Question 1

The length of a rectangle is halved, while its breadth is tripled. What is the percentage change in area?

Accepted Answer

Let original length = x and original breadth = y. Original area = xy. New length = x . 2 New breadth = 3y. New area = ❨ x x 3y ❩ = 3 xy. 2 2 ∴ Increase % = ❨ 1 xy x 1 x 100 ❩% = 50%. 2 xy

Question 2

A towel, when bleached, was found to have lost 20% of its length and 10% of its breadth. The percentage of decrease in area is:

Accepted Answer

Let original length = x and original breadth = y. Decrease in area = xy - ❨ 80 x x 90 y ❩ 100 100 = ❨ xy - 18 xy ❩ 25 = 7 xy. 25 ∴ Decrease % = ❨ 7 xy x 1 x 100 ❩% = 28%. 25 xy

Question 3

The length of a rectangular plot is 20 metres more than its breadth. If the cost of fencing the plot @ 26.50 per metre is Rs. 5300, what is the length of the plot in metres?

Accepted Answer

Let breadth = x metres. Then, length = (x + 20) metres. Perimeter = ❨ 5300 ❩ m = 200 m. 26.50 ∴ 2[(x + 20) + x] = 200 ⟹ 2x + 20 = 100 ⟹ 2x = 80 ⟹ x = 40. Hence, length = x + 20 = 60 m.

Question 4

The diagonal of the floor of a rectangular closet is 7½ feet. The shorter side of the closet is 4½ feet. What is the area of the closet in square feet?

Accepted Answer

Other side = ❨ 15 ❩ 2 - ❨ 9 ❩ 2 2 2 ft = 225 - 81 4 4 ft = 144 4 ft = 6 ft. ∴ Area of closet = (6 x 4.5) sq. ft = 27 sq. ft.

Question 5

A rectangular field is to be fenced on three sides leaving a side of 20 feet uncovered. If the area of the field is 680 sq. feet, how many feet of fencing will be required?

Accepted Answer

We have: l = 20 ft and lb = 680 sq. ft. So, b = 34 ft. ∴ Length of fencing = (l + 2b) = (20 + 68) ft = 88 ft.

Question 6

An error 2% in excess is made while measuring the side of a square. The percentage of error in the calculated area of the square is:

Accepted Answer

100 cm is read as 102 cm. ∴ A1 = (100 x 100) cm2 and A2 (102 x 102) cm2. (A2 - A1) = [(102)2 - (100)2] = (102 + 100) x (102 - 100) = 404 cm2. ∴ Percentage error = ❨ 404 x 100 ❩% = 4.04% 100 x 100

Question 7

A man walked diagonally across a square lot. Approximately, what was the percent saved by not walking along the edges?

Accepted Answer

Let the side of the square(ABCD) be x metres. Then, AB + BC = 2x metres. AC = √2x = (1.41x) m. Saving on 2x metres = (0.59x) m. Saving % = ❨ 0.59x x 100 ❩% = 30% (approx.) 2x

Question 8

The ratio between the length and the breadth of a rectangular park is 3 : 2. If a man cycling along the boundary of the park at the speed of 12 km/hr completes one round in 8 minutes, then the area of the park (in sq. m) is:

Accepted Answer

Perimeter = Distance covered in 8 min. = ❨ 12000 x 8 ❩m = 1600 m. 60 Let length = 3x metres and breadth = 2x metres. Then, 2(3x + 2x) = 1600 or x = 160. ∴ Length = 480 m and Breadth = 320 m. ∴ Area = (480 x 320) m2 = 153600 m2.

Question 9

The diagonal of a rectangle is √41 cm and its area is 20 sq. cm. The perimeter of the rectangle must be:

Accepted Answer

√l2 + b2 = √41. Also, lb = 20. (l + b)2 = (l2 + b2) + 2lb = 41 + 40 = 81 ⟹ (l + b) = 9. ∴ Perimeter = 2(l + b) = 18 cm.

Question 10

A rectangular park 60 m long and 40 m wide has two concrete crossroads running in the middle of the park and rest of the park has been used as a lawn. If the area of the lawn is 2109 sq. m, then what is the width of the road?

Accepted Answer

Area of the park = (60 x 40) m2 = 2400 m2. Area of the lawn = 2109 m2. ∴ Area of the crossroads = (2400 - 2109) m2 = 291 m2. Let the width of the road be x metres. Then, 60x + 40x - x2 = 291 ⟹ x2 - 100x + 291 = 0 ⟹ (x - 97)(x - 3) = 0 ⟹ x = 3.

Question 11

What is the least number of squares tiles required to pave the floor of a room 15 m 17 cm long and 9 m 2 cm broad?

Accepted Answer

Length of largest tile = H.C.F. of 1517 cm and 902 cm = 41 cm. Area of each tile = (41 x 41) cm2. ∴ Required number of tiles = ❨ 1517 x 902 ❩ = 814. 41 x 41

Question 12

The ratio between the perimeter and the breadth of a rectangle is 5 : 1. If the area of the rectangle is 216 sq. cm, what is the length of the rectangle?

Accepted Answer

2(l + b) = 5 b 1 ⟹ 2l + 2b = 5b ⟹ 3b = 2l b = 2 l 3 Then, Area = 216 cm2 ⟹ l x b = 216 ⟹l x 2 l = 216 3 ⟹ l2 = 324 ⟹ l = 18 cm.

Question 13

The difference between the length and breadth of a rectangle is 23 m. If its perimeter is 206 m, then its area is:

Accepted Answer

We have: (l - b) = 23 and 2(l + b) = 206 or (l + b) = 103. Solving the two equations, we get: l = 63 and b = 40. ∴ Area = (l x b) = (63 x 40) m2 = 2520 m2.

Question 14

A tank is 25 m long, 12 m wide and 6 m deep. The cost of plastering its walls and bottom at 75 paise per sq. m, is:

Accepted Answer

Area to be plastered = [2(l + b) x h] + (l x b) = {[2(25 + 12) x 6] + (25 x 12)} m2 = (444 + 300) m2 = 744 m2. ∴ Cost of plastering = Rs. ❨ 744 x 75 ❩ = Rs. 558. 100

Question 15

The percentage increase in the area of a rectangle, if each of its sides is increased by 20% is:

Accepted Answer

Let original length = x metres and original breadth = y metres. Original area = (xy) m2. New length = ❨ 120 x ❩m = ❨ 6 x ❩m. 100 5 New breadth = ❨ 120 y ❩m = ❨ 6 y ❩m. 100 5 New Area = ❨ 6 x x 6 y ❩m2 = ❨ 36 xy ❩m2. 5 5 25 The difference between the original area = xy and new-area 36/25 xy is = (36/25)xy - xy = xy(36/25 - 1) = xy(11/25) or (11/25)xy ∴ Increase % = ❨ 11 xy x 1 x 100 ❩% = 44%. 25 xy

Question 16

Area of a square is 1/2 hectare. The diagonal of the square is?

Accepted Answer

Area = 1/2 hectare = 10000 / 2 m2 = 5000 m2 Again Area = 1/2 x (Diagonal)2 So 1/2 x (Diagonal)2 = 5000m2 ⇒ Diagonal2= 10000 ∴ Diagonal = 100

Question 17

The base of right-angled triangle is 5 meters and hypotenuse is 13 meters . Its area will be?

Accepted Answer

Altitude = √(13)2 - (5)2 = √144 = 12m ∴ Area of the triangle = (5 x 12 ) / 2 m2 = 30m2

Question 18

The length of a rectangle is double while its breadth is halved. What is the percentage change in area?

Accepted Answer

Let length = l and breadth = b Then, area = lb New length = 2l And new breadth = b/2 ∴ New area = ( 2l ) x (b/2) = lb So, there is no change in area .

Question 19

The cost of carpeting a room 15 meters long with a carpet 75 cm wide at 30 paisa per meter is Rs. 36. The breadth of the room is?

Accepted Answer

Length of carpet = Total Cost / Rate = 3600 / 30 = 120 m Area of carpet = (120 x 75) / 100 m2 = 90 m2 ∴ Area of the room = 90 m2 Breadth of the room = Area /Length = 90 / 15 m = 6m

Question 20

If the side of a square is increased by 25%, then how much percent does its area get increased?

Accepted Answer

Let area 100 m2 Then, side = 10 m New side = 125 % of 10 = (125/100) x 10 = 12.5 m New area = 12.5 x 12.5 m2 =(12.5)2 sq. m ∴ Increase in area = (12.5)2 - (10)2 m2 = 22.5 x 2.5 m2 =56.25 m2 % Increase = 56.25 %

Area Questions