Area- Aptitude Questions and Answers

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

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

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

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

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

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%

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

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

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

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

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

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

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

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

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.

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.

Distance between two corners of hall
= √(Length² + Breadth²)
= √(40² + 30²) = √(1600 + 900) = 50 m