Mathematics – Class 8 – Chapter 11- Mensuration – Exercise 11.1 – NCERT Exercise Solution

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

Mathematics - Class 8 - Chapter 11 - Exercise 11.1 - Mensuration - NCERT Exercise Solution

Solution:

Given,

side of a square (s) = 60 m

Length of rectangular field, l = 80 m

Now,

According to question:

Perimeter of rectangular field = Perimeter of square field

By using formula:

2 (l + b) = 4 × Side

2(80+b) = 4×60

160+2b = 240

b = 40

Breadth of the rectangle is 40 m.

Now, Area of Square field

= (side)2

= (60)2 = 3600 m2

And Area of Rectangular field

= length × breadth = 80×40

= 3200 m2

Hence, area of square field is larger.

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.

Mathematics - Class 8 - Chapter 11 - Exercise 11.1 - Mensuration - NCERT Exercise Solution

Solution:

From given figure:

Side (s) of a square plot = 25 m

By formula:

Area of square plot = square of a side = (s)2

                (25)2 = 625

Therefore, the area of a square plot is 625 m2

From figure,

length of the house = 20 m and

Breadth of the house = 15 m

Now,

Area of the house = length × breadth

= 20×15 = 300 m2

Now,

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

325 m2 = 625 – 300

Cost of developing the garden per sq. m is Rs. 55

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

                                                                   = Rs. 17,875

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

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]

Mathematics - Class 8 - Chapter 11 - Exercise 11.1 - Mensuration - NCERT Exercise Solution

Solution:

According to given figure:

Total length = 20 m

Diameter of semi-circle = 7 m

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

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

Breadth of the rectangular field = 7 m

Now,

Area of rectangular field = length × breadth

                                        = 13×7

                                       = 91m2

Area of 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

So, Perimeter of two semi circles = 2πr

                                                     = 2 × (22/7) × 3.5

                                                     = 22 m

And Perimeter of garden = 22 + 13 + 13

                                         = 48 m

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]

Solution:

Given in question:

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

So, 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, no. required tiles to cover the floor = 45000

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.

Mathematics - Class 8 - Chapter 11 - Exercise 11.1 - Mensuration - NCERT Exercise Solution
(c)

Solution:

(a)

We know that radius = diameter/2

                                  = 2.8/2 cm = 1.4 cm

Now,

Circumference of semi-circle = πr

                                               = (22/7) × 1.4 = 4.4

So, circumference of semi-circle is 4.4 cm

Hence, total distance covered by the ant= circumference of semi-circle + diameter

= 4.4+2.8 = 7.2 cm

(b) Given, diameter of semi-circle = 2.8 cm

So,

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

Circumference of semi-circle = π r

= (22/7) × 1.4 = 4.4 cm

Hence, 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 = πr

= (22/7)×1.4

= 4.4 cm

Hence, total distance covered by the ant = 2+2+4.4 = 8.4 cm

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

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