Question 1

A square, a circle and equilateral triangle have same perimeter. Consider the following statements. I. The area of square is greater than the area of the triangle. II. The area of circle is less then the area of triangle. Which of the statement is/a…

Accepted Answer

Let the radius of circle is 'r' and a side of a square is 'a', then given condition 2πr = 4a ⇒ a = πr/2 ∴ Area of square = (πr/2)2 = π2 /4r2 = 9.86r2/4 = 2.46r2 and area of circle = πr2 = 3.14;r2 and let the side of equilateral triangle is x. Then, given condition, 3x = 2πr ⇒ x = 2πr/3 ∴ Area of equilateral triangle = √3/4 x 2 = √3/4 x 4π2r2/9 = π2/3√3r2 = 1.89r2 Hence, Area of circle > Area of square > Area of equilateral triangle.

Question 2

If the area of an equilateral triangle is x and its perimeter is y, then which one of the following is correct?

Accepted Answer

Area of equilateral triangle = √3a2/4 = x ......(i) And perimeter = 3a = y ⇒ a = y/3 ....(ii) Now, Putting the value of a from Eq. (ii) in Eq. (i). we get √3 (y/3)2/4 = x ⇒ x = √3 x y2/36 ⇒ x = y2/3√3x = y2/12√3 12√3 x = y2 On squaring both sides, we get y4 = 432x2

Question 3

The area of circle inscribed in an equilateral triangle is 154 sq cm. What is the perimeter of the triangle?

Accepted Answer

We know that, the radius of a circle inscribed in a equilateral triangle = a/[2√3] Where, a be the length of the side of an equilateral triangle. Given that, area of a circle inscribed in an equilateral tringle = 154 cm2 ∴ π(a/2√3)2 = 154 ⇒ (a/2√3)2 = 154 x (7/22) = (7)a2 ⇒ a = 42√3 cm Perimeter of an equilateral triangle = 3a = 3(14√3) = 42√3 cm

Question 4

The length of a rectangle is twice its breadth. If its length is decreased half of the 10 cm and the breadth is increased by half of the 10 cm cm, the area of the rectangle is increased by 5 sq cm, more than 70 sq cm. Find the length of the rectangl…

Accepted Answer

Given that, l = 2b [Here l = length and b = breadth] Decrease in length = Half of the 10 cm = 10/2 = 5 cm Increase in breadth = Half of the 10 cm = 10/2 = 5 cm Increase in the area = (70 + 5) = 75 sq cm According to the question, (l - 5) (b + 5) = lb + 75 ⇒ (2b - 5) (b + 5) = 2b2 + 75 [since l = 2b] ⇒ 5b - 25 = 75 ⇒ 5b = 100 ∴ b = 100/ 5 = 20 ∴ l = 2b = 2 x 20 = 40 cm

Question 5

The area of rectangle is 1.8 times the area of a square. The length of the rectangle is 5 times the breadth. The side of the square is 20 cm. What is the perimeter of the rectangle?

Accepted Answer

Area of square = (Side)2 = 202 = 400 sq cm ∴ Area of rectangle = 1.8 x 400 = 720 sq cm Let length and breadth of rectangle be 5k and k respectively. Then, according to the question, 5k x k = 720 ⇒ 5k2 = 720 ⇒ k2 = 720/5 = 144 ∴ k = √144 = 12 cm Perimeter of rectangle = 2(5k + k) = 12k = 12 x12 = 144 cm

Question 6

How many circular plates of diameter d be taken out of a square plate of side 2d with minimum loss of material?

Accepted Answer

Area of square plate = (Side)2 = (2d)2 = 4d2 Area of circular plate = π (d/2)2 = πd2/4 ∵ Number of square plates = [(4d2)/4] / [(πd2)/4] = (4 x 4)/π ≈ 5 Since, nearest integer value is 5.

Question 7

One diagonal of a rhombus is 60% of the other diagonal. Then, area of the rhombus is how many times the square of the length of the larger diagonal?

Accepted Answer

Let one diagonal be k. Then, other diagonal = (60k/100) = 3k/5 cm Area of rhombus =(1/2) x k x (3k/5) = (3/10) = 3/10 (square of longer diagonal) Hence, area of rhombus is 3/10 times.

Question 8

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%.

Question 9

If the diagonal of a rectangle is 17cm long and its perimeter is 46 cm. Find the area of the rectangle.

Accepted Answer

let length = x and breadth = y then 2(x+y) = 46 => x+y = 23 x²+y² = 17² = 289 now (x+y)² = 23² => x²+y²+2xy= 529 => 289+ 2xy = 529 => xy = 120 area = xy = 120 sq.cm

Question 10

If the radius of a circle is decreased by 50% , find the percentage decrease in its area.

Accepted Answer

let original radius = r and new radius = (50/100) r = r/2 original area = πr 2 and new area = π r / 2 2 decrease in area = 3 πr 2 / 4 * 1 πr 2 *100 = 75%

Question 11

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

Accepted Answer

l 2 + b 2 = 41 (or) l 2 + b 2 = 41 Also, l b = 20 l + b 2 = l 2 + b 2 + 2 l b = 41 + 40 = 81 (l + b) = 9. Perimeter = 2(l + b) = 18 cm.

Question 12

A parallelogram has sides 30m and 14m and one of its diagonals is 40m long. Then its area is

Accepted Answer

let ABCD be the given parallelogram area of parallelogram ABCD = 2 x (area of triangle ABC) now a = 30m, b = 14m and c = 40m s=1/2 x (30+14+40) = 42 Area of triangle ABC = s s - a s - b s - c = 42 12 28 2 = 168sq m area of parallelogram ABCD = 2 x 168 = 336 sq m

Question 13

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

Accepted Answer

Let original length = x metres and original breadth = y metres. Original area = xy sq.m Increased length = 120 100 and Increased breadth = 120 100 New area = 120 100 x * 120 100 y = 36 25 x y m 2 The difference between the Original area and New area is: 36 25 x y - x y 11 25 x y Increase % = 11 25 x y x y * 100 = 44%

Question 14

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

Accepted Answer

let the side of the square be x meters length of two sides = 2x meters diagonal = 2 x = 1.414x m saving on 2x meters = .59x m saving % = 0 . 59 x 2 x * 100 % = 30% (approx)

Question 15

A wire can be bent in the form of a circle of radius 56cm. If it is bent in the form of a square, then its area will be

Accepted Answer

length of wire = 2 πr = 2 x (22/7 ) x 56 = 352 cm side of the square = 352/4 = 88cm area of the square = 88 x 88 = 7744sq cm

Question 16

The ratio of the area of a square to that of the square drawn on its diagonal is

Accepted Answer

ratio = a 2 2 a 2

Question 17

The area of the largest circle that can be drawn inside a rectangle with sides 18cm by 14cm is

Accepted Answer

The diameter is equal to the shortest side of the rectangle. So radius= 14/2 = 7cm. Therefore, a r e a o f c i r c l e = πr 2 = 22 7 × 49 = 154 cm 2

Question 18

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

Accepted Answer

let ABC be the isosceles triangle, the AD be the altitude Let AB = AC = x then BC= 32-2x [because parameter = 2 (side) + Base] since in an isoceles triange the altitude bisects the base so BD = DC = 16-x In a 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 = 12 cm Hence, required area = 1 2 * B C * A D = 1 2 * 12 * 10 = 60 sq cm

Question 19

The base of a triangular field is three times its altitude. If the cost of cultivating the field at Rs. 24.68 per hectare be Rs. 333.18, find its base and height.

Accepted Answer

Area of the field = Total cost/rate = (333.18/25.6) = 13.5 hectares 13 . 5 * 10000 m 2 = 135000 m 2 Let altitude = x metres and base = 3x metres. Then, 1 2 * 3 x * x = 135000 ⇒ x 2 = 90000 ⇒ x = 300 Base = 900 m and Altitude = 300 m.

Question 20

A tank is 25m long 12m wide and 6m 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 × h + l × b = 2 25 + 12 × 6 + 25 × 12 = 744 s q . m Cost of plastering = 744 × 75 100 = R s . 558

Area Questions