NCERT Solutions for Class 8 Maths - (Ex-11.1, 11.2) - Mensuration

Exercise 11.1 

Question 1: 

A square and a rectangular field with measurements as given in the figure have the same perimeter. Which field has a larger area? 

Answer 1: 

Given:

The side of a square = 60 m

And the length of rectangular field = 80 m

According to question,

Perimeter of rectangular field = Perimeter of square field

=> 2 (/+b) = 4x side

=> 2(80 + b) = 4x60

=>  160 + 2b = 240

=> 2b = 240 -160

=>  2b = 80

=> b = 40 m

Now, Area of Square field = (Side)² = (60)² = 3600 m²

And Area of Rectangular field = length x breadth = 80 x 40 = 3200 m²

Hence, area of square field is larger.

Question 2: 

Mrs. Kaushik has a square plot with the measurement as shown in the figure. She wants to construct a house in the middle of the plot. A garden is developed around the house. Find the total cost of developing a garden around the house at the rate of Rs. 55 per m².

Answer 2: 

Side of a square plot = 25 m

∴ Area of square plot = (Side)² = (25)² = 625 m²

Length of the house = 20 m and

Breadth of the house = 15 m

∴ Area of the house = length x breadth = 20 x 15 = 300

Area of garden = Area of square plot - Area of house = 625 - 300 = 325 m²

∴ Cost of developing the garden per sq. m = Rs. 55

∴ Cost of developing the garden 325 sq. m

= Rs. 55x325 = Rs. 17,875

Hence total cost of developing a garden around is Rs. 17,875.

Question 3: 

The shape of a garden is rectangular in the middle and semi-circular at the ends as shown in the diagram. Find the area and the perimeter of this garden [Length of rectangle is 20 – (3.5 + 3.5 meters)

Answer 3: 

Given: Total length = 20 m

Diameter of semi circle = 7 m

Radius of semi circle = 7/2 = 3.5m

Length of rectangular field = 20 - (3.5 + 3.5) = 20 - 7 = 13 m

Breadth of the rectangular field = 7 m

Area of rectangular field = Ixb = 13x7 = 91 m²

Area of two semi circles = 

Area of garden = 91+38.5 = 129.5 m²

Now,  Perimeter of two semi circles = 2 x 22/7 x 3.5 = 22m

And

Perimeter of garden = 22 + 13 + 13 = 48 m

Question 4: 

A flooring tile has the shape of a parallelogram whose base is 24 cm and the corresponding height is 10 cm. How many such tiles are required to cover a floor of area 1080 m²? [If required you can split the tiles in whatever way you want to fill up the corners] 

Answer 4: 

Given:

Base of flooring tile = 24 cm = 0,24 m

Corresponding height of a flooring tile = 10 cm = 0,10 m

Now , Area of flooring tile = Base x Altitude = 0.24 x 0.10 = 0.024 m²

Number of tyles required = area of floor/ area of one tile

1080 / 0.024 = 45000 tiles

Hence 45000 dies are required to cover the floor.

Question 5: 

An ant is moving around a few food pieces of different shapes scattered on the floor. For which food-piece would the ant have to take a longer round? Remember, circumference of a circle can be obtained by using the expression c = 2πr where r is the radius of the circle. 

Answer 5: 

(a) 

Circumference of semi circle = πr = 22/7 x 1.4 = 4.4 cm

Total distance covered by the ant = Circumference of semi circle + Diameter

= 4.4 + 2.8 = 7.2 cm

(b) 

Circumference of semi circle = πr = 22/7 x 1.4 = 4.4 cm

Total distance covered by the ant = 1.5 + 2.8 + 1.5 + 4.4 = 10.2 cm

(c)

Circumference of semi circle = πr = 22/7 x 1.4 = 4.4 cm

Total distance covered by the ant = 2 + 2 +4.4 = 8.8cm

Hence for figure (b) food piece, the ant would take a longer round.

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 1: 

Here one parallel side of the trapezium (a) = 1 m

And second side (b) = 1.2 m and height (A) = 0.8 m

Area of top surface of the table = 1/2 (a + b) x h

= 1/2 x (1+1.2) x 0.8 

= 1/2 x 2.2 x 0.8  = 0.88 m²

Hence, the surface area of the table is 0.88 m².

Question 2: 

The area of a trapezium is 34 cm² 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 2: 

Let the length of the other parallel side be b

Length of one parallel side (a) = 10 am and height (h) =4 cm

Area of trapezium = 1/2 (a + b)xh

=>  34 = 1/2 (10 + b)x 4

=>  34 = (10 + b)x 2

34 = 20 + 2b

=>  34 - 20 = 2b

=>  14 = 2 b

=>   7 = b

=>  b = 7

Hence, the another required parallel side is 7 cm.

Question 3: 

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

Answer 3: 

Given: BC - 4B m, CD = 17 m, AD = 40 m and perimeter - 120 m

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

=> 120 =AB +48 + 17 + 40

=> 120 =AB = 105

=> AB = 120- 105 = 15 m

Question 4: 

The diagonal of a 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 4: 

Here h₁ = 13 m, h₂ = 8 m and AC = 24 m

Area of quadrilateral ABCD = Area of  ΔABC + Area of ΔADC

Hence,  the area of the field is 252 m²

Question 5: 

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

Answer 5: 

Question 6: 

Find the area of a rhombus whose side is 6 cm and whose altitude is 4 cm. If one of the diagonals is 8 cm long, find the length of the other diagonal.

Answer 6: 

Question 7: 

The floor of a building consists of 3000 tiles which are rhombus shaped and each of its diagonals are 45 cm and 30 cm in length. Find the total cost of polishing the floor, if the cost per m² is Rs. 4.

Answer 7: 

Here, d₁ = 45 cm and d₂ = 30 cm

∵ Area of one tile = 1/2 d₁d₂= 1/2 x45x30 = 675 cm²

∵ Area of 3000 tiles = 675 x 3000 = 2025000 cm²

= 2025000 / 10000 = 202.50 m²

Cost of polishing the floor per sq. meter = Rs. 4

Cost of polishing the floor per 202.50 sq. meter = 4 x 202.50 = Rs. 810

Hence, the total cost of polishing the floor is Rs. 810.

Question 8: 

Mohan 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 m² and the perpendicular distance between the two parallel sides is 100 m, find the length of the side along the river. 

Answer 8: 

Given: Perpendicular distance (A) = 100 m

Area of the trapezium shaped field = 10500 m²

Let side along the road be x m and side along the river = 2x m

Area of trapeziums = 1/2 (a+b) x h

Hence, the side along the river = 2x = 2 x 70 = 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 9: 

Given: Octagon having eight equal sides, each 5 m.

Construction: Divided the octagon in 3 figures, two trapeziums whose parallel and perpendicular sides are 11m and 4 m respectively and third figure is rectangle having length and breadth 11m and 5 m respectively.

Now

Area of two trapeziums = 2 x 1/2 (a+b) x h

= 2 x 1(11 + 5)x4 = 4x16 = 64m²

And

Area of rectangle = length x breadth = 11 x5 = 55m²

Total area of octagon = 64 + 55 = 119 m²

Question 10: 

There is a pentagonal shaped park as shown in the figure. 

For finding its area Jyoti 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 10: 

First way: By Jyoti’s diagram,

Area of pentagon = Area of trapezium ABCP + Area of trapezium AEDP

Second way: By Kavita's diagram

Here, a perpendicular AM drawn to BE,

AM = 30- 15 = 15 m

Area of pentagon = Area of ABE + Area of square BCDE

= ½ x l5x 15+ 15x15

= 112.5 + 225,0

= 337.5 m²

Hence, total area of pentagon shaped park = 337.5 m².

Question 11: 

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

Answer 11: 

Here two of given figures (1) and (11) are similar in dimensions.
And also figures [III] and ([V] are similar in dimensions.

Area of figure [I] = Area of trapezium = 1/2(a+b)xh

Also, Area of figure [II] = 96 cm²

Area of figure [III] = Area of trapezium = 1/2(a+b)xh

Also, Area of figure [IV] = 80 cm²