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 m^{2}

and area of rectangle = l × b = 80 m × 40 m

= 3200 m^{2}

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 m^{2}.

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 m^{2}

Hence the total cost of developing garden

= 325 m^{2} × 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 cm^{2}

Area of circle (two semicircle)

= πr^{2} = \(\frac{22}{7}\) × 3.5 × 3.5

= 38.5 cm^{2}

Hence, area of garden

= 38.5 cm^{2} + 91 cm^{2}

= 129.5 cm^{2}.

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 m^{2} ? (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 cm^{2}

Area of parallelogram

= 1080 m^{2}

= 1080 × 100 × 100 × cm^{2}

= 10800000cm^{2}

∴ 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.