# Area Aptitude Test Paper 7

31) One of the four angles of a rhombus is 90 degrees. If each side of the rhombus is 20 cm, what will be the length of the longer diagonal?

1. 20 √2 cm
2. 25 √2 cm
3. 25 √2 cm
4. 30 √2 cm

Explanation:

A rhombus with one of its angle 90 degrees is a square. So, both the diagonals are of equal length.

Now, Diagonal of a square= √2 × side

Given Side = 20 cm

So, Diagonal = 20 √2 cm

32) What is the area of the largest circle that can be drawn inside a square of side 20 cm?

1. 90 π cm2
2. 100 π cm2
3. 110 π cm2
4. 95 π cm2

Explanation:

The diameter of the largest circle that can be drawn inside a square = length of the side of the square

As per the question: The side of the square = 20 cm.

So, radius of the circle = 20/2 = 10 cm

Area of a circle = π × radius2

The area of the largest circle = π × 102 = 100 π cm2

33) If the side of a square is equal to the diameter of a circle, what is the area of the square if the area of the circle is 81π sq. cm?

1. 350 sq. cm.
2. 384 sq. cm.
3. 324 sq. cm.
4. 456 sq. cm.

Explanation:

Area of the circle = 81π sq. cm.

Diameter of circle = 9 *2= 18 cm

Now, Diameter of circle = Side of the square = 18 cm

So, Area of the square = Side2 = 182 = 324 sq. cm.

34) The perimeter of a circle and an equilateral triangle are equal. Find the area of the equilateral triangle if the area of the circle is 141π.

1. 271.34 sq. cm.
2. 281.34 sq. cm.
3. 261.34 sq. cm.
4. 251.34 sq. cm.

Explanation:

The perimeter of a circle and an equilateral triangle are equal:

Let the length of each side of equilateral triangle = A

As per the questions, the perimeters are equal.

So, 2 π r = 3 A

Area of circle is given = 141 π

So, π r2 = 144 π

r = 12

Thus, 2 π * 12 = 3 A

A = 24 π /3

The area of a equilateral triangle = (√3/4) * side2

= (√3/4) * (24 π /3) 2

= (√3/4) * 8 π * 8 π

= 0.43 * 25.12 * 25.12

= 271.34 sq. cm.

35) The diameter of a circle is increased by 100%. What is the percentage increase in area?

1. 250 %
2. 200 %
3. 400 %
4. 300 %

Explanation:

Let the diameter = d

Original area = π * (d/2) 2 = π d2/4

New Area = π * (2d/2) 2

= π (2d/2) (2d/2)

= π d2

Increase in area = (π d2 – π d2/4) = 3 π d2 / 4

Percentage increase = (increase in area/ original area) * 100

= (3 π d2 / 4) * (4/ π d2) * 100

= 300 %