# Mensuration

Quantitative Aptitude Questions and Answers section on “Mensuration” with solution and explanation for competitive examinations such as CAT, MBA, SSC, Bank PO, Bank Clerical and other examinations.
1.

If the length of the diagonal AC of a square ABCD is 5.2 cm, then the area of the square is :
[A]10.00 sq.cm
[B]12.62 sq.cm
[C]13.52 sq.cm
[D]15.12 sq.cm

13.52 sq.cm
Side of square = $\frac{Diagonal}{\sqrt{2}}$
∴ Area = $\frac{Diagonal^{2}}{2}$
$=\frac{\left ( 5.2 \right )^{2}}{2}$ $= \frac{27.04}{2}$
= 13.52 sq.cm
Hence option [C] is the right answer.

2.

The length of the diagonal of a square is ‘a’ cm. Which of the following represents the area of the square (in sq.cm) ?
[A]$\frac{a^{2}}{4}$
[B]$\frac{a}{\sqrt{2}}$
[C]2a
[D]$\frac{a^{2}}{2}$

$\frac{a^{2}}{2}$
$Side = \frac{Diagonal}{\sqrt{2}}$ $= \frac{a}{\sqrt{2}}$
$\therefore Area = (Side)^{2}$
$= \left ( \frac{a}{\sqrt{2}} \right )^{2} sq.cm = \frac{a^{2}}{2} sq.cm$
Hence option [D] is the right answer.

3.

The diagonal of a square is $4\sqrt{2}$cm. The diagonal of another square whose area is double that of the first square is :
[A]$\sqrt{32} cm$
[B]8 cm
[C]16 cm
[D]$8\sqrt{2} cm$

8 cm
Side of the first square = $\frac{1}{\sqrt{2}}\times 4\sqrt{2} = 4 cm$
Its area = $(4)^{2} = 16$
∴ Area of second square = 2$\times$16 = 32 sq.cm
∴ Required diagonal = $\sqrt{2}\times 4\sqrt{2}= 8 cm$
Hence option [D] is the right answer.

4.

The diffrence of the areas of two squares drawn on two line segments of different lengths is 32 sq.cm. Find the length of the greater line segment if one is longer than the other by 2 cm.
[A]16 cm
[B]11 cm
[C]9 cm
[D]7 cm

9 cm
Let the length of the smaller line segment = x cm.
The length of largest line segment = (x+2) cm.
According to the question $\left ( x+2 \right )^{2}-x^{2}=32$
$=>x^{2}+4x+4-x^{2}=32$
$=>4x=32-4=28$
$=>x=\frac{28}{4}=7$
Therefore The required lenght = x+2 = 7+2 = 9 cm.
Hence option [D] is the right answer.

5.

If the diagonals of two squares are in the ratio of 2 : 5, there area will be in the ratio of
[A]4 : 25
[B]4 : 5
[C]2 : 5
[D]$\sqrt{2} : \sqrt{5}$

4 : 25
Let diagonals be 2x and 5x.
$\therefore \frac{A1}{A2} = \frac{\frac{1}{2}\times (2x)^{2}}{\frac{1}{2}\times (5x)^{2}} = \frac{4}{25}$
=> 4 : 25
Hence option [A] is the right answer.

6.

The diagonal of a square A is (a+b). The diagonal of a square whose are is twice the area of square A, is
[A]$\sqrt{2}(a-b)$
[B]$\sqrt{2}(a+b)$
[C]$2(a+b)^{2}$
[D]$2(a+b)$

$\mathbf{\sqrt{2}(a+b)}$
Area of the square A = $\frac{\left ( diagonal \right )^{2}}{2}= \frac{\left ( a+b \right )^{2}}{2}$
Area of the new square
$= \frac{\left ( a+b \right )^{2}}{2}\times 2 = \left ( a+b \right )^{2}$
$=> side = (a+b)$
$\therefore Diagonal = \sqrt{2}\times side$
$= \sqrt{2}(a+b)$
Hence option [B] is the right answer.

7.

The ratio of the area of a square to that of the square drawn on its diagonal is :
[A]1 : 3
[B]1 : 5
[C]1 : 2
[D]1 : 6

1 : 2
Let the side of square be a unit.
Area of this square = $a^{2}$
The diagonal of square = $\sqrt{2}a$
∴ Area of square = $2a^{2}$
∴ Required Ratio = $a^{2} : 2a^{2}$
$= 1 : 2$
Hence option [C] is the right answer.

8.

The length of diagonal of a square is $15\sqrt{2}cm$. Its area is :
[A]250 sq.cm.
[B]225 sq.cm.
[C]112.5 sq.cm.
[D]450 sq.cm.

225 sq.cm.
Diagonal of square = $\sqrt{2}\times side$
$\therefore\sqrt{2}\times side = 15\sqrt{2}$
$=> Side = \frac{15\sqrt{2}}{\sqrt{2}} = 15$
∴ Area of square = $(side)^{2}$
= 15 $\times$ 15 = 225 sq.cm.
Hence option [B] is the right answer.

9.

The length of a plot is five times its breadth. A playground measuring 245 square meters occupies half of the total area of the plot. What is the length of the plot?
[A]$5\sqrt{2} m$
[B]$175\sqrt{2} m$
[C]$35\sqrt{2} m$
[D]$490 m$

$\mathbf{35\sqrt{2} m}$
Let breadth of plot = x m
∴ lenght = 5x m
As per the question,
$\frac{5x^{2}}{2} = 245$
$=> x^{2} = \frac{245\times 2}{5} = 98$
$=> x = 7 \sqrt{2} m$
$\therefore length = 5\times 7\sqrt{2} = 35\sqrt{2}m$
Hence option [C] is the right answer.

10.

The perimeter of five squares are 24 cm, 32 cm, 40 cm, 76 cm and 80 cm respectively. The perimeter of another square equal in area to sum of the ares of these squares is:
[A]62 cm
[B]124 cm
[C]31 cm
[D]961 cm

124 cm
Side of the squares are 6 cm, 8 cm, 10 cm, 19 cm and 20 cm respectively.
Sum of their areas $=\left ( 6^{2}+8^{2}+10^{2}+19^{2}+20^{2} \right ) cm^{2}$
$= (36 + 64 + 100 + 361 + 400) cm^{2}$
$= 961 cm^{2}$
∴ Area of largest other square $= 961 cm^{2}$
= Its side $= \sqrt{961} = 31 cm$
∴ Required perimeter = 4 $\times$ 31 = 124 cm.
Hence option [B] is the right answer.