Mensuration - Solutions 1
CBSE Class –VIII Mathematics
NCERT Solutions
CHAPTER - 11
Mensuration (Ex. 11.1)
NCERT Solutions
CHAPTER - 11
Mensuration (Ex. 11.1)
1. A square and a rectangular field with measurements as given in the figure have the same perimeter.
![](https://media-mycbseguide.s3.amazonaws.com/images/static/ncert/08/maths/11_1/image001.jpg)
Which field has a larger area?
![](https://media-mycbseguide.s3.amazonaws.com/images/static/ncert/08/maths/11_1/image001.jpg)
Which field has a larger area?
Ans. Given: The side of a square = 60 m and the length of rectangular field = 80 m
According to question,
Perimeter of rectangular file = Perimeter of square field
![](https://media-mycbseguide.s3.amazonaws.com/images/static/ncert/08/maths/11_1/image002.png)
![](https://media-mycbseguide.s3.amazonaws.com/images/static/ncert/08/maths/11_1/image002.png)
![](https://media-mycbseguide.s3.amazonaws.com/images/static/ncert/08/maths/11_1/image005.png)
![](https://media-mycbseguide.s3.amazonaws.com/images/static/ncert/08/maths/11_1/image002.png)
![](https://media-mycbseguide.s3.amazonaws.com/images/static/ncert/08/maths/11_1/image002.png)
![](https://media-mycbseguide.s3.amazonaws.com/images/static/ncert/08/maths/11_1/image002.png)
![](https://media-mycbseguide.s3.amazonaws.com/images/static/ncert/08/maths/11_1/image002.png)
Hence, the breadth of the rectangular field is 40 m.
Now, Area of Square field= (Side)2
= (60)2 sq.m = 3600 sq.m
= (60)2 sq.m = 3600 sq.m
Area of Rectangular field = (length
breadth)
= 80
40 sq. m = 3200 sq. m
![](https://media-mycbseguide.s3.amazonaws.com/images/static/ncert/08/maths/11_1/image004.png)
= 80
![](https://media-mycbseguide.s3.amazonaws.com/images/static/ncert/08/maths/11_1/image004.png)
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.
![](https://media-mycbseguide.s3.amazonaws.com/images/static/ncert/08/maths/11_1/image013.jpg)
Ans. Side of a square plot = 25 m
![](https://media-mycbseguide.s3.amazonaws.com/images/static/ncert/08/maths/11_1/image013.jpg)
Ans. Side of a square plot = 25 m
![](https://media-mycbseguide.s3.amazonaws.com/images/static/ncert/08/maths/11_1/image014.png)
Length and Breadth of the house is 20 m and 15 m respectively
![](https://media-mycbseguide.s3.amazonaws.com/images/static/ncert/08/maths/11_1/image014.png)
= 20
15 = 300 m2
![](https://media-mycbseguide.s3.amazonaws.com/images/static/ncert/08/maths/11_1/image004.png)
Area of garden = Area of square plot – Area of house
= (625 – 300) = 325 m2
![](https://media-mycbseguide.s3.amazonaws.com/images/static/ncert/08/maths/11_1/image017.png)
![](https://media-mycbseguide.s3.amazonaws.com/images/static/ncert/08/maths/11_1/image014.png)
![](https://media-mycbseguide.s3.amazonaws.com/images/static/ncert/08/maths/11_1/image004.png)
= 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]
[Length of rectangle is 20 – (3.5 + 3.5 meters]
![](https://media-mycbseguide.s3.amazonaws.com/images/static/ncert/08/maths/11_1/image018.png)
Ans. Given: Total length of the diagram = 20 m
Diameter of semi circle on both the ends = 7 m
![](https://media-mycbseguide.s3.amazonaws.com/images/static/ncert/08/maths/11_1/image014.png)
Length of rectangular field = [Total length - (radius of semicircle on both side)]
={20 – (3.5 + 3.5)}
= 20 – 7 = 13 m
Breadth of the rectangular field = 7 m
![](https://media-mycbseguide.s3.amazonaws.com/images/static/ncert/08/maths/11_1/image014.png)
= (13
7)
91 ![](https://media-mycbseguide.s3.amazonaws.com/images/static/ncert/08/maths/11_1/image012.png)
![](https://media-mycbseguide.s3.amazonaws.com/images/static/ncert/08/maths/11_1/image004.png)
![](https://media-mycbseguide.s3.amazonaws.com/images/static/ncert/08/maths/11_1/image002.png)
![](https://media-mycbseguide.s3.amazonaws.com/images/static/ncert/08/maths/11_1/image012.png)
Area of two semi circles = ![](https://media-mycbseguide.s3.amazonaws.com/images/static/ncert/08/maths/11_1/image021.png)
![](https://media-mycbseguide.s3.amazonaws.com/images/static/ncert/08/maths/11_1/image021.png)
=
= 38.5 m2
![](https://media-mycbseguide.s3.amazonaws.com/images/static/ncert/08/maths/11_1/image022.png)
Total Area of garden = (91 + 38.5)
129.5 m2
![](https://media-mycbseguide.s3.amazonaws.com/images/static/ncert/08/maths/11_1/image002.png)
Perimeter of two semi circles = ![](https://media-mycbseguide.s3.amazonaws.com/images/static/ncert/08/maths/11_1/image023.png)
![](https://media-mycbseguide.s3.amazonaws.com/images/static/ncert/08/maths/11_1/image023.png)
= 22 m
Hence, Perimeter of garden = (22 + 13 + 13)m = 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
? [If required you can split the tiles in whatever way you want to fill up the corners]
![](https://media-mycbseguide.s3.amazonaws.com/images/static/ncert/08/maths/11_1/image024.png)
Ans. Base of flooring tile = 24 cm
0.24 m
![](https://media-mycbseguide.s3.amazonaws.com/images/static/ncert/08/maths/11_1/image002.png)
height of a flooring tile = 10 cm
0.10 m [1cm = 1/100 m]
![](https://media-mycbseguide.s3.amazonaws.com/images/static/ncert/08/maths/11_1/image002.png)
Now, Area of flooring tile= Base
Altitude
![](https://media-mycbseguide.s3.amazonaws.com/images/static/ncert/08/maths/11_1/image004.png)
= 0.24
0.10 sq. m
![](https://media-mycbseguide.s3.amazonaws.com/images/static/ncert/08/maths/11_1/image004.png)
= 0.024 m2
![](https://media-mycbseguide.s3.amazonaws.com/images/static/ncert/08/maths/11_1/image014.png)
![](https://media-mycbseguide.s3.amazonaws.com/images/static/ncert/08/maths/11_1/image025.png)
= ![](https://media-mycbseguide.s3.amazonaws.com/images/static/ncert/08/maths/11_1/image026.png)
![](https://media-mycbseguide.s3.amazonaws.com/images/static/ncert/08/maths/11_1/image026.png)
= 45000 tiles
Hence 45000 tiles are required to cover the floor.
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
where
is the radius of the circle.
![](https://media-mycbseguide.s3.amazonaws.com/images/static/ncert/08/maths/11_1/image027.png)
![](https://media-mycbseguide.s3.amazonaws.com/images/static/ncert/08/maths/11_1/image028.png)
![](https://media-mycbseguide.s3.amazonaws.com/images/static/ncert/08/maths/11_1/image029.png)
Ans. (a) Radius = ![](https://media-mycbseguide.s3.amazonaws.com/images/static/ncert/08/maths/11_1/image030.png)
![](https://media-mycbseguide.s3.amazonaws.com/images/static/ncert/08/maths/11_1/image030.png)
= 1.4 cm
Circumference of semi circle = ![](https://media-mycbseguide.s3.amazonaws.com/images/static/ncert/08/maths/11_1/image031.png)
![](https://media-mycbseguide.s3.amazonaws.com/images/static/ncert/08/maths/11_1/image031.png)
=
4.4 cm
![](https://media-mycbseguide.s3.amazonaws.com/images/static/ncert/08/maths/11_1/image032.png)
![](https://media-mycbseguide.s3.amazonaws.com/images/static/ncert/08/maths/11_1/image002.png)
Total distance covered by the ant= (Circumference of semi circle + Diameter)![](https://media-mycbseguide.s3.amazonaws.com/images/static/ncert/08/maths/11_1/image033.jpg)
![](https://media-mycbseguide.s3.amazonaws.com/images/static/ncert/08/maths/11_1/image033.jpg)
=( 4.4 + 2.8 )cm
= 7.2 cm
(b) Diameter of semi circle = 2.8 cm
![](https://media-mycbseguide.s3.amazonaws.com/images/static/ncert/08/maths/11_1/image034.jpg)
Radius =
= 1.4 cm
![](https://media-mycbseguide.s3.amazonaws.com/images/static/ncert/08/maths/11_1/image034.jpg)
Radius =
![](https://media-mycbseguide.s3.amazonaws.com/images/static/ncert/08/maths/11_1/image035.png)
Circumference of semi circle = ![](https://media-mycbseguide.s3.amazonaws.com/images/static/ncert/08/maths/11_1/image031.png)
![](https://media-mycbseguide.s3.amazonaws.com/images/static/ncert/08/maths/11_1/image031.png)
=
4.4 cm
![](https://media-mycbseguide.s3.amazonaws.com/images/static/ncert/08/maths/11_1/image032.png)
![](https://media-mycbseguide.s3.amazonaws.com/images/static/ncert/08/maths/11_1/image002.png)
Total distance covered by the ant= (1.5 + 2.8 + 1.5 + 4.4)
10.2 cm
![](https://media-mycbseguide.s3.amazonaws.com/images/static/ncert/08/maths/11_1/image002.png)
(c) Diameter of semi circle = 2.8 cm
![](https://media-mycbseguide.s3.amazonaws.com/images/static/ncert/08/maths/11_1/image036.jpg)
Radius = ![](https://media-mycbseguide.s3.amazonaws.com/images/static/ncert/08/maths/11_1/image035.png)
![](https://media-mycbseguide.s3.amazonaws.com/images/static/ncert/08/maths/11_1/image035.png)
= 1.4 cm
Circumference of semi circle = ![](https://media-mycbseguide.s3.amazonaws.com/images/static/ncert/08/maths/11_1/image031.png)
![](https://media-mycbseguide.s3.amazonaws.com/images/static/ncert/08/maths/11_1/image031.png)
= ![](https://media-mycbseguide.s3.amazonaws.com/images/static/ncert/08/maths/11_1/image032.png)
4.4 cm
![](https://media-mycbseguide.s3.amazonaws.com/images/static/ncert/08/maths/11_1/image032.png)
![](https://media-mycbseguide.s3.amazonaws.com/images/static/ncert/08/maths/11_1/image002.png)
Total distance covered by the ant= (2 + 2 + 4.4) = 8.4 cm
Hence for figure (b) food piece, the ant would take a longer round.