Question 1

If the ratio of the area of two square is 9 : 1 the ratio of their perimeters is?

Accepted Answer

Let the area of square be (9x)2 m2 and (x2) m2 Then, their sides are (3x) m and x metres respectively ∴ Ratio of their perimeters = 12x / 4x =3:1

Question 2

The length of a rectangular field is increased by 60 % . By what percent would the width have to be decreased to maintain the same area?

Accepted Answer

Let length = L and breadth = B Let , New breadth = Z Then, New length = ( 160 / 100) L. = 8L / 5 ∴ 8L / 5 x Z = LB or Z = 5B/8 Decrease in breadth = (B-5B/8) = 3B/8 ∴ Decrease in percent = (3B/8 x1/B ) x 100 % = 371/2%

Question 3

The area of a rectangle is thrice that of square. Length of the rectangle is 40 cm and the breadth of the rectangle is ( 3 / 2 ) times that of the side of the square. The side of the square in cm is?

Accepted Answer

Let the side of the square = y cm Then, breadth of the rectangle = 3y/2 cm ∴ Area of rectangle = (40 x 3y/2) cm2 = 60y cm2 ∴ 60y = 3y2 ∴ y = 20 Hence, the side of the square = 20 cm

Question 4

If the radius of a circle be reduced by 50%. Its area is reduced by?

Accepted Answer

Original area = π x (r/2)2 = πr2/4 Reduction in area = π r2 - 3π r2/4 ∴ Reduction per cent = [ 3πr2/4 x 4/(πr2) x 100 ] % = 75%

Question 5

The radius of a circle has been reduced from 9 cm to 7 cm . the appropriate percentage decrease in area is?

Accepted Answer

Original area = (22/7) x 9 x 9 cm2 New area = (22/7) x 7 x 7 cm2 ∴ Decrease = 22/7 x [(9)2 -(7)2] cm2 =(22/7) x 16 x 2 cm2 Decrease percent = [(22/7 x 16 x 2) /( 7/22 x 9 x 9)] x 100 % = 39.5 %

Question 6

The difference between the circumference and the radius of a circle is 37 cm. The area of a circle is?

Accepted Answer

∵ 2πr - r = 37 ⇒ [(2 x 22/7) -1]r= 37 ⇒ 37r / 7= 37 ∴ r = 7 So, area of the circle =(22/7) x 7 x 7 cm2 = 154 cm2

Question 7

The area of a circular field is 13.86 hectares . The cost of fencing it at the rate of 20 paisa per meter is?

Accepted Answer

∵ 22/7 x r2 = 13.86 x 10000 ⇒ r2 = (13.86 x 10000 x 7) / 22 ∴ r = 210 m ⇒ Circumference = [2 x (22/7) x 210 ] m = 1320 m Cost of fencing = Rs. (1320 x 20)/100 = Rs . 264

Question 8

The area of circle is 38.5 sq. cm. Its circumference is?

Accepted Answer

∵ ( 22 x r2) / 7 = 38.5 ⇒ r2 =(38.5 x 7) /22 ∴ r = 3.5 cm ∴ Circumference = 2 x (22/7) x 3.5 cm = 22 cm

Question 9

If the circumference of a circle is the 352 meters, then its area in m square is?

Accepted Answer

circumference = 2πr = 2 x (22 / 7) x r = 352 ⇒ r = (352 x 7) / (22 x 2) =56 cm Now area = πr2 = (22 / 7) x 56 x 56 m2 = 9856 m2

Question 10

Area of a square with side x is equal to the area of a triangle with base x. The altitude of the triangle is?

Accepted Answer

Let h be the altitude of triangle. So area of triangle = (1/2)xh area of square = x2 From question area of square = area of triangle x2 = (1/2)xh ∴ h = (2x2)/x = 2x

Question 11

In a triangle ABC, BC = 5 cm AC = 12 cm and AB = 13 cm. The length of a altitude drawn from B on AC is?

Accepted Answer

∵ s= (13 + 5 + 12) / 2 cm = 15cm s-a = 2 cm, s-b = 10 cm and s-c = 3 cm ⇒ Area = √ (15 x 2 x 10 x 3 ) cm2 = 30 cm2 ⇒ (12 x h) / 2 = 30 ∴ h = 5 cm

Question 12

The perimeter of an isosceles triangle is equal to 14 cm . The lateral side is to the base in the ratio 5 : 4 the area of the triangle is?

Accepted Answer

Let lateral side = (5y) cm and base = (4y) cm ∴ perimeter = 5y + 5y + 4y = 14 ⇒y = 1 So, the sides are 5 cm , 5 cm and 4 cm Now s= 1/2 (5 + 5 + 4) cm = 7 cm (s-a) = 2 cm (s-b) = 2 cm and (s-c) = 3 cm ∴ Required Area = √ (7 x 2 x 2 x 3) cm2 =2√21 cm2

Question 13

The sides of a triangular board are 13 meters, 14 meters and 15 meters. The cost of paining it at the rate of Rs. 8.75 per m 2 is?

Accepted Answer

s = (13 + 14 + 15 ) / 2 = 21, s-a = 8 , s-b = 7, s-c= 6 ∴ Area to be painted = √ [s(s-a) (s-b) (s-c)] =√ [21 x 8 x 7 x 6] m2 = 84 m2 ∴ Cost of painting = Rs. (84 x 8.75) = Rs. 735

Question 14

If the diagonal of a square is double, how does the area of the square change?

Accepted Answer

Ratio of the areas = area of original square / area of new square = [ d2 / 2 ] / [ (2d)2 / 2 ] = 1/4 ∴ New area becomes 4 fold.

Question 15

A circle and a square have same area. The ratio of the side of the square and the radius of the circle is?

Accepted Answer

∵ Area of square = Area of circle ⇒ x2 = πr2 ⇒ x/r = √π = √ π : 1

Question 16

The radius of the wheel of a vehicle is 70 cm. The wheel makes 10 revolutions in 5 seconds. The speed of the vehicle is?

Accepted Answer

Circumference = 2πr = 2 x (22 / 7) x 70 cm = 440 cm Distance travelled in 10 revolutions = 440 x 10 cm = 4400 cm = 44 m ∴ Speed = distance / time = 44 / 5 m/sec = (44 / 5) x (18 / 5) km/hr = 31.68 km/hr

Question 17

A circular disc of area 0.49 π square meters, rolls down a length of 1.76 km. The number of revolutions it makes is?

Accepted Answer

Area = πr2 = 0.49π ⇒ r = 0.7 m Number of revolutions = total distance / circumference = (1.76 x 1000) / (2 x 22/7 x 0.7) = 400

Question 18

The Number of rounds that a wheel of diameter 7/11 m will make in going 4 km, is?

Accepted Answer

Number of round = total distance / circumference of wheel. = (4 x 1000) / πd = (4 x 1000) / ( 22/7 x 7/11) = 2000

Question 19

A wheel makes 1000 revolutions in covering a distance of 88 km. the diameter of the wheel is?

Accepted Answer

Distance covered in one revolution = total distance travelled / total number of revolution. = ( 88 x 1000) / 1000 m = 88 m We know that the distance covered in one revolution = circumference of the wheel. ⇒ πd = 88 ⇒ 22d / 7 = 88 ∴ d = 28 m

Question 20

The diameter of a wheel is 63 cms. Distance travelled by the wheel in 100 revolutions is?

Accepted Answer

Distance travelled in 1 revolution = circumference of the wheel = πd = ( 22/7 ) x 63 = 198 cm So the distance travelled in 100 revolutions = 100 x distance travelled in 1 revolution = 100 x 198 cm = 19800 cm = 198 m

Area Questions