Area of a Trapezium and a Polygon

Given:

Sides of the triangle = 12 cm, 15 cm

Height of the triangle = 10 cm

As we know, the area of the triangle = $\frac{1}{2}$ x side x height

= $\frac{1}{2}$ x 12 x 10 cm²

= $\frac{1}{2}$ x 120 cm²

= 60 cm²

When, side of the triangle = 15 cm

Let the height of the triangle = h cm

As we know, the area of the triangle = $\frac{1}{2}$ x side x height

= $\frac{1}{2}$ x 15 x h ²

[∵ Both areas remain same since they represent the area of the same triangle.]

⇒ $\frac{1}{2}$ x 15 x h = 60

⇒ 15 x h = 2 x 60

⇒ 15 x h = 120

⇒ h = $\frac{120}{15}$

⇒ h = 8 cm

Hence, the height of the triangle corresponding to 15 cm side is 8 cm.

Hence, option 3 is the correct option.

Given:

Base of the triangle = 8 cm

Hypotenuse of the triangle = 10 cm

Let h be the height of the triangle.

In a right-angled triangle, using the Pythagorean theorem,

Base² + Height² = Hypotenuse²

⇒ (8)² + h² = (10)²

⇒ 64 + h² = 100

⇒ h² = 100 - 64

⇒ h² = 36

⇒ h = $\sqrt{36}$

⇒ h = 6 cm

As we know, the area of the triangle = $\frac{1}{2}$ x side x height

= $\frac{1}{2}$ x 8 x 6 cm²

= $\frac{1}{2}$ x 48 cm²

= 24 cm²

Hence, option 4 is the correct option.

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

$$\begin{aligned}∵ s = \frac{a + b + c}{2}\\[1em] = \frac{4 + 3 + 5}{2}\\[1em] = \frac{12}{2}\\[1em] = 6\end{aligned}$$

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

= $\sqrt{6(6 - 4)(6 - 3)(6 - 5)}$ cm²

= $\sqrt{6 \times 2 \times 3 \times 1}$ cm²

= $\sqrt{36}$ cm²

= 6 cm²

Hence, option 4 is the correct option.

Given:

Altitude of the triangle = 4 cm

Base of the triangle = 2 cm more than its height

= 2 + 4 cm

= 6 cm

∵ As we know, the area of the triangle = $\frac{1}{2}$ x side x height

= $\frac{1}{2}$ x 6 x 4 cm²

= $\frac{1}{2}$ x 24 cm²

= 12 cm²

Hence, option 2 is the correct option.

Given:

Side of the equilateral triangle = 6 cm

∵ As we know, the area of the equilateral triangle = $\frac{\sqrt{3}}{4} \times \text{side}^2$

= $\frac{\sqrt{3}}{4} \times 6^2$ cm²

= $\frac{\sqrt{3}}{4} \times 36$ cm²

= $9\sqrt{3}$ cm²

Hence, option 2 is the correct option.

Let a = 10 cm, b = 24 cm and c = 26 cm.

$$\begin{aligned}∵ s = \frac{a + b + c}{2}\\[1em] = \frac{10 + 24 + 26}{2}\\[1em] = \frac{60}{2}\\[1em] = 30\end{aligned}$$

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

= $\sqrt{30(30 - 10)(30 - 24)(30 - 26)}$ cm²

= $\sqrt{30 \times 20 \times 6 \times 4}$ cm²

= $\sqrt{14,400}$ cm²

= 120 cm²

Hence, area of triangle = 120 cm².

(i) Given:

Side of the triangle = 6 cm

Height of the triangle = 4 cm

As we know, the area of the triangle = $\frac{1}{2}$ x side x height

= $\frac{1}{2}$ x 6 x 4 cm²

= $\frac{1}{2}$ x 24 cm²

= 12 cm²

Hence, area of triangle = 12 cm².

(ii) When, the side of the triangle = 8 cm

Let the height of the triangle be h cm

As we know area of triangle = $\frac{1}{2}$ x side x height

= $\frac{1}{2}$ x 8 x h ²

[∵ Both areas remain same since they represent the area of the same triangle.]

⇒ $\frac{1}{2}$ x 8 x h = 12

⇒ 4 x h = 12

⇒ h = $\frac{12}{4}$

⇒ h = 3 cm

Hence, height of the triangle corresponding to 8 cm side is 3 cm.

(i) Let a = 16 cm, b = 12 cm and c = 20 cm.

$$\begin{aligned}∵ s = \frac{a + b + c}{2}\\[1em] = \frac{16 + 12 + 20}{2}\\[1em] = \frac{48}{2}\\[1em] = 24\end{aligned}$$

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

= $\sqrt{24(24 - 16)(24 - 12)(24 - 20)}$ cm²

= $\sqrt{24 \times 8 \times 12 \times 4}$ cm²

= $\sqrt{9,216}$ cm²

= 96 cm²

Hence, area of triangle = 96 cm².

(ii) When, the largest side of the triangle = 20 cm

Let the height of the triangle = h cm

As we know, the area of the triangle = $\frac{1}{2}$ x side x height

= $\frac{1}{2}$ x 20 x h

[∵ Both areas remain same since they represent the area of the same triangle.]

⇒ $\frac{1}{2}$ x 20 x h = 96

⇒ 10 x h = 96

⇒ h = $\frac{96}{10}$

⇒ h = 9.6

Hence, height of the triangle corresponding to 20 cm side is 9.6 cm.

(iii) When, the smallest side of the triangle = 12 cm

Let the height of the triangle = h cm

As we know, the area of the triangle = $\frac{1}{2}$ x side x height

= $\frac{1}{2}$ x 12 x h

[∵ Both areas remain same since they represent the area of the same triangle.]

⇒ $\frac{1}{2}$ x 12 x h = 96

⇒ 6 x h = 96

⇒ h = $\frac{96}{6}$

⇒ h = 16 cm

Hence, height of the triangle corresponding to 12 cm side is 16 cm.

Given:

The base and the height of a triangle are in the ratio 4 : 5.

The area of the triangle = 40 m²

Let the base of the triangle be 4x m.

And, the height of the triangle be 5x m.

As we know, the area of a triangle = $\frac{1}{2} \times$ side $\times$ height

⇒ $\frac{1}{2}\times$ 4x $\times$ 5x = 40 m²

⇒ $\frac{1}{2}\times$ 20x² = 40 m²

⇒ 10x² = 40 m²

⇒ x² = $\frac{40}{10}$ m²

⇒ x² = 4 m²

⇒ x = $\sqrt{4}$ m

⇒ x = 2 m

So, base of the triangle = 4x

= 4 x 2 m

= 8 m

And, height of the triangle = 5x

= 5 x 2 m

= 10 m

Hence, base and height of the triangle are 8 m and 10 m, respectively.

Given:

The base and the height of a triangle are in the ratio 5 : 3.

The area of the triangle = 67.5 m²

Let the base of the triangle = 5x m

And, the height of the triangle = 3x m

As we know, the area of the triangle = $\frac{1}{2}$ x side x height

⇒ $\frac{1}{2}$ x 5x x 3x = 67.5 m²

⇒ $\frac{1}{2}$ x 15x² = 67.5 m²

⇒ 15x² = 2 $\times$ 67.5 m²

⇒ 15x² = 135 m²

⇒ x² = $\frac{135}{15}$ m²

⇒ x² = 9 m²

⇒ x = $\sqrt{9}$ m

⇒ x = 3 m

So, base of the triangle = 5x

= 5 x 3 m

= 15 m

And, height of the triangle = 3x

= 3 x 3 m

= 9 m

Hence, base and height of the triangle are 15 m and 9 m, respectively.

Let the side of the equilateral triangle be a.

It is given that the area of an equilateral triangle is $144\sqrt{3}\text{ cm}^2$.

∵ As we know, the area of the equilateral triangle = $\frac{\sqrt{3}}{4} \times \text{side}^2$

⇒ $\frac{\sqrt{3}}{4} \times a^2 = 144\sqrt{3}\text{ cm}^2$

⇒ $\frac{\cancel{\sqrt{3}}}{4} \times a^2 = 144\cancel{\sqrt{3}}\text{ cm}^2$

⇒ $\frac{1}{4} \times a^2 = 144\text{ cm}^2$

⇒ $a^2 = 4 \times 144\text{ cm}^2$

⇒ $a^2 = 576\text{ cm}^2$

⇒ $a = \sqrt{576}\text{ cm}^2$

⇒ a = 24 cm

Now, perimeter of equilateral triangle = 3 x side

= 3 x 24 cm

= 72 cm

Hence, the perimeter of the triangle is 72 cm.

Let the side of the equilateral triangle be a.

∵ As we know, the area of the equilateral triangle = $\frac{\sqrt{3}}{4} \times \text{ side}^2$

And, perimeter of equilateral triangle = 3 x side

It is given that the area of an equilateral triangle is numerically equal to its perimeter.

⇒ $\frac{\sqrt{3}}{4}a^2 = 3a$

⇒ $\sqrt{3}a^2 = 4 \times 3a$

⇒ $\sqrt{3}a^2 = 12a$

⇒ $\sqrt{3}a = 12$

⇒ $a = \frac{12}{\sqrt{3}}$

⇒ $a = \frac{12 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}}$

⇒ $a = \frac{12\sqrt{3}}{3}$

⇒ $a = 4\sqrt{3}$

As we know perimeter of equilateral triangle = 3 x side

= 3 x $4\sqrt{3}$ unit

= 12 $\sqrt{3}$ unit

= 12 x 1.74 unit

= 20.78 unit

Hence, the perimeter of the triangle is 20.78 unit.

Join BD and the field is divided into two triangular fields.

Triangle ABD is a right angled triangle.

AB = 18 m

AD = 24 m

Let BD be h m.

By using the Pythagorean theorem,

Base² + Height² = Hypotenuse²

⇒ (18)² + (24)² = h²

⇒ 324 + 576 = h²

⇒ h² = 900

⇒ h = $\sqrt{900}$

⇒ h = 30 m

For the right-angled triangle ABD,

As we know, the area of a triangle = $\frac{1}{2}$ x base x height

= $\frac{1}{2}$ x 18 x 24 m²

= $\frac{1}{2}$ x 432 m²

= 216 m²

For triangle BCD,

Let a = 40 m, b = 50 m and c = 30 m.

$$\begin{aligned}s = \frac{a + b + c}{2}\\[1em] = \frac{40 + 50 + 30}{2}\\[1em] = \frac{120}{2}\\[1em] = 60\end{aligned}$$

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

= $\sqrt{60(60 - 40)(60 - 50)(60 - 30)}$ m²

= $\sqrt{60 \times 20 \times 10 \times 30}$ m²

= $\sqrt{3,60,000}$ m²

= 600 m²

Area of rectangular field = Area of triangle ABD + Area of triangle BCD

= 216 + 600 m²

= 816 m²

Hence, the area of the field is 816 m².

Given:

Perimeter of the triangle = 96 cm

The lengths of the sides of a triangle are in the ratio 4 : 5 : 3.

Let the length of the sides of the triangle be 4a, 5a and 3a.

∵ Perimeter of triangle = sum of sides of the triangle.

⇒ 4a + 5a + 3a = 96 cm

⇒ 12a = 96 cm

⇒ a = $\frac{96}{12}$ cm

⇒ a = 8 cm

So, the lengths of the sides are 4a, 5a and 3a.

= 4 x 8, 5 x 8 and 3 x 8

= 32, 40 and 24

Let a = 32 m, b = 40 m and c = 24 m.

$$\begin{aligned}∵ s = \frac{a + b + c}{2}\\[1em] = \frac{32 + 40 + 24}{2}\\[1em] = \frac{96}{2}\\[1em] = 48\end{aligned}$$

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

= $\sqrt{48(48 - 32)(48 - 40)(48 - 24)}$ cm²

= $\sqrt{48 \times 16 \times 8 \times 24}$ cm²

= $\sqrt{1,47,456}$ cm²

= 384 cm²

Hence, the area of the triangle is 384 cm².

Given:

One of the equal sides of an isosceles triangle = 13 cm

Perimeter of the triangle = 50 cm

Let a be the length of the third side of the triangle.

∵ Perimeter of triangle = sum of sides

⇒ 13 + 13 + a = 50 cm

⇒ 26 + a = 50 cm

⇒ a = 50 - 26 cm

⇒ a = 24 cm

Let a = 13 m, b = 13 m and c = 24 m.

$$\begin{aligned}∵ s = \frac{a + b + c}{2}\\[1em] = \frac{13 + 13 + 24}{2}\\[1em] = \frac{50}{2}\\[1em] = 25\end{aligned}$$

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

= $\sqrt{25(25 - 13)(25 - 13)(25 - 24)}$ cm²

= $\sqrt{25 \times 12 \times 12 \times 1}$ cm²

= $\sqrt{3,600}$ cm²

= 60 cm²

Hence, the area of the triangle is 60 cm².

Given:

Ratio of altitude and base of the field = 6 : 5

Total cost = ₹ 49,57,200

Rate = ₹ 36,720 per hectare

As we know that area of the triangular field x rate = total cost

⇒ Area of the triangular field x $\frac{₹ 36,720}{1,00,00}$ = ₹ 49,57,200

⇒ Area of the triangular field = ₹ $\frac{49,57,200}{36,720} \times 10,000$

⇒ Area of the triangular field = ₹ $135 \times 10,000$

⇒ Area of the triangular field = 13,50,000 m²

Let the altitude be 6a and the base be 5a.

As we know, the area of a triangle = $\frac{1}{2}$ x base x height

⇒ 13,50,000 = $\frac{1}{2}$ x 6a x 5a

⇒ 13,50,000 = $\frac{1}{2}$ x 30a²

⇒ 13,50,000 = 15a²

⇒ a² = $\frac{13,50,000}{15}$

⇒ a² = 90000

⇒ a = $\sqrt{90000}$

⇒ a = 300

So, the altitude = 6a

= 6 x 300

= 1800 m

And, the base = 5a

= 5 x 300

= 1500 m

Hence, the dimensions of the field are 1800 m and 1500 m.

Given:

Base of the triangle = 24 cm

Hypotenuse of the triangle = 40 cm

Let h be the height of the triangle.

In a right-angled triangle, using the Pythagorean theorem,

Base² + Height² = Hypotenuse²

⇒ (24)² + h² = (40)²

⇒ 576 + h² = 1600

⇒ h² = 1600 - 576

⇒ h² = 1024

⇒ h = $\sqrt{1024}$

⇒ h = 32 cm

As we know, the area of a triangle = $\frac{1}{2}$ x side x height

= $\frac{1}{2}$ x 24 x 32 cm²

= $\frac{1}{2}$ x 768 cm²

= 384 cm²

Hence, the area of the triangle is 384 cm².

(i) Given:

In triangle ABC,

Base of the triangle = 24 cm

Height of the triangle = 7 cm

Let h be the hypotenuse of the triangle.

In a right-angled triangle, using the Pythagorean theorem,

Base² + Height² = Hypotenuse²

⇒ (24)² + (7)² = h²

⇒ 576 + 49 = h²

⇒ h² = 625

⇒ h = $\sqrt{625}$

⇒ h = 25 cm

Hence, the length of AC = 25 cm.

(ii) As we know, the area of a triangle = $\frac{1}{2}$ x side x height

= $\frac{1}{2}$ x 24 x 7 cm²

= $\frac{1}{2}$ x 168 cm²

= 84 cm²

Hence, the area of the triangle = 84 cm².

(iii) The base of the triangle = 25 cm

Let the height of the triangle be h.

As we know, the area of a triangle = $\frac{1}{2}$ x side x height

[∵ Both areas remain same since they represent the area of the same triangle.]

⇒ $\frac{1}{2}$ x 25 x h = 84

⇒ 25 x h = 2 x 84

⇒ 25 x h = 168

⇒ h = $\frac{168}{25}$

⇒ h = 6.7 cm

Hence, the length of BD = 6.7 cm.

Given:

The length and the breadth of a rectangle are in the ratio 9 : 5.

The area of the rectangle is 180 m².

Let the length of the rectangle be 9a and the breadth of the rectangle be 5a.

As we know, the area of the rectangle = length x breadth

⇒ 180 m² = 9a x 5a

⇒ 180 m² = 45a²

⇒ a² = $\frac{180}{45}$ m²

⇒ a² = 4 m²

⇒ a = $\sqrt{4}$ m

⇒ a = 2 m

So, the length of the rectangle is 9a.

= 9 x 2 m

= 18 m

Hence, option 4 is the correct option.

Given:

The adjacent sides of the rectangle are 16 m and 9 m.

As we know, the area of the rectangle = length x breadth

= 16 x 9 m²

= 144 m²

SInce the area of the rectangle is equal to the area of the square,

And, the area of the square = side²

⇒ side² = 144 m²

⇒ side = $\sqrt{144}$ m

⇒ side = 12 m

Hence, option 3 is the correct option.

Given:

The perimeter of the square is 16 cm.

As we know, the perimeter of a square = 4 x side.

⇒ 4 x side = 16 cm

⇒ side = $\frac{16}{4}$ cm

⇒ side = 4 cm

And, the area of the square = side²

= (4)² cm²

= 16 cm²

Hence, option 1 is the correct option.

Given:

The area of the square is 169 m².

Let the side of the square be x m.

As we know, the area of a square = side²

⇒ x² = 169 m²

⇒ x = $\sqrt{169}$ m

⇒ x = 13 m

And, the perimeter of the square = 4 x side

= 4 x 13 m

= 52 m

Hence, option 1 is the correct option.

Given:

The diagonal of the square is 10 cm.

Let each side of the square be a cm.

Then, by using the Pythagoras theorem,

⇒ Diagonal² = Side² + Side²

⇒ (10)² = a² + a²

⇒ 100 = 2a²

⇒ a² = $\frac{100}{2}$

⇒ a² = 50

⇒ a = $\sqrt{50}$

⇒ a = 5 $\sqrt{2}$

As we know, the area of a square = side²

= (5 $\sqrt{2}$) ² cm²

= 25 x 2 cm²

= 50 cm²

Hence, option 2 is the correct option.

Given:

The breadth of the rectangle is 8 cm.

The area of the rectangle is 120 cm².

Let the length of the rectangle be l.

As we know, the area of the rectangle = length x breadth

⇒ l x 8 = 120

⇒ l = $\frac{120}{8}$

⇒ l = 15 cm

The perimeter of the rectangle = 2(length + breadth)

= 2 (15 + 8)

= 2 x 23

= 46 cm

Hence, the length of the rectangle is 15 cm, and the perimeter is 46cm.

(i) Given:

The perimeter of the rectangle is 46 m.

The length of the rectangle is 15 m.

Let the breadth of the rectangle be b m.

As we know, the perimeter of the rectangle = 2(length + breadth)

⇒ 2(15 + b) = 46

⇒ 15 + b = $\frac{46}{2}$

⇒ 15 + b = 23

⇒ b = 23 - 15

⇒ b = 8 m

Hence, the breadth of the rectangle is 8 m.

(ii) The area of the rectangle = length x breadth

= 15 x 8 m²

= 120 m²

Hence, the area of the rectangle is 120 m².

(iii) By using Pythagoras theorem,

⇒ Diagonal² = Length² + Breadth²

⇒ Diagonal² = 15² + 8²

⇒ Diagonal² = 225 + 64

⇒ Diagonal² = 289

⇒ Diagonal = $\sqrt{289}$

⇒ Diagonal = 17 m

Hence, the diagonal of the rectangle is 17 m.

Given:

The diagonal of the rectangle is 34 cm.

The breadth of the rectangle is 16 cm.

Let the length of the rectangle be l cm.

By using Pythagoras theorem,

⇒ Diagonal² = Length² + Breadth²

⇒ (34)² = l² + (16)²

⇒ 1,156 = l² + 256

⇒ l² = 1,156 - 256

⇒ l² = 900

⇒ l = $\sqrt{900}$

⇒ l = 30 cm

Hence, the length of the rectangle is 30 cm.

(ii) As we know, the area of the rectangle = length x breadth

= 30 x 16 cm²

= 480 cm²

Hence, the area of the rectangle is 480 cm².

Given:

The area of a small rectangular plot is 84 m².

The difference between its length and the breadth is 5 m.

Let the length of the rectangle be l m.

So, breadth of the rectangle = l - 5

As we know, the area of the rectangle = length x breadth

⇒ l x (l - 5) = 84

⇒ l² - 5l = 84

⇒ l² - 5l - 84 = 0

⇒ l² - (12 - 7)l - 84 = 0

⇒ l² - 12l + 7l - 84 = 0

⇒ l(l - 12) + 7(l - 12) = 0

⇒ (l - 12)(l + 7) = 0

⇒ l = 12 or - 7

Since length cannot be negative, the length will be 12 m.

So, breadth of the rectangle = (l - 5)

= (12 - 5) m

= 7 m

And, perimeter of the rectangle = 2(length + breadth)

= 2(12 + 7) m

= 2 x 19 m

= 38 m

Hence, the perimeter of the rectangle is 38 m.

Given:

Diagonal of the square is 15 m.

Let each side of the square be a m.

Then, by using Pythagoras theorem,

⇒ Diagonal² = Side² + Side²

⇒ (15)² = a² + a²

⇒ 225 = 2a²

⇒ a² = $\frac{225}{2}$

⇒ a² = 112.5

⇒ a = $\sqrt{112.5}$

⇒ a = 10.6

As we know that perimeter of the square = 4 x side

= 4 x 10.6 m

= 42.4 m

Hence, the side of square is 10.6 m and its perimeter is 42.4 m.

Given:

The length of the rectangle is 16 cm.

The perimeter of the rectangle = The perimeter of a square.

The side of the square is 12.5 cm.

As we know, the perimeter of the square = 4 x side

= 4 x 12.5 cm

= 50 cm

Let the breadth of the rectangle be b cm.

And, the perimeter of the rectangle = 2 (length + breadth)

⇒ 2 (16 + b) = 50

⇒ 16 + b = $\frac{50}{2}$

⇒ 16 + b = 25

⇒ b = 25 - 16

⇒ b = 9 cm

Now, area of the rectangle = length x breadth

= 16 x 9 cm²

= 144 cm²

Hence, the area of the rectangle is 144 cm².

Given:

The perimeter of a square = The area of a square.

Let the side of the square be a unit.

As we know, the perimeter of the square = 4 x side

= 4a

And, the area of the square = side²

= a²

Thus, 4a = a²

⇒ a = 4 unit

Now, the area of the square = a²

= 4²

= 16 sq. units

Hence, the area of the square is 16 sq. units.

(i) Given:

Each side of a rectangle is doubled.

Let the length and breadth of the rectangle be l and b.

Thus, the new length and breadth of the rectangle are 2l and 2b.

As we know, the perimeter of a rectangle = 2 (length + breadth)

The perimeter of the original rectangle = 2 (l + b)

The perimeter of the new rectangle = 2 (2l + 2b)

= 4 (l + b)

Thus, the ratio of the perimeters of the original rectangle and the resulting rectangle = $\frac{2 (l + b)}{4 (l + b)}$

= $\frac{2 \cancel{(l + b)}}{4 \cancel{(l + b)}}$

= $\frac{2}{4}$

= $\frac{1}{2}$

Hence, the ratio of the perimeters of the original rectangle and the resulting rectangle is $\frac{1}{2}$.

(ii) As we know, the area of a rectangle = length x breadth

The area of the original rectangle = l x b

= lb

The area of the new rectangle = 2l x 2b

= 4 lb

Thus, the ratio of the areas of the original rectangle and the resulting rectangle = $\frac{lb}{4lb}$

= $\frac{\cancel{lb}}{4 \cancel{lb}}$

= $\frac{1}{4}$

Hence, the ratio of areas of the original rectangle and the resulting rectangle is $\frac{1}{4}$.

(i) The area of the shaded portion = Area of the rectangle - Area of the square.

It is given that the length and breadth of the rectangle area 3.2 m and 1.8 m.

As we know, the area of the rectangle = length x breadth

= 3.2 x 1.8 m²

= 5.76 m²

The side of the square is 1.4 m.

And, the area of the square = side²

= 1.4² m²

= 1.96 m²

Thus, the area of the shaded portion = 5.76 - 1.96 m²

= 3.8 m²

Hence, the area of the shaded portion is 3.8 m².

(ii) The area of the shaded portion = Area of the square - Area of the rectangle.

It is given that the side of the square is 6 m.

And, the area of the square = side²

= 6² m²

= 36 m²

The length and breadth of the rectangle are 4.8 m and 3.6 m, respectively.

As we know, the area of the rectangle = length x breadth

= 4.8 x 3.6 m²

= 17.28 m²

Area of the shaded portion = 36 - 17.28 m²

= 18.72 m²

Hence, the area of the shaded portion is 18.72 m².

Given:

The side of the square field is 21 m.

The width of the path is 3 m.

Area of the path = Area of bigger square field - Area of smaller square field

The side of the bigger square = 21 m + 3 m + 3m

= 27 m

As we know, the area of square = side²

⇒ Area of the bigger square = 27² m²

= 729 m²

⇒ Area of the smaller square = 21² m²

= 441 m²

So, the area of the path = 729 - 441 m²

= 288 m²

Hence, the area of the path is 288 m².

Given:

The length of the rectangular field is 30 m.

The breadth of the rectangular field is 27 m.

The width of the path is 2.5 m.

Area of the path = Area of bigger rectangular field - Area of smaller rectangular field

The length of the smaller rectangular field = 30 m - 2.5 m - 2.5 m

= 25 m

The breadth of the smaller rectangular field = 27 m - 2.5 m - 2.5 m

= 22 m

As we know, the area of a rectangle = length x breadth

⇒ Area of the bigger rectangular field = 30 x 27 m²

= 810 m²

⇒ Area of the smaller square = 25 x 22 m²

= 550 m²

So, the area of the path = 810 - 550 m²

= 260 m²

Hence, the area of the path is 260 m².

(i) Given:

Length of the floor = 18 m

Breadth of the floor = 13.5 m

As we know, the area of the floor = length x breadth

= 18 x 13.5 m²

= 243 m²

Side of the square tile = 25 cm

= $\frac{25}{100}$ m

= $\frac{1}{4}$ m

As we know, the area of a tile = side²

= $\Big(\frac{1}{4}\Big)$² m²

= $\frac{1}{16}$ m²

Area of square tiles x Number of tiles = Area of the floor

⇒ Number of tiles = $\frac{\text{Area of the floor}}{\text{Area of square tiles}}$

⇒ Number of tiles = $\frac{243}{\frac{1}{16}}$

⇒ Number of tiles = 243 x 16

⇒ Number of tiles = 3,888

Rate per tile = ₹ 6 per tile

Total cost = ₹ 3,888 x 6

= ₹ 23,328

Hence, the number of tiles is 3,888 and the cost of the tiles is ₹ 23,328.

(ii) Width of the margin on each side = 1.5 m

Length of the inner floor = 18 m - 1.5 m - 1.5 m

= 18 - 3 m

= 15 m

Breadth of the inner floor = 13.5 m - 1.5 m - 1.5 m

= 13.5 - 3 m

= 10.5 m

As we know, the area of the floor = length x breadth

= 15 x 10.5 m²

= 157.5 m²

Side of the square tile = 25 cm

= $\frac{25}{100}$ m

= $\frac{1}{4}$ m

As we know, the area of a tile = side²

= $\Big(\frac{1}{4}\Big)$² m²

= $\frac{1}{16}$ m²

Area of square tiles x Number of tiles = Area of the floor

⇒ Number of tiles = $\frac{\text{Area of the floor}}{\text{Area of square tiles}}$

⇒ Number of tiles = $\frac{157.5}{\frac{1}{16}}$

⇒ Number of tiles = 157.5 x 16

⇒ Number of tiles = 2,520

Rate of tiles = ₹ 6 per tile

Total cost = ₹ 2,520 x 6

= ₹ 15,120

Hence, the number of tiles is 2,520 and the cost of the tiles is ₹ 15,120.

Given:

Length of the rectangular field = 30 m

Width of the rectangular field = 22 m

Width of each road = 2.5 m

Area of crossroad IJKL (parallel to the width of field) = 2.5 x 22 m²

= 55 m²

Area of crossroad EFGH (parallel to the length of field) = 2.5 x 30 m²

= 75 m²

Side of the square MNOP = 2.5

As we know, the area of the square = side²

Area of the square field MNOP (area common to both roads) = 2.5² m²

= 6.25 m²

Now, area of crossroad = Area of cross road EFGH + Area of cross road IJKL - Area of square field MNOP

= 75 m² + 55 m² - 6.25 m²

= 130 m² - 6.25 m²

= 123.75 m²

Hence, the area of the crossroads is 123.75 m².

Given:

The length and the breadth of a rectangular field are in the ratio 5 : 4.

The area of the rectangular field = 3380 m²

Rate of fencing = ₹ 75 per meter

Let the length of the rectangular field be 5a and the breadth be 4a.

As we know, the area of the rectangle = length x breadth

⇒ 5a x 4a = 3380 m²

⇒ 20a² = 3380 m²

⇒ a² = $\frac{3380}{20}$ m²

⇒ a² = 169 m²

⇒ a = $\sqrt{169}$ m

⇒ a = 13 m

So, the length of the rectangular field = 5a

= 5 x 13 m

= 65 m

And, the breadth of the rectangular field = 4a.

= 4 x 13 m

= 52 m

Perimeter of the rectangle = 2(l + b)

= 2(65 + 52)

= 2 x 117

= 234 m

Total cost of fencing = Rate of fencing x Perimeter of fencing

= ₹ 75 x 234

= ₹ 17,550

Hence, the cost of fencing is ₹ 17,550.

(i) Given:

The length and the breadth of a conference hall are in the ratio 7 : 4.

The perimeter of the hall = 110 m

Let the length of the conference hall be 7a and the breadth be 4a.

As we know, the perimeter of the hall = 2(length + breadth)

⇒ 2(7a + 4a) = 110

⇒ 2 x 11a = 110

⇒ 22a = 110

⇒ a = $\frac{110}{22}$

⇒ a = 5

So, the length of the hall = 7a

= 7 x 5 m

= 35 m

And, the breadth of the hall = 4a

= 4 x 5 m

= 20 m

Area of the hall = length x breadth

= 35 x 20 m²

= 700 m²

Hence, the area of the conference hall is 700 m².

(ii) Given:

Length of the tile = 25 cm

= $\frac{25}{100}$ m

= $\frac{1}{4}$ m

Breadth of the tile = 20 cm

= $\frac{20}{100}$ m

= $\frac{1}{5}$ m

Area of the tile = length x breadth

= $\frac{1}{4} \times \frac{1}{5}$ m²

= $\frac{1}{20}$ m²

Area of hall = Area of tiles x Number of tiles

⇒ 700 = $\frac{1}{20}$ x Number of tiles

⇒ Number of tiles = 700 x 20

⇒ Number of tiles = 14,000

Hence, the number of tiles = 14,000.

(iii) Cost of tiles = ₹ 1,400 per 100 tiles

Total cost = $\frac{14,000 \times 1400}{100}$

= ₹ 1,96,000

Hence, the cost of tiles = ₹ 1,96,000.

Given:

The lengths of the parallel sides of the trapezium are 12 cm and 10 cm.

The distance between the parallel sides is 10 cm.

As we know, the area of a trapezium = $\frac{1}{2}$ x (sum of parallel sides) x height

= $\frac{1}{2}$ x (12 + 10) x 10 cm²

= $\frac{1}{2}$ x 22 x 10 cm²

= $\frac{1}{2}$ x 220 cm²

= 110 cm²

Hence, option 2 is the correct option.

Given:

Two parallel sides of a trapezium are in the ratio 2 : 3.

The distance between the parallel sides = 12 cm.

The area of the trapezium = 240 cm².

Let the parallel sides of a trapezium be 2a and 3a.

As we know, the area of a trapezium = $\frac{1}{2}$ x (sum of parallel sides) x height

⇒ $\frac{1}{2}$ x (2a + 3a) x 12 cm² = 240 cm²

⇒ $\frac{1}{2}$ x 5a x 12 = 240

⇒ $\frac{1}{2}$ x 60a = 240

⇒ 30a = 240

⇒ a = $\frac{240}{30}$

⇒ a = 8

So, the length of its larger parallel side = 3a

= 3 x 8 cm

= 24 cm

Hence, option 3 is the correct option.

Given:

The base of the parallelogram = 12 cm.

The height of the parallelogram = 6 cm.

As we know, the area of parallelogram = base x height

The area of the parallelogram using the longer side:

= 12 x 6 cm²

= 72 cm².

The base of the parallelogram = 9 cm.

Let the height of the parallelogram be b cm.

The area of the parallelogram using the shorter side:

= 9 x b cm²

Since the area is the same, we can equate both areas:

⇒ 9 x b cm² = 72 cm²

⇒ b = $\frac{72}{9}$ cm

⇒ b = 8 cm

Hence, option 3 is the correct option.

Given:

The base of the rhombus = 9 cm.

The height of the rhombus = 6 cm.

As we know, the area of a rhombus = base x height

= 9 x 6 cm²

= 54 cm²

Hence, option 1 is the correct option.

Given:

One diagonal of a rhombus = 15 cm.

Area = 60 cm²

Let the other diagonal of the rhombus be d.

As we know, the area of a rhombus = $\frac{1}{2}$ x product of its diagonals

⇒ $\frac{1}{2}$ x (15 x d) = 60

⇒ $\frac{15}{2}$ x d = 60

⇒ d = $\frac{60 \times 2}{15}$ cm

⇒ d = $\frac{120}{15}$ cm

⇒ d = 8 cm

Hence, option 2 is the correct option.

Given:

DC = 6.4 m

AD = 80 cm = $\frac{80}{100}$ m = 0.8 m

BC = 2.4 m

Area of the cross-section = Area of trapezium ABCD

= $\frac{1}{2}$ x (sum of parallel sides) x height

= $\frac{1}{2}$ x (0.8 + 2.4) x 6.4

= $\frac{1}{2}$ x 3.2 x 6.4

= $\frac{1}{2}$ x 20.48

= 10.24 m²

Hence, the area of cross-section is 10.24 m².

Given:

The parallel sides of a trapezium are in the ratio 3 : 4.

The distance between the parallel sides = 9 dm

The area = 126 dm²

Let the parallel sides of trapezium be 3a and 4a.

As we know, the area of a trapezium = $\frac{1}{2}$ x (sum of parallel side) x height

⇒ $\frac{1}{2}$ x (3a + 4a) x 9 = 126

⇒ $\frac{1}{2}$ x 7a x 9 = 126

⇒ $\frac{1}{2}$ x 63a = 126

⇒ 63a = 126 x 2

⇒ 63a = 252

⇒ a = $\frac{252}{63}$

⇒ a = 4

So, the parallel sides of the trapezium are:

For 3a:

= 3 x 4 dm = 12 dm

And, for 4a:

= 4 x 4 dm = 16 dm

Hence, the lengths of its parallel sides are 12 dm and 16 dm.

Given:

The two parallel sides and the distance between them are in the ratio 3 : 4 : 2.

The area of the trapezium is 175 cm².

So, the two parallel sides of the trapezium are 3a and 4a.

Distance between them is 2a.

As we know, the area of a trapezium = $\frac{1}{2}$ x (sum of parallel side) x height

⇒ $\frac{1}{2}$ x (3a + 4a) x 2a = 175

⇒ $\frac{1}{2}$ x 7a x 2a = 175

⇒ $\frac{1}{2}$ x 14a² = 175

⇒ 7a² = 175

⇒ a² = $\frac{175}{7}$

⇒ a² = 25

⇒ a = $\sqrt{25}$

⇒ a = 5

So, height of the trapezium = 2a

= 2 x 5 cm

= 10 cm

Hence, the height of the trapezium is 10 cm.

Let the parallelogram be ABCD as shown in the figure below:

For triangle ABC,

AB = 21 cm, BC = 28 cm and AC = 35 cm.

$$\begin{aligned}∵ s = \frac{a + b + c}{2}\\[1em] = \frac{21 + 28 + 35}{2}\\[1em] = \frac{84}{2}\\[1em] = 42\end{aligned}$$

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

= $\sqrt{42(42 - 21)(42 - 28)(42 - 35)}$ cm²

= $\sqrt{42 \times 21 \times 14 \times 7}$ cm²

= $\sqrt{86,436}$ cm²

= 294 cm²

∵ Diagonal of parallelogram divides it into two equal parts.

So, area of parallelogram ABCD = 2 x area of triangle ABC.

= 2 x 294 cm²

= 588 cm²

Hence, the area of the parallelogram is 588 cm².

(i) Given:

The diagonals of a rhombus are 18 cm and 24 cm.

As we know, the area of rhombus = $\frac{1}{2}$ x product of its diagonal

= $\frac{1}{2}$ x 18 x 24 cm²

= $\frac{1}{2}$ x 432 cm²

= 216 cm²

Hence, the area of the rhombus is 216 cm².

(ii) AC = 18 cm

Then, OA = OC = $\frac{18}{2}$ = 9 cm

And, BD = 24 cm

Then, OB = OD = $\frac{24}{2}$ = 12 cm

Since the diagonals of a rhombus bisect at 90°.

Applying pythagoras theorem in triangle AOB, we get:

AB² = OA² + OB²

⇒ AB² = (9)² + (12)²

⇒ AB² = 81 + 144

⇒ AB² = 225

⇒ AB = $\sqrt{225}$

⇒ AB = 15 cm

Hence, the length of each side of the rhombus is 15 cm.

(iii) As we know, the perimeter of the rhombus = 4 x side

= 4 x 15 cm

= 60 cm

Hence, the perimeter of the rhombus is 60 cm.

(i) Given:

The perimeter of a rhombus = 40 cm.

One diagonal = 16 cm.

As we know, the perimeter of the rhombus = 4 x side

⇒ 4 x side = 40 cm

⇒ Side = $\frac{40}{4}$ cm

⇒ Side = 10 cm

AB = 10 cm

AC = 16 cm

Then, OA = OC = $\frac{16}{2}$ = 8 cm

Let OB be a cm.

Since the diagonal of a rhombus bisect at 90°.

Applying pythagoras theorem in triangle AOB, we get:

AB² = OA² + OB²

⇒ (10)² = (8)² + a²

⇒ 100 = 64 + a²

⇒ a² = 100 - 64

⇒ a² = 36

⇒ a = $\sqrt{36}$

⇒ a = 6

So, OB = 6 cm

BD = 2 x 6 cm

= 12 cm

Hence, the length of other diagonal is 12 cm.

(ii) As we know that area of rhombus = $\frac{1}{2}$ x product of its diagonal

= $\frac{1}{2}$ x 16 x 12 cm²

= $\frac{1}{2}$ x 192 cm²

= 96 cm²

Hence, the area of the rhombus is 96 cm².

Given:

The length of the diagonals of a rhombus is in the ratio 4 : 3.

The area = 384 cm²

Let the length of the diagonals be 4a and 3a.

As we know, the area of rhombus = $\frac{1}{2}$ x product of its diagonal

⇒ $\frac{1}{2}$ x 4a x 3a = 384

⇒ $\frac{1}{2}$ x 12a² = 384

⇒ 6a² = 384

⇒ a² = $\frac{384}{6}$

⇒ a² = 64

⇒ a = $\sqrt{64}$

⇒ a = 8

The length of the diagonals be 4a and 3a.

= 4 x 8 cm and 3 x 8 cm

= 32 cm and 24 cm

AC = 32 cm

Then, OA = OC = $\frac{32}{2}$ = 16 cm

And, BD = 24 cm

Then, OB = OD = $\frac{24}{2}$ = 12 cm

Since the diagonal of a rhombus bisect at 90°.

Applying pythagoras theorem in ΔAOB, we get:

AB² = OA² + OB²

⇒ AB² = (16)² + (12)²

⇒ AB² = 256 + 144

⇒ AB² = 400

⇒ AB = $\sqrt{400}$

⇒ AB = 20 cm

Hence, the length of its side is 20 cm.

Given:

Side of iron sheet (AB) = 10 m

Diagonal of the sheet (AC) = 16 m

Join BD.

Then, OA = OC = $\frac{AC}{2}$ = $\frac{16}{2}$ = 8 m

Since the diagonals of a rhombus bisect each other at 90°, applying the pythagoras theorem in Δ AOB, we get:

AB² = OA² + OB²

⇒ (10)² = (8)² + OB²

⇒ 100 = 64 + OB²

⇒ OB² = 100 - 64

⇒ OB² = 36

⇒ OB = $\sqrt{36}$

⇒ OB = 6 m

Thus, BD = 2 x OB = 2 x 6 m = 12 m

The area of rhombus = $\frac{1}{2}$ x product of diagonals

= $\frac{1}{2}$ x 16 x 12 m²

= $\frac{1}{2}$ x 192 m²

= 96 m²

The rate of painting is ₹ 6 per m².

Therefore, the total cost of painting is:

Total cost = 2 x area of sheet x rate of painting

= 2 x 96 x 6

= 192 x 6

= ₹ 1,152

We also know that the area of the rhombus = base x height

⇒ 96 = 10 x height

⇒ height = $\frac{96}{10}$

⇒ height = 9.6 m

Hence, the total cost of painting is ₹ 1,152 and the distance between the opposite sides of the rhombus is 9.6 m.

Given:

The area of a trapezium = 279 sq. cm

The distance between its two parallel sides = 18 cm.

One of the parallel sides is longer than the other side by 5 cm.

Let one of the parallel sides be a. Then, the other parallel side is a - 5.

As we know, the area of a trapezium = $\frac{1}{2}$ (sum of parallel sides) x distance between the parallel sides

Substituting the given values:

⇒ 279 = $\frac{1}{2}$ (a + a - 5) x 18

⇒ 279 = $\frac{1}{2}$ (2a - 5) x 18

⇒ 279 = $\frac{1}{2}$ (36a - 90)

⇒ 279 = 18a - 45

⇒ 18a = 279 + 45

⇒ 18a = 324

⇒ a = $\frac{324}{18}$

⇒ a = 18 cm

So, the parallel sides are 18 cm and (18 - 5) = 13 cm.

Hence, the lengths of the parallel sides are 18 cm and 13 cm, respectively.

Given:

The area of a rhombus = The area of a triangle

The base of the triangle = 24 cm

The altitude of the triangle = 16 cm

One of the diagonals of the rhombus = 19.2 cm

As we know, the area of a triangle = $\frac{1}{2}$ x base x altitude

= $\frac{1}{2}$ x 24 x 16 cm²

= $\frac{1}{2}$ x 384 cm²

= 192 cm²

Let the length of the other diagonal of the rhombus be d.

The area of the rhombus = $\frac{1}{2}$ x (product of diagonals)

⇒ $\frac{1}{2}$ x (19.2 x d) = 192

⇒ 9.6 x d = 192

⇒ d = $\frac{192}{9.6}$

⇒ d = 20 cm

Hence, the length of the other diagonal is 20 cm.

Given:

AB = 18 cm

∠B = ∠C = 90°

CD = 12 cm

AD = 10 cm

Draw DL perpendicular to AB.

Since, AL = AB - DC

= 18 - 12

= 6 cm

Also, DL = BC (Both are perpendiculars)

Using pythagoras theorem,

AD² = AL² + LD²

⇒ 10² = 6² + LD²

⇒ 100 = 36 + LD²

⇒ LD² = 100 - 36

⇒ LD² = 64

⇒ LD = $\sqrt{64}$

⇒ LD = 8 cm

As we know, the area of a trapezium = $\frac{1}{2}$ (sum of parallel sides) x distance betweem the parallel sides.

= $\frac{1}{2}$ (18 + 12) x 8

= $\frac{1}{2}$ x 30 x 8

= $\frac{1}{2}$ x 240

= 120 cm²

Hence, the area of the trapezium is 120 cm².

Given:

The circumference of a circle = The area of a circle

Let r be the radius of the circle.

As we know, the circumference of the circle = 2πr

And, the area of the circle = πr²

So,

⇒ 2πr = πr²

⇒ 2r = r²

⇒ 2 = r

Hence, the radius of the circle is 2 units.

Hence, option 2 is the correct option.

Given:

The perimeter of a square = The circumference of a circle

Side of the square = 11 cm

As we know, the perimeter of a square = 4 x side

= 4 x 11 cm

= 44 cm

Let r be the radius of the circle.

And, the circumference of the circle = 2πr

⇒ 2πr = 44

Substitute π = $\frac{22}{7}$ :

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

So, the diameter of the circle = 2 x radius

= 2 x 7 cm

= 14 cm

Hence, option 1 is the correct option.

Given:

The circumference of a circle = The sum of the circumferences of two circles with radii 10 cm and 12 cm.

As we know, the circumference of a circle = 2πr

For r = 10 cm:

Circumference of the circle = 2 x $\frac{22}{7}$ x 10

= $\frac{44}{7}$ x 10

= $\frac{440}{7}$

For r = 12 cm

Circumference of the circle = 2 x $\frac{22}{7}$ x 12

= $\frac{44}{7}$ x 12

= $\frac{528}{7}$

Let R be the radius of the larger circle.

The circumference of the larger circle =

$$\begin{aligned}\frac{440}{7} + \frac{528}{7}\\[1em] ⇒ 2πR = \frac{440 + 528}{7}\\[1em] ⇒ 2 \times \frac{22}{7} \times R = \frac{440 + 528}{7}\\[1em] ⇒ \frac{44}{7} \times R = \frac{968}{7}\\[1em] ⇒ R = \frac{968 \times 7}{7 \times 44}\\[1em] ⇒ R = \frac{968 \times \cancel{7}}{\cancel{7} \times 44}\\[1em] ⇒ R = \frac{968}{44}\\[1em] ⇒ R = 22 \text{ cm}\end{aligned}$$

Hence, option 3 is the correct option.

Given:

The diameter of a circular wheel = 56 cm

Total number of rounds = 100

Since diameter = 2 x Radius

⇒ Radius = $\frac{\text{Diameter}}{2}$

⇒ Radius = $\frac{56}{2}$

= 28 cm

As we know, the circumference of a circle = 2πr

$$\begin{aligned}= 2 \times \frac{22}{7} \times 28\\[1em] = \frac{44}{7} \times 28\\[1em] = \frac{1,232}{7}\\[1em] = 176 \text{ cm}\end{aligned}$$

Total distance = circumference of the circle x number of rounds

= 176 x 100

= 17600 cm

Hence, option 2 is the correct option.

Given:

Radius of circular wheel = 3.5 cm

Total rounds = 600

Total time = 48 seconds

As we know, the circumference of a circle = 2πr

$$\begin{aligned}= 2 \times \frac{22}{7} \times 3.5\\[1em] = \frac{44}{7} \times 3.5\\[1em] = \frac{154}{7}\\[1em] = 22 \text{ cm}\end{aligned}$$

Total distance = The circumference of the circle x number of rounds

= 22 x 600

= 13200 cm

And, Speed = $\frac{\text{Distance}}{\text{Time}}$

= $\frac{13200}{48}$

= 275 cm/s

Hence, option 1 is the correct option.

(i) Let r be the radius of the circle.

The circumference of a circle = 132 cm

As we know, the circumference of the circle = 2πr

$$\begin{aligned}⇒ 2 \times \frac{22}{7} \times r = 132\\[1em] ⇒ \frac{44}{7} \times r = 132\\[1em] ⇒ r = \frac{132 \times 7}{44}\\[1em] ⇒ r = \frac{924}{44}\\[1em] ⇒ r = 21 \text{ cm}\end{aligned}$$

And, area of the circle = πr²

$$\begin{aligned}= \frac{22}{7} \times 21^2\\[1em] = \frac{22}{7} \times 441\\[1em] = \frac{9,702}{7}\\[1em] = 1,386 \text{ cm}^2\end{aligned}$$

Hence, the radius of the circle is 21 cm and the area is 1,386 cm².

(ii) Let r be the radius of the circle.

The circumference of a circle = 22 cm

As we know, the circumference of the circle = 2πr

$$\begin{aligned}⇒ 2 \times \frac{22}{7} \times r = 22\\[1em] ⇒ \frac{44}{7} \times r = 22\\[1em] ⇒ r = \frac{22 \times 7}{44}\\[1em] ⇒ r = \frac{154}{44}\\[1em] ⇒ r = 3.5 \text{ m}\end{aligned}$$

And, area of the circle = πr²

$$\begin{aligned}= \frac{22}{7} \times 3.5^2\\[1em] = \frac{22}{7} \times 12.25\\[1em] = \frac{269.50}{7}\\[1em] = 38.5 \text{ m}^2\end{aligned}$$

Hence, the radius of the circle is 3.5 m and the area is 38.5 m².

(i) Let r be the radius of the circle.

The area of circle = 154 cm²

As we know, the area of the circle = πr²

$$\begin{aligned}⇒ \frac{22}{7} \times r^2 = 154\\[1em] ⇒ r^2 = \frac{154 \times 7}{22}\\[1em] ⇒ r^2 = \frac{1,078}{22}\\[1em] ⇒ r^2 = 49\\[1em] ⇒ r = \sqrt{49}\\[1em] ⇒ r = 7 \text{ cm}\end{aligned}$$

Circumference of the circle = 2πr

$$\begin{aligned}= 2 \times \frac{22}{7} \times 7\\[1em] = \frac{44}{7} \times 7\\[1em] = \frac{308}{7}\\[1em] = 44 \text{ cm}\end{aligned}$$

Hence, the radius of the circle is 7 cm and the circumference is 44 cm.

(ii) Let r be the radius of the circle.

The area of circle = 6.16 m²

As we know, the area of the circle = πr²

$$\begin{aligned}⇒ \frac{22}{7} \times r^2 = 6.16\\[1em] ⇒ r^2 = \frac{6.16 \times 7}{22}\\[1em] ⇒ r^2 = \frac{43.12}{22}\\[1em] ⇒ r^2 = 1.96\\[1em] ⇒ r = \sqrt{1.96}\\[1em] ⇒ r = 1.4 \text{ m}\end{aligned}$$

Circumference of the circle = 2πr

$$\begin{aligned}= 2 \times \frac{22}{7} \times 1.4\\[1em] = \frac{44}{7} \times 1.4\\[1em] = \frac{61.6}{7}\\[1em] = 8.8 \text{ m}\end{aligned}$$

Hence, the radius of the circle is 1.4 m and the circumference is 8.8 m.

Given:

The circumference of a circular table = 88 m

Let r be the radius of the circle.

As we know, the circumference of the circle = 2πr

$$\begin{aligned}⇒ 2 \times \frac{22}{7} \times r = 88\\[1em] ⇒ \frac{44}{7} \times r = 88\\[1em] ⇒ r = \frac{88 \times 7}{44}\\[1em] ⇒ r = \frac{616}{44}\\[1em] ⇒ r = 14 \text{ m}\end{aligned}$$

The area of the circle = πr²

$$\begin{aligned}= \frac{22}{7} \times 14^2\\[1em] = \frac{22}{7} \times 196\\[1em] = \frac{4,312}{7}\\[1em] = 616 \text{ m}^2\end{aligned}$$

Hence, the area of the circular table is 616 m².

Given:

The area of a circle = 1386 sq. cm

Let r be the radius of the circle.

As we know, the area of the circle = πr²

$$\begin{aligned}⇒ \frac{22}{7} \times r^2 = 1386\\[1em] ⇒ r^2 = \frac{1386 \times 7}{22}\\[1em] ⇒ r^2 = \frac{9,702}{22}\\[1em] ⇒ r^2 = 441\\[1em] ⇒ r = \sqrt{441}\\[1em] ⇒ r = 21 \text{ cm}\end{aligned}$$

Circumference of the circle = 2πr

$$\begin{aligned}= 2 \times \frac{22}{7} \times 21\\[1em] = \frac{44}{7} \times 21\\[1em] = \frac{324}{7}\\[1em] = 132 \text{ cm}\end{aligned}$$

Hence, the circumference of the circle is 132 cm.

Given:

Radius of outer circle = 9 cm

Radius of inner circle = 5 cm

Area of circular ring = Area of outer circle - Area of inner circle

As we know, the area of the circle = πr²

Area of circular ring =

$$\begin{aligned}\frac{22}{7} \times 9^2 - \frac{22}{7} \times 5^2\\[1em] = \frac{22}{7} \times 81 - \frac{22}{7} \times 25\\[1em] = \frac{1,782}{7} - \frac{550}{7}\\[1em] = \frac{1,782 - 550}{7}\\[1em] = \frac{1,232}{7}\\[1em] = 176 \text{ cm}^2\end{aligned}$$

Hence, the area of the circular ring is 176 cm².

(i) Given:

Radius of outer circle = 70 m

Radius of inner circle = 63 m

Area of circular track = Area of outer circle - Area of inner circle

As we know, the area of the circle = πr²

Area of circular ring =

$$\begin{aligned}\frac{22}{7} \times 70^2 - \frac{22}{7} \times 63^2\\[1em] = \frac{22}{7} \times 4,900 - \frac{22}{7} \times 3,969\\[1em] = \frac{1,07,800}{7} - \frac{87,318}{7}\\[1em] = \frac{1,07,800 - 87,318}{7}\\[1em] = \frac{20,482}{7}\\[1em] = 2,926 \text{ m}^2\end{aligned}$$

Hence, the area of the circular track is 2,926 m².

(ii) The difference between the lengths of the two circumferences of the track = Circumference of outer circle - Circumference of inner circle

As we know, the circumference of the circle = 2πr

The difference between the lengths of the two circumferences of the track =

$$\begin{aligned}2 \times \frac{22}{7} \times 70 - 2 \times \frac{22}{7} \times 63\\[1em] = \frac{44}{7} \times 70 - \frac{44}{7} \times 63\\[1em] = \frac{3080}{7} - \frac{2772}{7}\\[1em] = \frac{3080 - 2772}{7}\\[1em] = \frac{308}{7}\\[1em] = 44 \text{ m}\end{aligned}$$

Hence, the difference between the lengths of the two circumferences of the track is 44 m.

Given:

Radius of outer circular field = 105 m

Radius of inner circular field = 105 - 5 m

= 100 m

Area of circular ring = Area of outer circular field - Area of inner circular field

As we know, the area of the circle = πr²

Area of circular ring =

$$\begin{aligned}\frac{22}{7} \times 105^2 - \frac{22}{7} \times 100^2\\[1em] = \frac{22}{7} \times 11,025 - \frac{22}{7} \times 10,000\\[1em] = \frac{2,42,500}{7} - \frac{2,20,000}{7}\\[1em] = \frac{2,42,500 - 2,20,000}{7}\\[1em] = \frac{22,550}{7}\\[1em] = 3221\frac{3}{7} \text{ m}^2\end{aligned}$$

Hence, the area of circular field is $3221\frac{3}{7}$ m².

Given:

Area of the square = 484 cm²

Let s be the side of the square.

As we know, the area of the square = side²

⇒ s² = 484

⇒ s = $\sqrt{484}$

⇒ s = 22

Hence, one side of square is 22 cm.

(ii) Total length of the wire = perimeter of the square

As we know, the perimeter of the square = 4 x side

= 4 x 22

= 88 cm

Hence, the length of the wire is 88 cm.

(iii) Perimeter of the square = circumference of the circle

Let r be the radius of the circle.

⇒ 2πr = 88 cm

$$\begin{aligned}⇒ 2 \times \frac{22}{7} \times r = 88\\[1em] ⇒ \frac{44}{7} \times r = 88\\[1em] ⇒ r = \frac{7 \times 88}{44}\\[1em] ⇒ r = \frac{616}{44}\\[1em] ⇒ r = 14 \text{ cm}\end{aligned}$$

Area of the circle = πr²

$$\begin{aligned}= \frac{22}{7} \times 14^2\\[1em] = \frac{22}{7} \times 196\\[1em] = \frac{4312}{7}\\[1em] = 616 \text{ cm}^2\end{aligned}$$

Hence, the largest area that can be enclosed when the same wire is bent to form a circle is 616 cm².

Given:

Area of the square = 196 cm²

Let s be the side of the square.

As we know, the area of the square = side²

⇒ s² = 196

⇒ s = $\sqrt{196}$

⇒ s = 14

As we know, the perimeter of the square = 4 x side

= 4 x 14

= 56 cm

Perimeter of the square = circumference of the circle

Let r be the radius of the circle.

⇒ 2πr = 56

$$\begin{aligned}⇒ 2 \times \frac{22}{7} \times r = 56\\[1em] ⇒ \frac{44}{7} \times r = 56\\[1em] ⇒ r = \frac{7 \times 56}{44}\\[1em] ⇒ r = \frac{392}{44}\\[1em] ⇒ r = 8\frac{10}{11} \text{ cm}\end{aligned}$$

Area of the circle = πr²

$$\begin{aligned}= \frac{22}{7} \times \Big(8\frac{10}{11}\Big)^2\\[1em] = \frac{22}{7} \times \Big(\frac{98}{11}\Big)^2\\[1em] = \frac{22}{7} \times \Big(\frac{9604}{121}\Big)\\[1em] = \frac{211,288}{847}\\[1em] = 249.45 \text{ cm}^2\end{aligned}$$

Hence, the area of the circle is 249.45 cm².

(i) Given:

The radius of a circular wheel = 42 cm = 0.42 m

As we know, the circumference of the circle = 2πr

$$\begin{aligned}2 \times \frac{22}{7} \times 0.42\\[1em] = \frac{44}{7} \times 0.42\\[1em] = \frac{18.48}{7}\\[1em] = 2.64 \text{ m}\end{aligned}$$

Hence, the distance traveled by wheel in 1 revolution is 2.64 m.

(ii) Distance traveled by wheel in 50 revolutions = Circumference of the wheel x 50

= 2.64 x 50 m

= 132 m

Hence, the distance traveled by wheel in 50 revolutions is 132 m.

(ii) Distance traveled by wheel in 200 revolutions = Circumference of the wheel x 200

= 2.64 x 200 m

= 528 m

Hence, the distance traveled by wheel in 200 revolutions is 528 m.

(i)Given:

Diameter of wheel = 0.70 m

Radius = $\frac{\text{Diameter}}{2}$

= $\frac{0.70}{2}$ m

= 0.35 m

As we know, the circumference of the circle = 2πr

$$\begin{aligned}2 \times \frac{22}{7} \times 0.35\\[1em] = \frac{44}{7} \times 0.35\\[1em] = \frac{15.40}{7}\\[1em] = 2.2 \text{ m}\end{aligned}$$

Distance traveled by wheel in 500 revolutions = Circumference of the wheel x 500

= 2.2 x 500 m

= 1100 m

Speed = $\frac{\text{Distance}}{\text{Time}}$

Time taken by wheel = 5 min = 5 x 60 sec = 300 sec.

$$\begin{aligned}\text{Speed} = \frac{1100}{300}\\[1em] = 3\frac{2}{3} \text{ m}/\text{s}\end{aligned}$$

Hence, the speed = $3\frac{2}{3}$ m/s.

(ii) Speed = $3\frac{2}{3}$ m/s

$$\begin{aligned}= 3\frac{2}{3} \times \frac{3600}{1000} \text{ km}/\text{hr}\\[1em] = \frac{11}{3} \times \frac{3600}{1000} \text{ km}/\text{hr}\\[1em] = \frac{39600}{3000} \text{ km}/\text{hr}\\[1em] = 13.2 \text{ km}/\text{hr}\end{aligned}$$

Hence, the speed = 13.2 km/hr.

Given:

Diameter of wheel = 56 cm

Diameter of wheel in Kilometre = 0.00056 km

Radius = $\frac{\text{Diameter}}{2}$

= $\frac{0.00056}{2}$

= 0.00028 km

As we know, the circumference of the circle = 2πr

$$\begin{aligned}= 2 \times \frac{22}{7} \times 0.00028\\[1em] = \frac{44}{7} \times 0.00028\\[1em] = \frac{0.01232}{7}\\[1em] = 0.00176 \text{ km}\end{aligned}$$

Distance traveled by wheel in 45 revolutions = Circumference of the wheel x 45

= 0.00176 x 45 km

= 0.0792 km

And, speed = $\frac{\text{Distance}}{\text{Time}}$

It is given that time taken by wheel = 10 sec = $\frac{10}{3600}$.

$$\begin{aligned}\text{Speed} = \frac{0.0792 \times 3600}{10}\\[1em] = \frac{285.12}{10}\\[1em] = 28.512 \text{ km}/\text{hr}\end{aligned}$$

Hence, the speed = 28.512 km/hr.

(i) Given:

Diameter of roller = 1.4 m

Radius = $\frac{\text{Diameter}}{2}$

= $\frac{1.4}{2}$ m

= 0.7 m

As we know, the circumference of the circle = 2πr

$$\begin{aligned}2 \times \frac{22}{7} \times 0.7\\[1em] = \frac{44}{7} \times 0.7\\[1em] = \frac{30.8}{7}\\[1em] = 4.4 \text{ m}\end{aligned}$$

Hence, the circumference of the roller is 4.4 m.

(ii) Let the roller make a revolutions while traveling 61.6 m.

Distance travelled by roller in a revolutions = Circumference of the wheel x a

⇒ 61.6 = 4.4 x a

⇒ a = $\frac{61.6}{4.4}$

⇒ a = 14

Hence, the roller makes 14 revolutions while traveling 61.6 m.

Given:

Length of circumference = Sum of the lengths of the circumferences with radii 15 cm and 13 cm

Let r be the radius of the circle.

As we know, the circumference of the circle = 2πr

So,

⇒ 2πr = 2π x 15 + 2π x 13

$$\begin{aligned}⇒ 2 \times \frac{22}{7} \times r = 2 \times \frac{22}{7} \times 15 + 2 \times \frac{22}{7} \times 13\\[1em] ⇒ \frac{44}{7} \times r = \frac{44}{7} \times 15 + \frac{44}{7} \times 13\\[1em] ⇒ \frac{44}{7} \times r = \frac{660}{7} + \frac{572}{7}\\[1em] ⇒ \frac{44}{7} \times r = \frac{660 + 572}{7}\\[1em] ⇒ \frac{44}{7} \times r = \frac{1,232}{7}\\[1em] ⇒ r = \frac{1,232 \times 7}{7 \times 44}\\[1em] ⇒ r = \frac{1,232 \times \cancel{7}}{\cancel{7} \times 44}\\[1em] ⇒ r = \frac{1,232}{44}\\[1em] ⇒ r = 28 \text{ cm}\end{aligned}$$

So, radius of the circle = 28 cm

And, area of the circle = πr²

$$\begin{aligned}= \frac{22}{7} \times 28^2\\[1em] = \frac{22}{7} \times 784\\[1em] = \frac{17,248}{7}\\[1em] = 2,464 \text{ cm}^2\end{aligned}$$

Hence, the area of the circle is 2,464 cm².

Given:

Total length of the wire = 108 cm

Let r be the radius of the circle.

Total length of wire = Circumference of the semicircle + diameter

As we know, the circumference of the semicircle = πr

$$\begin{aligned}⇒ 108 = \frac{22}{7} \times r + 2r\\[1em] ⇒ 108 = \frac{22}{7}r + \frac{2}{1}r\\[1em] ⇒ 108 = \frac{22}{7}r + \frac{2 \times 7}{1 \times 7}r\\[1em] ⇒ 108 = \frac{22}{7}r + \frac{14}{7}r\\[1em] ⇒ 108 = \frac{22 + 14}{7}r\\[1em] ⇒ 108 = \frac{36}{7}r\\[1em] ⇒ r = \frac{7 \times 108}{36}\\[1em] ⇒ r = \frac{756}{36}\\[1em] ⇒ r = 21 \text{ cm}\end{aligned}$$

And, area of the semicircle = $\frac{1}{2}$πr²

$$\begin{aligned}= \frac{1}{2} \times \frac{22}{7} \times 21^2\\[1em] = \frac{22}{14} \times 441\\[1em] = \frac{9,702}{14}\\[1em] = 693 \text{ cm}^2\end{aligned}$$

Hence, the radius of the circle is 21 cm and the area is 693 cm².

Given:

BC = 14 cm

AB = 14 + 14 + 14 + 7

= 49 cm

Area of shaded portion = Area of rectangle ABCD - (3 x Area of circle + Area of semicircle)

As we know, the area of rectangle = length x breadth

= 49 x 14 cm²

= 686 cm²

Diameter of the circle = 14 cm

Radius of the circle = $\frac{14}{2}$ cm = 7 cm

Area of the circle = πr²

$$\begin{aligned}\text{Area of the circle} = \frac{22}{7} \times 7^2\\[1em] = \frac{22}{7} \times 49\\[1em] = \frac{1078}{7}\\[1em] = 154 \text{ cm}^2\end{aligned}$$

Area of 3 circles = 3 x 154 cm² = 462 cm²

Area of the semicircle = $\frac{1}{2}$πr²

$$\begin{aligned}= \frac{1}{2} \times \frac{22}{7} \times 7^2\\[1em] = \frac{22}{14} \times 49\\[1em] = \frac{1078}{14}\\[1em] = 77 \text{ cm}^2\end{aligned}$$

⇒ Area of shaded portion = 686 - (462 + 77) cm²

= 686 - 539 cm²

= 147 cm²

Hence, the area of the shaded portion is 147 cm².

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

$s = \frac{a + b + c}{2}$

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

When each side of a triangle is doubled, then 2a, 2b and 2c are sides of new triangle.

$$\begin{aligned}s' = \frac{a + b + c}{2}\\[1em] s' = \frac{2a + 2b + 2c}{2}\\[1em] s' = \frac{2(a + b + c)}{2}\\[1em] s' = 2\frac{(a + b + c)}{2}\\[1em] s' = 2s\end{aligned}$$

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

= $\sqrt{2s(2s - 2a)(2s - 2b)(2s - 2c)}$

= $\sqrt{2s2(s - a)2(s - b)2(s - c)}$

= $\sqrt{16s(s - a)(s - b)(s - c)}$

= 4 $\sqrt{s(s - a)(s - b)(s - c)}$

Ratio of original triangle to new triangle = $\frac{\sqrt{s(s - a)(s - b)(s - c)}}{4\sqrt{s(s - a)(s - b)(s - c)}}$

= $\frac{1}{4}$

Hence, option 3 is the correct option.

Given:

The perimeter of a rectangle = The perimeter of a square

Each side of the square = 30 cm

Length of the rectangle = 40 cm

Let b be the breadth of the rectangle.

As we know that perimeter of the rectangle = 2(length + breadth)

And, perimeter of the square = 4 x side

⇒ 2(40 + b) = 4 x 30

⇒ 80 + 2b = 120

⇒ 2b = 120 - 80

⇒ 2b = 40

⇒ b = $\frac{40}{2}$

⇒ b = 20

Hence, option 4 is the correct option.

Given:

The lengths of the parallel sides of the trapezium are 6 unit and 14 unit.

The area of a trapezium = 80 square unit

Let d be the distance between parallel sides.

As we know, the area of a trapezium = $\frac{1}{2}$ x (sum of parallel sides) x height

⇒ $\frac{1}{2}$ x (6 + 14) x d = 80

⇒ $\frac{1}{2}$ x 20 x d = 80

⇒ 10 x d = 80

⇒ d = $\frac{80}{10}$

⇒ d = 8 unit

Hence, option 2 is the correct option.

Given:

Radius of outer circle = 10 cm

Radius of inner circle = 8 cm

Area of circular ring = Area of outer circle - Area of inner circle

As we know, the area of the circle = πr²

⇒ Area of circular ring = π x (10)² - π x (8)²

= 100π - 64π

= 36π cm²

Hence, option 1 is the correct option.

Given:

The adjacent sides of a rectangle are 20 cm and 24 cm.

The perimeter of rectangle = The circumference of a circle

Let r be the radius of the circle.

As we know, the perimeter of the rectangle = 2(length + breadth)

And, the circumference of the circle = 2πr

⇒ 2(20 + 24) = 2 x $\frac{22}{7}$ x r

⇒ 2 x 44 = $\frac{44}{7}$ x r

⇒ 88 = $\frac{44}{7}$ x r

⇒ r = $\frac{7 \times 88}{44}$

⇒ r = $\frac{616}{44}$

⇒ r = 14 cm

Hence, option 1 is the correct option.

Given,

AB (base) = 12 cm, perpendicular dropped from D to AB (height) = 9 cm.

By formula,

The area of parallelogram = Base x Height

= 12 x 9

= 108 cm².

∴ Statement 1 is false, and statement 2 is true.

Hence, option 4 is the correct option.

Given,

The diagonal of the rectangle = 17 m

The breadth of the rectangle = 8 m

Using Pythagoras theorem in rectangle,

⇒ Diagonal² = Length² + Breadth²

⇒ 17² = l² + 8²

⇒ 289 = l² + 64

⇒ l² = 289 - 64

⇒ l² = 225

⇒ l = $\sqrt{225}$

⇒ l = 15 m.

By formula,

Area of rectangle = l x b = 15 x 8 = 120 m².

Half of the area of rectangle = $\frac{120}{2}$ = 60 m².

So, assertion (A) is false.

ABCD is a rectangle. BD is a diagonal.

In Δ ABD and Δ BDC,

⇒ AB = DC (Opposite sides of rectangle are equal)

⇒ AD = BC (Opposite sides of rectangle are equal)

⇒ BD = BD (Common)

∴ Δ ABD ≅ Δ BDC (By SSS congruency criterion)

Thus the diagonal of every rectangle, divide into two congruent right-triangles.

So, reason (R) is true.

∴ A is false, but R is true.

Hence, option 4 is the correct option.

Given, height of the equilateral triangle = 2 $\sqrt{3}$ cm

By formula,

The area of an equilateral triangle = $\frac{1}{\sqrt{3}}$ x (Height)²

So, reason (R) is true.

Substituting values we get :

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

So, assertion (A) is true.

∴ Both A and R are correct, and R is the correct explanation for A.

Hence, option 1 is the correct option.

Given, the diagonal of a square = $9\sqrt{2}$ cm

By formula,

The area of a square = $\frac{1}{2}$ x d², where d is the length of the diagonal.

So, reason (R) is true.

Given,

The diagonal of a square is $9\sqrt{2}$ cm.

$$\begin{aligned}\text{Area of square } = \frac{1}{2} \times (9\sqrt{2})^2 \\[1em] = \frac{1}{2} \times 81 \times 2 \\[1em] = 81 \text{ cm}^2.\end{aligned}$$

So, assertion (A) is true.

∴ Both A and R are correct, and R is the correct explanation for A.

Hence, option 1 is the correct option.

Given,

Base = 10 cm

Height = 5 cm

The side parallel to the given base = 6 cm

By formula,

Area of trapezium = $\frac{1}{2}$ x (sum of parallel sides) x height.

$$\begin{aligned}\text{Area of given trapezium} = \frac{1}{2} \times (10 + 6) \times 5 \\[1em] = \frac{1}{2} \times 16 \times 5 \\[1em] = 8 \times 5 \\[1em] = 40 \text{ cm}^2.\end{aligned}$$

∴ A is true, but R is false.

Hence, option 3 is the correct option.

The perimeter of a trapezium = 52 cm.

The altitude = 8 cm

Perimeter of the trapezium = AB + BC + CD + DA

⇒ 52 = AB + 10 + CD + 10

⇒ 52 = AB + CD + 20

⇒ AB + CD = 52 - 20

⇒ AB + CD = 32

⇒ Sum of parallel sides = 32

As we know, the area of a trapezium = $\frac{1}{2}$ x (sum of parallel sides) x height

= $\frac{1}{2}$ x 32 x 8 cm²

= $\frac{1}{2}$ x 256 cm²

= 128 cm²

Hence, the area of the trapezium is 128 cm².

Given:

Total length of garden = 20 m

Radius of the circle = $\frac{7}{2}$ m

Length of rectangle = total length - 2 x radius of the circle

= 20 - 2 x $\frac{7}{2}$ m

= 20 - 7 m

= 13 m

Perimeter of the figure = l + l + 2 x $\frac{1}{2}$ x circumference of semicircle

As we know that circumference of the circle = 2πr

= 13 + 13 + 2 x $\frac{1}{2}$ x 2πr

= 26 + 2 x $\frac{22}{7} \times \frac{7}{2}$

= 26 + $\cancel{2} \times \frac{22}{\cancel{7}} \times \frac{\cancel{7}}{\cancel{2}}$

= 26 + 22

= 48 m

Area of the figure = Area of rectangle + 2 x Area of semicircle

As we know that area of rectangle = length x breadth

And, area of the circle = πr²

So, area

$$\begin{aligned}= 13 \times 7 + \frac{22}{7} \times \big( \frac{7}{2} \big)^2 \\[1em] = 91 + \frac{22}{7} \times \frac{49}{4} \\[1em] = 91 + \frac{22 \times 49}{7 \times 4} \\[1em] = 91 + \frac{1078}{28} \\[1em] = 91 + 38.5 \\[1em] = 129.5 \text{ m}^2\end{aligned}$$

Hence, the area of the garden is 129.5 m² and the perimeter is 48 m.

(i) Given :

AB = 13 cm

AC = 10 cm

Then, OA = OC = $\frac{10}{2}$ = 5 cm

Since the diagonals of a rhombus bisect at 90°.

Applying pythagoras theorem in triangle AOB, we get:

AB² = OA² + OB²

⇒ (13)² = (5)² + OB²

⇒ 169 = 25 + OB²

⇒ OB² = 169 - 25

⇒ OB² = 144

⇒ OB = $\sqrt{144}$

⇒ OB = 12

BD = 2 x OB

= 2 x 12 cm

= 24 cm

Hence, the length of other diagonal is 24 cm.

(ii) As we know, the area of rhombus = $\frac{1}{2}$ x product of its diagonal

= $\frac{1}{2}$ x 10 x 24 cm²

= $\frac{1}{2}$ x 240 cm²

= 120 cm²

Hence, the area of the rhombus is 120 cm².

Given:

Length of rectangle = 20 cm

Breadth of rectangle = 10 cm

Height of triangle = 8 cm

Hypotenuse of triangle = 10 cm

Applying pythagoras theorem in triangle COB, we get:

BC² = OB² + OC²

⇒ (10)² = OB² + (8)²

⇒ 100 = OB² + 64

⇒ OB² = 100 - 64

⇒ OB² = 36

⇒ OB = $\sqrt{36}$

⇒ OB = 6 cm

Area of shaded figure = area of rectangle - area of triangle

As we know that area of rectangle = length x breadth

And, area of the triangle = $\frac{1}{2}$ x base x height

So, Area

$$\begin{aligned}= 20 \times 10 - \frac{1}{2} \times 6 \times 8\\[1em] = 200 - \frac{1}{2} \times 48\\[1em] = 200 - 24\\[1em] = 176 \text{ cm}^2\end{aligned}$$

Hence, the area of the shaded portion is 176 cm².

Let r be the radius of the circle.

The circumference of a circle = 264 cm

As we know, the circumference of the circle = 2πr

$$\begin{aligned}⇒ 2 \times \frac{22}{7} \times r = 264\\[1em] ⇒ \frac{44}{7} \times r = 264\\[1em] ⇒ r = \frac{264 \times 7}{44}\\[1em] ⇒ r = \frac{1848}{44}\\[1em] ⇒ r = 42 \text{ cm}\end{aligned}$$

And, area of the circle = πr²

$$\begin{aligned}= \frac{22}{7} \times 42^2\\[1em] = \frac{22}{7} \times 1764\\[1em] = \frac{38808}{7}\\[1em] = 5544 \text{ cm}^2\end{aligned}$$

Hence, the area of the circle is 5,544 cm².

Let r be the radius of the circle.

The area of circle = 1386 cm²

As we know, the area of the circle = πr²

$$\begin{aligned}⇒ \frac{22}{7} \times r^2 = 1386\\[1em] ⇒ r^2 = \frac{1386 \times 7}{22}\\[1em] ⇒ r^2 = \frac{9702}{22}\\[1em] ⇒ r^2 = 441\\[1em] ⇒ r = \sqrt{441}\\[1em] ⇒ r = 21 \text{ cm}\end{aligned}$$

Circumference of the circle = 2πr

$$\begin{aligned}= 2 \times \frac{22}{7} \times 21\\[1em] = \frac{44}{7} \times 21\\[1em] = \frac{924}{7}\\[1em] = 132 \text{ cm}\end{aligned}$$

Hence, the circumference of the circle is 132 cm.

Given:

Area of the square = 484 cm²

Let s be the side of the square.

As we know, the area of the square = side²

⇒ s² = 484

⇒ s = $\sqrt{484}$

⇒ s = 22 cm

Total length of the wire = Perimeter of the square

As we know, the perimeter of the square = 4 x side

= 4 x 22

= 88 cm

Perimeter of the square = Circumference of the circle

Let r be the radius of the circle.

⇒ 2πr = 88 cm

$$\begin{aligned}⇒ 2 \times \frac{22}{7} \times r = 88\\[1em] ⇒ \frac{44}{7} \times r = 88\\[1em] ⇒ r = \frac{7 \times 88}{44}\\[1em] ⇒ r = \frac{616}{44}\\[1em] ⇒ r = 14 \text{ cm}\end{aligned}$$

Area of the circle = πr²

$$\begin{aligned}= \frac{22}{7} \times 14^2\\[1em] = \frac{22}{7} \times 196\\[1em] = \frac{4312}{7}\\[1em] = 616 \text{ cm}^2\end{aligned}$$

Hence, the area of the circle is 616 cm².

Given:

The perimeters of two squares are 120 cm and 64 cm.

Let a be the side of the square.

As we know that perimeter of the square = 4 x side

⇒ 4 x a = 120

⇒ a = $\frac{120}{4}$

⇒ a = 30 cm

The perimeter of square is 64 cm.

Let b be the side of the square.

As we know that perimeter of the square = 4 x side

⇒ 4 x b = 64

⇒ a = $\frac{64}{4}$

⇒ a = 16 cm

It is given that area of third square = the sum of the areas of these two squares

Let R be the side of the square.

As we know that area of the square = side²

⇒ R² = 30² + 16²

⇒ R² = 900 + 256

⇒ R² = 1156

⇒ R = $\sqrt{1156}$

⇒ R = 34 cm

Perimeter of third square = 4 x 34 cm

= 136 cm

Hence, the perimeter of the square is 136 cm.

Given:

Area of square = Half the sum of areas of three squares with sides 3 cm, 4 cm and 5 cm.

Let R be the side of the square.

As we know that area of the square = side²

$$\begin{aligned}⇒ R^2 = \frac{1}{2} \times (3^2 + 4^2 + 5^2)\\[1em] ⇒ R^2 = \frac{1}{2} \times (9 + 16 + 25)\\[1em] ⇒ R^2 = \frac{1}{2} \times 50\\[1em] ⇒ R^2 = 25\\[1em] ⇒ R = \sqrt{25}\\[1em] ⇒ R = 5 \text{ cm}\end{aligned}$$

And, perimeter of the square = 4 x side

Perimeter of the square = 4 x 5 cm

= 20 cm

Hence, the perimeter of the square is 20 cm.

Given:

Diameter of wheel = 28 cm

Radius = $\frac{\text{Diameter}}{2}$

= $\frac{28}{2}$ cm

= 14 cm

As we know, the circumference of the circle = 2πr

$$\begin{aligned}2 \times \frac{22}{7} \times 14\\[1em] = \frac{44}{7} \times 14\\[1em] = \frac{616}{7}\\[1em] = 88 \text{ cm}\end{aligned}$$

Let a be the number of revolutions.

Total distance = 11 km

= 11 x 1,00,000 cm

= 11,00,000 cm

Distance traveled by wheel in a revolutions = circumference of the wheel x a

⇒ 11,00,000 = 88 x a

⇒ a = $\frac{11,00,000}{88}$

⇒ a = 12,500 revolutions

Hence, the wheel rotates 12,500 times during a journey of 11 km.