Area- Aptitude Questions and Answers - in Hindi

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There is a path made around a circular park. If the difference of external and internal perimeters is 132 meter, then the width of the path is:


किसी वृत्तकार उद्यान के चारों ओर एक समान चौड़ाई का एक पथ बना हुआ है । यदि इस वृत्ताकार पथ की आंतरिक तथा बाह्य‌ परिधियों का अंतर 132 मीटर है, तो पथ की चौड़ाई है:

A)
22 m

B)
20 m

C)
21 m

D)
24 m


Correct Answer :

21 m


Explanation :

Let the inner and outer radii be r and R meters.
then width of the path => 2(pi)R-2(pi)r = 132m
there pi =22/7
So R-r= 132*7/2*22 =21m

therefore width of the path is R-r=21m

The base of right-angled triangle is 5 meters and hypotenuse is 13 meters . Its area will be ?

A)
25 m2

B)
28 m2

C)
30 m2

D)
24 m2


Correct Answer :

30 m2


Explanation :

Hypotenuse = √(base2+height2)
height2 = Hypotenuse2 - base2
height = √(132 - 52)
⇒ √144 ⇒ 12

Area of the triangle = 1/2(base*height)
=1/2(5 * 12 ) m2
= 30m2

If the side of a square is increased by 25%, then how much percent does its area get increased ?

A)

125


B)

156.25


C)

50


D)

56.25



Correct Answer :

56.25


Explanation :

Let side of square is 100 m2 Then, side = 10 m
And, Area = length of a side2 (10)2 m2

increased side by 25 Then, new side = 125 % of 10 ⇒ (125/100) x 10 ⇒ 12.5 m
And,new area = length of a side2 ⇒ (12.5)2 sq. m

Increase in area = (12.5)2 - (10)2 m2
= 156.25 - 100 m2
= 56.25 m2
% Increase = 56.25

if the side of a square be increased by 4 cms. The area increased by 60 sq. cms . The side of the square is ?

A)

12 cm


B)

13 cm


C)

14 cm


D)

5.5 cm



Correct Answer :

5.5 cm


Explanation :

Let each side = x cm
Then, (x + 4 )2 - x2 = 60
⇒ x 2 + 8x + 16 - x2 = 60
⇒ x = 5.5 cm

If the diameters of a circle is increased by 100% . Its area is increased by ?

A)

100%


B)

200%


C)

300%


D)

400%



Correct Answer :

300%


Explanation :

Area of circle = πd2/4

diameters increased by 100%, then new area = π(2d)2 /4 => πd2
Then, New area = (πd2 - πd2/4) => 3πd2/4

increase percent =[(3πd2/4) / ( πd2/4) ]*100 %
= [(3πd2/4) x (4/πd2)]*100 %
= 300%

Find the area of a triangle whose sides measure 8 cm, 10 cm and 12 cm?

A)

8√63 sq cm


B)

5√63 sq cm


C)

6√53 sq cm


D)

7√93 sq cm



Correct Answer :

5√63 sq cm


Explanation :

Area of triangle = √[s(s-a)(s-b)(s-c)] there s = (a+b+c)/2

let, a = 8 cm, b = 10 cm and c = 12 cm
then,s = (a + b + c)/2
= (8 + 10 + 12)/2
= 15 cm

so,Area of triangle = √s(s - a) (s - b) (s - c)
= √[15 * (15 - 8) * (15 - 10) * (15 - 12)]
= √1575
= √[25 x 63]
= 5√63 cm

One side of a parallelogram is 8.06 cm and its perpendicular distance from opposite side is 2.08 cm. What is the approximate area of the parallelogram?

A)

12.56 cm2


B)

14.56 cm2


C)

16.76 cm2


D)

22.56 cm2



Correct Answer :

16.76 cm2


Explanation :

Area of parallelogram = base x height
= 8.06 x 2.08 = 16.76 cm2

Find the area of a rectangle having 15m length and 8m breadth?

A)

120 sq m


B)

111 sq m


C)

115 sq m


D)

125 sq m



Correct Answer :

120 sq m


Explanation :

Area of Rectangle= Length x Breadth
= 15 x 8 = 120 sq m

The length of the side of a square is represented by x+2. The length of the side of an equilateral triangle is 2x. If the square and the equilateral triangle have equal perimeter, then the value of x is _______?

A)

4


B)

5


C)

6


D)

8



Correct Answer :

4


Explanation :

Since the side of the square is x + 2, its perimeter = 4 (x + 2) = 4x + 8
Since the side of the equilateral triangle is 2x, its perimeter = 3 * 2x = 6x
Also, the perimeters of both are equal.
4x + 8 = 6x
2x = 8
x = 4

The perimeters of two squares are 160cm and 164cm. Find the perimeter of a third square whose area is equal to the difference of the areas of the two squares(in cm).

A)
81

B)
60

C)
36

D)
9


Correct Answer :

36


Explanation :

Perimeter of Square = 4 * length of a side
Area of Square= length of a side2

The perimeters of two squares are 160cm and 164cm ,
So the side of first square = 160/4 = 40 cm
side of second square = 164/4 = 41 cm

Area of third squre = 412 - 402 = 81 sq cm
Then, side of third square = √81 = 9 cm

required perimeter = 4*9 = 36 cm

The sides of a triangle are in the ratio 3:4:5. The perimeter of the triangle is 24cm. The area (in cm2) of the triangle is:

A)
39

B)
26

C)
24

D)
32


Correct Answer :

24


Explanation :

Perimeter = (a+b+c) ,there a,b,c are the sides of a triangle
Area = 1/2(base*height) = √[s(s-a)(s-b)(s-c)] there s = (a+b+c)/2

let's the sides of triangle are 3 cm, 4 cm, and 5 cm in length , then
Perimeter = 3 cm + 4 cm + 5 cm =12 cm

So Area = √[s(s-a)(s-b)(s-c)] there s = (a+b+c)/2 = 12/2 =6
Area = √[6(6-3)(6-4)(6-5)] =√[6*3*2*1] =6

hence Area may be 6 or multiple of 6 that is 6, 12 , 24 ...because sides are given in ratio

In a rhombus of side 10 cm, one of diagonals is 16 cm long. The length of In a rhombus of side 10 cm, one of diagonals is 16 cm long. The length of the second diagonal is ?

A)
16 cm

B)
12 cm

C)
18 cm

D)
20 cm


Correct Answer :

12 cm


Explanation :

The diagonals meet in the middle at a right angle.
All sides of rhombus are equal i.e. 10cm
(Side)2=(half of one diagonal)2+(half of second diagonal)2
by using this formula
(10)2=(8)2+(half of second diagonal)2
(half of second diagonal)2 = 36
half of second diagonal=6
second diagonal=12cm

One diagonal and perimeter of a rhombus are 24 cm and 52 cm respectively. The area of rhombus will be?

A)
60 cm2

B)
120 cm2

C)
100 cm2

D)
30 cm2


Correct Answer :

120 cm2


Explanation :


Side of roumbus =52/4 = 13 cm

Area of the roumbus = (product of diagonals)/2

Let ABCD is rhombus, in which Side of roumbus =AB=13cm , BD=24cm and AC=? cm
the digonal of rhoumbus bisect each other at right angle

in the right angel AOB
OA2+OB2=AB2
OA2+(BD/2)2=AB2
OA2+(12)2=(13)2
OA2= (13)2 - (12)2
OA = √(169-144) =√25 = 5
Hence the lenght of other dignoal is AC=2*OA= 2*5= 10cm
Thus,Area of the roumbus = (product of diagonals)/2
=> ( 24 * 10 ) /2 => 120 cm2

ABCD is a square of side X cm. Its side is increased by 30%. What is the change in its area?


ABCD एक वर्ग है जिसकी भुजा 'X' सेमी हे। यदि इसकी एक भुजा 30% की वृद्धि कर दे तो इसके क्षेत्र में क्या परिवर्तन होगा?

A)
60%

B)
62.5%

C)
63%

D)
69%


Correct Answer :

69%


Explanation :

Let side = X cm
then Area = length of a side2 ⇒ (X)2 cm2

increased side by 30 then, new side = 130 % of X ⇒ (130/100) *X ⇒ 1.3X cm
And,new area = length of a side2 ⇒ (1.3)2 sq. cm

Increase in area = (1.3*X)2 - (X)2 cm2
= 1.69*X2 - X2
= .69*X2
% Increase = 69

The sides of a triangle a, b, c are such that 2a = 3b = 4c and its perimeter is 208 cm. What is the length of the longest side?

A)
48cm

B)
96cm

C)
64cm

D)
54cm


Correct Answer :

96cm


Explanation :

Let 2a = 3b = 4c = x
So, a = x/2 b = x/3 c = x/4
The L.C.M of 2, 3 and 4 is 12
Therefore, a : b : c = x/2 × 12 : x/3 × 12 : x/4 = 12
= 6x : 4x : 3x
= 6 : 4 : 3
Therefore, a : b : c = 6 : 4 : 3

the perimeter of a triangle is sum of its side
Lets take a variable x for the sides,
hence the sides will be 6x,4x,3x respectively.

Perimeter = 6x+4x+3x
208 = 6x+4x+3x = 13x
13x = 208
=> x = 208/13
x=16

Length of the longest side = 6x = 6 x 16 = 96.

What is the longest side of a rectangle which has a perimeter of 70 units and an area of 276 square units?

A)
12 units

B)
18 units

C)
23 units

D)
36 units


Correct Answer :

23 units


Explanation :

let Length =L and Width=W

Perimeter = 2(L + W) = 70
L + W = 35
L = 35 - W

Area = WL = 276
W(35 - W) = 276
35W - W2 = 276
0 = W2 - 35W + 276
0 = (W - 23)(W - 12)
W = {12, 23}
L = {23, 12}

Therefore, the sides of the rectangle are 23 and 12
The longest side of the rectangle is 23 units.

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