These NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.2 Questions and Answers are prepared by our highly skilled subject experts.

## NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Exercise 11.2

Question 1.

The shape of the top surface of a table is a trapezium. Find its area if its parallel sides are 1 m and 1.2 m and perpendicular distance between them is 0.8 m.

Answer:

Here, = a = 1.2m b = 1m and h = 0.8m

Area of a trapezium = \(\frac{1}{2}\) × h × (a + b)

= \(\frac{1}{2}\) × 0.8 × (1.2 + 1)m^{2} = 0.4 × 2.2 = 0.88 m^{2}

Question 2.

The area of trapezium is 34 cm^{2} and the length of one of the parallel sides is 10 cm and its height is 4 cm. Find the length of the other parallel side.

Answer:

Let the length of the other parallel side be ‘X’, height of a trapezium = 4 m. Length of one parallel side = 10 cm

Area of a trapezium = 34 cm^{2}

\(\frac{1}{2}\) × h × (a + b) = 34

\(\frac{1}{2}\) × 4 × (10 + x) = 34

2 (10 + x) = 34

20 + 2x = 34

2x = 34 – 20

2x = 14

x = \(\frac{14}{2}\) = 7 cm

Hence, length of the other side = 7 cm

Question 3.

Length of the fence of a trapezium shaped field ABCD is 120 m. If BC = 48 m, CD = 17m and AD = 40 m, find the area of this field. Side AB is perpendicular to the parallel sides AD and BC.

Answer:

Given BC = 48m,CD = 17mandAD = 40m

Perimeter of a trapezium = 120 m

AB + BC + CD + DA = 120

AB + 48 + 17 + 40 = 120

AB + 105 = 120

AB = 120 – 105 = 15 m

AB is the height of the trapezium.

Area of a trapezium = \(\frac{1}{2}\) × h × (a + b)

= \(\frac{1}{2}\) × 15 × (48 + 40) m^{2}

= \(\frac{1}{2}\) × 15 × 88 m^{2} = 15 × 44 m^{2} = 660 m^{2}

Area of the given trapezium = 660 m^{2}

Question 4.

The diagonal of quadrilateral shaped field is 24 m and the perpendiculars dropped on it from the remaining opposite vertices are 8 m and 13 m. Find the area of the field.

Answer:

In the given quadrilateral, d = 24 m, h_{1} = 13 m and h_{2} = 8 m

Area of the quadrilateral = \(\frac{1}{2}\) × d × (h_{1} + h_{2})

= \(\frac{1}{2}\) × 24 × (13 + 8) m^{2}

= 12 × 21 m^{2} = 252 m^{2}

Question 5.

The diagonals of a rhombus are 7.5 cm and 12 cm. Find its area.

Answer:

Here d_{1} = 7.5 cm, d_{2} = 12 cm

Area of the rhombus = \(\frac{1}{2}\) × d_{1} × d_{2}

= \(\frac{1}{2}\) × 7.5 × 12 cm^{2} = 7.5 × 6 cm^{2} = 45 cm^{2}

Question 6.

Find the area of a rhombus whose side is 5 cm and whose altitude is 4.8 cm. If one of its diagonals is 8 cm long, find the length of the other diagonal.

Answer

We know that rhombus is also a parallelogram.

Area of the rhombus = Area of a parallelogram = bh = 5 × 4.8 cm^{2} × = 24 cm^{2}

Let the other diagonal be ‘x’

Area of a rhombus = 24 cm^{2}

\(\frac{1}{2}\) × d_{1} × d_{2} = 24

\(\frac{1}{2}\) × 8 × x = 24

4 × x = 24

x = \(\frac{24}{4}\) = 6 cm

∴ The required diagonal = 6 cm

Question 7.

The floor of a building consists of 3000 tiles which are rhombus shaped and each of its diagonal are 45 cm and 30 cm length.

Find the total cost of polishing the floor if the cost per m^{2} is ₹ 4.

Answer:

Tiles are in the shape of a rhombus.

Here d_{1} = 45 cm and d_{2} = 30 cm

Area of the rhombus (one tile) = \(\frac{1}{2}\) × d_{1} × d_{2}

= \(\frac{1}{2}\) × 45 × 30 cm^{2} = 45 × 15 = 675 cm^{2}

Total number of tiles = 3000

Area of the floor = 3000 × 675 cm^{2}

= \(\frac{3000 \times 675}{100 \times 100}\)m^{2} = \(\frac{2025}{10}\)m^{2}

Cost of polishing the floor = \(\frac{2025}{10}\) × 4

= 405 × 2 = ₹ 810

Question 8.

wants to buy a trapezium shaped field. Its side along the river is parallel to and twice the side along the road. If the area of this field is 10500 m^{2} and the perpendicular distance between the two parallel sides is 100 m, find the length of the side along the river.

Answer:

Let the length of the side be ‘x’ (along the river)

Length of the other side = \(\frac{1}{2}\) × x m (along the road)

Perpendicular distance between two parallel sides = 100 m

Area of the trapezium field = 10500 m^{2}

\(\frac{1}{2}\) × h (a + b) = 10500

\(\frac{1}{2}\) × 100 (x + \(\frac{1}{2}\)x) = 10500

\(\frac{1}{2}\) × 100 × \(\frac{3 x}{2}\) = 10500

\(\frac{300}{4}\) x = 10500

x = \(\frac{10500}{4}\) × 4 = 35 × 4m = 140 m

∴ Length of the side along the river = 140 m

Question 9.

Top surface of a raised platform is in the shape of a regular octagon as shown in the figure. Find the area of the octagonal surface.

Answer:

The regular octagon can be divided into two trapeziums and one rectangle.

Area of the octagonal surface

= Area of rectangular portion + 2 (Area of trapezoidal portion)

= 11 × 5 + 2[\(\frac { 1 }{ 2 }\) × 4(5 + 11)]

= 55 + 2 × \(\frac { 1 }{ 2 }\) × 4 × 16 m

= 55 + 64 m^{2} = 119 m^{2}

∴ Area of the octagonal surface =119 cm^{2}

Question 10.

There is a pentagonal shaped park as shown in the figure. For finding its area, Joyti and Kavita divided it in two different ways.

Find the area of this park using both ways. Can you suggest some other way of finding its area?

Answer:

For Joyti’s diagram

The given shape is divided into two

congruent trapeziums

Here a = 30m, b = 15m (parallel sides),

h = \(\frac { 15 }{ 2 }\) m.

Area of a trapezium = \(\frac { 1 }{ 2 }\) × h(a + b)squnits.

Area of one trapezium

= \(\frac { 1 }{ 2 }\) × \(\frac { 15 }{ 2 }\) × (30 + 15) m^{2}

= \(\frac { 1 }{ 2 }\) × \(\frac { 15}{ 2 }\) × 45 m^{2} = \(\frac { 675 }{ 4 }\)^{2}

Area of the pentagonal shape = 2 \(\frac { 675}{ 4 }\)

= \(\frac { 675}{ 2 }\)m^{2} = 337.5 m^{2}

For Kavita’s diagram

The given shape is divided into a square and a triangle.

Here, side of a square is 15 m; height of the triangle is30-15 = 15m

Area of the pentagonal shape = Area of the square + Area of the triangle

= 15 × 15 + \(\frac { 1 }{ 2 }\) × 15 × 15 m^{2}

= 225 + \(\frac { 225 }{ 2 }\)m^{2}

= 225 + 112.5 m^{2}

= 337.5 m^{2}

Another method

Divide the pentagonal shape into three triangles of side 15 m and height 15 m. Area of the pentagonal shape

= 3 × Area of the triangle

= 3 × \(\frac { 1 }{ 2 }\) × 15 × 15 = 337.5 m^{2}

Question 11.

Diagram of the adjacent picture frame has outer dimensions = 24 cm × 28 cm and inner dimensions 16 cm × 20 cm. Find the area of each section of the frame, if the width of each section is same.

Answer:

In trapezium (1) parallel sides are 24 cm and 16 cm.

Height = \(\frac{28-20}{2}\) = 4 cm

Areaofitstrapezium(1) = \(\frac{1}{2}\) × 4 × (24 + 16)

= 2 × 40 = 80 cm^{2}

In trapezium (2) parallel sides are 28 cm and 20 cm.

Height = \(\frac{24-16}{2}=\frac{8}{2}\) = 4 cm

Area of the trapezium (2)

= \(\frac{1}{2}\) × 4 × (28 + 20)cm^{2} = 2 × 48cm^{2} = 96cm^{2}

Area of each section of the frame is 80 cm^{2}, 96 cm^{2}, 80 cm^{2} and 96 cm^{2}

Note: [Area of 1 section = Area of 3rd section and Area of 2nd section = Area of 4th section]