Question 1

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 2

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 3

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 4

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

Accepted Answer

πr 2 = 26 . 64

Question 5

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 6

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 7

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 8

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 9

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 10

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

Question 11

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 12

If a square and a rhombus stand on the same base, then the ratio of the areas of the square and the rhombus is:

Accepted Answer

A square and a rhombus on the same base are equal in area

Question 13

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 14

The length of a rope in meter by which a cow must be tethered in order that she may be able to graze an area of 2826 sq.m is ?

Accepted Answer

The cow can graze an area of 2826 sq.m i.e it is able to cover area of circle of 2826 sq.m Therefore, πr 2 =2826 r 2 =(2826×7/22) = 899.18 = 900 r = 30 m. Therefore, the radius of the circle implies the length of the rope = 30 mts.

Question 15

An order was placed for the supply of a carpet whose breadth was 6 m and length was 1.44 times the breadth. What be the cost of a carpet whose length and breadth are 40% more and 25% more respectively than the first carpet. Given that the ratio of c…

Accepted Answer

Length of the first carpet = (1.44)(6) = 8.64 cm Area of the second carpet = 8.64(1 + 40/100) 6 (1 + 25/100) = 51.84(1.4)(5/4) sq m = (12.96)(7) sq m Cost of the second carpet = (45)(12.96 x 7) = 315 (13 - 0.04) = 4095 - 12.6 = Rs. 4082.40

Question 16

In a rectangle, length is three times its breadth. If the length and the breadth of the rectangle are increased by 30% and 100% respectively, then its perimeter increases by

Accepted Answer

Let the breadth of the rectangle = x Length of the the rectangle will be = 3 times of breadth = 3x So the initial perimeter = 2(length + breadth) = 2(x + 3x) = 8x New breadth after increase = x + 10x/100 = 1.1x New length after increase = 3x + 30*3x/100 = 3.9x New perimeter = 2(1.1x + 3.9x) = 10x Percentage change in perimeter = ( 10x-8x)*100/8x = 25%

Question 17

A rectangular grassy plot 110m by 65cm has a gravel path .5cm wide all round it on the inside. Find the cost of gravelling the path at 80 paise per sq.mt

Accepted Answer

area of theplot = 110 * 65 = 7150 sq.m area of the plot excluding the path = (110-5)* (65-5) = 6300 sq.m area of the path = 7150 - 6300 =850 sq.m cost of gravelling the path = 850 x (80/100) = 680 Rs

Question 18

The ratio of bases of two triangles is x:y and that of their areas is a:b. Then the ratio of their corresponding altitudes wll be

Accepted Answer

a : b =(1/2 * x * H) : (1/2 * y * h) bxH = ayh => H/h =ay/bx Hence H:h = ay:bx

Question 19

If each side of a square is increased by 25%, find the percentage change in its area.

Accepted Answer

Let each side of the square be a. Then, area = . a 2 New side = 125 a 100 = 5 a 4 . New area = 5 a 4 2 = 25 a 2 16 Increase in area = 25 a 2 16 - a 2 = 9 a 2 16 Increase% = 9 a 2 16 * 1 a 2 * 100 % = 56.25%.

Area Questions