HBSE 8th Class Maths Solutions Chapter 11 Mensuration Ex 11.1

Haryana State Board HBSE 8th Class Maths Solutions Chapter 11 Mensuration Ex 11.1 Textbook Exercise Questions and Answers.

Haryana Board 8th Class Maths Solutions Chapter 11 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 ?

Solution:
(a) Perimeter of square
= 4 × side
= 4 × 60 = 240 m

(b) Perimeter of rectangle = 2 (l + b)
240 m = 2(80 + b)
\(\frac{240}{2}\) = 80 + b
120 = 80 + b
120 – 80 = b
40 = b
∴ b = 40 m
∴ Area of square = (side)² = (60)²
60 × 60 = 3600 m²
and area of rectangle = l × b = 80 m × 40 m
= 3200 m²
Hence square field is larger than rectangle.

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.

Solution:
Area of garden
= Area of square plot – Area of middle plot (Rectangular plot)
= (side)² – (l × b)
= (25)² – (20 × 15)
= 625 – 300 = 325 m²
Hence the total cost of developing garden
= 325 m² × Rs. 55 = Rs. 17875.

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) metres].

Solution:
Length of rectangle
= 20 – (3.5 + 3.5)
= 20 – 7 = 13 m
Breadth of rectangle = 7 m
Radius of semi-circle = 3.5 cm
Two semicircle = one circle
Hence perimeter (or circumference) of circle
= 2pr = 2 × \(\frac{22}{7}\) × 3.5 = 22 cm
Perimeter of rectangle
= 2 (l + b)
= 2 × (13 + 7)
= 2 × 20 = 40 cm
Hence, perimeter of this garden
=(22 cm + 40 cm) = 62 cm
Now, area of rectangle
= l × b = 13 × 7 = 91 cm²
Area of circle (two semicircle)
= πr² = \(\frac{22}{7}\) × 3.5 × 3.5
= 38.5 cm²
Hence, area of garden
= 38.5 cm² + 91 cm²
= 129.5 cm².

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).
Solution:
Area of parallelogram tiles
= base × height
= b × h
= 24 × 10 = 240 cm²
Area of parallelogram
= 1080 m²
= 1080 × 100 × 100 × cm²
= 10800000cm²
∴ Number of tiles
= \(\frac{Area of floor}{7}\)
= \(\frac{10800000}{240}\) = 45000.

Question 5.
An ant is moving around a few food

Solution:
(a) Circumference (perimeter) of semicircle
= \(\frac{2 \pi r}{2}\) = \(\frac{2 \times 22 \times 1.4}{2 \times 7}\)
(d = 2.8 cm, r = 1.4 cm)
= 4.4 cm
Perimeter = 4.4 cm + 2.8 cm = 7.2 cm.

(b) Circumference of semicircle + Perimeter of square
= 4.4 cm + 1.5 cm + 2.8 cm + 1.5 cm
= 10.2 cm.

(c) Circumference of semicircle + Perimeter of triangle
= 4.4 cm + 2 cm + 2 cm
= 8.4 cm
Hence (b) food piece would the ant have to take a longer round.