You can Download KSEEB Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.1 Questions and Answers helps you to revise the complete syllabus.

## KSEEB Solutions for 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:

side of square = 60 m

length of rectangle = 80 cm

Perimeter of square = 4 × side

= 4 × 60

= 240 m

Perimeter of rectangle = 2 (l + b) = 2(80 + b)

Perimeter of rectangle = perimeter of square

⇒ 2(l + b) = 4 × side

⇒ 2(80 + b) = 4 × 60

⇒ 80 + b = \(\frac{4 \times 60}{2}\)

⇒ 80 + b = 120

⇒ b = 120 – 80

⇒ b = 40 m

Area of square = (side)^{2}

= (60)^{2}

= 60 × 60

= 3600 m^{2}

Area of rectangle = l × b

= 80 × 40

= 3200 m^{2}

The area of the square is larger than the area of a 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 25 m Rs. 55 per m^{2}.

Solution:

Length of house = 20 m

Breadth of house = 15 m

Length of garden = 25 m

Breadth of garden = 25 m

Area of garden = Area of plot – area of house

= 25 × 25 – 20 × 15

= 625 – 300

= 325 m^{2}

Cost of developing the garden = 325 × 55 = Rs. 17875

Question 3.

The shape of a garden is rectangular in the middle and semicircular 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].

Solution:

Length of rectangle = 20 – (3.5 + 3.5)

= 20 – 7

= 13 m

Breadth of rectangle = 7 m

Radius of semi circle = \(\frac{7}{2}\) m

Perimeter of garden = 2 circle inference of semi circle + 2 (length of rectangle)

= \(\frac{2 \times 2 \pi r}{2}\) + 2l

= 2πr + 2l

= \(2 \times \frac{22}{7} \times \frac{7}{2}+2 \times 13\)

= 22 + 26

= 48 m

Area of garden = 2(Area of semicircle) + Area of rectangle

= \(2 \times \frac{\pi r^{2}}{2}+l \times b\)

= πr^{2} + lb

= \(\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2}+13 \times 7\)

= \(\frac{77}{2}\) + 91

= \(\frac{77+182}{2}\)

= \(\frac{259}{2}\)

= 129.5 m^{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:

Base of tile = 24 cm

Height of tile = 10 cm

Area of tile = 24 × 10 = 240 cm^{2}

Area of floor = 1080 m^{2} = 1080 × 100 × 100 cm^{2}

Number of tiles = \(\frac{\text { Area of floor }}{\text { Area of tile }}\)

= \(\frac{1080 \times 100 \times 100}{240}\)

= 45000

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, the circumference of a circle can be obtained by using the expression c = 2πr, where r is the radius of the circle.

Solution:

(a) Circumference of semicircle = \(\frac{2 \pi r}{2}\) = πr

= \(\frac{22}{7} \times 1.4\)

= 22 × 0.2

= 4.4 cm

Perimeter of semicircle = πr + 2r

= 4.4 + 2 × 1.4

= 4.4 + 2.8

= 7.2 cm

(b) Perimeter of figure = Perimeter of semicirlce + Perimeter of rectangle

= \(\frac{2 \pi r}{2}\) + (2 × b + l)

= πr + 2l + b

= \(\frac{22}{7}\) × 1.4 + 2 × 1.5 + 2.8

= 22 × 0.2 + 3 + 2.8

= 4.4 + 3 + 2.8

= 10.2 cm

(c) Perimeter of figure = Perimeter of semi-circle + 2l

= \(\frac{2 \pi r}{2}\) + 2l

= πr + 2l

= \(\frac{22}{7}\) × 1.4 + 2 × 2

= 4.4 + 4

= 8.4 cm

Hence, for (b) food piece, and would have to take a longer round.