm.

From figure,

Area of △ABC = $\frac{1}{2} \times$ base × height

Substituting values we get,

⇒ $\frac{1}{2} \times$ AB × CD = 35

⇒ $\frac{1}{2} \times$ AB × 5 = 35

⇒ AB = $\frac{35 \times 2}{5}$ = 14 cm.

From figure,

AB is the diameter of semicircle.

Radius = $\frac{\text{Diameter}}{2} = \frac{14}{2}$ = 7 cm.

Area of semi-circle = $\frac{πr^2}{2} = \frac{22}{7} \times (7)^2 \times \frac{1}{2}$

$= \frac{22}{7} \times 49 \times \frac{1}{2}$

= 11 × 7

= 77 cm².

Area of shaded region = Area of semi-circle - Area of △ABC

= 77 - 35

= 42 cm².

Hence, area of shaded region = 42 cm².

Area of quadrant = $\frac{πr^2}{4}$

$$\begin{aligned}= \frac{3.14 \times 10^2}{4} \\[1em] = \frac{314}{4} \\[1em] = 78.5 \text{ m}^2.\end{aligned}$$

Area of triangle AOB = $\frac{1}{2} \times OB \times AO$

$$\begin{aligned}= \frac{1}{2} \times 10 \times 10 \\[1em] = \frac{1}{2} \times 100 \\[1em] = 50 \text{ m}^2\end{aligned}$$

Area of shaded region = Area of quadrant - Area of triangle

= 78.5 - 50

= 28.5 ≈ 29 m².

Hence, area of shaded region = 29 m².

Area of quadrant = $\frac{πr^2}{4}$

$$\begin{aligned}= \frac{\frac{22}{7} \times 7^2}{4} \\[1em] = \frac{154}{4} \\[1em] = 38.5 \text{ cm}^2.\end{aligned}$$

Area of triangle AOD = $\frac{1}{2} \times OA \times OD$

$$\begin{aligned}= \frac{1}{2} \times 7 \times 4 \\[1em] = \frac{1}{2} \times 28 \\[1em] = 14 \text{ cm}^2\end{aligned}$$

Area of shaded region = Area of quadrant - Area of triangle

= 38.5 - 14

= 24.5 cm².

Hence, area of shaded region = 24.5 cm².

From figure,

Let ABCD be the rectangular piece.

Area of rectangular piece = 30 × 21 = 630 cm²

From figure,

Radius of the biggest circle that can be cut from the rectangular piece = $\frac{21}{2}$ = 10.5 cm

Area of circle = πr²

= $\frac{22}{7} \times (10.5)^2$

= $\frac{22}{7} \times 110.25$

= 22 × 15.75

= 346.5 cm².

Area of paper left = Area of rectangular piece - Area of circle

= 630 - 346.5

= 283.5 cm².

Hence, the area of the biggest circle that can be cut from the rectangular piece = 346.5 cm² and area of remaining paper = 283.5 cm².

Let ABCD be a rectangle with AB = 4 cm.

From figure,

Diameter of circle, AC = AO + OC = 2.5 + 2.5 = 5 cm.

In right angle triangle ABC,

⇒ AC² = AB² + BC²

⇒ 5² = 4² + BC²

⇒ BC² = 5² - 4²

⇒ BC² = 25 - 16 = 9

⇒ BC = $\sqrt{9}$ = 3 cm.

By formula,

Area of rectangle = l × b

= AB × BC

= 4 × 3

= 12 cm².

Hence, area of rectangle = 12 cm².

From figure,

O is the center. In right angle triangle ABC,

Using pythagoras theorem,

⇒ AC² = AB² + BC²

⇒ AC² = 5² + 12²

⇒ AC² = 25 + 144 = 169

⇒ AC = $\sqrt{169}$ = 13 cm.

From figure,

AC is the diameter and OA is the radius = $\frac{13}{2}$ = 6.5 cm.

Area of circle = πr²

= 3.142 × (6.5)²

= 3.142 × 42.25

= 132.75 cm².

Area of rectangle = l × b

= 12 × 5 = 60 cm².

Area of shaded region = Area of circle - Area of rectangle

= 132.75 - 60

= 72.75 cm².

Hence, area of shaded region = 72.75 cm².

By formula,

Area of △ABC = $\frac{1}{2}$ × base × height

= $\frac{1}{2}$ × BC × AB

= $\frac{1}{2}$ × 7 × 7

= $\frac{49}{2}$ = 24.5 cm².

In right angle triangle,

Using pythagoras theorem,

⇒ AC² = AB² + BC²

⇒ AC² = 7² + 7²

⇒ AC² = 49 + 49 = 98

⇒ AC = $\sqrt{98} = 7\sqrt{2}$ cm.

From figure,

Radius of semi-circle (r) = $\frac{AC}{2} = \frac{7\sqrt{2}}{2}$

By formula,

Area of semi-circle = $\frac{πr^2}{2}$

$$\begin{aligned}= \frac{1}{2} \times \frac{22}{7} \times (\frac{7\sqrt{2}}{2})^2 \\[1em] = \frac{1}{2} \times \frac{22}{7} \times \frac{98}{4} \\[1em] = \frac{2156}{56} = 38.5 \text{ cm}^2.\end{aligned}$$

Area of the shaded region = Area of the semi-circle – Area of △ABC

= 38.5 - 24.5

= 14 cm².

Hence, area of the shaded region = 14 cm².

By formula,

Perimeter = 2πr

$$\begin{aligned}\Rightarrow 2πr = 660 \\[1em] \Rightarrow πr = 330 \\[1em] \Rightarrow r = \frac{330}{π} = \frac{330}{\frac{22}{7}} \\[1em] \Rightarrow r = \frac{330 \times 7}{22} \\[1em] \Rightarrow r = \frac{30 \times 7}{2} \\[1em] \Rightarrow r = 15 \times 7 = 105 \text{ m}.\end{aligned}$$

From figure,

BD = BO + OD = r + r = 2r = 2 × 105 = 210 m.

Let O be the center of the circle and ABCD be square of side x metres.

Area of square = (side)² = x².

In right angle triangle BCD,

⇒ BD² = BC² + CD²

⇒ 210² = x² + x²

⇒ 2x² = 44100

⇒ x² = $\frac{44100}{2}$ = 22050 m².

Hence, area of square = 22050 m².

(i) Let side of the square be 2a units.

From figure,

AD = Diameter of incircle.

Radius of incircle (r) = $\frac{\text{Diameter}}{2} = \frac{2a}{2}$ = a units.

In right angle triangle ABC,

Using pythagoras theorem,

AC² = AB² + BC²

AC² = (2a)² + (2a)²

AC² = 4a² + 4a²

AC² = 8a²

AC = $\sqrt{8a^2} = 2\sqrt{2}a$ units.

From figure,

AC is the diameter of circumcircle and AO is radius.

AO (R) = $\frac{\text{Diameter}}{2} = \frac{2\sqrt{2}a}{2} = \sqrt{2}$a units.

$$\begin{aligned}\text{Ratio between circumference } = \frac{\text{Circumference of incircle}}{\text{Circumference of circumcircle}} \\[1em] = \frac{πr}{πR} \\[1em] = \frac{r}{R} \\[1em] = \frac{a}{\sqrt{2}a} \\[1em] = \frac{1}{\sqrt{2}} = 1 : \sqrt{2}.\end{aligned}$$

Hence, ratio between circumferences = $1 : \sqrt{2}$.

(ii)

$$\begin{aligned}\text{Ratio between areas } = \frac{\text{Area of incircle}}{\text{Area of circumcircle}} \\[1em] = \frac{πr^2}{πR^2} \\[1em] = \frac{r^2}{R^2} \\[1em] = \frac{a^2}{(\sqrt{2}a)^2} \\[1em] = \frac{a^2}{2a^2} = 1 : 2.\end{aligned}$$

Hence, ratio between areas = 1 : 2.

Given,

PQ = 200 m and PT = 70 m.

(i) By formula,

Area of rectangle PQST = l × b

= 200 × 70

= 14000 m².

Radius of each semi-circular part on either side of rectangle = $\frac{PT}{2} = \frac{70}{2}$ = 35 m.

Area of both semi-circular parts = 2 × $\frac{πr^2}{2}$ = πr²

= $\frac{22}{7} \times 35 \times 35$

= 22 × 5 × 35

= 3850 m².

So, the total area of grassed enclosure = 14000 + 3850 = 17850 m².

Hence, area of glassed enclosure = 17850 m².

(ii) Given,

Width of track around the enclosure = 7 m.

From figure,

AB = PQ = 200 m

ED = ST = 200 m

EA = PT + ET + AP = 70 + 7 + 7 = 84 m

BD = DS + QS + BQ = 70 + 7 + 7 = 84 m

Outer radius of semi-circle (R) = $\frac{EA}{2} = \frac{84}{2}$ = 42 m.

Circumference of both semi-circular part = πR + πR = 2πR.

= $2 \times \frac{22}{7} \times 42$

= 264 m.

From figure,

Outer perimeter = Circumference of both semi-circular part + ED + AB

= 264 + 200 + 200

= 664 m.

Hence, perimeter of outer track ABCDEF = 664 m.

Given,

Perimeter of inside semi-circular track = 312 m.

⇒ 90 + πr + 90 + πr = 312

⇒ 2πr + 180 = 312

⇒ 2πr = 312 - 180

⇒ 2πr = 132

⇒ πr = $\frac{132}{2}$

⇒ πr = 66

⇒ r = $\frac{66}{π} = \frac{66}{\frac{22}{7}} = \frac{66 \times 7}{22} = 21$ m.

So, length of AB = 2r = 2 × 21 = 42 m.

Since, width of track = 2 m.

So, HE = GF = 42 + 2 + 2 = 46 m.

Radius of outer semi-circle (R) = $\frac{46}{2}$ = 23 m.

From figure,

Area of track = Area of outer semi-circle (with diametre HE) + Area of outer semi-circle( with diametre GF) + Area of outer rectangle (EFGH) - [Area of inner semi-circle (with diameter AB) + Area of inner semi-circle (with diameter DC) + Area of inner rectangle ABCD]

$$\begin{aligned}= \frac{πR^2}{2} + \frac{πR^2}{2} + HG \times HE - (\frac{πr^2}{2} + \frac{πr^2}{2} + AB \times BC) \\[1em] = πR^2 + 90 \times 46 - (πr^2 + 42 \times 90) \\[1em] = πR^2 - πr^2 + 90 \times 46 - 90 \times 42 \\[1em] = π(R^2 - r^2) + 90 \times (46 - 42) \\[1em] = \frac{22}{7} \times (23^2 - 21^2) + 90 \times 4 \\[1em] = \frac{22}{7} \times (529 - 441) + 360 \\[1em] = \frac{22 \times 88}{7} + 360 \\[1em] = \frac{1936}{7} + 360 \\[1em] = \frac{1936 + 2520}{7} \\[1em] = \frac{4456}{7} \\[1em] = 636\frac{4}{7} \text{ m}^2.\end{aligned}$$

Hence, area of semi-circular track = $636\frac{4}{7}$ m².

Radius of circle with center A = AC = 8 cm.

Area of circle with center A = πr²

= $\frac{22}{7} \times 8^2 = \frac{22 \times 64}{7}$

= $\frac{1408}{7}$ = 201.14 cm².

From figure,

BC = AC - AB = 8 - 3 = 5 cm.

Area of circle with center B = πr²

= $\frac{22}{7} \times 5^2 = \frac{22 \times 25}{7}$

= $\frac{550}{7}$ = 78.57 cm².

Area of shaded region = Area of circle with center A - Area of circle with center B

= 201.14 - 78.57

= 122.57 cm².

Hence, area of shaded region = 122.57 cm².

Given,

Radius of each quadrant = 7 cm

From figure,

Area of shaded region = Area of square – 4 area of each quadrant

= (side)² – 4 × $\frac{πr^2}{4}$

= 14² – $4 \times \frac{22}{7} \times 7 \times 7 \times \frac{1}{4}$

= 196 – 154

= 42 cm².

Hence, area of shaded region = 42 cm².

Area of lawn ABCD = (side)²

= (56)² = 3136 m².

In square,

Length of diagonal (AC) = $\sqrt{2}$ side = $56\sqrt{2}$ m.

Since, diagonals of square are equal and bisect each other,

∴ AO = OC = OD = OB.

Radius of quadrant ODC = Radius of quadrant OAB = $\frac{AC}{2} = \frac{56\sqrt{2}}{2} = 28\sqrt{2}$ m.

Area of quadrant ODC = Area of quadrant OAB = $\frac{1}{4}πr^2$

$$\begin{aligned}= \frac{1}{4} \times \frac{22}{7} \times 28\sqrt{2} \times 28\sqrt{2} \\[1em] = 22 \times 28 \times 2 \\[1em] = 1232 \text{ m}^2.\end{aligned}$$

Since, diagonals of square divide it into four equal triangles.

Area of △ODC = Area of △OAB = $\frac{\text{Area of square}}{4} = \frac{3136}{4}$ = 784 m².

From figure,

Area of flower bed = Area of quadrant ODC - Area of △ODC

= 1232 - 784 = 448 m².

Since there are two flower beds so area = 2 × 448 = 896 m².

Area of lawn and flower beds = 3136 + 896 = 4032 m².

Hence, sum of the areas of the lawn and the flower beds = 4032 m².

Given, length of each side of square = 20 cm.

We know that,

Length of diagonal of a square = $\sqrt{2}$ side = $20\sqrt{2}$ cm.

∴ OB = $20\sqrt{2}$ cm.

From figure,

OB is the radius of the quadrant OPBQ.

Area of quadrant OPBQ = $\frac{1}{4}πr^2$

$$\begin{aligned}= \frac{1}{4} \times 3.14 \times 20\sqrt{2} \times 20\sqrt{2} \\[1em] = 3.14 \times 200 \\[1em] = 628 \text{ cm}^2.\end{aligned}$$

Area of square OABC = (side)²

= (20)² = 400 cm².

Area of shaded region = Area of quadrant OPBQ - Area of square OABC

= 628 - 400

= 228 cm².

Hence, area of shaded region = 228 cm².

Since, ABCD is a rectangle.

AD = BC = 7 cm.

CD = AB = 14 cm.

From figure,

AD and BC are diameters of smaller circles.

Radius = $\frac{7}{2}$ = 3.5 cm.

Area of each small semi-circle = $\frac{πr^2}{2}$

= $\frac{22}{7} \times (3.5)^2 \times \frac{1}{2}$

= 19.25 cm².

From figure,

CD is the diameter of the larger semi-circle.

Radius = $\frac{14}{2}$ = 7 cm.

Area of larger semi-circle = $\frac{πr^2}{2}$

= $\frac{22}{7} \times (7)^2 \times \frac{1}{2}$

= 77 cm².

Area of rectangle ABCD = AB × BC

= 14 × 7

= 98 cm².

From figure,

Area of shaded region = Area of rectangle ABCD + 2 × Area of each smaller semi-circle - Area of larger semi-circle

= 98 + (2 × 19.25) - 77

= 98 + 38.5 - 77

= 59.5 cm².

Hence, area of shaded region = 59.5 cm².

We know that,

Angle in semi-circle = 90°.

∴ ∠A = 90°.

In right angle △ABC

Using Pythagoras theorem,

⇒ BC² = AC² + AB²

⇒ BC² = 24² + 7²

⇒ BC² = (576 + 49) = 625

⇒ BC = $\sqrt{625}$ = 25 cm.

From figure,

Radius of circle (OB) = $\frac{BC}{2} = \frac{25}{2}$ = 12.5 cm.

By formula,

Area of △ABC = $\frac{1}{2}$ × AB × AC

= $\frac{1}{2}$ × 7 × 24

= 84 cm².

Area of circle = πr²

= 3.14 × 12.5 × 12.5

= 490.63 cm².

Area of quadrant COD = $\frac{1}{4}πr^2 = \frac{1}{4} \times 490.63$ = 122.66 cm².

Area of shaded portion = Area of circle – (Area of △ABC + Area of quadrant COD)

= 490.63 – (84 + 122.66)

= 490.63 – 206.66

= 283.97 cm².

Hence, area of shaded portion = 283.97 cm².

Let r cm be the radius of each circle.

From figure,

⇒ r + r = AD

⇒ 2r = 14

⇒ r = 7 cm.

Hence, radius of each circle = 7 cm.

Area of each circle = πr²

= $\frac{22}{7} \times (7)^2$

= $\frac{22}{7} \times 7 \times 7$

= 22 x 7

= 154 cm².

Area of 4 circles = 4 × 154 = 616 cm².

From figure,

Radius of each quadrant in square ABCD = 7 cm.

Area of each quadrant = $\frac{1}{4}πr^2$

= $\frac{1}{4} \times \frac{22}{7} \times 7^2$

= 38.5 cm².

Area of 4 quadrants = 4 × 38.5 = 154 cm².

Area of square = (side)²

= (14)² = 196 cm².

Area of shaded region = (Area of 4 circles - Area of 4 quadrants) + (Area of square - Area of 4 quadrants)

= (616 - 154) + (196 - 154)

= 462 + 42

= 504 cm².

Hence, area of shaded region = 504 cm².

From figure,

Diameter of big semi-circle = 10 cm.

Radius of big semi-circle (R) = $\frac{10}{2}$ = 5 cm,

Diameter of small semi-circle = 5 cm.

Radius of each smaller semi-circle (r) = $\frac{5}{2}$ = 2.5 cm.

(i) Length of boundary = Circumference of bigger semi-circle + 2 x circumference of smaller semi-circles

= πR + πr + πr

= π(R + 2r)

= 3.14(5 + 2 × $\frac{5}{2}$)

= 3.14(5 + 5)

= 3.14 × 10

= 31.4 cm.

Hence, length of boundary = 31.4 cm.

(ii) From figure,

Area of shaded region = Area of bigger semi-circle + Area of one smaller semi-circle – Area of other smaller semi-circle

$$\begin{aligned}= \frac{πR^2}{2} + \frac{πr^2}{2} - \frac{πr^2}{2} \\[1em] = \frac{πR^2}{2} \\[1em] = \frac{3.14 \times 5^2}{2} \\[1em] = \frac{3.14 \times 25}{2} \\[1em] = 1.57 \times 25 \\[1em] = 39.25 \text{ cm}^2.\end{aligned}$$

Hence, area of shaded region = 39.25 cm².

Let r₁ = 5 cm, r₂ = 3 cm and r₃ = 2 cm.

Perimeter of shaded region = Circumference of largest semi-circle + Circumference of smaller semi-circle + Circumference of smallest semi-circle

= πr₁ + πr₂ + πr₃

= π(5 + 3 + 2)

= 10π

= 10 x 3.14

= 31.4 cm.

Area of shaded region = Area of largest semi-circle - Area of smaller semi-circle + Area of smallest semi-circle

$$\begin{aligned}= \frac{πr_1^2}{2} - \frac{πr_2^2}{2} + \frac{πr_3^2}{2} \\[1em] = π\Big(\frac{r_1^2}{2} - \frac{r_2^2}{2} + \frac{r_3^2}{2}\Big) \\[1em] = 3.14\Big(\frac{5^2}{2} - \frac{3^2}{2} + \frac{2^2}{2}\Big) \\[1em] = 3.14\Big(\frac{25}{2} - \frac{9}{2} + \frac{4}{2}\Big) \\[1em] = 3.14\Big(\frac{25 - 9 + 4}{2} \Big) \\[1em] = 3.14\Big(\frac{20}{2} \Big) \\[1em] = 3.14\times 10 \\[1em] = 31.4 \text{ cm}^2.\end{aligned}$$

Hence, perimeter of shaded region = 31.4 cm and area of shaded region = 31.4 cm².

We know that,

Side of square ABCD = 4 cm

Radius of each quadrant circle (r) = 1 cm

Given,

Diameter of circle in the center = 2 cm

∴ Radius of circle in center (r₁) = $\frac{2}{2}$ = 1 cm.

From figure,

Area of shaded region = Area of square – Area of 4 quadrants – Area of circle at center

$$\begin{aligned}= \text{ side}^2 - 4 \times \frac{πr^2}{4} - πr_1^2 \\[1em] = 4^2 - πr^2 - πr_1^2 \\[1em] = 16 - 3.14(1)^2 - 3.14(1)^2 \\[1em] = 16 - 3.14 - 3.14 \\[1em] = 16 - 6.28 \\[1em] = 9.72 \text{ cm}^2.\end{aligned}$$

Hence, area of shaded region = 9.72 cm².

Area of rectangle = AB × BC = 14 × 7 = 98 cm².

Since, ABCD is a rectangle.

∴ CD = AB = 14 cm.

BFEC is a quadrant of radius, BC = r₁ = 7 cm.

∴ CE = 7 cm.

From figure,

DE = CD - CE = 14 - 7 = 7 cm.

From figure,

DE is the diameter of semi-circle DGE.

So, radius (r) = $\frac{DE}{2} = \frac{7}{2}$ = 3.5 cm.

Area of semi-circle = $\frac{πr^2}{2}$

$$\begin{aligned}= \frac{\frac{22}{7} \times (3.5)^2}{2} \\[1em] = \frac{22 \times 12.25}{14} \\[1em] = \frac{269.5}{14} \\[1em] = 19.25 \text{ cm}^2.\end{aligned}$$

Area of quadrant BFEC = $\frac{πr_1^2}{4}$

$$\begin{aligned}= \frac{\frac{22}{7} \times 7^2}{4} \\[1em] = \frac{22 \times 7}{4} \\[1em] = \frac{154}{4} = 38.5 \text{ cm}^2.\end{aligned}$$

Area of remaining piece of rectangle = Area of rectangle - Area of semicircle DGE - Area of quadrant BFEC

= 98 - 19.25 - 38.5

= 40.25 cm².

Hence, area of remaining piece of rectangle = 40.25 cm².

From figure,

∠ECF = ∠BCD = 90°.

Area of right angle △ECF = $\frac{1}{2} \times$ CF × EC

= $\frac{1}{2} \times 6 \times 6$

= 18 cm².

Area of quadrant BCD = $\frac{πr^2}{4}$

$$\begin{aligned}= \frac{\frac{22}{7} \times 42^2}{4} \\[1em] = \frac{22 \times 42 \times 42}{7 \times 4} \\[1em] = \frac{38808}{28} \\[1em] = 1386 \text{ cm}^2.\end{aligned}$$

Area of shaded region = Area of quadrant BCD + Area of right angle △ECF

= 18 + 1386 = 1404 cm².

Hence, area of shaded region = 1404 cm².

(i) From figure,

OA = $\frac{AD}{2} = \frac{14}{2}$ = 7 cm.

OB = OA - AB = 7 - 3.5 = 3.5 cm.

Radius of smallest semi-circle = $\frac{3.5}{2}$ = 1.75 cm.

Circumference of semi-circle = πr.

Length of boundary = Circumference of largest semi-circle + Circumference of smaller semi-circle + 2 × Circumference of smallest semi-circle

= 7π + 3.5π + (2 x 1.75π)

= 7π + 3.5π + 3.5π

= 14π

= $14 \times \frac{22}{7}$

= 2 × 22

= 44 cm.

Hence, length of boundary = 44 cm.

(ii) Area of shaded region = Area of large semi-circle + Area of smaller semi-circle - 2 × Area of smallest semi-circle

$$\begin{aligned}= \frac{π(7)^2}{2} + \frac{π(3.5)^2}{2} - 2 \times \frac{π(1.75)^2}{2} \\[1em] = \frac{49π}{2} + \frac{12.25π}{2} - \frac{6.125π}{2} \\[1em] = \frac{49π + 12.25π - 6.125π}{2} \\[1em] = \frac{55.125π}{2} \\[1em] = \frac{55.125 \times \frac{22}{7}}{2} \\[1em] = \frac{1212.75}{14} \\[1em] = 86.625 \text{ cm}^2.\end{aligned}$$

Hence, area of shaded region = 86.625 cm².

From figure,

Radius of quadrant = BC = 3.5 cm.

EC = BC = 3.5 cm (As both equal to radius of quadrant BFEC)

By formula,

Area of trapezium = $\frac{1}{2}$ × (Sum of || sides) × distance between them

= $\frac{1}{2}$ × (AB + DC) × BC

= $\frac{1}{2}$ × (AB + EC + DE) × BC

= $\frac{1}{2}$ × (3.5 + 3.5 + 2) × 3.5

= $\frac{1}{2}$ × 9 × 3.5

= 4.5 × 3.5

= 15.75 cm².

So, the area of quadrant BCEF = $\frac{πr^2}{4}$

$$\begin{aligned}= \frac{\frac{22}{7} \times (3.5)^2}{4} \\[1em] = \frac{\frac{22}{7} \times 12.25}{4} \\[1em] = \frac{22 \times 12.25}{4 \times 7} \\[1em] = \frac{269.5}{28} \\[1em] = 9.625 \text{ cm}^2\end{aligned}$$

Area of shaded portion = Area of trapezium - Area of quadrant

= 15.75 – 9.625 = 6.125 cm².

Hence, area of shaded portion = 6.125 cm².

In right angle △ABC,

Using Pythagoras theorem,

⇒ AC² = AB² + BC²

⇒ AC² = 28² + 21²

⇒ AC² = 784 + 441

⇒ AC² = 1225

⇒ AC = $\sqrt{1225}$ = 35 cm.

Radius of semi-circle (R) = $\frac{AC}{2} = \frac{35}{2}$ = 17.5 cm.

Radius of quadrant (r) = BC = 21 cm.

From figure,

Area of shaded region = Area of △ABC + Area of semi-circle – Area of quadrant

$$\begin{aligned}= \frac{1}{2} \times BC \times AB + \frac{πR^2}{2} - \frac{πr^2}{4}\\[1em] = \frac{1}{2} \times 21 \times 28 + \frac{\frac{22}{7} \times (17.5)^2}{2} - \frac{\frac{22}{7} \times (21)^2}{4} \\[1em] = 21 \times 14 + \frac{22 \times 306.25}{7 \times 2} - \frac{22 \times 441}{7 \times 4} \\[1em] = 294 + \frac{6737.5}{14} - \frac{9702}{28} \\[1em] = 294 + 481.25 - 346.5 \\[1em] = 428.75 \text{ cm}^2.\end{aligned}$$

Hence, area of shaded region = 428.75 cm².

We know that

△ABC is an equilateral triangle of side 8 cm

A, B, C are the centres of three circular arcs of equal radius

Radius = $\frac{8}{2}$ = 4 cm

By formula,

Area of △ABC = $\frac{\sqrt{3}}{4} \text{ side}^2$

= $\frac{\sqrt{3}}{4}$ × 8 × 8

= $\frac{\sqrt{3}}{4} \times 64$

= $16\sqrt{3}$

= 16 × 1.732

= 27.712 cm².

So, the area of 3 equal sectors of 60° whose radius is 4 cm = 3 × πr² × $\frac{60}{360}$

= 3 × 3.142 × 4 × 4 × $\frac{1}{6}$

= 3.142 × 8

= 25.136 cm².

Area of shaded region = Area of equilateral triangle - Area of 3 sectors

= 27.712 – 25.136

= 2.576 ≈ 2.58 cm².

Hence, area of shaded region = 2.58 cm².

(i) Side of the regular hexagon (a) = $2\sqrt{3}$ cm.

Now since a regular hexagon has 6 sides, hence we can say the central angle of a hexagon = $\frac{360°}{6}$ = 60°.

From figure,

∠AOB = 60°.

Now in the ΔAOB,

Since the total sum of the angles of a triangle is equal to 180°

∴ ∠AOB + ∠OAB + ∠OBA = 180° .......(1)

Since, OA = OB = radius of circle,

So, ∠OAB = ∠OBA = x (let) (Angles opposite to equal sides are equal).

Substituting value in equation 1 we get,

⇒ 60° + x + x = 180°

⇒ 2x + 60° = 180°

⇒ 2x = 120°

⇒ x = 60°.

Now since all the angles of the triangle are equal hence we can say the triangle is an equilateral triangle. So, all the sides of the triangle will also be equal.

∴ AO = BO = AB = $2\sqrt{3}$ cm.

Draw perpendicular from O to AB.

From figure,

AT = BT = $\frac{2\sqrt{3}}{2} = \sqrt{3}$ cm. (As altitude and median are same in equilateral triangle.)

From figure,

In right angle triangle OAT,

$$\begin{aligned}\Rightarrow OA^2 = OT^2 + AT^2 \\[1em] \Rightarrow (2\sqrt{3})^2 = OT^2 + (\sqrt{3})^2 \\[1em] \Rightarrow 12 = OT^2 + 3 \\[1em] \Rightarrow OT^2 = 12 - 3 \\[1em] \Rightarrow OT^2 = 9 \\[1em] \Rightarrow OT = \sqrt{9} = 3 \text{ cm}.\end{aligned}$$

Hence, radius of circle = OT = 3 cm.

We know the circumference of a circle is given by the formula,

Circumference = 2πr

= $2 \times \frac{22}{7} \times 3$

= $\frac{132}{7}$ cm.

Hence, circumference of inscribed circle = $\frac{132}{7}$ cm.

(ii) Area of inscribed circle = πr²

= $\frac{22}{7} \times (3)^2$

= $\frac{22 \times 9}{7} = \frac{198}{7}$ cm².

Hence, area of inscribed circle = $\frac{198}{7}$ cm².

Given,

Radius of the circle = 10 cm

Angle at the centre subtended by a chord AB = 90°.

We know that,

Area of sector OACB = $πr^2 \times \frac{90}{360}$

= 3.14 × 10 × 10 × $\frac{1}{4}$

= $\frac{314}{4}$

= 78.5 cm².

In right angle △OAB,

Area of △OAB = $\frac{1}{2}$ × OA × OB

= $\frac{1}{2}$ × 10 × 10

= 50 cm².

Area of minor segment = Area of sector OACB – Area of △OAB

= 78.5 - 50

= 28.5 cm².

Area of circle = πr²

= 3.14 × 10 × 10

= 314 cm².

Area of major segment = Area of circle – Area of minor segment

= 314 - 28.5

= 285.5 cm².

Hence, area of sector OACB = 78.5 cm² and area of major segment = 285.5 cm².

Given,

Length of edge of cube (a) = 7 cm

By formula,

Surface area of cube = 6a²

= 6 × (7)²

= 6 × 7 × 7

= 294 cm².

Volume of cube = a³

= (7)³

= 7 × 7 × 7

= 343 cm³.

Hence, surface area and volume of cube = 294 cm² and 343 cm³ respectively.

In a rectangular solid,

Let l = 5 m, b = 4 m and h = 3 m

By formula,

Surface area of rectangular solid = 2(lb + bh + lh)

= 2(5 × 4 + 4 × 3 + 5 × 3)

= 2(20 + 12 + 15)

= 2 × 47

= 94 m²

Volume of rectangular solid = l × b × h

= 5 × 4 × 3

= 60 m³.

Diagonal of cuboid = $\sqrt{l^2 + b^2 + h^2}$

$$\begin{aligned}= \sqrt{5^2 + 4^2 + 3^2} \\[1em] = \sqrt{25 + 16 + 9} \\[1em] = \sqrt{50} \\[1em] = 5\sqrt{2} \\[1em] = 5 \times 1.414 \\[1em] = 7.07 \text{ m}.\end{aligned}$$

Hence, surface area = 94 m², volume = 60 m³ and the length of diagonal is 7.07 m.

Given,

Length of rectangular solid = 25 cm

Breadth of rectangular solid = 20 cm

Volume of rectangular solid = 7000 cm³

Let height of rectangular solid = h cm.

By formula,

Volume = l × b × h

Substituting the values we get,

⇒ 7000 = 25 × 20 × h

⇒ 500 × h = 7000

⇒ h = $\frac{7000}{500}$

⇒ h = 14 cm.

Hence, height of rectangular solid is 14 cm.

The given dimensions of class room are

Length (l) = 10 m

Breadth (b) = 6 m

Height (h) = 4 m

We know that,

Floor area of class room = l × b = 10 × 6 = 60 m².

Given,

One student needs 1.5 m² floor area.

So, the number of students = $\frac{60}{1.5}$ = 40.

By formula,

Volume of class room = l × b × h

= 10 × 6 × 4

= 240 m³.

Cubic metres of air for each student = $\frac{\text{ Vol. of classroom }}{\text{ No. of students }}$

= $\frac{240}{40}$

= 6 m³.

Hence, the classroom can accommodate 40 students and each student will have 6 m³ of air.

(a) Given,

Volume of cuboid = 1440 cm³

Height of cuboid = 10 cm.

Given, cross section is a square.

∴ length = breadth = x cm (let).

By formula,

Volume of cuboid = area of square × height

Substituting the values we get,

⇒ 1440 = $x$ × $x$ × 10

⇒ $x^2$ = $\frac{1440}{10}$

⇒ $x^2$ = 144

⇒ $x$ = $\sqrt{144}$ = 12 cm.

Hence, the side of square is 12 cm.

Given,

Perimeter of one face of a cube = 20 cm

Perimeter of one face of a cube = 4 × side (As it is a square)

⇒ 20 = 4 × side

⇒ Side = $\frac{20}{4}$ = 5 cm

Surface area of cube = 6(side)²

= 6(5)²

= 6 × 5 × 5 = 150 cm²

Volume of cube = side × side × side

= 5 × 5 × 5

= 125 cm³.

Hence, surface area of cube = 150 cm² and volume = 125 cm³.

By formula,

Surface area of the cuboid (box) = 2(lh + bh + hl)

= 2(80 × 40 + 40 × 20 + 20 × 80)

= 2(3200 + 800 + 1600)

= 2 × 5600

= 11200 cm².

Area of square sheet = (side)²

= 40²

= 1600 cm²

Let no. of sheets required to cover be n.

So, total area of square sheets = 1600n cm².

In order to cover the box with square sheets, their areas must be equal.

∴ 1600n = 11200

⇒ n = $\frac{11200}{1600}$ = 7.

Hence, 7 square sheets will be required.

Given,

Volume of a cuboid = 3600 cm³

Height of cuboid = 12 cm

Cross section is a rectangle with length and breadth in ratio 4 : 3.

Let length = 4x cm and breadth = 3x cm.

By formula,

Volume of cuboid = length × breadth × height

⇒ 3600 = 4x × 3x × 12

⇒ 144x² = 3600

⇒ x² = $\frac{3600}{144}$

⇒ x² = 25

⇒ x = $\sqrt{25}$ = 5 cm.

So.

Length of rectangle = 4x = 4 × 5 = 20 cm

Breadth of rectangle = 3x = 3 × 5 = 15 cm

Perimeter of the cross section = 2(l + b)

= 2(20 + 15)

= 2 × 35

= 70 cm.

Hence, perimeter of cross-section = 70 cm.

Given,

Volume of a cube = 729 cm³.

By formula,

Volume of a cube = (side)³

∴ (side)³ = 729

⇒ (side)³ = (9)³

⇒ side = 9 cm

By formula,

Surface area of cube = 6(side)²

= 6 × (9)²

= 6 × 9 × 9

= 486 cm².

So the length of a diagonal = $\sqrt{3}$ × side

= $\sqrt{3}$ × 9

= 1.732 × 9

= 15.57 cm.

Hence, surface area = 486 cm² and length of diagonal = 15.57 cm.

The longest rod which can be kept inside a rectangular box will be equal to the diagonal of the box.

Let h cm be the inner height of box.

By formula,

Length of diagonal = $\sqrt{l^2 + b^2 + h^2}$

∴ 17 = $\sqrt{12^2 + 8^2 + h^2}$

Squaring both sides,

⇒ 17² = 12² + 8² + h²

⇒ 289 = 144 + 64 + h²

⇒ 289 = 208 + h²

⇒ h² = 289 - 208

⇒ h² = 81

⇒ h = $\sqrt{81}$ = 9 cm.

Hence, the inner height of rectangular box is 9 cm.

Given,

Inner length of rectangular box = 90 cm

Inner breadth of rectangular box = 80 cm

Inner height of rectangular box = 70 cm

We know that

Capacity of rectangular box = volume of rectangular box = l × b × h

= 90 × 80 × 70

= 504000 cm³.

Area of tin foil = Surface area of box = 2(lb + bh + lh)

= 2(90 × 80 + 80 × 70 + 90 × 70)

= 2(7200 + 5600 + 6300)

= 2 × 19100

= 38200 cm².

Hence, capacity of box = 504000 cm³ and area of tin-foil needed = 38200 cm².

By formula,

Volume of cuboidal box = l × b × h

Substituting values we get,

Volume of box = 20 cm × 16 cm × 24 cm = 7680 cm³

By formula,

Volume of cube = (side)³

Substituting values we get,

Volume of cubes = 4 cm × 4 cm × 4 cm = 64 cm³

Let no. of cubes be n.

So, total volume of cubes = 64n cm³.

In order to fill the box with cubes,

Volume of cubes = Volume of box

∴ 64n = 7680

⇒ n = $\frac{7680}{64}$ = 120.

Hence, 120 cubes can be put into the box.

Since the height of the box is 7 cm, so only 3 cubes can be put height-wise. (If we put 4 cubes, the height becomes 8 cm which is more than the height of the box.)

Height of 3 cubes = 3 x 2 = 6 cm

∴ We will only consider height of box up to 6 cm for placing the cubes inside it.

By formula,

Volume of cuboidal box = l × b × h

Substituting values we get,

Volume of box = 10 cm × 8 cm × 6 cm = 480 cm³

By formula,

Volume of cube = (side)³

Substituting values we get,

Volume of cubes = 2 cm × 2 cm × 2 cm = 8 cm³

Let no. of cubes be n.

So, total volume of cubes = 8n cm³.

In order to fill the box with cubes,

Volume of cubes = Volume of box

∴ 8n = 480

⇒ n = $\frac{480}{8}$ = 60.

Hence, 60 cubes can be put into the box.

Given,

Cost of 1 m³ wood = ₹ 25,000

Cost of a solid cubical block = ₹ 18,225

As we know that, cost of a solid cubical block = volume of block x cost of 1 m³ wood

$$\begin{aligned}\text{Volume of block }= \frac{\text{Cost of a solid cubical block}}{\text{Cost of wood per m}^3}\\[1em] = \frac{18225}{25000}\\[1em] = \frac{18225}{25000}\\[1em] = \frac{729}{1000}\\[1em] = 0.729\end{aligned}$$

By formula,

Volume of cuboidal block = (side)³

∴ (side)³ = 0.729 m³

$$\begin{aligned}\Rightarrow \text{(side)}^3 = \frac{729}{1000}\\[1em] \Rightarrow \text{side} = \sqrt[3]{\frac{729}{1000}}\\[1em]\end{aligned}$$

On factorising 729 and 1000 we get,

$$\begin{aligned}\Rightarrow \text{side} = \sqrt[3]{\frac{3 \times 3 \times 3 \times 3 \times 3 \times 3}{2 \times 2 \times 2 \times 5 \times 5 \times 5}}\\[1em] \Rightarrow \text{side} = {\frac{3 \times 3}{2 \times 5}}\\[1em] \Rightarrow \text{side} = {\frac{9}{10}}\\[1em] \Rightarrow \text{side} = 0.9\end{aligned}$$

Hence, volume of block = 0.729 m³ and the length of one edge of the block is 0.9 m.

Let rise in height of water be h cm.

The volume of water rising will be equal to volume of cube.

∴ 15 × 12 × h = 11 × 11 × 11

180h = 1331

h = $\frac{1331}{180}$ = 7.39 cm.

Hence, the rise in the water level is 7.39 cm.

Given, the base of rectangular container is a square

∴ l = 6 cm and b = 6 cm

When a cube is placed in it, water rises to top i.e. through height 1 cm and 2 cm³ of water overflows.

We know that,

Volume of cube = Volume of water displaced

= 6 × 6 × 1 + 2

= 36 + 2

= 38 cm³

Hence, volume of cube = 38 cm³.

From figure,

length of cuboid = 12 + 12 = 24 cm

breadth of cuboid = 12 cm

height of cuboid = 12 cm

By formula,

Total surface area of cuboid = 2(lb + bh + hl)

= 2(24 × 12 + 12 × 12 + 12 × 24)

= 2(288 + 144 + 288)

= 2 × 720

= 1440 cm².

Hence, surface area of resulting cuboid = 1440 cm².

Side of a cube = 6 cm

Volume of cube = (side)³ = 6³ = 216 cm³.

Since, the same metal is melted and casted into cuboid so,

Volume of cuboid = Volume of cube.

By formula,

Volume of cuboid = l × b × h

Substituting values we get,

⇒ 216 = 9 × 8 × h

⇒ h = $\frac{216}{72}$ = 3.

Hence, height of cuboid = 3 cm.

Given,

Area of playground = 4800 m²

We can write it as,

l × b = 4800

Given,

Depth of level = 1 cm

∴ h = 1 cm = $\frac{1}{100}$ m

By formula,

Volume of gravel needed = l × b × h

= 4800 × $\frac{1}{100}$

= 48 m³

Cost of gravel = ₹260 per cubic metre

So the total cost = 260 × 48 = ₹12480

Hence, cost of covering the playground with 1 cm deep gravel = ₹12480.

From figure,

ABCD is a field.

Volume of the earth dug out = 6 × 4 × 3 = 72 m³

Area of field ABCD = AB × BC = 18 × 30 = 540 m².

Area of pit EFGH = EF × FG = 4 × 6 = 24 m².

Area of remaining field = 540 - 24 = 516 m².

Let h metres is the level raised over the field uniformly.

Volume of rise in level = Volume of earth dug out

∴ Area of remaining field × h = 72

516h = 72

h = $\frac{72}{516} = 0.1395$ m = 13.95 cm

Hence, the level of the remaining field has been raised by 13.95 cm.

Let ABCD be the rectangular plot.

Given,

Length of plot (l) = 24 m

Width of plot (b) = 20 m

So, the area of plot = l × b = 24 × 20 = 480 m².

We know that,

Side of cubical pit = 4 m

Volume of each pit = 4³ = 64 m³.

Volume of 4 pits at the corners = 4 × 64 = 256 m³.

Surface area of each pit = 4(side)²

= 4 × 4²

= 64 m²

So, the area of remaining plot = 480 – 64 = 416 m².

Let height of soil be h meters spread over remaining land.

So, the volume of soil raised = Volume of soil dug

∴ 416 × h = 256

h = $\frac{256}{416} = \frac{8}{13}$ metre.

Hence, remaining plot gets raised by $\frac{8}{13}$ metre.

Given,

Inner dimensions of wooden box are 2 m, 1.2 m and 0.75 m.

Thickness of the wood = 2.5 cm = $\frac{2.5}{100}$ = 0.025 m.

So the external dimensions of wooden box are,

⇒ (2 + 2 × 0.025), (1.2 + 2 × 0.025), (0.75 + 2 × 0.025)

⇒ (2 + 0.05), (1.2 + 0.05), (0.75 + 0.5)

⇒ 2.05 m, 1.25 m, 0.80 m.

By formula,

Volume of solid = External volume of box – Internal volume of box

Substituting the values we get,

Volume of solid = (2.05 × 1.25 × 0.80) – (2 × 1.2 × 0.75)

= 2.05 – 1.80

= 0.25 m³.

Cost of box = Cost per m³ × Volume of box

= ₹ 22,000 × 0.25

= ₹ 22,000 x $\frac{25}{100}$

= ₹ 5,500.

Hence, cost of wood required to make the box is ₹ 5,500.

Given,

Internal edge of cubical wooden box = 1 m

Thickness of wood = 5 cm = $\frac{5}{100}$ m = 0.05 m

We know that,

External length = (1 + 0.05 × 2) = 1.1 m

External Breadth = (1 + 0.05 × 2) = 1.1 m

External Height = (1 + 0.05) = 1.05 m [Since, box is open at top]

By formula,

Volume of the wood used = Outer volume – Inner volume

Substituting the values we get,

Volume of the wood used = 1.1 × 1.1 × 1.05 – 1 × 1 × 1

= 1.2705 – 1

= 0.2705 m³

Given,

Cost of 1 m³ of wood = ₹ 28,800

Cost of the wood required to make the box = Cost of wood per m³ × Volume of wood

= ₹ 28,800 × 0.2705

= ₹ 7,790.40

≈ ₹ 7,800 (to nearest hundred)

Hence, cost of wood required to make the box = ₹ 7,800.

Given,

Side of square brass plate = $x$ cm

Here, l = $x$ cm and b = $x$ cm

Thickness of plate = 1 mm = $\frac{1}{10}$ = 0.1 cm

We know that,

Volume of the plate = l × b × h

Substituting the values

= $x$ × $x$ × 0.1

= $0.1x^2$ cm³

Given,

Weight of 1 cm³ of brass = 8.4 g

∴ Weight of $0.1x^2$ cm³ of brass = 8.4 x $0.1x^2$ g

But weight of square brass plate is given as 4725 g

$$\begin{aligned}\therefore 8.4 \times 0.1x^2 = 4725 \\[1em] \Rightarrow x^2 = \frac{4725}{8.4 \times 0.1} \\[1em] \Rightarrow x^2 = \frac{472500}{84 \times 1} \\[1em] \Rightarrow x^2 = 5625 \\[1em] \Rightarrow x = \sqrt{5625} = 75.\end{aligned}$$

Hence, x = 75.

Since, cubes are melted and recasted into a single cube.

∴ Total volume of three cubes = Volume of new single cube

Substituting values we get,

$$\begin{aligned}\Rightarrow x^3 + 8^3 + 10^3 = 12^3 \\[1em] \Rightarrow x^3 + 512 + 1000 = 1728 \\[1em] \Rightarrow x^3 + 1512 = 1728 \\[1em] \Rightarrow x^3 = 216 \\[1em] \Rightarrow x^3 = 6^3 \\[1em] \Rightarrow x = 6.\end{aligned}$$

Hence, x = 6.

It is given that

Area of cross-section of pipe = 3.5 cm²

Speed of water = 40 cm/s

Length of water column in 1 sec = 40 cm

We know that,

Volume of water flowing in 1 second = Area of cross section × length

= 3.5 × 40

= 140 cm³.

So, the volume of water flowing in 1 minute i.e., 60 sec = 140 × 60 = 8400 cm³.

1 cm³ = $\frac{1}{1000}$ litre

∴ 8400 cm³ = $8400 \times \frac{1}{1000}$ = 8.4 litres.

Hence, 8.4 litres of water is delivered by the pipe in one minute.

(a) From figure,

The line AB divides the figure, into two cuboids vertical and horizontal.

Volume of solid = Volume of vertical cuboid + Volume of horizontal cuboid

= 4 × 2 × 6 + 4 × 4 × 2

= 48 + 32

= 80 cm³

Hence, volume of solid = 80 cm³.

Figure (ii) is a trapezium with parallel sides 2 m and 3.5 m

(i) Area of cross section = $\frac{1}{2} \times$ (sum of parallel sides) × height

= $\frac{1}{2} \times$ (2 + 3.5) × 6

= $\frac{1}{2}$ × 5.5 × 6

= 5.5 × 3

= 16.5 m².

Hence, area of cross-section = 16.5 m².

(ii) Volume of concrete in the wall = Area of cross section × length

= 16.5 × 400

= 6600 m³.

Hence, volume of concrete wall = 6600 m³.

Figure (iii) is a trapezium with parallel sides 2 m and 3m.

By formula,

Area of cross section = $\frac{1}{2} \times$ (sum of parallel sides) × distance between them

= $\frac{1}{2} \times$ (2 + 3) × 40

= $\frac{1}{2}$ × 5 × 40

= 100 m².

So, the volume of water it will hold when full = area of cross section × width

= 100 × 10

= 1000 m³.

Hence, swimming pool will hold 1000 m³ of water.

Given,

Length of swimming pool = 50 m

Width of swimming pool = 15 m

Its shallow and deep ends are 1.5 m and 4.5 m deep

By formula,

Area of cross section of swimming pool = $\frac{1}{2} \times$ (sum of parallel sides) × length

= $\frac{1}{2} \times (1.5 + 4.5) \times 50$

= $\frac{1}{2}$ × 6 × 50

= 150 m².

Amount of water required to fill pool = Area of cross section × width

= 150 × 15

= 2250 m³.

Hence, amount of water required to fill pool 2250 m³.

By formula,

Area of triangle = $\frac{1}{2}$ × base × height.

Substituting values we get,

30 = $\frac{1}{2}$ × 10 × height

height = $\frac{30 \times 2}{10} = \frac{60}{10}$ = 6 cm.

Hence, Option 2 is the correct option.

By formula,

Perimeter of square = 4 × side

Substituting values we get,

80 = 4 × side

side = $\frac{80}{4}$ = 20 cm.

By formula,

Area of square = side × side = 20 × 20 = 400 cm².

Hence, Option 3 is the correct option.

By formula,

Area of parallelogram = base × height

Substituting values we get,

⇒ 48 = base × 6

⇒ base = $\frac{48}{6}$ = 8 cm.

Hence, Option 1 is the correct option.

r = $\frac{\text{Diameter}}{2} = \frac{d}{2}$.

By formula,

Area of circle = πr² = $π \times \Big(\frac{d}{2}\Big)^2 = \frac{πd^2}{4}.$

Hence, Option 3 is the correct option.

By formula,

Area of trapezium = $\frac{1}{2}$ × sum of parallel sides × distance between them

Substituting values we get,

⇒ 64 = $\frac{1}{2}$ × sum of parallel sides × 8

⇒ 64 = 4 × sum of parallel sides

⇒ sum of parallel sides = $\frac{64}{4}$ = 16 cm.

Hence, Option 4 is the correct option.

By formula,

Area of rhombus = $\frac{1}{2}$ × d₁ × d₂

Substituting values we get,

⇒ Area of rhombus = $\frac{1}{2}$ × 8 × 6 = 24 cm².

Hence, Option 2 is the correct option.

Let diagonals be d₁ and d₂.

By formula,

Area of rhombus = $\frac{1}{2}$ × d₁ × d₂

If doubled, diagonals = 2d₁ and 2d₂.

Area of new rhombus = $\frac{1}{2}$ × 2d₁ × 2d₂

= 4 × $\frac{1}{2}$ × d₁ × d₂

= 4 × Area of rhombus.

Hence, Option 3 is the correct option.

Let ABCD be the quadrilateral with diagonal BD.

Figure unavailable in the source: If the length of a diagonal of a quadrilateral is 10 cm and lengths of the perpendiculars on it from opposite vertices are 4 cm and 6 cm, then area of quadrilateral is? Mensuration, ML Aggarwal Understanding Mathematics Solutions ICSE Class…

Let AM and CN be the perpendiculars from A and C on diagonal BD.

From figure,

BD divides quadrilateral in two triangles.

Area of △ABD = $\frac{1}{2}$ × base × height

= $\frac{1}{2}$ × BD × AM

= $\frac{1}{2}$ × 10 × 4

= 20 cm².

Area of △BCD = $\frac{1}{2}$ × base × height

= $\frac{1}{2}$ × BD × CN

= $\frac{1}{2}$ × 10 × 6

= 30 cm².

Area of quadrilateral = Area of △ABD + Area of △BCD

= 20 + 30 = 50 cm².

Hence, Option 3 is the correct option.

Let diagonals be d₁ and d₂.

By formula,

Area of rhombus = $\frac{1}{2}$ × d₁ × d₂

Substituting values we get,

⇒ 90 = $\frac{1}{2}$ × 10 × d₂

⇒ d₂ = $\frac{90 \times 2}{10}$

⇒ d₂ = 18 cm.

Hence, Option 1 is the correct option.

Perimeter of quadrant = $\frac{2πr}{4} + 2r$

$$\begin{aligned}= \frac{2 \times \frac{22}{7} \times 7}{4} + 2 \times 7 \\[1em] = \frac{44}{4} + 14 \\[1em] = 11 + 14 \\[1em] = 25 \text{ cm}.\end{aligned}$$

Hence, Option 3 is the correct option.

Area of square OABC = (7)² = 49 cm².

Area of quadrant OAC = $\frac{πr^2}{4}$

$$\begin{aligned}= \frac{\frac{22}{7} \times (7)^2}{4} \\[1em] = \frac{22 \times 7}{4} \\[1em] = \frac{154}{4}\\[1em] = 38.5 \text{ cm}^2.\end{aligned}$$

Area of shaded region = Area of square OABC - Area of quadrant OAC

= 49 - 38.5 = 10.5 cm².

Hence, Option 1 is the correct option.

From figure,

Let length = 14 cm and breadth = 10 cm.

Diameter of semi-circle = 14 cm and radius = 7 cm.

Perimeter of shaded region = length + breadth + breadth + πr.

= 14 + 10 + 10 + $\frac{22}{7} \times 7$

= 34 + 22 = 56 cm.

Hence, Option 2 is the correct option.

Area of shaded region = Area of rectangle + Area of semi-circle

= length × breadth + $\frac{πr^2}{2}$

= 14 × 10 + $\frac{\frac{22}{7} \times (7)^2}{2}$

= 140 + $\frac{154}{2}$

= 140 + 77

= 217 cm².

Hence, Option 4 is the correct option.

From figure,

Radius of larger circle (R) = 14 cm.

Area of shaded region = Area of larger circle - Area of 1st smaller circle + Area of 2nd smaller circle .......(1)

From figure,

Diameter of both the smaller semi-circles = 14 cm.

∴ Radius = 7 cm and area of both the circle are equal.

∴ Area of shaded region = Area of larger circle = $\frac{πR^2}{2}$

$$\begin{aligned}= \frac{1}{2} \times \frac{22}{7} \times (14)^2 \\[1em] = \frac{11}{7} \times 196 \\[1em] = 11 \times 28 \\[1em] = 308 \text{ cm}^2.\end{aligned}$$

Hence, Option 4 is the correct option.

Radius of larger semi-circle (R) = 14 cm and radius of smaller semi-circle (r) = 7 cm.

From figure,

Perimeter of shaded region = Circumference of larger semi-circle + 2 × Circumference of smaller semi-circle

= πR + 2πr

= $\frac{22}{7} \times 14 + 2 \times \frac{22}{7} \times 7$

= 44 + 44 = 88 cm.

Hence, Option 2 is the correct option.

Area of right angle triangle ABC = $\frac{1}{2}$ × base × height

= $\frac{1}{2}$ × AB × BC

= $\frac{1}{2}$ × 12 × 5

= 30 cm².

From figure,

AB = 12 cm is the diameter of circle.

Radius = $\frac{AB}{2}$ = 6 cm.

Area of semi-circle = $\frac{πr^2}{2} = \frac{π(6)^2}{2} = \frac{36π}{2}$ = 18π cm².

Area of shaded region = Area of right angle triangle ABC + Area of semi-circle

= (30 + 18π) cm².

Hence, Option 3 is the correct option.

In right angle triangle ABC,

⇒ AC² = AB² + BC²

⇒ AC² = (12)² + (5)²

⇒ AC² = 144 + 25

⇒ AC² = 169

⇒ AC = $\sqrt{169}$ = 13 cm.

From figure,

radius of semi-circle (r) = $\frac{AB}{2} = \frac{12}{2}$ = 6 cm.

Perimeter of shaded region = AC + CB + Circumference of semi-circle

= 13 + 5 + πr

= (18 + 6π) cm.

Hence, Option 4 is the correct option.

By formula,

Volume of cube = (Side)³

∴ (Side)³ = 729

Side = $\sqrt[3]{729}$ = 9 cm.

⇒ Surface area of cube = 6(side)²

= 6(9)²

= 6 × 81 = 486 cm².

Hence, Option 1 is the correct option.

Let side of cube = x cm.

Given,

Surface area of cube = 96 cm²

By formula,

Surface area of a cube = 6x²

∴ 6x² = 96

⇒ x² = 16

⇒ x = $\sqrt{16}$ = 4 cm.

Volume of cube = (side)³

= (x)³

= 4³ = 64 cm³.

Hence, Option 3 is the correct option.

The longest pole in a cuboid is equal to the diagonal of cuboid.

Diagonal of cuboid = $\sqrt{l^2 + b^2 + h^2}$

$$\begin{aligned}= \sqrt{10^2 + 10^2 + 5^2} \\[1em] = \sqrt{100 + 100 + 25} \\[1em] = \sqrt{225} \\[1em] = 15 \text{ m}.\end{aligned}$$

Hence, Option 1 is the correct option.

By formula,

Lateral surface area of cube = 4(side)²

⇒ 4(side)² = 256

⇒ (side)² = 64

⇒ side = $\sqrt{64}$ = 8 m.

Volume of cube = (side)³

= 8³ = 512 m³.

Hence, Option 1 is the correct option.

Each face of cube is a square. Let length of each side = x cm.

Given, perimeter = 40 cm.

∴ 4x = 40 cm

⇒ x = 10 cm.

There are 12 edges in a cube.

Sum of edges = 12 × 10 = 120 cm.

Hence, Option 2 is the correct option.

Let l, b and h be the length, breadth and height of the cuboid container.

Volume = l × b × h = lbh.

If they are doubled then,

New Volume = 2l × 2b × 2h = 8lbh

Hence, volume becomes 8 times.

So, capacity becomes 8 times.

So, container can hold 50 × 8 = 400 boxes.

Hence, Option 3 is the correct option.

Volume of plank = 4 m × 50 cm × 20 cm

= 4 m × 0.50 m × 0.20 m

= 0.4 m³.

Volume of pit = 16 m × 12 m × 4 m

= 768 m³.

No. of planks that can be stored in pit = $\frac{\text{Volume of pit}}{\text{Volume of plank}} = \frac{768}{0.4}$ = 1920.

Hence, Option 2 is the correct option.

Given that the circumference is 10π cm.

⇒ 2πr = 10π

⇒ r = $\frac{10π}{2π}$

⇒ r = 5 cm

Now, calculate the area using this radius,

⇒ A = πr²

= π.5²

= 25π cm².

∴ Statement 1 is true.

We know that,

Circumference of the circle = 2πr

2πr × π = 2π²r ≠ πr².

∴ Statement 2 is false.

∴ Statement 1 is true, and Statement 2 is false.

Hence, option 3 is the correct option.

Given, sides of a triangle are 9 cm, 12 cm and 15 cm.

A scalene triangle is a triangle in which all three sides have different lengths.

The given side lengths are 9 cm, 12 cm, and 15 cm. All these lengths are distinct.

Therefore, it is a scalene triangle.

If the square of the longest side equals the sum of the squares of the other two sides, then it's a right-angled triangle.

The longest side is 15 cm.

Let's check: 15² = 9² + 12²

⇒ 225 = 81 + 144

⇒ 225 = 225

Since the equality holds, the triangle is a right-angled triangle.

∴ Assertion (A) is true.

By formula,

Area of a right triangle = $\frac{1}{2}$ x base x height.

∴ Reason (R) is true.

∴ Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct reason (or explanation) for Assertion (A).

Hence, option 4 is the correct option.

With the help of heron's formula, we can find the area of any triangle and not particularly scalene triangle.

∴ Assertion (A) is false.

If ABC is a triangle with side a, b and c respectively, then its area = $\sqrt{s(s - a)(s - b)(s - c)}$, where s = $\frac{a + b + c}{2}$.

This is the correct definition and formula for Heron's formula. The variable 's' represents the semi-perimeter of the triangle.

∴ Reason (R) is true.

∴ Assertion (A) is false, Reason (R) is true.

Hence, option 2 is the correct option.

Given: length, breadth and diagonal of cuboid = 4 m, 3 m and 13 m.

By formula,

Length of diagonal of a cuboid = $\sqrt{l^2 + b^2 + h^2}$

∴ Reason (R) is true.

Substituting the values, we get

$$\begin{aligned}\Rightarrow 13 = \sqrt{3^2 + 4^2 + h^2}\\[1em] \Rightarrow 13^2 = 3^2 + 4^2 + h^2\\[1em] \Rightarrow 169 = 9 + 16 + h^2\\[1em] \Rightarrow 169 = 25 + h^2\\[1em] \Rightarrow h^2 = 169 - 25\\[1em] \Rightarrow h^2 = 144\\[1em] \Rightarrow h = \sqrt{144}\\[1em] \Rightarrow h = 12\end{aligned}$$

Volume of cuboid = l x b x h

= 3 x 4 x 12

= 144 m³.

∴ Assertion (A) is true.

∴ Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct reason for Assertion (A).

Hence, option 3 is the correct option.

From figure,

Area of △AOB = $\frac{1}{2}$ × base × height

= $\frac{1}{2}$ × AO × OB

= $\frac{1}{2}$ × 12 × 5

= 30 cm².

In right angle triangle AOB,

Using pythagoras theorem,

⇒ AB² = AO² + OB²

⇒ AB² = 12² + 5²

⇒ AB² = 144 + 25

⇒ AB² = 169

⇒ AB = $\sqrt{169}$ = 13 cm.

In △ABC,

Let a = BC = 14 cm, b = AC = 15 cm and c = AB = 13 cm.

s = $\frac{a + b + c}{2} = \frac{14 + 15 + 13}{2} = \frac{42}{2}$ = 21 cm.

By Heron's formula,

Area = $\sqrt{s(s - a)(s - b)(s- c)}$

Substituting values we get,

$$\begin{aligned}= \sqrt{21(21 - 14)(21 - 15)(21 - 13)} \\[1em] = \sqrt{21 \times 7 \times 6 \times 8} \\[1em] = \sqrt{7056} \\[1em] = 84 \text{ cm}^2.\end{aligned}$$

Area of shaded region = Area of △ABC - Area of △AOB

= 84 - 30 = 54 cm².

Hence, area of shaded region = 54 cm².

Let length of square be x cm.

New length = (x + 3) cm.

According to the question,

⇒ (x + 3)(x + 3) = 121

⇒ x² + 3x + 3x + 9 = 121

⇒ x² + 6x = 112

⇒ x² + 6x - 112 = 0

⇒ x² + 14x - 8x - 112 = 0

⇒ x(x + 14) - 8(x + 14) = 0

⇒ (x - 8)(x + 14) = 0

⇒ x - 8 = 0 or x + 14 = 0

⇒ x = 8 or x = -14.

Since, length cannot be negative.

∴ x ≠ -14.

Perimeter of original square = 4 × side = 4x

= 4 × 8 = 32 cm.

Hence, perimeter of original square = 32 cm.

From figure,

Area of shaded region = Area of square ABCD - Area of △EFG - Area of △HIJ ........(1)

In △EFG and △HIJ,

FG = HI

IJ = EF

∠HIJ = ∠GFE (Both equal to 90°).

Hence, △EFG ≅ △HIJ by SAS axiom.

∴ Area of △EFG = Area of △HIJ.

Area of △EFG = $\frac{1}{2}$ × base × height

= $\frac{1}{2}$ × EF × FG

= $\frac{1}{2}$ × 5 × 6 = 15 cm².

Area of △HIJ = Area of △EFG = 15 cm²

Area of square ABCD = (side)² = (9)² = 81 cm².

Substituting values in (1) we get,

Area of shaded region = 81 - 15 - 15 = 51 cm².

Hence, area of shaded region = 51 cm².

In right angle △ABD,

⇒ BD² = AB² + AD²

⇒ BD² = 6² + 8²

⇒ BD² = 36 + 64

⇒ BD² = 100

⇒ BD = $\sqrt{100}$ = 10 cm.

In right angle △BDC,

⇒ BC² = BD² + DC²

⇒ 26² = 10² + DC²

⇒ DC² = 676 - 100

⇒ DC² = 576

⇒ DC = $\sqrt{576}$ = 24 cm.

Area of △ABD = $\frac{1}{2}$ × base × height

= $\frac{1}{2}$ × AD × AB

= $\frac{1}{2}$ × 8 × 6

= 24 cm².

Area of △BDC = $\frac{1}{2}$ × base × height

= $\frac{1}{2}$ × DC × BD

= $\frac{1}{2}$ × 24 × 10

= 120 cm².

Area of quadrilateral ABCD = Area of △ABD + Area of △BDC

= 24 + 120 = 144 cm².

Hence, area of quadrilateral ABCD = 144 cm².

The points are labelled as shown in the figure below:

Area of right angle △AEH = $\frac{1}{2}$ × base × height

= $\frac{1}{2}$ × AH × AE

= $\frac{1}{2}$ × 5 × 5

= 12.5 m².

Area of right angle △EBF = $\frac{1}{2}$ × base × height

= $\frac{1}{2}$ × BF × EB

= $\frac{1}{2}$ × 5 × 7

= 17.5 m².

Area of trapezium GDCF = $\frac{1}{2}$ × (sum of parallel sides) × distance between them

= $\frac{1}{2}$ × (GD + FC) × DC

= $\frac{1}{2}$ × (3 + 7) × 12

= 60 m².

Area of square ABCD = (side)² = (12)² = 144 m².

From figure,

Area of shaded region = Area of square ABCD - (Area of right angle △AEH + Area of right angle △EBF + Area of trapezium GDCF)

= 144 - (12.5 + 17.5 + 60)

= 144 - 90 = 54 m².

Hence, area of shaded region = 54 m².

The points are labelled as shown in the figure below:

In △ABK,

a = BK = 1 cm, b = AK = 5 cm and c = AB = 5 cm.

s = $\frac{a + b + c}{2} = \frac{1 + 5 + 5}{2} = \frac{11}{2}$ = 5.5 cm.

Area of △ABK = $\sqrt{s(s - a)(s - b)(s - c)}$

Substituting values we get,

$$\begin{aligned}= \sqrt{5.5 \times (5.5 - 1)(5.5 - 5)(5.5 - 5)} \\[1em] = \sqrt{5.5 \times 4.5 \times 0.5 \times 0.5} \\[1em] = \sqrt{6.1875} \\[1em] = 2.5 \text{ cm}.\end{aligned}$$

Area of △KIJ = $\frac{1}{2} \times$ base × height

= $\frac{1}{2}$ × KI × KJ

= $\frac{1}{2}$ × 1.5 × 6

= 4.5 cm².

Area of △CBD = $\frac{1}{2} \times$ base × height

= $\frac{1}{2}$ × BD × CB

= $\frac{1}{2}$ × 1.5 × 6

= 4.5 cm².

Area of rectangle BKHE = length × breadth

= BE × EH

= 6.5 × 1 = 6.5 cm².

Let EL and HM be perpendicular to LM.

Since, EFGH is an isosceles trapezium so,

FL = MG = $\frac{2 - 1}{2} = \frac{1}{2}$ = 0.5 cm.

In right angle △EFL,

⇒ EF² = EL² + FL²

⇒ 1² = EL² + (0.5)²

⇒ EL² = 1² - (0.5)²

⇒ EL² = 1 - 0.25

⇒ EL² = 0.75

⇒ EL = $\sqrt{0.75}$ = 0.87 cm.

Area of trapezium EFGH = $\frac{1}{2}$ × (sum of parallel sides) × distance between them

= $\frac{1}{2}$ × (EH + FG) × EL

= $\frac{1}{2}$ × (1 + 2) × 0.87

= 1.3 cm².

Area of chart paper used = Area of aeroplane = Area of △ABK + Area of △KIJ + Area of △CBD + Area of rectangle BKHE + Area of trapezium EFGH

= 2.5 + 4.5 + 4.5 + 6.5 + 1.3

= 19.3 cm².

Hence, area of chart paper used = 19.3 cm².

By formula,

Area of circle = πr²

⇒ πr² = 78.5

⇒ 3.14r² = 78.5

⇒ r² = $\frac{78.5}{3.14}$ = 25

⇒ r = $\sqrt{25}$ = 5 cm.

Circumference = 2πr = 2 × 3.14 × 5 = 31.4 cm.

Hence, circumference = 31.4 cm.

Let radius of the circle be r cm.

Given,

Area of the circle = 154 cm²

πr² = 154

$$\begin{aligned}\Rightarrow \frac{22}{7} \times r^2 = 154 \\[1em] \Rightarrow r^2 = \frac{154 \times 7}{22} \\[1em] \Rightarrow r^2 = 49 \\[1em] \Rightarrow r = \sqrt{49} = 7 \text{ cm}.\end{aligned}$$

The biggest circle that can be cut from a square has the diameter equal to the side of the square.

Side = 2r = 2 × 7 = 14 cm.

Area of cardboard = (side)² = 14² = 196 cm².

Hence, area of cardboard = 196 cm².

Length of sheet = 2 m = 200 cm,

Breadth of sheet = 1.5 m = 150 cm.

Given,

Radius of each circle = 5 cm,

Diameter of each circle = 2 × 5 = 10 cm.

No. of circles lengthwise to be cut = $\frac{\text{Length of sheet}}{\text{Diameter of each circle}} = \frac{200}{10}$ = 20.

No. of circles breadthwise to be cut = $\frac{\text{Breadth of sheet}}{\text{Diameter of each circle}} = \frac{150}{10}$ = 15.

Total no. of circles that can be cut = 20 × 15 = 300.

Area of paper wasted = Area of sheet - Area of circles

= l × b - 300 × πr²

= 200 × 150 - 300 × 3.14 × (5)²

= 30000 - 23550

= 6450 cm².

Hence, total no. of circles that can be cut = 300 and area of paper wasted = 6450 cm².

Diameter of semi-circular protractor = 14 cm

Radius = $\frac{14}{2}$ = 7 cm.

Perimeter of semi-circular protractor = πr + 2r = $\frac{22}{7} \times 7 + 2 \times 7$ = 22 + 14 = 36 cm.

Hence, perimeter of semi-circular protractor = 36 cm.

Given,

Circumference of circular park = 88 m

Let radius of circular park be r meters.

2πr = 88

$$\begin{aligned}\Rightarrow 2 \times \frac{22}{7} \times r = 88 \\[1em] \Rightarrow r = \frac{88 \times 7}{2 \times 22} \\[1em] \Rightarrow r = 14 \text{ m}.\end{aligned}$$

From figure,

Radius of outer circle (R) = Radius of circular park + Width of road = 14 + 3.5 = 17.5 m

Area of road = Area of outer circle - Area of circular park

= πR² - πr²

= π(17.5)² - π(14)²

= π[306.25 - 196]

= 110.25π

= $110.25 \times \frac{22}{7}$ = 346.5 m².

Cost of paving the road = Area of road × Rate = 346.5 × 60 = ₹ 20790.

Hence, cost of paving road = ₹ 20790.

From figure,

Length of inner rectangle ABCD = 140 m

Breadth of inner rectangle ABCD = 42 m

Diameter of inner semi-circle = 42 m

Radius of inner semi-circle (r) = $\frac{42}{2}$ = 21 m.

Length of outer rectangle = Length of inner rectangle = 140 m

Breadth of outer rectangle = Breadth of inner rectangle + 2 × Width of outer track = 42 + 2 × 3.5 = 42 + 7 = 49 m.

Diameter of outer semi-circle = 49 m

Radius of outer semi-circle (R) = $\frac{49}{2}$ = 24.5 m.

Area of track = Area of outer rectangle - Area of inner rectangle + 2(Area of outer semi-circle - Area of inner semi-circle)

$$\begin{aligned}= 140 × 49 - 140 × 42 + 2(\frac{πR^2}{2} - \frac{πr^2}{2}) \\[1em] = 140(49 - 42) + 2π \times \frac{1}{2}(R^2 - r^2) \\[1em] = 140 × 7 + π[(24.5)^2 - (21)^2] \\[1em] = 980 + π[600.25 - 441] \\[1em] = 980 + \frac{22}{7} \times 159.25 \\[1em] = 980 + 500.5 \\[1em] = 1480.5 \text{ m}^2.\end{aligned}$$

Hence, area of track = 1480.5 m².

Angle in semi-circle = 90°

∴ ∠ACB = 90°

Using pythagoras theorem,

⇒ AB² = AC² + BC²

⇒ AB² = 12² + 16²

⇒ AB² = 144 + 256

⇒ AB² = 400

⇒ AB = $\sqrt{400}$ = 20 cm.

AB is the diameter of circle,

∴ Radius = $\frac{\text{AB}}{2}$ = $\frac{20}{2}$ = 10 cm.

Area of shaded region = Area of semi-circle - Area of triangle

$$\begin{aligned}= \frac{πr^2}{2} - \frac{1}{2} \times \text{ base × height } \\[1em] = \frac{3.142 \times 10^2}{2} - \frac{1}{2} \times AC \times BC \\[1em] = \frac{3.142 \times 100}{2} - \frac{1}{2} \times 12 \times 16 \\[1em] = \frac{314.2}{2} - 96 \\[1em] = 157.1 - 96 \\[1em] = 61.1 \text{ cm}^2.\end{aligned}$$

Perimeter of shaded region = Circumference of semi-circle + AC + CB

= πr + 12 + 16

= 3.142 × 10 + 28

= 31.42 + 28

= 59.42 cm.

Hence, perimeter of shaded region = 59.42 cm and area of semi-circle = 61.1 cm².

Perimeter of quarter of circle = $\frac{2πr}{4} + 2r = \frac{πr}{2} + 2r$

Substituting values we get,

$$\begin{aligned}= \frac{\frac{22}{7} \times 3.5}{2} + 2 \times 3.5 \\[1em] = \frac{22 \times 0.5}{2} + 7 \\[1em] = \frac{11}{2} + 7 \\[1em] = 12.5 \text{ cm}.\end{aligned}$$

Hence, perimeter of quarter of circle = 12.5 cm.

(i) Let O be the center of the circle.

B be the mid-point of side of square.

From figure,

OB = 2 + 1 = 3 cm

AB = 1 cm

Using Pythagoras theorem,

OA = $\sqrt{OB^2 + AB^2}$

OA = $\sqrt{3^2 + 1^2} = \sqrt{9 + 1} = \sqrt{10}$.

So, the radius of the circle = $\sqrt{10}$ cm.

Hence, the radius of circle = $\sqrt{10}$ cm.

(ii) We know that,

Area of the circle = πr².

= 3.14 × $\sqrt{10}^2$

= 3.14 × 10

= 31.4 cm².

Area of 5 square of side 2 cm each = 2² × 5

= 4 × 5

= 20 cm².

So, the area of shaded portion = 31.4 – 20 = 11.4 cm².

Hence, area of shaded portion = 11.4 cm².