Question 1

In two triangles, the ratio of the areas is 4 : 3 and the ratio of their heights is 3 : 4. Find the ratio of their bases.

Accepted Answer

Let the bases of the two triangles be x and y and their heights be 3h and 4h respectively. Then, 1 2 × x × 3 h 1 2 × y × 4 h = 4 3 ⇒ x y = 4 3 × 4 3 = 16 9 Required ratio = 16 : 9.

Question 2

The area of a circular field is 13.86 hectares. Find the cost of fencing it at the rate of Rs. 4.40 per metre.

Accepted Answer

Area = (13.86 x 10000) sq.m = 138600 sq.m πR 2 = 138600 ⇒ R 2 = 138600 × 7 22 ⇒ R = 210 m Circumference = 2 πR 2 = 2 × 22 7 × 210 = 1320 m Cost of fencing = Rs. (1320 x 4.40) = Rs. 5808.

Question 3

A semicircular shaped window has diameter of 63cm. Its perimeter equals

Accepted Answer

perimeter of window = π r+2r = [(22/7) x (63/2) + 63] = 99+63 = 162 cm

Question 4

A triangle and a parallelogram are constructed on the same base such that their areas are equal. If the altitude of the parallelogram is 100 m , then the altitude of the triangle is

Accepted Answer

Let the triangle and parallelogram have common base b, let the Altitude of triangle is h1 and of parallelogram is h2(which is equal to 100 m), then Area of triangle = 1 2 × b × h 1 Area of rectangle = b*h2 As per Given, 1 2 * b * h 1 = b * h 2 1 2 * b * h 1 = b * 100 h1=200

Question 5

What is the area of an obtuse angled triangle given below with two sides are 8 and 12 and the angle included between two sides is 150 deg?

Accepted Answer

We know that, The area of a triangle with two sides given and included angle A = 1/2 x product of sides x Sin(angle) Here the two sides are 8 & 12 Angle = 150 Area = 1/2 x 8 x 12 x sin150 Sin(150) = sin(90+60) = cos(60) = 1/2 A = 48 x 1/2 = 24 Area of the given triangle = 24 sq units.

Question 6

A sector of 120 degrees, cut out from a circle, has an area of 66/7 sq cm. Find the radius of the circle.

Accepted Answer

πr = 66 7 r =3

Question 7

The length of the room is 5.5m and width is 3.75m. Find the cost of paving the floor by slabs at the rate of Rs.800 per sq meter

Accepted Answer

l=5.5m w=3.75m area of the floor = 5.5 x 3.75 = 20.625 sq m cost of paving = 800 x 20.625 = Rs. 16500

Question 8

What is the length of the longest pole which can be kept in a room 12 m long, 4 m broad and 3 m high ?

Accepted Answer

The length of the longest pole which can be kept in a room 12 m long, 4 m broad and 3 m high is => d 2 = 12 2 + 4 2 + 3 2 = 13 mts.

Question 9

The diagonals of two squares are in the ratio of 2 : 5. Find the ratio of their areas.

Accepted Answer

Let the diagonals of the squares be 2x and 5x respectively. Ratio of their areas = 1 2 * 2 x 2 : 1 2 * 5 x 2 = 4 : 25

Question 10

An equilateral triangle is described on the diagonal of a square. What is the ratio of the area of the triangle to that of the square?

Accepted Answer

area of a square = a² sq cm length of the diagonal = 2 a cm area of equilateral triangle with side 2 a = 3 4 * 2 a 2 required ratio = 3 4 * 2 a 2 : a 2 = 3 : 2

Question 11

A took 15 seconds to cross a rectangular field diagonally walking at the rate of 52 m/min and B took the same time to cross the same field along its sides walking at the rate of 68m/min. The area of the field is?

Accepted Answer

length of the diagonal = 52 x (15/60) =13m sum of length and breadth = 68 x (15/60) = 17m l 2 + b 2 = 13 o r l + b = 17 area =lb = 1/2[(2lb)] = 1/2[(l+b)² - (l²+b²)] = 1/2[17² -169] =1/2 x 120 = 60 sq.m

Question 12

The circumference of a circle, whose area is 24.64 sq.m

Accepted Answer

πr 2 = 26 . 64

Question 13

In an old Palace there are 50 stairs to go on top floor. Each is 20 cm high, 30 cm wide and 60 cm long. If all the stairs are re-plastered at the rate of Rs. 4 per 100 sq.cm. What is the expenditure ?

Accepted Answer

Total are of all stairs = 50 (20x60 + 30x60) = 50 (1200 + 1800) = 50 x 3000 = 150000 sq cm. Given rate of 100 sq cm = Rs. 4 Then for 150000 sq. cm = ? => 4 x 150000/100 = Rs. 6000

Question 14

The length of a rectangle is 10 cm more than the breadth. Find the diagonal length of rectangle if perimeter is 44 cm ?

Accepted Answer

Let the length of the rectangle = L cm Then the breadth of the rectangle = (L - 10) cm Perimeter of a rectangle = 2(L + B) cm => 44 = 2(L + L - 10) => 44 = 4L - 20 => 4L = 64 => L = 16 cm => B = L - 10 = 6 cm Diagonal = L 2 + B 2 = 16 2 + 6 2 = 17.08 cm

Question 15

The diagonal of the floor of a rectangular closet is 7 12 feet. The shorter side of the closet is 4 12 feet. What is the area of the closet in square feet?

Accepted Answer

Other side = d 2 - s 2 = 225 4 - 81 4 = 144 4 = 6 ft Area of closet = (6 x 4.5) sq. ft = 27 sq. ft.

Question 16

The altitude drawn to the base of an isosceles triangle is 8 cm and the perimeter is 32 cm. Find the area of the triangle.

Accepted Answer

Let ABC be the isosceles triangle and AD be the altitude Let AB = AC = x. Then, BC = (32 - 2x). Since, in an isosceles triangle, the altitude bisects the base, so BD = DC = (16 - x). In triangle ADC, A C 2 = A D 2 + D C 2 ⇒ x 2 = 8 2 + 16 - x 2 ⇒ x = 10 BC = (32- 2x) = (32 - 20) cm = 12 cm. Hence, required area = 1 2 * B C * A D = 1 2 * 12 * 8 = 48 c m 2

Question 17

In measuring the sides of a rectangle, one side is taken 5% in excess and the other 4% in deficit. Find the error percent in the area, calculate from the those measurements.

Accepted Answer

let x and y be the sides of the rectangle then correct area = 105 100 x × 96 100 y Error% = 504 500 x y - x y = 4 500 x y % 4 500 x y × 1 x y × 100 = 4 5 = 0 . 8

Question 18

Four circular cardboard pieces, each of radius 7cm are placed in such a way that each piece touches two other pieces. The area of the space encosed by the four pieces is

Accepted Answer

required area = [14 x 14 - (4 x 1/4 x 22/7 x 7 x 7) ]sq.cm = 196 - 154 = 42 sq.cm.

Question 19

A rectangular plot measuring 90 meters by 50 meters is to be enclosed by wire fencing. If the poles of the fence are kept 5 meters apart. How many poles will be needed?

Accepted Answer

perimeter of the plot = 2(90+50) = 280m no of poles =280/5 =56m

Question 20

A large field of 700 hectares is divided into two parts. The difference of the areas of the two parts is one-fifth of the average of the two areas. What is the area of the smaller part in hectares?

Accepted Answer

Let the areas of the two parts be x and (700-x) hectares therefore, x - 700 - x = 1 5 x + 700 - x 2 ⇒ 2 x - 700 = 70 ⇒ x = 385 So, the two parts are 385 and 315. Hence, Area of the smaller = 315 hectares

Area Questions