Mensuration

Given,

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

Area of the field = 14580 m²

Let the length of the field be 9a metres and the breadth of the field be 5a metres.

By formula,

Area of a rectangle = length × breadth

⇒ 9a × 5a = 14580

⇒ 45a² = 14580

⇒ a² = $\frac{14580}{45}$

⇒ a² = 324

⇒ a = $\sqrt{324}$

⇒ a = 18

So, length of the field = 9a = 9 × 18 m = 162 m

And, breadth of the field = 5a = 5 × 18 m = 90 m

Length of the fence required = Perimeter of the field

= 2(length + breadth)

= 2(162 + 90) m

= 2 × 252 m

= 504 m

Cost of fencing = Length of the fence × Rate of fencing

= 504 m × ₹325/m

= ₹1,63,800

Hence, the cost of surrounding the field with a fence is ₹1,63,800.

Given,

Length of the rectangle = 16 m

Breadth of the rectangle = 9 m

By formula,

Area of a rectangle = length × breadth

= 16 × 9 m²

= 144 m²

Let the side of the square be a metres.

It is given that the area of the square is equal to the area of the rectangle.

⇒ a² = 144

⇒ a = $\sqrt{144}$

⇒ a = 12 m

Perimeter of the rectangle = 2(length + breadth)

= 2(16 + 9) m

= 2 × 25 m

= 50 m

Perimeter of the square = 4 × side

= 4 × 12 m

= 48 m

Difference of the perimeters = 50 - 48 m = 2 m

Hence, a side of the square is 12 m and the perimeter of the rectangle exceeds the perimeter of the square by 2 m.

Given,

Two adjacent sides of the parallelogram are 24 cm and 18 cm.

Distance between the longer sides = 12 cm

By formula,

Area of a parallelogram = base × corresponding height

Taking 24 cm as the base,

Area of the parallelogram = 24 × 12 cm²

= 288 cm²

Let the distance between the shorter sides be h cm.

Taking 18 cm as the base,

Area of the parallelogram = 18 × h cm²

⇒ 18 × h = 288

⇒ h = $\frac{288}{18}$

⇒ h = 16 cm

Hence, the distance between the shorter sides is 16 cm.

Given,

Side of the square plot = 24 m

Length of the house = 18 m

Breadth of the house = 12 m

By formula,

Area of a square = side²

Area of the square plot = (24)² m²

= 576 m²

By formula,

Area of a rectangle = length × breadth

Area covered by the house = 18 × 12 m²

= 216 m²

Area of the garden = Area of the plot - Area covered by the house

= 576 - 216 m²

= 360 m²

Cost of developing the garden = Area of the garden × Rate

= 360 m² × ₹200/m²

= ₹72,000

Hence, the total cost of developing the garden around the house is ₹72,000.

Given,

Base of the parallelogram shaped tile = 18 cm

Corresponding height of the tile = 6 cm

By formula,

Area of a parallelogram = base × height

Area of one tile = 18 × 6 cm²

= 108 cm²

Area of the floor = 540 m²

Converting the area of the floor into square centimetres,

Area of the floor = 540 × 10000 cm²

= 5400000 cm²

$$\begin{aligned}\text{ Number of tiles required } = \frac{\text{Area of the floor}}{\text{Area of one tile}} \\[1em] = \frac{5400000}{108}\\[1em] = 50000\end{aligned}$$

Hence, 50000 tiles are required to cover the floor.

The ant has to move along the boundary of each food piece. So, we find the perimeter of each of the three shapes.

(a) The boundary of this food piece consists of a semicircular arc of diameter 2.8 cm and a straight side of length 2.8 cm.

Radius of the semicircle = $\frac{2.8}{2}$ cm = 1.4 cm

Length of the semicircular arc = πr

= $\frac{22}{7}$ x 1.4 cm

= 4.4 cm

Perimeter of the food piece = 4.4 + 2.8 cm

= 7.2 cm

(b) The boundary of this food piece consists of a semicircular arc of diameter 2.8 cm, two straight sides of length 1.5 cm each and a straight side of length 2.8 cm.

Length of the semicircular arc = 4.4 cm

Perimeter of the food piece = 4.4 + 1.5 + 1.5 + 2.8 cm

= 10.2 cm

(c) The boundary of this food piece consists of a semicircular arc of diameter 2.8 cm and two straight sides of length 2 cm each.

Length of the semicircular arc = 4.4 cm

Perimeter of the food piece = 4.4 + 2 + 2 cm

= 8.4 cm

On comparing the three perimeters, 10.2 cm is the greatest.

Hence, the ant has to take a longer round for the food piece (b).

Given,

Area enclosed between the concentric circles = 770 cm²

Radius of the outer circle (R) = 21 cm

Let the radius of the inner circle be r cm.

Area enclosed between the circles = Area of the outer circle - Area of the inner circle

⇒ πR² - πr² = 770

⇒ π(R² - r²) = 770

⇒ $\frac{22}{7}$ × (21² - r²) = 770

⇒ 21² - r² = $\frac{770 \times 7}{22}$

⇒ 441 - r² = 245

⇒ r² = 441 - 245

⇒ r² = 196

⇒ r = $\sqrt{196}$

⇒ r = 14

Hence, the radius of the inner circle is 14 cm.

Given,

Area enclosed by the wire when bent in the form of a square = 121 cm²

Let the side of the square be a cm.

By formula,

Area of a square = side²

⇒ a² = 121

⇒ a = $\sqrt{121}$

⇒ a = 11 cm

Length of the wire = Perimeter of the square

= 4 × side

= 4 × 11 cm

= 44 cm

When the same wire is bent into the form of a circle, the length of the wire becomes the circumference of the circle.

Let the radius of the circle be r cm.

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

By formula,

Area of a circle = πr²

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

Hence, the area of the circle is 154 cm².

(i) From the figure, ΔABC is a right-angled triangle, right angled at A.

Base of the triangle, AB = 3 cm

Height of the triangle, AC = 4 cm

By formula,

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

= $\frac{1}{2}$ x 3 x 4 cm²

= $\frac{1}{2}$ x 12 cm²

= 6 cm²

Hence, the area of ΔABC is 6 cm².

(ii) In right-angled triangle ABC, BC is the hypotenuse.

By Pythagoras theorem,

Base² + Height² = Hypotenuse²

⇒ AB² + AC² = BC²

⇒ (3)² + (4)² = BC²

⇒ 9 + 16 = BC²

⇒ BC² = 25

⇒ BC = $\sqrt{25}$

⇒ BC = 5 cm

Hence, the length of BC is 5 cm.

(iii) Let the length of the altitude AD from A to BC be h cm.

Taking BC as the base,

Area of ΔABC = $\frac{1}{2}$ × BC × AD

[∵ Both the products give the area of the same triangle.]

⇒ $\frac{1}{2}$ x 5 x h = 6

⇒ 5 × h = 2 × 6

⇒ 5 × h = 12

⇒ h = $\frac{12}{5}$

⇒ h = 2.4 cm

Hence, the length of the altitude from A to BC is 2.4 cm.

Given,

Length of the rectangular garden = 80 m

Breadth of the rectangular garden = 40 m

Width of each cross-path = 2.5 m

(i) The two cross-paths are rectangles which cross each other, and the square where they cross is common to both of them.

Area of the path along the length = 80 × 2.5 m²

= 200 m²

Area of the path along the breadth = 40 × 2.5 m²

= 100 m²

Area of the common square portion = 2.5 × 2.5 m²

= 6.25 m²

Area of the cross-paths = 200 + 100 - 6.25 m²

= 293.75 m²

Hence, the area of the cross-paths is 293.75 m².

(ii) Area of the garden = length × breadth

= 80 × 40 m²

= 3200 m²

Area of the unshaded portion = Area of the garden - Area of the cross-paths

= 3200 - 293.75 m²

= 2906.25 m²

Hence, the area of the unshaded portion is 2906.25 m².

Given,

Length of rectangle ABCD = 18 cm

Breadth of rectangle ABCD = 12 cm

By formula,

Area of a rectangle = length × breadth

Area of rectangle ABCD = 18 × 12 cm²

= 216 cm²

The three unshaded triangles are right-angled triangles.

By formula,

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

Area of the upper left triangle = $\frac{1}{2}$ × 10 × 12 cm²

= 60 cm²

Area of the upper right triangle = $\frac{1}{2}$ × (18 - 10) × (12 - 7) cm²

= $\frac{1}{2}$ × 8 × 5 cm²

= 20 cm²

Area of the lower triangle = $\frac{1}{2}$ × 18 × 7 cm²

= 63 cm²

Area of the shaded region = Area of rectangle ABCD - Sum of the areas of the three triangles

= 216 - (60 + 20 + 63) cm²

= 216 - 143 cm²

= 73 cm²

Hence, the area of the shaded region is 73 cm².

Given,

Area of the square grassy lawn ABCD = 729 m²

Area of the path = 295 m²

(i) Let the side of the square lawn be a metres.

By formula,

Area of a square = side²

⇒ a² = 729

⇒ a = $\sqrt{729}$

⇒ a = 27 m

Area of the square field enclosing the lawn and the path = Area of the lawn + Area of the path

= 729 + 295 m²

= 1024 m²

Let the side of this bigger square field be b metres.

⇒ b² = 1024

⇒ b = $\sqrt{1024}$

⇒ b = 32 m

The path of uniform width runs all around the lawn, so it lies on both sides of the lawn.

Width of the path = $\frac{32 - 27}{2}$ m

= $\frac{5}{2}$ m

= 2.5 m

Hence, the width of the path is 2.5 m.

(ii) Length of the boundary of the square field = Perimeter of the bigger square

= 4 × side

= 4 × 32 m

= 128 m

Hence, the length of the boundary of the square field enclosing the lawn and the path is 128 m.

Given,

Each side of the rhombus = 13 cm

One diagonal of the rhombus = 10 cm

Let ABCD be the rhombus in which AB = BC = CD = DA = 13 cm and AC = 10 cm. Let its diagonals meet at O.

As we know, the diagonals of a rhombus bisect each other at right angles.

So, OA = $\frac{1}{2}$ × AC

= $\frac{1}{2}$ × 10 cm

= 5 cm

(i) In right-angled triangle AOB, AB is the hypotenuse.

By Pythagoras theorem,

OA² + OB² = AB²

⇒ (5)² + OB² = (13)²

⇒ 25 + OB² = 169

⇒ OB² = 169 - 25

⇒ OB² = 144

⇒ OB = $\sqrt{144}$

⇒ OB = 12 cm

Length of the other diagonal, BD = 2 × OB

= 2 × 12 cm

= 24 cm

Hence, the length of the other diagonal is 24 cm.

(ii) By formula,

Area of a rhombus = $\frac{1}{2}$ × product of its diagonals

= $\frac{1}{2}$ × 10 × 24 cm²

= $\frac{1}{2}$ × 240 cm²

= 120 cm²

Hence, the area of the rhombus is 120 cm².

Given,

The cross-section ABCD of the swimming pool is a trapezium.

Width AB = 14 m

Depth at the shallow end = 1.5 m

Depth at the deep end = 8 m

The two depths are the parallel sides of the trapezium and the width AB is the distance between them.

By formula,

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

= $\frac{1}{2}$ × (1.5 + 8) × 14 m²

= $\frac{1}{2}$ × 9.5 × 14 m²

= 9.5 × 7 m²

= 66.5 m²

Hence, the area of the cross-section of the swimming pool is 66.5 m².

Given,

Area of the trapezium = 360 m²

Distance between the two parallel sides = 20 m

One of the parallel sides = 25 m

Let the other parallel side be a metres.

By formula,

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

⇒ $\frac{1}{2}$ × (25 + a) × 20 = 360

⇒ (25 + a) × 10 = 360

⇒ 25 + a = $\frac{360}{10}$

⇒ 25 + a = 36

⇒ a = 36 - 25

⇒ a = 11 m

Hence, the other parallel side of the trapezium is 11 m.

Given,

Side of the rhombus = 8 cm

Altitude of the rhombus = 5.25 cm

One diagonal of the rhombus = 15 cm

A rhombus is a parallelogram whose all sides are equal.

By formula,

Area of a parallelogram = base × corresponding height

Area of the rhombus = 8 × 5.25 cm²

= 42 cm²

Let the length of the other diagonal be d cm.

By formula,

Area of a rhombus = $\frac{1}{2}$ × product of its diagonals

⇒ $\frac{1}{2}$ × 15 × d = 42

⇒ 15 × d = 2 × 42

⇒ 15 × d = 84

⇒ d = $\frac{84}{15}$

⇒ d = 5.6 cm

Hence, the area of the rhombus is 42 cm² and the length of the other diagonal is 5.6 cm.

From the given diagram,

ED = 7 m, AC = 13 m and the distance between ED and AC is 6.5 m.

(i) In trapezium ACDE, ED and AC are the parallel sides and the distance between them is 6.5 m.

By formula,

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

= $\frac{1}{2}$ × (7 + 13) × 6.5 m²

= $\frac{1}{2}$ × 20 × 6.5 m²

= 10 × 6.5 m²

= 65 m²

Hence, the area of trapezium ACDE is 65 m².

(ii) In parallelogram ABDE, AB is the base and the distance between AB and ED is the corresponding height.

Since ABDE is a parallelogram, AB = ED = 7 m.

By formula,

Area of a parallelogram = base × corresponding height

= 7 × 6.5 m²

= 45.5 m²

Hence, the area of parallelogram ABDE is 45.5 m².

(iii) Area of triangle BCD = Area of trapezium ACDE - Area of parallelogram ABDE

= 65 - 45.5 m²

= 19.5 m²

Hence, the area of triangle BCD is 19.5 m².

Given,

Base of the triangle = 24.8 cm

Corresponding altitude of the triangle = 16.5 cm

One diagonal of the rhombus = 22 cm

By formula,

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

= $\frac{1}{2}$ × 24.8 × 16.5 cm²

= 12.4 × 16.5 cm²

= 204.6 cm²

It is given that the area of the rhombus is equal to the area of the triangle.

So, area of the rhombus = 204.6 cm²

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

By formula,

Area of a rhombus = $\frac{1}{2}$ × product of its diagonals

⇒ $\frac{1}{2}$ × 22 × d = 204.6

⇒ 11 × d = 204.6

⇒ d = $\frac{204.6}{11}$

⇒ d = 18.6 cm

Hence, the length of the other diagonal of the rhombus is 18.6 cm.

Given,

Perimeter of the trapezium = 52 cm

Each non-parallel side = 10 cm

Altitude of the trapezium = 8 cm

By formula,

Perimeter of a trapezium = sum of parallel sides + sum of non-parallel sides

⇒ 52 = sum of parallel sides + (10 + 10)

⇒ 52 = sum of parallel sides + 20

⇒ Sum of parallel sides = 52 - 20

⇒ Sum of parallel sides = 32 cm

By formula,

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

= $\frac{1}{2}$ × 32 × 8 cm²

= 16 × 8 cm²

= 128 cm²

Hence, the area of the trapezium is 128 cm².

Given,

Area of the trapezium = 540 cm²

The ratio of the parallel sides = 7 : 5

Distance between the parallel sides = 18 cm

Let the parallel sides of the trapezium be 7a cm and 5a cm.

By formula,

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

$$\begin{aligned}\Rightarrow \frac{1}{2} \times (7a + 5a) \times 18 = 540 \\[1em] \Rightarrow \frac{1}{2} \times 12a \times 18 = 540 \\[1em] \Rightarrow 108a = 540 \\[1em] \Rightarrow a = \frac{540}{108} \\[1em] \Rightarrow a = 5\end{aligned}$$

So, one parallel side = 7a = 7 × 5 cm = 35 cm

And, the other parallel side = 5a = 5 × 5 cm = 25 cm

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

The given shape can be divided into two rectangles and a right-angled triangle by drawing two vertical lines, as shown below.

From the figure, the total height of the shape = 5 + 4 = 9 cm

The width of the shape at the bottom = BE = 7 cm

Width of the third portion = 7 - 2 - 3 = 2 cm

By formula,

Area of a rectangle = length × breadth

Area of the Rectangle ABCH = AH x AB = 2 × 5 cm²

= 10 cm²

As, GC = GH + HC = 4 + 5 = 9

Area of the Rectangle GCDF = GF x GC = 3 × 9 cm²

= 27 cm²

By formula,

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

Since DE = BE - BC - CD = 7 - 2 - 3 = 2 and FD = GC = 9

Area of the right-angled triangle FDE = $\frac{1}{2}$ x DE x FD

= $\frac{1}{2}$ × 2 × 9 cm²

= 9 cm²

Area enclosed by the given shape = 10 + 27 + 9 cm²

= 46 cm²

Hence, the area enclosed by the given shape is 46 cm².

The given shape can be divided into two rectangles and a parallelogram, as shown below.

From the figure, the total height of the shape = 9 cm, the total width of the shape = 9 cm, the width of each vertical rectangle = 2 cm and the height of each inner notch = 7 cm

By formula,

Area of a rectangle = length × breadth

Area of Rectangle ABCD = AB x AD = 2 × 9 cm²

= 18 cm²

Area of Rectangle EFGH = GH x FH = 2 × 9 cm²

= 18 cm²

⇒ Area of the two vertical rectangles = 2 × 18 cm²

= 36 cm²

From figure,

CI = CB - IB = 9 - 7 = 2 cm

The distance between these two vertical sides (BE) = 9 - 2 - 2 = 5 cm

By formula,

Area of a parallelogram = base × corresponding height

Area of the parallelogram IEJC = IC x BE = 2 × 5 cm²

= 10 cm²

Area enclosed by the given shape = 36 + 10 cm²

= 46 cm²

Hence, the area enclosed by the given shape is 46 cm².

From the given sketch,

DC = 15 cm, AB = 40 cm, DB = 41 cm and ∠DAB = 90°.

(i) In right-angled triangle DAB, DB is the hypotenuse.

By Pythagoras theorem,

AD² + AB² = DB²

⇒ AD² + (40)² = (41)²

⇒ AD² + 1600 = 1681

⇒ AD² = 1681 - 1600

⇒ AD² = 81

⇒ AD = $\sqrt{81}$

⇒ AD = 9 cm

Hence, the length AD is 9 cm.

(ii) In trapezium ABCD, DC and AB are the parallel sides and AD is the distance between them.

By formula,

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

= $\frac{1}{2}$ × (15 + 40) × 9 cm²

= $\frac{1}{2}$ × 55 × 9 cm²

= $\frac{495}{2}$ cm²

= 247.5 cm²

Hence, the area of trapezium ABCD is 247.5 cm².

(iii) By formula,

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

Area of triangle ABD = $\frac{1}{2}$ × AB × AD

= $\frac{1}{2}$ × 40 × 9 cm²

= 20 × 9 cm²

= 180 cm²

Area of triangle BCD = Area of trapezium ABCD - Area of triangle ABD

= 247.5 - 180 cm²

= 67.5 cm²

Hence, the area of triangle BCD is 67.5 cm².

Given,

Outer dimensions of the frame = 28 cm × 32 cm

Inner dimensions of the frame = 20 cm × 24 cm

The width of each section is the same.

Width of each section = $\frac{28 - 20}{2}$ cm

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

= 4 cm

Each section of the frame is a trapezium.

The upper and the lower sections are trapeziums whose parallel sides are 28 cm and 20 cm and the distance between them is 4 cm.

By formula,

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

Area of the upper section = $\frac{1}{2}$ × (28 + 20) × 4 cm²

= $\frac{1}{2}$ × 48 × 4 cm²

= 96 cm²

Area of the lower section = 96 cm²

The left and the right sections are trapeziums whose parallel sides are 32 cm and 24 cm and the distance between them is 4 cm.

Area of the left section = $\frac{1}{2}$ × (32 + 24) × 4 cm²

= $\frac{1}{2}$ × 56 × 4 cm²

= 112 cm²

Area of the right section = 112 cm²

Hence, the areas of the four sections of the frame are 112 cm², 96 cm², 112 cm² and 96 cm².

From the given figure,

AB = 6 cm, AD = 8 cm, BC = 26 cm, ∠BAD = 90° and ∠BDC = 90°.

In right-angled triangle BAD, BD is the hypotenuse.

By Pythagoras theorem,

AB² + AD² = BD²

⇒ (6)² + (8)² = BD²

⇒ 36 + 64 = BD²

⇒ BD² = 100

⇒ BD = $\sqrt{100}$

⇒ BD = 10 cm

In right-angled triangle BDC, BC is the hypotenuse.

By Pythagoras theorem,

BD² + DC² = BC²

⇒ (10)² + DC² = (26)²

⇒ 100 + DC² = 676

⇒ DC² = 676 - 100

⇒ DC² = 576

⇒ DC = $\sqrt{576}$

⇒ DC = 24 cm

By formula,

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

Area of triangle BAD = $\frac{1}{2}$ × 6 × 8 cm²

= 24 cm²

Area of triangle BDC = $\frac{1}{2}$ × 10 × 24 cm²

= 120 cm²

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

= 24 + 120 cm²

= 144 cm²

Hence, the area of the quadrilateral ABCD is 144 cm².

The octagonal surface can be divided into a rectangle in the middle and two trapeziums, one at the top and one at the bottom, as shown in the figure.

The middle portion is a rectangle of length = DC = 15 m and breadth = CF = 8 m.

By formula,

Area of a rectangle = length × breadth

Area of the Rectangle DEFC = DC x CF = 15 × 8 m²

= 120 m²

Since the octagon is regular, each of its sides is 8 m.

The upper portion is a trapezium whose parallel sides are 8 m and 15 m and the distance between them is 6 m.

By formula,

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

Area of the Trapezium ABCD = $\frac{1}{2}$ × (AB + DC) x AO

= $\frac{1}{2}$ × (8 + 15) × 6 m²

= $\frac{1}{2}$ × 23 × 6 m²

= 23 × 3 m²

= 69 m²

Area of the Trapezium EFGH = Area of the Trapezium ABCD = 69 m²

Area of the octagonal surface = Area of the Rectangle DEFC + Area of the Trapezium EFGH + Area of the Trapezium ABCD

= 120 + 69 + 69 m²

= 258 m²

Hence, the area of the octagonal surface is 258 m².

Given,

ABCD is a rectangle of size 18 cm by 10 cm.

In ΔBEC, ∠E = 90° and EC = 8 cm.

Since BC is a side of the rectangle, BC = 10 cm.

In right-angled triangle BEC, BC is the hypotenuse.

By Pythagoras theorem,

BE² + EC² = BC²

⇒ BE² + (8)² = (10)²

⇒ BE² + 64 = 100

⇒ BE² = 100 - 64

⇒ BE² = 36

⇒ BE = $\sqrt{36}$

⇒ BE = 6 cm

By formula,

Area of a rectangle = length × breadth

Area of rectangle ABCD = 18 × 10 cm²

= 180 cm²

By formula,

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

Area of triangle BEC = $\frac{1}{2}$ × 8 × 6 cm²

= 24 cm²

Area of pentagon ABECD = Area of rectangle ABCD - Area of triangle BEC

= 180 - 24 cm²

= 156 cm²

Hence, the area enclosed by the pentagon ABECD is 156 cm².

Given,

PS = 11 cm, PY = 9 cm, PX = 8 cm, PW = 5 cm, PV = 3 cm, QV = 5 cm, UW = 4 cm, RX = 6 cm and TY = 2 cm

The diagonal PS divides the polygon PQRSTU into two parts. QV, RX, UW and TY are perpendiculars drawn on PS.

From the figure,

VX = PX - PV = 8 - 3 = 5 cm

XS = PS - PX = 11 - 8 = 3 cm

WY = PY - PW = 9 - 5 = 4 cm

YS = PS - PY = 11 - 9 = 2 cm

Area of the portion above PS

Area of triangle PQV = $\frac{1}{2}$ × PV × QV

= $\frac{1}{2}$ × 3 × 5 cm²

= 7.5 cm²

Area of trapezium VQRX = $\frac{1}{2}$ × (QV + RX) × VX

= $\frac{1}{2}$ × (5 + 6) × 5 cm²

= $\frac{1}{2}$ × 11 × 5 cm²

= 27.5 cm²

Area of triangle RXS = $\frac{1}{2}$ × XS × RX

= $\frac{1}{2}$ × 3 × 6 cm²

= 9 cm²

Area of the portion below PS

Area of triangle PUW = $\frac{1}{2}$ × PW × UW

= $\frac{1}{2}$ × 5 × 4 cm²

= 10 cm²

Area of trapezium WUTY = $\frac{1}{2}$ × (UW + TY) × WY

= $\frac{1}{2}$ × (4 + 2) × 4 cm²

= $\frac{1}{2}$ × 6 × 4 cm²

= 12 cm²

Area of triangle TYS = $\frac{1}{2}$ × YS × TY

= $\frac{1}{2}$ × 2 × 2 cm²

= 2 cm²

Area of polygon PQRSTU = Area of triangle PQV + Area of trapezium VQRX + Area of triangle RXS + Area of triangle PUW + Area of trapezium WUTY + Area of triangle TYS

= 7.5 + 27.5 + 9 + 10 + 12 + 2 cm²

= 68 cm²

Hence, the area of the polygon PQRSTU is 68 cm².

Given,

Volume of the cube = 343 cm³

Let the length of an edge of the cube be a cm.

By formula,

Volume of a cube = (edge)³

⇒ a³ = 343

⇒ a = $\sqrt[3]{343}$

⇒ a = 7 cm

Hence, the length of an edge of the cube is 7 cm.

By formula,

Volume of a cuboid = length × breadth × height

(i) Volume of the cuboid = 90 cm³, breadth = 5 cm and height = 3 cm

Let the length of the cuboid be l cm.

⇒ l × 5 × 3 = 90

⇒ 15 × l = 90

⇒ l = $\frac{90}{15}$

⇒ l = 6 cm

Hence, the length of the cuboid is 6 cm.

(ii) Length = 15 cm, breadth = 8 cm and height = 7 cm

Volume of the cuboid = 15 × 8 × 7 cm³

= 840 cm³

Hence, the volume of the cuboid is 840 cm³.

(iii) Volume of the cuboid = 62.5 m³, length = 10 m and breadth = 5 m

Let the height of the cuboid be h m.

⇒ 10 × 5 × h = 62.5

⇒ 50 × h = 62.5

⇒ h = $\frac{62.5}{50}$

⇒ h = 1.25 m

Hence, the height of the cuboid is 1.25 m.

Given,

Volume of the cuboid = 312 cm³

Base area of the cuboid = 26 cm²

Let the height of the cuboid be h cm.

By formula,

Volume of a cuboid = base area × height

⇒ 26 × h = 312

⇒ h = $\frac{312}{26}$

⇒ h = 12 cm

Hence, the height of the cuboid is 12 cm.

Given,

Dimensions of the godown = 55 m × 45 m × 30 m

Volume of one box = 1.25 m³

By formula,

Volume of a cuboid = length × breadth × height

Volume of the godown = 55 × 45 × 30 m³

= 74250 m³

$$\begin{aligned}\text{ Number of boxes } = \frac{\text{Volume of the godown}}{\text{Volume of one box}} \\[1em] = \frac{74250}{1.25} \\[1em] = \frac{7425000}{125} \\[1em] = 59400\end{aligned}$$

Hence, 59400 cuboidal boxes can be stored in the godown.

Given,

Length of the rectangular pit = 1.4 m = 140 cm

Breadth of the rectangular pit = 90 cm

Depth of the rectangular pit = 70 cm

Number of bricks = 1000

Base of each brick = 21 cm × 10.5 cm

By formula,

Volume of a cuboid = length × breadth × height

Volume of the earth dug out = 140 × 90 × 70 cm³

= 882000 cm³

Volume of one brick = $\frac{\text{Volume of the earth dug out}}{\text{Number of bricks}}$

= $\frac{882000}{1000}$ cm³

= 882 cm³

Let the height of each brick be h cm.

Volume of one brick = base area × height

⇒ 21 × 10.5 × h = 882

⇒ 220.5 × h = 882

⇒ h = $\frac{882}{220.5}$

⇒ h = 4 cm

Hence, the height of each brick is 4 cm.

Let the edge of the cube be a units.

By formula,

Volume of a cube = (edge)³

Volume of the given cube = a³ cubic units

When each edge is tripled, the edge of the new cube = 3a units

Volume of the new cube = (3a)³

= 27a³ cubic units

= 27 × Volume of the given cube

Hence, the volume of the cube becomes 27 times its original volume.

Given,

Radius of the cylindrical milk tank = 1.4 m

Height of the cylindrical milk tank = 8 m

By formula,

Volume of a cylinder = πr²h

$$\begin{aligned}= \frac{22}{7} \times (1.4)^2 \times 8\\[1em] = \frac{22}{7} \times 1.96 \times 8\\[1em] = 22 \times 0.28 \times 8\\[1em] = 49.28 \text{ m}^3\end{aligned}$$

So, volume of the tank = 49.28 m³

Converting the volume into litres,

Quantity of milk = 49.28 × 1000 litres

[∵ 1 m³ = 1000 litres]

= 49280 litres

Hence, 49280 litres of milk can be stored in the tank.

Given,

External dimensions of the closed box = 84 cm × 75 cm × 64 cm

Thickness of the wood = 2 cm

Since the box is closed, the wood is present on both the opposite faces in each direction. So, each external dimension decreases by 2 × 2 = 4 cm.

Internal length = 84 - 4 cm = 80 cm

Internal breadth = 75 - 4 cm = 71 cm

Internal height = 64 - 4 cm = 60 cm

By formula,

Volume of a cuboid = length × breadth × height

External volume of the box = 84 × 75 × 64 cm³

= 403200 cm³

Internal volume of the box = 80 × 71 × 60 cm³

= 340800 cm³

Volume of the wood = External volume - Internal volume

= 403200 - 340800 cm³

= 62400 cm³

Hence, the volume of the wood required to make the box is 62400 cm³.

Given,

The diameters of the two cylindrical jars are in the ratio 3 : 4.

Both the jars contain the same amount of milk.

Let the diameters of the two jars be 3a and 4a.

By formula,

Radius of a circle = $\frac{\text{Diameter}}{2}$

Radius of the first jar (r₁) = $\frac{3a}{2}$

Radius of the second jar (r₂) = $\frac{4a}{2}$ = 2a

Let the heights of the two jars be h₁ and h₂ respectively.

By formula,

Volume of a cylinder = πr²h

Since both the jars contain the same amount of milk, their volumes are equal.

$$\begin{aligned}\Rightarrow \pi \times \left(\frac{3a}{2}\right)^2 \times h_1 = \pi \times (2a)^2 \times h_2\\[1em] \Rightarrow \frac{9a^2}{4} \times h_1 = 4a^2 \times h_2\\[1em] \Rightarrow \frac{h_1}{h_2} = 4a^2 \times \frac{4}{9a^2}\\[1em] \Rightarrow \frac{h_1}{h_2} = \frac{16}{9}\end{aligned}$$

Hence, the ratio of the heights of the two jars is 16 : 9.

Let the radius of the base of the original cylinder be r and its height be h.

By formula,

Volume of a cylinder = πr²h

Volume of the original cylinder = πr²h

For the new cylinder, the radius is halved and the height is doubled.

Radius of the new cylinder = $\frac{r}{2}$

Height of the new cylinder = 2h

$$\begin{aligned}\text{Volume of the new cylinder} = \pi \times \left(\frac{r}{2}\right)^2 \times 2h\\[1em] = \pi \times \frac{r^2}{4} \times 2h\\[1em] = \frac{\pi r^2 h}{2}\end{aligned}$$

Ratio of the volumes = $\frac{\text{Volume of the new cylinder}}{\text{Volume of the original cylinder}}$

$$\begin{aligned}= \frac{\pi r^2 h}{2} \div \pi r^2 h\\[1em] = \frac{\pi r^2 h}{2} \times \frac{1}{\pi r^2 h}\\[1em] = \frac{1}{2}\end{aligned}$$

Hence, the ratio of the volume of the new cylinder to that of the original cylinder is 1 : 2.

Given,

Size of the rectangular piece of tin = 30 cm × 18 cm

Case 1: Rolled along its length (30 cm)

Here, the length of the sheet becomes the circumference of the base of the cylinder and the breadth becomes its height.

Let the radius of the base of this cylinder be r₁ cm.

⇒ 2πr₁ = 30

⇒ r₁ = $\frac{30}{2\pi}$

Height of this cylinder, h₁ = 18 cm

Case 2: Rolled along its breadth (18 cm)

Here, the breadth of the sheet becomes the circumference of the base of the cylinder and the length becomes its height.

Let the radius of the base of this cylinder be r₂ cm.

⇒ 2πr₂ = 18

⇒ r₂ = $\frac{18}{2\pi}$

Height of this cylinder, h₂ = 30 cm

By formula,

Volume of a cylinder = πr²h

$$\begin{aligned}\text{Ratio of the volumes} = \frac{\pi r_1^2 h_1}{\pi r_2^2 h_2}\\[1em] = \frac{\pi \times \left(\frac{30}{2\pi}\right)^2 \times 18}{\pi \times \left(\frac{18}{2\pi}\right)^2 \times 30}\\[1em] = \frac{30^2 \times 18}{18^2 \times 30}\\[1em] = \frac{30}{18}\\[1em] = \frac{5}{3}\end{aligned}$$

Hence, the ratio of the volumes of the two cylinders so formed is 5 : 3.

Given,

Internal diameter of the cylindrical pipe = 7 cm

Speed of the water = 5 m per sec

Radius of the pipe = $\frac{7}{2}$ cm = 3.5 cm

(i) Length of the column of water discharged in one second = 5 m = 500 cm

Length of the column of water discharged in one minute = 500 × 60 cm

= 30000 cm

By formula,

Volume of a cylinder = πr²h

$$\begin{aligned}\text{Volume of water discharged in one minute} = \frac{22}{7} \times (3.5)^2 \times 30000\\[1em] = \frac{22}{7} \times 12.25 \times 30000\\[1em] = 22 \times 1.75 \times 30000\\[1em] = 1155000 \text{ cm}^3\end{aligned}$$

So, volume of water discharged in one minute = 1155000 cm³

Converting the volume into litres,

Volume of water = $\frac{1155000}{1000}$ litres

[∵ 1 litre = 1000 cm³]

= 1155 litres

Hence, 1155 litres of water is discharged by the pipe in one minute.

(ii) Dimensions of the rectangular tank = 4 m × 3 m × 2.31 m

= 400 cm × 300 cm × 231 cm

By formula,

Volume of a cuboid = length × breadth × height

Volume of the tank = 400 × 300 × 231 cm³

= 27720000 cm³

Time taken to fill the tank = $\frac{\text{Volume of the tank}}{\text{Volume of water discharged in one minute}}$

= $\frac{27720000}{1155000}$ minutes

= 24 minutes

Hence, the pipe would take 24 minutes to fill the empty rectangular tank.

Given,

Radius of the first vessel = 15 cm and its height = 40 cm

Radius of the second vessel = 20 cm and its height = 45 cm

Height of the third vessel = 30 cm

By formula,

Volume of a cylinder = πr²h

Volume of the first vessel = π × (15)² × 40

= 9000π cm³

Volume of the second vessel = π × (20)² × 45

= 18000π cm³

Total volume of the milk = 9000π + 18000π cm³

= 27000π cm³

Let the radius of the third cylindrical vessel be r cm.

Since the third vessel just contains all the milk,

⇒ π × r² × 30 = 27000π

⇒ r² = $\frac{27000\pi}{30\pi}$

⇒ r² = 900

⇒ r = $\sqrt{900}$

⇒ r = 30

Hence, the radius of the other cylindrical vessel is 30 cm.

Given,

Height of the wooden pole = 7 m

Diameter of the wooden pole = 20 cm

Weight of the wood = 225 kg per m³

Radius of the pole = $\frac{20}{2}$ cm = 10 cm = 0.1 m

By formula,

Volume of a cylinder = πr²h

$$\begin{aligned}= \frac{22}{7} \times (0.1)^2 \times 7\\[1em] = \frac{22}{\cancel{7}} \times 0.01 \times \cancel{7}\\[1em] = 22 \times 0.01\\[1em] = 0.22 \text{ m}^3\end{aligned}$$

So, volume of the wooden pole = 0.22 m³

Weight of the pole = Volume of the pole × Weight of 1 m³ of wood

= 0.22 × 225 kg

= 49.5 kg

Hence, the weight of the wooden pole is 49.5 kg.

Given,

Length of the wooden cuboid = 30 cm

Side of the square cross-section = 14 cm

The cylinder of maximum volume that can be cut from this cuboid has its height equal to the length of the cuboid and the diameter of its base equal to the side of the square cross-section.

Height of the cylinder = 30 cm

Diameter of the base of the cylinder = 14 cm

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

By formula,

Volume of a cylinder = πr²h

$$\begin{aligned}= \frac{22}{7} \times 7^2 \times 30\\[1em] = \frac{22}{\cancel{7}} \times \cancel{7} \times 7 \times 30\\[1em] = 22 \times 7 \times 30\\[1em] = 4620 \text{ cm}^3\end{aligned}$$

So, volume of the cylinder = 4620 cm³

By formula,

Volume of a cuboid = length × breadth × height

Volume of the wooden cuboid = 14 × 14 × 30 cm³

= 5880 cm³

Volume of the wood wasted = Volume of the cuboid - Volume of the cylinder

= 5880 - 4620 cm³

= 1260 cm³

Hence, the volume of the cylinder is 4620 cm³ and the volume of the wood wasted is 1260 cm³.

Given,

Surface area of the cube = 384 cm²

(i) Let the length of an edge of the cube be a cm.

By formula,

Surface area of a cube = 6 × (edge)²

⇒ 6 × a² = 384

⇒ a² = $\frac{384}{6}$

⇒ a² = 64

⇒ a = $\sqrt{64}$

⇒ a = 8 cm

Hence, the length of an edge of the cube is 8 cm.

(ii) By formula,

Volume of a cube = (edge)³

= (8)³ cm³

= 512 cm³

Hence, the volume of the cube is 512 cm³.

Given,

Radius of the solid cylinder = 5 cm

Height of the solid cylinder = 10 cm

By formula,

Total surface area of a cylinder = 2πr(r + h)

= 2 × π × 5 × (5 + 10) cm²

= 2 × π × 5 × 15 cm²

= 150π cm²

Hence, the total surface area of the solid cylinder is 150π cm².

Given,

External measures of the cuboidal aquarium = 70 cm × 28 cm × 35 cm

Here, length = 70 cm, breadth = 28 cm and height = 35 cm.

The paper is needed for the base, the two side faces and the back face.

By formula,

Area of a rectangle = length × breadth

Area of the base = 70 × 28 cm²

= 1960 cm²

Area of the two side faces = 2 × (28 × 35) cm²

= 2 × 980 cm²

= 1960 cm²

Area of the back face = 70 × 35 cm²

= 2450 cm²

Area of the paper needed = 1960 + 1960 + 2450 cm²

= 6370 cm²

Hence, the area of the paper needed is 6370 cm².

Given,

Internal dimensions of the rectangular hall = 15 m × 12 m × 4 m

Here, length = 15 m, breadth = 12 m and height = 4 m.

By formula,

Area of four walls = 2(length + breadth) × height

= 2(15 + 12) × 4 m²

= 2 × 27 × 4 m²

= 216 m²

Area of 4 windows = 4 × (2 × 1.5) m²

= 4 × 3 m²

= 12 m²

Area of 2 doors = 2 × (1.5 × 2.5) m²

= 2 × 3.75 m²

= 7.5 m²

Area to be white washed = Area of four walls - Area of windows - Area of doors

= 216 - 12 - 7.5 m²

= 196.5 m²

Cost of white washing the four walls = Area to be white washed × Rate

= 196.5 m² × ₹50/m²

= ₹9,825

Area of the ceiling = length × breadth

= 15 × 12 m²

= 180 m²

Area to be white washed including the ceiling = 196.5 + 180 m²

= 376.5 m²

Cost of white washing including the ceiling = 376.5 m² × ₹50/m²

= ₹18,825

Hence, the cost of white washing all four walls is ₹9,825 and the cost of white washing including the ceiling is ₹18,825.

Given,

Length of the swimming pool = 50 m

Breadth of the swimming pool = 30 m

Depth of the swimming pool = 2.5 m

By formula,

Area of a rectangle = length × breadth

Area of the floor = 50 × 30 m²

= 1500 m²

By formula,

Area of four walls = 2(length + breadth) × height

Area of the walls = 2(50 + 30) × 2.5 m²

= 2 × 80 × 2.5 m²

= 400 m²

Total area to be cemented = 1500 + 400 m²

= 1900 m²

Cost of cementing = Total area × Rate

= 1900 m² × ₹120/m²

= ₹2,28,000

Hence, the cost of cementing the floor and the walls of the swimming pool is ₹2,28,000.

Given,

Perimeter of the floor of the rectangular hall = 236 m

Height of the hall = 4.5 m

By formula,

Area of four walls = perimeter of the floor × height

= 236 × 4.5 m²

= 1062 m²

Cost of painting the four walls = Area of four walls × Rate

= 1062 m² × ₹42/m²

= ₹44,604

Hence, the cost of painting the four walls of the hall is ₹44,604.

Given,

Length of the cuboidal fish tank = 30 cm

Breadth of the cuboidal fish tank = 20 cm

Height of the cuboidal fish tank = 20 cm

The tank is three-quarters full of water.

Height of the water in the tank = $\frac{3}{4}$ x 20 cm

= 15 cm

The water touches the base of the tank and the four walls up to a height of 15 cm.

By formula,

Area of a rectangle = length × breadth

Area of the base = 30 × 20 cm²

= 600 cm²

By formula,

Area of four walls = 2(length + breadth) × height

Area of the walls in contact with water = 2(30 + 20) × 15 cm²

= 2 × 50 × 15 cm²

= 1500 cm²

Area of the tank in contact with water = 600 + 1500 cm²

= 2100 cm²

Hence, the area of the tank which is in contact with water is 2100 cm².

Given,

Volume of the cuboid = 448 cm³

Height of the cuboid = 7 cm

The base of the cuboid is a square.

(i) Let a side of the square base be a cm.

By formula,

Volume of a cuboid = base area × height

⇒ a² × 7 = 448

⇒ a² = $\frac{448}{7}$

⇒ a² = 64

⇒ a = $\sqrt{64}$

⇒ a = 8 cm

Hence, a side of the square base is 8 cm.

(ii) Here, length = 8 cm, breadth = 8 cm and height = 7 cm.

By formula,

Surface area of a cuboid = 2(lb + bh + hl)

= 2(8 × 8 + 8 × 7 + 7 × 8) cm²

= 2(64 + 56 + 56) cm²

= 2 × 176 cm²

= 352 cm²

Hence, the surface area of the cuboid is 352 cm².

Given,

The length, breadth and height of the rectangular solid are in the ratio 5 : 4 : 2.

Total surface area of the solid = 1216 cm²

Let the length, breadth and height of the solid be 5a cm, 4a cm and 2a cm respectively.

By formula,

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

⇒ 2(5a × 4a + 4a × 2a + 2a × 5a) = 1216

⇒ 2(20a² + 8a² + 10a²) = 1216

⇒ 2 × 38a² = 1216

⇒ 76a² = 1216

⇒ a² = $\frac{1216}{76}$

⇒ a² = 16

⇒ a = $\sqrt{16}$

⇒ a = 4

So, length of the solid = 5a = 5 × 4 cm = 20 cm

Breadth of the solid = 4a = 4 × 4 cm = 16 cm

Height of the solid = 2a = 2 × 4 cm = 8 cm

By formula,

Volume of a cuboid = length × breadth × height

= 20 × 16 × 8 cm³

= 2560 cm³

Hence, the volume of the rectangular solid is 2560 cm³.

Given,

Length of the rectangular room = 6 m

Breadth of the rectangular room = 5 m

Height of the rectangular room = 3.5 m

By formula,

Area of four walls = 2(length + breadth) × height

= 2(6 + 5) × 3.5 m²

= 2 × 11 × 3.5 m²

= 77 m²

Area of the ceiling = length × breadth

= 6 × 5 m²

= 30 m²

Area of 2 doors = 2 × (1.1 × 2) m²

= 2 × 2.2 m²

= 4.4 m²

Area of 3 windows = 3 × (1.5 × 1.4) m²

= 3 × 2.1 m²

= 6.3 m²

Area to be white washed = Area of four walls + Area of the ceiling - Area of doors - Area of windows

= 77 + 30 - 4.4 - 6.3 m²

= 107 - 10.7 m²

= 96.3 m²

Cost of white washing = Area to be white washed × Rate

= 96.3 m² × ₹40/m²

= ₹3,852

Hence, the cost of white washing the walls and the ceiling of the room is ₹3,852.

Given,

Dimensions of the cuboidal block of metal = 36 cm × 32 cm × 0.25 m

Edge of each cube = 4 cm

Converting the height into centimetres,

Height of the block = 0.25 m = 25 cm

(i) By formula,

Volume of a cuboid = length × breadth × height

Volume of the block = 36 × 32 × 25 cm³

= 28800 cm³

By formula,

Volume of a cube = (edge)³

Volume of one cube = (4)³ cm³

= 64 cm³

Number of cubes = $\frac{\text{Volume of the block}}{\text{Volume of one cube}}$

= $\frac{28800}{64}$

= 450

Hence, 450 such cubes can be made.

(ii) By formula,

Surface area of a cube = 6 × (edge)²

Surface area of one cube = 6 × (4)² cm²

= 6 × 16 cm²

= 96 cm²

Surface area of 450 cubes = 450 × 96 cm²

= 43200 cm²

Cost of silver coating = Total surface area × Rate

= 43200 cm² × ₹0.75/cm²

= ₹32,400

Hence, the cost of silver coating the surfaces of the cubes is ₹32,400.

Given,

Edges of the three cubes of silver = 3 cm, 4 cm and 5 cm

By formula,

Volume of a cube = (edge)³

Sum of the volumes of the three cubes = (3)³ + (4)³ + (5)³ cm³

= 27 + 64 + 125 cm³

= 216 cm³

Let the edge of the new cube be a cm.

Since the three cubes are melted and recast into a single cube, the volume remains the same.

⇒ a³ = 216

⇒ a = $\sqrt[3]{216}$

⇒ a = 6 cm

By formula,

Surface area of a cube = 6 × (edge)²

Surface area of the new cube = 6 × (6)² cm²

= 6 × 36 cm²

= 216 cm²

Cost of coating with gold = Surface area × Rate

= 216 cm² × ₹35/cm²

= ₹7,560

Hence, the cost of coating the surface of the new cube with gold is ₹7,560.

Given,

Curved surface area of the hollow cylinder = 4375 cm²

Width of the rectangular sheet = 35 cm

When the cylinder is cut along its height and opened out, the rectangular sheet so formed has the same area as the curved surface area of the cylinder.

Area of the rectangular sheet = 4375 cm²

Let the length of the rectangular sheet be l cm.

By formula,

Area of a rectangle = length × breadth

⇒ l × 35 = 4375

⇒ l = $\frac{4375}{35}$

⇒ l = 125 cm

By formula,

Perimeter of a rectangle = 2(length + breadth)

= 2(125 + 35) cm

= 2 × 160 cm

= 320 cm

Hence, the perimeter of the rectangular sheet is 320 cm.

Given,

Diameter of the road roller = 0.7 m

Width of the road roller = 1.2 m

Size of the playground = 120 m × 44 m

Radius of the road roller = $\frac{0.7}{2}$ m = 0.35 m

In one complete revolution, the roller levels an area equal to its curved surface area.

By formula,

Curved surface area of a cylinder = 2πrh

Curved surface area of the road roller = 2 × $\frac{22}{7}$ × 0.35 × 1.2 m²

= 2 × 22 × 0.05 × 1.2 m²

= 2.64 m²

So, area levelled in one revolution = 2.64 m²

By formula,

Area of a rectangle = length × breadth

Area of the playground = 120 × 44 m²

= 5280 m²

Number of revolutions = $\frac{\text{Area of the playground}}{\text{Area levelled in one revolution}}$

= $\frac{5280}{2.64}$

= 2000

Hence, the roller must take at least 2000 revolutions to level the playground.

Given,

Diameter of the base of the cylindrical container = 14 cm

Height of the cylindrical container = 20 cm

The label is placed 2 cm from the top and 2 cm from the bottom.

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

Height of the label = 20 - 2 - 2 cm = 16 cm

The label goes around the curved surface of the container, so its area is the curved surface area of a cylinder of radius 7 cm and height 16 cm.

By formula,

$$\begin{aligned}\text{Curved surface area of a cylinder} = 2\pi rh \\[1em] = 2 \times \frac{22}{7} \times 7 \times 16\\[1em] = 2 \times \frac{22}{\cancel{7}} \times \cancel{7} \times 16\\[1em] = 2 \times 22 \times 16\\[1em] = 704 \text{ cm}^2\end{aligned}$$

Hence, the area of the label is 704 cm².

Given,

Sum of the radius and the height of the cylinder = 37 cm

Total surface area of the cylinder = 1628 cm²

Let the radius of the cylinder be r cm and its height be h cm.

So, r + h = 37

By formula,

Total surface area of a cylinder = 2πr(r + h)

$$\begin{aligned}\Rightarrow 2 \times \frac{22}{7} \times r \times 37 = 1628\\[1em] \Rightarrow \frac{1628}{7} \times r = 1628\\[1em] \Rightarrow r = \frac{1628 \times 7}{1628}\\[1em] \Rightarrow r = 7 \text{ cm}\end{aligned}$$

So, radius of the cylinder = 7 cm

Since r + h = 37,

⇒ 7 + h = 37

⇒ h = 37 - 7

⇒ h = 30 cm

By formula,

Volume of a cylinder = πr²h

$$\begin{aligned}= \frac{22}{7} \times 7^2 \times 30\\[1em] = \frac{22}{\cancel{7}} \times \cancel{7} \times 7 \times 30\\[1em] = 22 \times 7 \times 30\\[1em] = 4620 \text{ cm}^3\end{aligned}$$

Hence, the height of the cylinder is 30 cm and its volume is 4620 cm³.

Given,

The ratio between the curved surface and the total surface of the cylinder = 1 : 2

Total surface area of the cylinder = 616 cm²

Let the radius of the cylinder be r cm and its height be h cm.

$$\begin{aligned}\Rightarrow \frac{\text{Curved surface area of the cylinder}}{\text{Total surface area of the cylinder}} = \frac{1}{2}\\[1em] \Rightarrow \frac{2\pi rh}{2\pi r(r + h)} = \frac{1}{2}\\[1em] \Rightarrow \frac{h}{r + h} = \frac{1}{2}\\[1em] \Rightarrow 2h = r + h\\[1em] \Rightarrow 2h - h = r\\[1em] \Rightarrow r = h\end{aligned}$$

Since the total surface area is 616 cm²,

$$\begin{aligned}\Rightarrow 2 \times \frac{22}{7} \times r \times (r + r) = 616\\[1em] \Rightarrow \frac{44}{7} \times r \times 2r = 616\\[1em] \Rightarrow \frac{88}{7} \times r^2 = 616\\[1em] \Rightarrow r^2 = \frac{616 \times 7}{88}\\[1em] \Rightarrow r^2 = 49\\[1em] \Rightarrow r = \sqrt{49}\\[1em] \Rightarrow r = 7 \text{ cm}\end{aligned}$$

So, radius of the cylinder = 7 cm and height of the cylinder = 7 cm

By formula,

Volume of a cylinder = πr²h

$$\begin{aligned}= \frac{22}{7} \times 7^2 \times 7\\[1em] = \frac{22}{\cancel{7}} \times \cancel{7} \times 7 \times 7\\[1em] = 22 \times 7 \times 7\\[1em] = 1078 \text{ cm}^3\end{aligned}$$

Hence, the volume of the cylinder is 1078 cm³.

Given,

Length of the metal pipe = 77 cm

Inner diameter of the cross section = 4 cm

Outer diameter of the cross section = 4.4 cm

Inner radius, r = $\frac{4}{2}$ cm = 2 cm

Outer radius, R = $\frac{4.4}{2}$ cm = 2.2 cm

(i) By formula,

Curved surface area of a cylinder = 2πrh

$$\begin{aligned}\text{Inner curved surface area} = 2 \times \frac{22}{7} \times 2 \times 77\\[1em] = 2 \times 22 \times 2 \times 11\\[1em] = 968 \text{ cm}^2\end{aligned}$$

Hence, the inner curved surface area of the pipe is 968 cm².

(ii) By formula,

Curved surface area of a cylinder = 2πRh

Outer curved surface area = 2 × $\frac{22}{7}$ × 2.2 × 77 cm²

= 2 × 22 × 2.2 × 11 cm²

= 1064.8 cm²

Hence, the outer curved surface area of the pipe is 1064.8 cm².

(iii) The two ends of the pipe are rings, each having outer radius 2.2 cm and inner radius 2 cm.

$$\begin{aligned}\text{ Area of one ring }= \pi R^2 - \pi r^2 \\[1em] = \pi (R^2 - r^2)\\[1em] = \frac{22}{7} \times \left[(2.2)^2 - 2^2\right]\\[1em] = \frac{22}{7} \times (4.84 - 4)\\[1em] = \frac{22}{7} \times 0.84\\[1em] = 22 \times 0.12\\[1em] = 2.64 \text{ cm}^2\end{aligned}$$

So, area of one ring = 2.64 cm²

Area of the two rings = 2 × 2.64 cm²

= 5.28 cm²

Total surface area of the pipe = Inner curved surface area + Outer curved surface area + Area of the two rings

= 968 + 1064.8 + 5.28 cm²

= 2038.08 cm²

Hence, the total surface area of the pipe is 2038.08 cm².

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

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

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

(iv) Area is measured in square units.

(v) Volume of a solid is the measurement of the space occupied by it.

(vi) Volume is measured in cubic units.

(vii) The volume of a unit cube is 1 cubic unit.

(viii) 1 litre = 1000 cm³

(ix) 1 m³ = 1000 litres

(x) Volume of a cuboid = area of base or length × breadth × height.

(xi) Cylinders in which line segment joining the centres of the circular faces is perpendicular to the base are called right circular cylinders.

(xii) Volume of a cylinder = area of base × height.

(xiii) Area of four walls = perimeter of floor × height of the room.

(xiv) Lateral surface area of a cube = 4 × (side)²

(xv) Total surface area of a cylinder of radius r and height h is 2πr(h + r).

(i) True

Reason:

The perimeter of a closed figure is the total length of its boundary. So, the perimeter of a rectangle is the sum of the lengths of its four sides.

(ii) True

Reason:

A diagonal of a quadrilateral divides it into two triangles. So, the area of a quadrilateral can be found by adding the areas of these two triangles.

(iii) False

Reason:

The perimeter (circumference) of a circle of radius r is 2πr. The expression πr² gives the area of the circle, not its perimeter.

(iv) False

Reason:

Volume of a cube = (side)³. The expression 6 × (side)² gives the total surface area of the cube.

(v) False

Reason:

1 m = 100 cm, so 1 m³ = 100 × 100 × 100 cm³ = 1000000 cm³.

(vi) True

Reason:

A cuboid has three pairs of identical opposite faces of areas lb, bh and hl. So, its total surface area = 2(lb + bh + hl).

(vii) False

Reason:

Volume is the amount of space occupied by a solid, whereas capacity is the amount of substance that the solid can hold. So, they are not the same.

(viii) True

Reason:

The total surface area of a cylinder is the sum of its lateral (curved) surface area and the areas of its two circular ends, that is, 2πrh + 2πr².

(ix) False

Reason:

A cube has six identical square faces, so its surface area = 6 × (side)². The expression 4 × (side)² gives the lateral surface area of the cube.

(x) True

Reason:

Lateral surface area of a cuboid = 2(l + b)h = perimeter of the base × height.

Given,

Area of the triangle = 30 cm²

Base of the triangle = 10 cm

Let the height of the triangle be h cm.

By formula,

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

⇒ $\frac{1}{2}$ x 10 x h = 30

⇒ 5 × h = 30

⇒ h = $\frac{30}{5}$

⇒ h = 6 cm

Hence, option 2 is the correct option.

Given,

Perimeter of the square = 80 cm

Let the side of the square be a cm.

By formula,

Perimeter of a square = 4 × side

⇒ 4 × a = 80

⇒ a = $\frac{80}{4}$

⇒ a = 20 cm

By formula,

Area of a square = side²

= (20)² cm²

= 400 cm²

Hence, option 3 is the correct option.

Given,

Area of the parallelogram = 48 cm²

Height of the parallelogram = 6 cm

Let the base of the parallelogram be b cm.

By formula,

Area of a parallelogram = base × height

⇒ b × 6 = 48

⇒ b = $\frac{48}{6}$

⇒ b = 8 cm

Hence, option 1 is the correct option.

Let d be the diameter of the circle.

By formula,

Radius of a circle = $\frac{\text{diameter}}{2}$

So, radius of the circle = $\frac{d}{2}$

By formula,

Area of a circle = πr²

$$\begin{aligned}= \pi \times \left(\frac{d}{2}\right)^2\\[1em] = \pi \times \frac{d^2}{4}\\[1em] = \frac{\pi d^2}{4}\end{aligned}$$

Hence, option 3 is the correct option.

Given,

Area of the trapezium = 64 cm²

Distance between the parallel sides = 8 cm

Let the sum of the parallel sides be a cm.

By formula,

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

⇒ $\frac{1}{2}$ × a × 8 = 64

⇒ 4 × a = 64

⇒ a = $\frac{64}{4}$

⇒ a = 16 cm

Hence, option 4 is the correct option.

Given,

Diagonals of the rhombus = 8 cm and 6 cm

By formula,

Area of a rhombus = $\frac{1}{2}$ × product of its diagonals

= $\frac{1}{2}$ × 8 × 6 cm²

= $\frac{1}{2}$ × 48 cm²

= 24 cm²

Hence, option 2 is the correct option.

Let the lengths of the diagonals of the rhombus be d₁ and d₂.

By formula,

Area of a rhombus = $\frac{1}{2}$ × product of its diagonals

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

When the lengths of the diagonals are doubled, the new diagonals are 2d₁ and 2d₂.

$$\begin{aligned}\text{Area of the new rhombus} = \frac{1}{2} \times 2d_1 \times 2d_2\\[1em] = 4 \times \frac{1}{2} \times d_1 \times d_2\\[1em] = 4 \times \text{Area of the given rhombus}\end{aligned}$$

Hence, option 3 is the correct option.

Given,

Length of a diagonal of the quadrilateral, d = 10 cm

Lengths of the perpendiculars on it from the opposite vertices, h₁ = 4 cm and h₂ = 6 cm

By formula,

Area of a quadrilateral = $\frac{1}{2}$ × d × (h₁ + h₂)

Area of the quadrilateral = $\frac{1}{2}$ × 10 × (4 + 6) cm²

= $\frac{1}{2}$ × 10 × 10 cm²

= 50 cm²

Hence, option 3 is the correct option.

Given,

Area of the rhombus = 90 cm²

Length of one diagonal = 10 cm

Let the length of the other diagonal be d cm.

By formula,

Area of a rhombus = $\frac{1}{2}$ × product of its diagonals

⇒ $\frac{1}{2}$ × 10 × d = 90

⇒ 5 × d = 90

⇒ d = $\frac{90}{5}$

⇒ d = 18 cm

Hence, option 1 is the correct option.

Given,

Volume of the cube = 729 cm³

Let the side of the cube be a cm.

By formula,

Volume of a cube = (side)³

⇒ a³ = 729

⇒ a = $\sqrt[3]{729}$

⇒ a = 9 cm

By formula,

Surface area of a cube = 6 × (side)²

= 6 × (9)² cm²

= 6 × 81 cm²

= 486 cm²

Hence, option 1 is the correct option.

Given,

Lateral surface area of the cube = 100 cm²

Let the side of the cube be a cm.

By formula,

Lateral surface area of a cube = 4 × (side)²

⇒ 4 × a² = 100

⇒ a² = $\frac{100}{4}$

⇒ a² = 25

⇒ a = $\sqrt{25}$

⇒ a = 5 cm

By formula,

Volume of a cube = (side)³

= (5)³ cm³

= 125 cm³

Hence, option 2 is the correct option.

Let the side of the original cube be a units.

By formula,

Volume of a cube = (side)³

Volume of the original cube = a³ cubic units

When the length of the side is doubled, the side of the new cube = 2a units

Volume of the new cube = (2a)³

= 8a³ cubic units

Ratio of the volumes = $\frac{\text{Volume of the new cube}}{\text{Volume of the original cube}}$

= $\frac{8a^3}{a^3}$

= 8

Hence, option 4 is the correct option.

Given,

Dimensions of the rectangular room = 10 m × 12 m × 9 m

Here, length = 10 m, breadth = 12 m and height = 9 m.

By formula,

Area of four walls = 2(length + breadth) × height

= 2(10 + 12) × 9 m²

= 2 × 22 × 9 m²

= 396 m²

Cost of painting the four walls = Area of four walls × Rate

= 396 m² × ₹80/m²

= ₹31,680

Hence, option 3 is the correct option.

Given,

Volume of the cylinder = 1848 cm³

Diameter of the base of the cylinder = 14 cm

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

Let the height of the cylinder be h cm.

By formula,

Volume of a cylinder = πr²h

$$\begin{aligned}\Rightarrow \frac{22}{7} \times 7^2 \times h = 1848\\[1em] \Rightarrow \frac{22}{\cancel{7}} \times \cancel{7} \times 7 \times h = 1848\\[1em] \Rightarrow 154 \times h = 1848\\[1em] \Rightarrow h = \frac{1848}{154}\\[1em] \Rightarrow h = 12 \text{ cm}\end{aligned}$$

Thus, the height of the cylinder is 12 cm.

Hence, option 1 is the correct option.

Let the radius of the original cylinder be r and its height be h.

By formula,

Volume of a cylinder = πr²h

Volume of the original cylinder = πr²h

For the new cylinder, the radius is doubled and the height is halved.

Radius of the new cylinder = 2r

Height of the new cylinder = $\frac{h}{2}$

$$\begin{aligned}\text{Volume of the new cylinder} = \pi \times (2r)^2 \times \frac{h}{2}\\[1em] = \pi \times 4r^2 \times \frac{h}{2}\\[1em] = 2\pi r^2 h\\[1em] = 2 \times \text{Volume of the original cylinder}\end{aligned}$$

Hence, option 2 is the correct option.

Statement I

Parallel sides of the isosceles trapezium = 8 cm and 10 cm

Distance between the parallel sides = 6 cm

By formula,

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

= $\frac{1}{2}$ × (8 + 10) × 6 cm²

= $\frac{1}{2}$ × 18 × 6 cm²

= 9 × 6 cm²

= 54 cm²

So, the area of the trapezium is 54 cm² and Statement I is true.

Statement II

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

This is the standard formula for the area of a trapezium, so Statement II is true.

Hence, option 3 is the correct option.

Statement I

By formula,

Curved surface area of a cylinder = 2πrh

And, total surface area of a cylinder = 2πrh + 2πr²

Since the radius r of a cylinder is always positive, 2πr² is always positive.

So, the total surface area of a cylinder is always greater than its curved surface area and Statement I is true.

Statement II

The expression 2πrh gives the curved surface area of a cylinder and not its total surface area.

The total surface area of a cylinder = 2πr(h + r)

So, Statement II is false.

Hence, option 1 is the correct option.

Statement I

Circumference of the base of the cylinder = 100 cm

Height of the cylinder = 10 cm

Let the radius of the base of the cylinder be r cm.

By formula,

Circumference of a circle = 2πr

⇒ 2πr = 100

⇒ r = $\frac{100}{2\pi}$

⇒ r = $\frac{50}{\pi}$

By formula,

Volume of a cylinder = πr²h

$$\begin{aligned}= \pi \times \left(\frac{50}{\pi}\right)^2 \times 10\\[1em] = \pi \times \frac{2500}{\pi^2} \times 10\\[1em] = \frac{25000}{\pi} \text{ cm}^3\end{aligned}$$

Taking $\pi = \frac{22}{7}$, the volume of the cylinder is about 7954.5 cm³, which is not equal to 1000 cm³.

So, Statement I is false.

Statement II

Volume of a cylinder = πr²h

This is the standard formula for the volume of a cylinder, so Statement II is true.

Hence, option 2 is the correct option.

Given,

Side of the square field = 65 m

Length of the rectangular field = 75 m

Both the fields have the same perimeter.

By formula,

Perimeter of a square = 4 × side

Perimeter of the square field = 4 × 65 m

= 260 m

Let the breadth of the rectangular field be b metres.

By formula,

Perimeter of a rectangle = 2(length + breadth)

⇒ 2(75 + b) = 260

⇒ 75 + b = $\frac{260}{2}$

⇒ 75 + b = 130

⇒ b = 130 - 75

⇒ b = 55 m

By formula,

Area of a square = side²

Area of the square field = (65)² m²

= 4225 m²

By formula,

Area of a rectangle = length × breadth

Area of the rectangular field = 75 × 55 m²

= 4125 m²

Since 4225 m² is greater than 4125 m², the square field has a larger area.

Difference of the areas = 4225 - 4125 m²

= 100 m²

Hence, the square field has a larger area, by 100 m².

Given,

Parallel sides of the trapezium shaped table top = 1.5 m and 2.5 m

Perpendicular distance between the parallel sides = 0.8 m

By formula,

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

= $\frac{1}{2}$ × (1.5 + 2.5) × 0.8 m²

= $\frac{1}{2}$ × 4 × 0.8 m²

= 2 × 0.8 m²

= 1.6 m²

Hence, the area of the table top is 1.6 m².

Given,

Length of the rectangular hall = 26 m

Breadth of the rectangular hall = 22 m

Area required by one student = 1.1 sq. m

By formula,

Area of a rectangle = length × breadth

Area of the hall = 26 × 22 m²

= 572 m²

$$\begin{aligned}\text{ Maximum number of students }= \frac{\text{Area of the hall}}{\text{Area required by one student}}\\[1em] = \frac{572}{1.1}\\[1em] = \frac{5720}{11}\\[1em] = 520\end{aligned}$$

Hence, the maximum number of students that can be seated in the hall is 520.

Given,

Total cost of fencing the square field = ₹9360

Rate of fencing = ₹78 per metre

Perimeter of the field = $\frac{\text{Total cost of fencing}}{\text{Rate of fencing}}$

= $\frac{9360}{78}$ m

= 120 m

Let the side of the square field be a metres.

By formula,

Perimeter of a square = 4 × side

⇒ 4 × a = 120

⇒ a = $\frac{120}{4}$

⇒ a = 30 m

By formula,

Area of a square = side²

= (30)² m²

= 900 m²

Cost of levelling the field = Area of the field × Rate of levelling

= 900 m² × ₹25/m²

= ₹22,500

Hence, the cost of levelling the field is ₹22,500.

(i) From the figure, the shaded portion consists of the central rectangle and the four small rectangles at the corners.

The outer rectangle is 30 cm long and 10 cm wide, and the central shaded rectangle is 18 cm long and 6 cm wide.

Length of each corner rectangle = $\frac{30 - 18}{2}$ cm

= $\frac{12}{2}$ cm

= 6 cm

Breadth of each corner rectangle = $\frac{10 - 6}{2}$ cm

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

= 2 cm

By formula,

Area of a rectangle = length × breadth

Area of the central rectangle = 18 × 6 cm²

= 108 cm²

Area of one corner rectangle = 6 × 2 cm²

= 12 cm²

Area of the four corner rectangles = 4 × 12 cm²

= 48 cm²

Area of the shaded portion = 108 + 48 cm²

= 156 cm²

Hence, the area of the shaded portion is 156 cm².

(ii) The given shaded figure can be divided into three portions, as shown below.

The first portion is a rectangle 4 cm long and 2 cm wide.

The second portion is a rectangle 6 cm long and 1 cm wide.

The third portion is a rectangle 4 cm long and 2 cm wide from which a rectangle 3 cm long and 1 cm wide has been removed.

By formula,

Area of a rectangle = length × breadth

Area of the first portion = 4 × 2 cm²

= 8 cm²

Area of the second portion = 6 × 1 cm²

= 6 cm²

Area of the third portion = (4 × 2) - (3 × 1) cm²

= 8 - 3 cm²

= 5 cm²

Area of the shaded portion = 8 + 6 + 5 cm²

= 19 cm²

Hence, the area of the shaded portion is 19 cm².

Given,

Area of the trapezium = 160 sq. cm

The lengths of the parallel sides are in the ratio 1 : 3.

Length of the smaller parallel side = 10 cm

Since the parallel sides are in the ratio 1 : 3,

Length of the larger parallel side = 3 × 10 cm

= 30 cm

Let the perpendicular distance between the parallel sides be h cm.

By formula,

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

⇒ $\frac{1}{2}$ x (10 + 30) x h = 160

⇒ $\frac{1}{2}$ x 40 x h = 160

⇒ 20 × h = 160

⇒ h = $\frac{160}{20}$

⇒ h = 8 cm

Hence, the perpendicular distance between the parallel sides is 8 cm.

Given,

Area of the trapezium = 729 cm²

Distance between the two parallel sides = 18 cm

One parallel side is 3 cm shorter than the other parallel side.

Let the longer parallel side be a cm.

Then, the shorter parallel side = (a - 3) cm

By formula,

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

⇒ $\frac{1}{2}$ x (a + a - 3) x 18 = 729

⇒ (2a - 3) × 9 = 729

⇒ 2a - 3 = $\frac{729}{9}$

⇒ 2a - 3 = 81

⇒ 2a = 81 + 3

⇒ 2a = 84

⇒ a = $\frac{84}{2}$

⇒ a = 42 cm

So, the longer parallel side = 42 cm

And, the shorter parallel side = 42 - 3 cm = 39 cm

Hence, the lengths of the parallel sides of the trapezium are 39 cm and 42 cm.

Given,

Side of the hexagon ABCDEF = 5 cm

From the figure, AC = FD = 8 cm and the distance between AC and FD is 5 cm.

The height of the hexagon, that is the distance between B and E is 11 cm.

First student's way:

The first student divides the hexagon into two triangles, ABC and FED, and a rectangle ACDF.

Total height = 11 cm

AF = 5 cm

From figure,

Height of each triangle = $\frac{11 - 5}{2}$ cm

= $\frac{6}{2}$ cm

= 3 cm

Height of triangle ABC = 3 cm

By formula,

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

Area of triangle ABC = $\frac{1}{2}$ x 8 x 3 cm²

= 12 cm²

Area of triangle FED = 12 cm²

By formula,

Area of a rectangle = length × breadth

Area of rectangle ACDF = 8 × 5 cm²

= 40 cm²

Area of the hexagon = 12 + 12 + 40 cm²

= 64 cm²

Second student's way:

The second student divides the hexagon into two congruent trapeziums, ABEF and BCDE, by joining B and E.

In trapezium ABEF, the parallel sides are AF and BE.

Length of AF = 5 cm and length of BE = 11 cm

Distance between AF and BE = $\frac{8}{2}$ cm

= 4 cm

By formula,

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

Area of trapezium ABEF = $\frac{1}{2}$ x (5 + 11) x 4 cm²

= $\frac{1}{2}$ x 16 x 4 cm²

= 32 cm²

Area of trapezium BCDE = 32 cm²

Area of the hexagon = 32 + 32 cm²

= 64 cm²

Hence, by both the ways the area of the hexagon is 64 cm².

From the given figure,

AC = 60 m, AF = 16 m, AH = 46 m, EF = 24 m, DH = 14 m and BG = 16 m

The diagonal AC divides the polygon into two parts. EF, DH and BG are perpendiculars drawn on AC.

FH = AH - AF = 46 - 16 = 30 m

HC = AC - AH = 60 - 46 = 14 m

Area of the portion above AC:

By formula,

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

Area of triangle AEF = $\frac{1}{2}$ x 16 x 24 m²

= 192 m²

By formula,

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

Area of trapezium FEDH = $\frac{1}{2}$ x (24 + 14) x 30 m²

= $\frac{1}{2}$ x 38 x 30 m²

= 570 m²

Area of triangle DHC = $\frac{1}{2}$ x 14 x 14 m²

= 98 m²

Area of the portion below AC:

Area of triangle ABC = $\frac{1}{2}$ x 60 x 16 m²

= 480 m²

Area of the polygon = 192 + 570 + 98 + 480 m²

= 1340 m²

Hence, the area of the polygon is 1340 m².

Given,

Diagonals of the rhombus = 16 m and 12 m

Let ABCD be the rhombus in which AC = 16 m and BD = 12 m. Let its diagonals meet at O.

(i) By formula,

Area of a rhombus = $\frac{1}{2}$ × product of its diagonals

= $\frac{1}{2}$ x 16 x 12 m²

= 8 × 12 m²

= 96 m²

Hence, the area of the rhombus is 96 m².

(ii) As we know, the diagonals of a rhombus bisect each other at right angles.

OA = $\frac{1}{2}$ x 16 m = 8 m

OB = $\frac{1}{2}$ x 12 m = 6 m

In right-angled triangle AOB, AB is the hypotenuse.

By Pythagoras theorem,

OA² + OB² = AB²

⇒ (8)² + (6)² = AB²

⇒ 64 + 36 = AB²

⇒ AB² = 100

⇒ AB = $\sqrt{100}$

⇒ AB = 10 m

Hence, the length of a side of the rhombus is 10 m.

(iii) By formula,

Perimeter of a rhombus = 4 × side

= 4 × 10 m

= 40 m

Hence, the perimeter of the rhombus is 40 m.

Given,

Area of the parallelogram = 98 cm²

One altitude is half of the corresponding base.

Let the base of the parallelogram be b cm.

Then, the corresponding altitude = $\frac{b}{2}$ cm

By formula,

Area of a parallelogram = base × corresponding altitude

⇒ b x $\frac{b}{2}$ = 98

⇒ b² = 2 × 98

⇒ b² = 196

⇒ b = $\sqrt{196}$

⇒ b = 14 cm

So, the altitude of the parallelogram = $\frac{14}{2}$ cm = 7 cm

Hence, the base of the parallelogram is 14 cm and its altitude is 7 cm.

Given,

Dimensions of the hall = 18 m × 15 m × 5 m

Area painted by one can of paint = 120 m²

Here, length = 18 m, breadth = 15 m and height = 5 m.

By formula,

Area of four walls = 2(length + breadth) × height

= 2(18 + 15) × 5 m²

= 2 × 33 × 5 m²

= 330 m²

Area of the ceiling = length × breadth

= 18 × 15 m²

= 270 m²

Total area to be painted = 330 + 270 m²

= 600 m²

Number of cans of paint = $\frac{\text{Total area to be painted}}{\text{Area painted by one can}}$

= $\frac{600}{120}$

= 5

Hence, she needs 5 cans of paint to paint the hall.

Given,

Size of the rectangular paper = 22 cm × 14 cm

Height of the cylinder formed = 14 cm

Since the height of the cylinder is 14 cm, the side of length 22 cm becomes the circumference of the base of the cylinder.

Let the radius of the base of the cylinder be r cm.

⇒ 2πr = 22

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

By formula,

Volume of a cylinder = πr²h

$$\begin{aligned}= \frac{22}{7} \times (3.5)^2 \times 14\\[1em] = \frac{22}{7} \times 12.25 \times 14\\[1em] = 22 \times 12.25 \times 2\\[1em] = 539 \text{ cm}^3\end{aligned}$$

Hence, the volume of the cylinder is 539 cm³.

Given,

Inner dimensions of the closed rectangular wooden box = 90 cm by 80 cm by 70 cm

Here, length = 90 cm, breadth = 80 cm and height = 70 cm.

By formula,

Volume of a cuboid = length × breadth × height

Capacity of the box = 90 × 80 × 70 cm³

= 504000 cm³

Since the box is closed, the tin foil is needed for all its six inner faces.

By formula,

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

Area of the tin foil = 2(90 × 80 + 80 × 70 + 70 × 90) cm²

= 2(7200 + 5600 + 6300) cm²

= 2 × 19100 cm²

= 38200 cm²

Hence, the capacity of the box is 504000 cm³ and the area of the tin foil needed to line its inner surface is 38200 cm².

Given,

Lateral surface area of the cuboid = 224 cm²

Height of the cuboid = 7 cm

The base of the cuboid is a square.

(i) Let the length of the side of the square base be a cm.

Since the base is a square, length = breadth = a cm.

By formula,

Lateral surface area of a cuboid = 2(length + breadth) × height

⇒ 2(a + a) × 7 = 224

⇒ 2 × 2a × 7 = 224

⇒ 28a = 224

⇒ a = $\frac{224}{28}$

⇒ a = 8 cm

Hence, the length of a side of the square base is 8 cm.

(ii) By formula,

Volume of a cuboid = length × breadth × height

= 8 × 8 × 7 cm³

= 448 cm³

Hence, the volume of the cuboid is 448 cm³.

Given,

Inner dimensions of the closed wooden box = 2 m by 1.2 m by 0.75 m

Thickness of the wood = 2.5 cm = 0.025 m

Since the box is closed, the wood is present on both the opposite faces in each direction. So, each inner dimension increases by 2 × 0.025 = 0.05 m.

Outer length = 2 + 0.05 m = 2.05 m

Outer breadth = 1.2 + 0.05 m = 1.25 m

Outer height = 0.75 + 0.05 m = 0.8 m

By formula,

Volume of a cuboid = length × breadth × height

Outer volume of the box = 2.05 × 1.25 × 0.8 m³

= 2.05 m³

Inner volume of the box = 2 × 1.2 × 0.75 m³

= 1.8 m³

Volume of the wood = Outer volume - Inner volume

= 2.05 - 1.8 m³

= 0.25 m³

Cost of the wood = Volume of the wood × Rate

= 0.25 m³ × ₹27000/m³

= ₹6,750

Hence, the cost of the wood required to make the box is ₹6,750.

Given,

Length of the petrol tank = 40 cm

Width of the petrol tank = 28 cm

Depth of the petrol tank = 25 cm

Average consumption of the car = 13.5 km per litre

By formula,

Volume of a cuboid = length × breadth × height

Volume of the petrol tank = 40 × 28 × 25 cm³

= 28000 cm³

Converting the volume into litres,

Capacity of the tank = $\frac{28000}{1000}$ litres

[∵ 1 litre = 1000 cm³]

= 28 litres

Distance travelled = Capacity of the tank × Average consumption

= 28 × 13.5 km

= 378 km

Hence, the car can travel 378 km with a full tank of petrol.

Given,

Diameter of the garden roller = 1.4 m

Length of the garden roller = 2 m

Radius of the roller = $\frac{1.4}{2}$ m = 0.7 m

In one complete revolution, the roller covers an area equal to its curved surface area.

By formula,

Curved surface area of a cylinder = 2πrh

$$\begin{aligned}= 2 \times \frac{22}{7} \times 0.7 \times 2\\[1em] = 2 \times 22 \times 0.1 \times 2\\[1em] = 8.8 \text{ m}^2\end{aligned}$$

So, area covered in one revolution = 8.8 m²

Area covered in 5 revolutions = 5 × 8.8 m²

= 44 m²

Hence, the roller will cover an area of 44 m² in 5 revolutions.

Given,

Capacity of the open cylindrical tank = 2079 m³

Diameter of the base of the tank = 21 m

Radius of the base = $\frac{21}{2}$ m = 10.5 m

Let the height of the tank be h m.

By formula,

Volume of a cylinder = πr²h

$$\begin{aligned}\Rightarrow \frac{22}{7} \times (10.5)^2 \times h = 2079\\[1em] \Rightarrow \frac{22}{7} \times 110.25 \times h = 2079\\[1em] \Rightarrow 346.5 \times h = 2079\\[1em] \Rightarrow h = \frac{2079}{346.5}\\[1em] \Rightarrow h = 6 \text{ m}\end{aligned}$$

So, height of the tank = 6 m

Since the tank is open at the top, its inner surface consists of the curved surface and the base.

By formula,

Curved surface area of a cylinder = 2πrh

$$\begin{aligned}= 2 \times \frac{22}{7} \times 10.5 \times 6\\[1em] = 2 \times 22 \times 1.5 \times 6\\[1em] = 396 \text{ m}^2\end{aligned}$$

So, curved surface area = 396 m²

By formula,

Area of a circle = πr²

$$\begin{aligned}\text{Area of the base} = \frac{22}{7} \times (10.5)^2\\[1em] = \frac{22}{7} \times 110.25\\[1em] = 22 \times 15.75\\[1em] = 346.5 \text{ m}^2\end{aligned}$$

So, area of the base = 346.5 m²

Inner surface area to be plastered = 396 + 346.5 m²

= 742.5 m²

Cost of plastering = Inner surface area × Rate

= 742.5 m² × ₹40/m²

= ₹29,700

Hence, the cost of plastering the inner surface of the tank is ₹29,700.

Given,

Height of the solid right circular cylinder = 1.21 m = 121 cm

Diameter of the cylinder = 28 cm

Number of equal solid cubes formed = 7

Radius of the cylinder = $\frac{28}{2}$ cm = 14 cm

By formula,

Volume of a cylinder = πr²h

$$\begin{aligned}= \frac{22}{7} \times 14^2 \times 121\\[1em] = \frac{22}{7} \times 196 \times 121\\[1em] = 22 \times 28 \times 121\\[1em] = 74536 \text{ cm}^3\end{aligned}$$

Since the cylinder is melted and recast into 7 equal cubes, the total volume remains the same.

Volume of one cube = $\frac{74536}{7}$ cm³

= 10648 cm³

Let the edge of each cube be a cm.

Volume of a cube = (edge)³

⇒ a³ = 10648

⇒ a = $\sqrt[3]{10648}$

⇒ a = 22 cm

Hence, the edge of each cube is 22 cm.

Given,

Depth of the well = 20 m

Diameter of the well = 2 m

Radius of the well = $\frac{2}{2}$ m = 1 m

(i) The soil to be dug out occupies a cylindrical space of radius 1 m and height 20 m.

By formula,

Volume of a cylinder = πr²h

$$\begin{aligned}= \frac{22}{7} \times 1^2 \times 20\\[1em] = \frac{22 \times 20}{7}\\[1em] = \frac{440}{7}\\[1em] = 62\frac{6}{7} \text{ m}^3\end{aligned}$$

Hence, $\mathbf{62\frac{6}{7}}$ cubic metres of soil must be dug out.

(ii) The inner curved surface of the well is the curved surface of the cylinder.

By formula,

Curved surface area of a cylinder = 2πrh

$$\begin{aligned}= 2 \times \frac{22}{7} \times 1 \times 20\\[1em] = \frac{880}{7} \text{ m}^2\end{aligned}$$

So, inner curved surface area = $\frac{880}{7}$ m²

Cost of plastering = Inner curved surface area × Rate

= $\frac{880}{7}$ m² × ₹200/m²

= ₹$\frac{176000}{7}$

= ₹25142.86 (approximately)

Rounding off to the nearest hundred rupees, the cost of plastering is ₹25,100.

Hence, the cost of plastering the inner curved surface of the well is ₹25,100.

Given,

The length of the room is 50% more than its breadth.

Cost of carpeting the room at ₹385 per m² = ₹9240

Cost of papering the walls at ₹33 per m² = ₹2145

The room has one door of dimensions 1 m × 2 m and two windows each of dimensions 1 m × 1.5 m.

Area of the floor = $\frac{\text{Cost of carpeting}}{\text{Rate of carpeting}}$

= $\frac{9240}{385}$ m²

= 24 m²

Let the breadth of the room be b metres.

Since the length is 50% more than the breadth,

Length of the room = $b + \frac{50}{100} \times b$

= $b + \frac{b}{2}$

= $\frac{3b}{2}$ metres

By formula,

Area of a rectangle = length × breadth

⇒ $\frac{3b}{2} \times b = 24$

⇒ 3b² = 48

⇒ b² = $\frac{48}{3}$

⇒ b² = 16

⇒ b = $\sqrt{16}$

⇒ b = 4 m

So, breadth of the room = 4 m

And, length of the room = $\frac{3 \times 4}{2}$ m = 6 m

Area of the walls papered = $\frac{\text{Cost of papering}}{\text{Rate of papering}}$

= $\frac{2145}{33}$ m²

= 65 m²

Area of the door = 1 × 2 m²

= 2 m²

Area of the two windows = 2 × (1 × 1.5) m²

= 3 m²

Let the height of the room be h metres.

By formula,

Area of four walls = 2(length + breadth) × height

Area of four walls - Area of the door - Area of the windows = Area of the walls papered

⇒ 2(6 + 4) × h - 2 - 3 = 65

⇒ 20h - 5 = 65

⇒ 20h = 65 + 5

⇒ 20h = 70

⇒ h = $\frac{70}{20}$

⇒ h = 3.5 m

Hence, the dimensions of the room are 6 m × 4 m × 3.5 m.

Given,

Length of the barrel of the fountain pen = 7 cm

Diameter of the barrel = 5 mm = 0.5 cm

A full barrel of ink is used up in writing 310 words.

Volume of the bottle of ink = one-fifth of a litre

Radius of the barrel = $\frac{0.5}{2}$ cm = 0.25 cm

By formula,

Volume of a cylinder = πr²h

$$\begin{aligned}= \frac{22}{7} \times (0.25)^2 \times 7\\[1em] = \frac{22}{\cancel{7}} \times 0.0625 \times \cancel{7}\\[1em] = 22 \times 0.0625\\[1em] = 1.375 \text{ cm}^3\end{aligned}$$

So, volume of ink in a full barrel = 1.375 cm³

Volume of ink in the bottle = $\frac{1}{5} \times 1000$ cm³

[∵ 1 litre = 1000 cm³]

= 200 cm³

Number of barrels that can be filled from the bottle = $\frac{200}{1.375}$

= $\frac{200000}{1375}$

= $\frac{1600}{11}$

Number of words = $\frac{1600}{11} \times 310$

= $\frac{496000}{11}$

= 45090.9 (approximately)

Rounding off to the nearest 100 words, the number of words is 45100.

Hence, about 45100 words would use up a bottle of ink containing one-fifth of a litre.

Given,

Height of the cylindrical jar = 20 cm

Internal diameter of the jar = 7 cm

Edge of the iron cube = 5 cm

Original height of the water = 12 cm

Radius of the jar = $\frac{7}{2}$ cm = 3.5 cm

By formula,

Volume of a cube = (edge)³

Volume of the iron cube = (5)³ cm³

= 125 cm³

When the cube is completely immersed, the volume of the water that rises is equal to the volume of the cube.

Let the rise in the level of water be h cm.

By formula,

Volume of a cylinder = πr²h

$$\begin{aligned}\Rightarrow \frac{22}{7} \times (3.5)^2 \times h = 125\\[1em] \Rightarrow \frac{22}{7} \times 12.25 \times h = 125\\[1em] \Rightarrow 38.5 \times h = 125\\[1em] \Rightarrow h = \frac{125}{38.5}\\[1em] \Rightarrow h = \frac{1250}{385}\\[1em] \Rightarrow h = \frac{250}{77}\\[1em] \Rightarrow h = 3\frac{19}{77} \text{ cm}\end{aligned}$$

Hence, the rise in the level of water is $\mathbf{3\frac{19}{77}}$ cm.

Given,

Size of the tin sheet = 42 cm by 30 cm

Side of each square cut off from the corners = 6 cm

When squares of side 6 cm are cut off from the four corners and the flaps are folded up, the side of each square becomes the height of the open box, and each dimension of the sheet decreases by 6 cm at both the ends.

Length of the box = 42 - 6 - 6 cm

= 30 cm

Breadth of the box = 30 - 6 - 6 cm

= 18 cm

Height of the box = 6 cm

By formula,

Volume of a cuboid = length × breadth × height

Capacity of the box = 30 × 18 × 6 cm³

= 3240 cm³

Hence, the capacity of the box is 3240 cm³.