Exercise 11.1 Exercise 11.2 Exercise 11.3 Exercise 11.4

# 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?

Sol :

Perimeter of square = 4 (Side of the square) = 4 (60 m) = 240 m

Perimeter of rectangle = 2 (Length + Breadth)

= 2 (80 m + Breadth)

= 160 m + 2 × Breadth

It is given that the perimeter of the square and the rectangle are the same.

160 m + 2 × Breadth = 240 m

Breadth of the rectangle = = 40 m

Area of square = (Side)^{2} = (60 m)^{2} = 3600 m^{2}

Area of rectangle = Length × Breadth = (80 × 40) m^{2} = 3200 m^{2}

Thus, the area of the square field is larger than the area of the rectangular field.

Question 2

Mrs. Kaushik has a square plot with the measurement as shown in the following 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}.

Sol :

Area of the square plot = (25 m)^{2} = 625 m^{2}

Area of the house = (15 m) × (20 m) =300 m^{2}

Area of the remaining portion = Area of square plot − Area of the house

= 625 m^{2} − 300 m^{2} = 325 m^{2}

The cost of developing the garden around the house is Rs 55 per m^{2}.

Total cost of developing the garden of area 325 m^{2} = Rs (55 × 325)

= Rs 17,875

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 the garden [Length of rectangle is 20 − (3.5 + 3.5) metres]

Sol :

Length of the rectangle = [20 − (3.5 + 3.5)] metres = 13 m

Circumference of 1 semi-circular part = π*r *

Circumference of both semi-circular parts = (2 × 11) m = 22 m

Perimeter of the garden = AB + Length of both semi-circular regions BC and

DA + CD

= 13 m + 22 m + 13 m = 48 m**

Area of the garden = Area of rectangle + 2 × Area of two semi-circular regions

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).

Sol :

Area of parallelogram = Base × Height

Hence, area of one tile = 24 cm × 10 cm = 240 cm^{2}

Required number of tiles =** **

= 45000 tiles

Thus, 45000 tiles are required to cover a floor of area 1080 m^{2}.

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* = 2*πr*, where *r* is the radius of the circle.

Sol :

(a)Radius (*r*) of semi-circular part =

Perimeter of the given figure = 2.8 cm + π*r*

(b)Radius (*r*) of semi-circular part =

Perimeter of the given figure = 1.5 cm + 2.8 cm + 1.5 cm +π (1.4 cm)

(c)Radius (*r*) of semi-circular part =

Perimeter of the figure(c) = 2 cm + π*r* + 2 cm

Thus, the ant will have to take a longer round for the food-piece (b), because the perimeter of the figure given in alternative (b) is the greatest among all.