Question 1

Angles of a quadrilateral are in the ratio 3 : 4 : 5 : 8. The smallest angle is

Accepted Answer

Let First angle = 3k Second angle = 4k Third angle = 5k and Fourth angle = 8k We know 3k + 4k + 5k + 8k = 360° ⇒ 20k = 360° ∴ k = 18 ° Measure of smallest angle = 3k =3 x 18 ° = 54 °

Question 2

Floor of a square room of side 10 m is to be completely covered with square tiles, each having length 50 cm. The smallest number of tiles needed is?

Accepted Answer

Area of square room = (10)2 = 100 sq m = 100 x (100)2 sq cm = 100 x 100 x 100 sq cm Now, area of tiles = (50)2 = 50 x 50 sq cm ∴ Number of tiles needed = (Area of square room) / (Area of tile) = (100 x 100 x 100) / (50 x 50) = 400 Hence, 400 tiles will be needed.

Question 3

The perimeter of an isosceles triangle is 26 cm while equal sides together measure 20 cm. The third side and each of the equal sides are respectively.

Accepted Answer

Let the third side be x. According to the question, x + 20 = 26 ⇒ x = 26 - 20 = 6 cm ∴ Each equal side = 20/2 = 10 cm

Question 4

The ratio of length and breadth of a rectangle is 5 : 3 .If length is 8 m more than breadth, find the area of the rectangle.

Accepted Answer

Let length of rectangle = 5k and breadth of rectangle = 3k According to the quecation, 5k - 3k = 8 ⇒ 2k =8 ∴ k = 4 ∴ Lenght = 5k = 5 x 4 = 20 m Breadth = 3k = 3 x 4 = 12 m ∴ Required area = Lenght x Breadth = 20 x 12 = 240 sq m

Question 5

The ratio between the lenght and the breadth of a rectangle is 2 : 1. If breadth is 5 cm less than the length, what will be the parimeter of the rectangle?

Accepted Answer

Let length = 2x and breadth = x According to the question, 2x - x = 5 ⇒ x = 5 ∴ Required perimeter = 2(2x + x) = 6x = 30 cm

Question 6

The perimeter of a rectangle having area equal to 144 cm 2 and sides in the ratio 4 : 9 is

Accepted Answer

Let l = 4k and b = 9k Area of rectangle = l x b 144 = 4k x 9k ⇒ k2 = 144/36 ⇒ k2 = 4 ∴ k = 2 ∴ l = 8 cm and b = 18 cm Perimeter of rectangle = 2(l + b) = 2(8 + 18) = 2 x 26 = 52 cm

Question 7

Area of rectangular field is 3584 m 2 and the length and the breadth are in the ratio 7 : 2, respectively. What is the perimeter of the rectangle?

Accepted Answer

Let length of the rectangular field = 7k m and breadth of the rectangular field = 2k m According to the question, Area of a rectangular field = Length x Breadth ⇒ 3584 = 7k x 2k ⇒ 14 x k2 = 3584 ⇒ k2 = 3584/14 = 256 ⇒ k2 = 256 = 16 m ∴ Length of rectangular field = 7k = 7 x 16 = 112 m And breadth of rectangular field = 2 x 16 = 32 m ∴ Perimeter of rectangle = 2(Length x Breadth) = 2(112 + 32) = 2 x 144 = 288 m

Question 8

The length and perimeter of a rectangle are the ratio of 5 : 18 What well be the ratio of its length and breadth?

Accepted Answer

According to the question, l/[2(l + b)] = 5/18 ⇒ 10l + 10b = 18l = 10b ⇒ l/b = 10/8 = 5/4 ∴ l : b = 5 : 4 Hence, ratio of length and breadth of a rectangle is 5 : 4

Question 9

Find the cost of carpeting a room 8 m long and 6 m broad with a carpet 75 cm wide at ₹ 20 per m.

Accepted Answer

Area of the carpet = Area of the room = 8 x 6 = 48 sq m Width of the carpet = 75/100 = 3/4 m Length of the carpet = 48 x (4/3) = 16 x 4 = 64 m ∴ Cost of carpeting = 64 x 20 = ₹ 1280

Question 10

A rectangle has 20 cm as its length and 200 sq cm as its area. If the area is increased by 1 1/ 5 time the original area by increase its length only then the perimeter of the rectangle so formed (in cm) is

Accepted Answer

l1 = 20 cm, A1 = 200 sq cm ∴ b1 = 200/20 = 10 cm Now, A2 = 200 x 6/5 = 240 sq cm b2 = 10 cm ∴ l2 = 240/10 = 24 cm ∴ Perimeter of new rectangle = 2(l2 + b2) = 2(24 + 10) = 2 x 34 = 68 cm

Question 11

The breadth of a rectangle is 25 m. The total cost of putting a grass bed on this field was ₹ 12375, at the rate of ₹ 15 per sq m. What is the length of the rectangular field?

Accepted Answer

Area to the rectangular field = 12375/15 = 825 sq m According to the question, (L x B) = 825 [L = length and B = breadth] ⇒ L x 25 = 825 ∴ L = 825/25 = 33 m

Question 12

What is the area of a square having perimeter 68 cm?

Accepted Answer

According to the question, 4a = 68 [ where a = side] ∴a = 68/4 = 17 cm ∴ Required area = a2 = (17)2 = 289 sq cm

Question 13

The perimeter of two squares are 68 cm. Find the perimeter of the third square whose area is equal to the different of the areas of these two squares.

Accepted Answer

a1 = 68/4 = 17 cm and a2 = 60/4 = 15 cm [ where a1 and a2 are sides] According to the question, Area of the third square = [(17)2 - (15)2 ] = (17 + 15) (17 - 15) = 32 x 2 = 64 sq cm Let a3 = Side of the third square. According to the question, (a3)2 = 64 sq cm ∴ a3 = √64 = 8 cm ∴ Perimeter of the third square = 4 x a3 = 4 x 8 = 32 cm.

Question 14

A wheel make 4000 revolution in moving a distance of 44 km. Find the radius of the wheel.

Accepted Answer

Distance covered in 1 revolution = (44 x 1000) / 4000 = 11 m According to the question, 2πr = 11 ⇒ (44/7) x r = 11 ∴ r = (11 x 7) / 44 = 1.75 m

Question 15

If the area of a semi-circle be 77 sq m, find its perimeter.

Accepted Answer

According to the question, Area of semi- circle = 77 m (1/2) x π x r2 = 77 ⇒ r2 = (77 x 2 x 7)/22 ∴ r = 7m ∴Circumference of semi- circle =πr + 2r = ( π + 2)r = [(22/7) + 2] x 7 = 36 m

Question 16

A railing of 288 m is required for fencing a semi-circular park. Find the area of the park. (π = 22/7)

Accepted Answer

Let the radius of the park be r, then πr + 2r = 288 (π + 2)r = 288 ⇒ [(22/7) + 2]r = 288 ⇒ r = (288 x 7)/36 = 56 ∴ Area of the park = (1/2)πr2 = (1/2) x (22/7) x 56 x 56 = 4928

Question 17

If the radius of a circle is increased by 6%, find the percentage increase in its area.

Accepted Answer

Given that, a = 6 According to the formula, Percentage increase in area = 2a + [a2/100]% = 2 x 6 + [36/100]% = (12 + 0.36)% = 12.36%

Question 18

The wheel of an engine turns 350 times round its axle to cover a distance of 1.76 km. The diameter of the wheel is

Accepted Answer

Distance covered in 1 round = (Total Distance) / (Total round) = (1.76 x 1000)/350 = 176/35 m ∴ 2πr = (1.76 x 100) / 35 cm 2π = Diameter = 17600 x 7/22 x 35 = 160 cm

Question 19

The ratio of the area of the circumcircle and the incircle of a square is

Accepted Answer

Ratio of the areas of the circumcircle and incircle of a square = [(Diagonal)2π] / [(Side)2π] = [(Side x √2)2] / (Side)2 = 2/1 or 2 : 1

Question 20

The radius of a circle is so increased that its circumference increased by 5%. The area of the circle, then increases by

Accepted Answer

Increase in circumference of circle = 5% ∴ Increase in radius is also 5%. Now, increase in area of circle = 2a + (a2/100) % Where, a = increase in radius= 2 x 5 + (5 x 5)/100 % = 10.25%

Area Questions