# GSEB Solutions Class 8 Maths Chapter 11 Mensuration Ex 11.1

Gujarat BoardĀ GSEB Textbook Solutions Class 8 Maths Chapter 11 Mensuration Ex 11.1 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 8 Maths Chapter 11 Mensuration Ex 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?
Solution:
(a) Side of the square = 60 m

ā“ Its perimeter = 4 Ć Side
= 4 v 60 m = 240 m
Area of the square = Side Ć Side
= 60 m Ć 60 m
= 3600 m

(b) āµ Perimeter of the rectangle = Perimeter of the given square
ā“ Perimeter of the rectangle = 240 m
or 2 Ć [Length + Breadth] = 240 m
or 2 Ć [80 m + Breadth] = 240 m
or 80 m + Breadth = $$\frac{240}{2}$$m = 120 m
ā“ Breadth = (120 – 80)m = 40 m
Now, Area of the rectangle = Length Ć Breadth
= 80 m Ć 40 m
= 3200 m2
Since, 3600 m2 Ć 3200 m2
ā“ Area of the square field (a) is greater.

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 ā¹ 55 per m2.

Solution:
āµ The given plot is a square with side as 25 m.
ā“ Area of the plot = Side Ć Side
= 25 m Ć 25 m
= 625 m2
āµ The constructed portion is a rectangle having length = 20 m and breadth = 15 m.
ā“ Area of the constructed portion
= 20 m Ć 15 m
= 300 m2
Now area of the garden
= [Total plot area] – [Total constructed area]
= (625 – 300) m2
= 325 m2
ā“ Cost of developing the garden
= ā¹ 55 Ć 325 = ā¹17,875

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? [Lenght of rectangle is 20 – (3.5 + 3.5) metres.]

Solution:
For the semi-circular part:
Diameter of the semi-circle = 7m
ā“ Radius of the semi-circle = $$\frac{7}{2}$$ = 3.5 m
Area of the 2 semi – circles = 2[$$\frac{1}{2}$$ Ļr2]

Also, perimeter of the 2 semicircles
= 2($$\frac{2Ļr}{2}$$) = 2Ļr = 2 Ć $$\frac{22}{7}$$ Ć $$\frac{35}{10}$$m = 22m

For rectangular part:
Length of the rectangle = 20 – (3.5 + 3.5)m
= 13 m
Breadth of the rectangle = 7 m
ā“ Area = Length Ć Breadth = 13 m Ć 7 m
= 91 m2
Perimeter = 2 Ć [Length + Breadth]
= 2 Ć [13 m + 0m] = 2 Ć 13 m = 26 m
Now, Area of the garden = (38.5 + 91)m2
= 129.5 m2
Perimeter of the garden = (38.5 + 91)m2 = 129.5 m2
Perimeter of the garden = 22 m + 26 m = 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 want to fill up the comers.)
Solution:
Area of a parallelogram
= Base Ć Corresponding height
Area of a tile = ($$\frac{24}{100}$$ Ć $$\frac{10}{100}$$)m2 = $$\frac{240}{10000}$$
m2 = 0.024 m2
ā“ Area of the floor = 1080 m2
Now, number of tiles =
= $$\frac{1080}{0.024}$$ = 45000 tiles

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 = 2Ļr, where r is the radius of the circle?

Solution:
(a) Diameter = 2.8 cm
ā Radius = $$\frac{2.8}{2}$$
= 1.4 cm
Perimeter of a circle = 2Ļr
ā Perimeter of a semicircle = $$\frac{2Ļr}{2}$$ = Ļr
ā“ Perimeter of the figure = Ļr + Diameter
= $$\frac{22}{7}$$ Ć 1.4 cm + 2.8 cm
= $$\frac{22}{7}$$ Ć $$\frac{14}{10}$$ cm + 2.8 cm
= 4.4 cm + 2.8 cm = 7.2 cm
Note: In this fugure, the diameter is also a part of the figure.

(b) Perimeter of the semi-circular part = Ļr

= $$\frac{22}{7}$$ Ć 1.4 cm = 4.4 cm
Perimeter of the remaining part
= 1.5 cm + 28 cm + 1.5 cm = 5.8 cm
ā“ Perimeter of the figure
= 4.4 cm + 5.8 cm = 10.2 cm

(c) Perimeter of the semi – circular part

ā“ Perimeter of the figure
= 4.4 cm + 2 cm + 2 cm = 8.4 cm
Since, 7.2 cm < 8.4 cm < 10.2 cm
ā“ Perimeter of figure ‘b’ has the longest round.
Note: In figures ‘b’ and ‘c’, the diameters are not part of the figures.