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NCERT · Class 8 · Maths

Chapter 11: Mensuration

Practice each question, then open the answer or worked solution.

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Exercise 11.1

5 questions
Q1Written answer

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

Ncert solution class 8 chapter 11-1

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Worked solution

Side of a square = 60 m (Given)

And the length of the rectangular field, l = 80 m (Given)

According to the question,

The perimeter of the rectangular field = Perimeter of the square field

2(l+b) = 4×Side (using formulas)

2(80+b) = 4×60

160+2b = 240

b = 40

The breadth of the rectangle is 40 m.

Now, the Area of the Square field

= (side)2

= (60)2 = 3600 m2

And the Area of the Rectangular field

= length×breadth = 80×40

= 3200 m2

Hence, the area of the square field is larger.

Q2Written answer

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.

Ncert solution class 8 chapter 11-2

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Worked solution

Side of the square plot = 25 m

Formula: Area of the square plot =  square of the side = (side)2

= (25)2 = 625

Therefore the area of the square plot is 625 m2

Length of the house = 20 m and

Ncert solution class 8 chapter 11-3

The breadth of the house = 15 m

Area of the house = length×breadth

= 20×15 = 300 m2

Area of the garden = Area of the square plot – Area of the house

= 625–300 = 325 m2

∵ The cost of developing the garden per sq. m is Rs. 55

The cost of developing the garden 325 sq. m = Rs. 55×325

= Rs. 17,875

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

Q3Written answer

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]

Ncert solution class 8 chapter 11-4

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Worked solution

Solution::

Given: Total length = 20 m

Diameter of the semi-circle = 7 m

Radius of the semi-circle = 7/2 = 3.5 m

Length of the rectangular field

= 20-(3.5+3.5) = 20-7 = 13 m

Breadth of the rectangular field = 7 m

Area of rectangular field = l×b

= 13×7= 91m2

Area of the two semi-circles = 2×(1/2)×π×r2

= 2×(1/2)×22/7×3.5×3.5

= 38.5 m2

Area of garden = 91+38.5 = 129.5 m2

Now, the perimeter of the two semi-circles =  2πr = 2×(22/7)×3.5  = 22 m

And the perimeter of the garden = 22+13+13

= 48 m. Answer

Q4Written answer

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]

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Worked solution

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×Altitude

= 0.24×0.10

= 0.024

Area of flooring tile is 0.024m2

Number of tiles required to cover the floor= Area of floor/Area of one tile =  1080/0.024

= 45000 tiles

Hence 45000 tiles are required to cover the floor.

Q5Written answer

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.

Ncert solution class 8 chapter 11-5

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Worked solution

(a) Radius =  Diameter/2 = 2.8/2 cm = 1.4 cm

Circumference of semi-circle =  πr

=  (22/7)×1.4 = 4.4

Circumference of the semi-circle is 4.4 cm

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

= 4.4+2.8 = 7.2 cm

(b) Diameter of semi-circle = 2.8 cm

Radius =  Diameter/2 = 2.8/2  = 1.4 cm

Circumference of semi-circle = r

= (22/7)×1.4 = 4.4 cm

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

(c) Diameter of semi-circle = 2.8 cm

Radius =  Diameter/2 = 2.8/2

= 1.4 cm

Circumference of semi-circle =  π r

=  (22/7)×1.4

= 4.4 cm

Total distance covered by the ant = 2+2+4.4 = 8.4 cm

After analyzing the results of three figures, we concluded that for figure (b) food piece, the ant would take a longer round.