CAT 2021

Slot 2

QA

Question & Solution

All the sides of the rhombus are equal.

The area of a rhombus is $12\ cm^2$

Considering d1 to be the length of the longer diagonal, d2 to be the length of the shorter diagonal.

The area of a rhombus is $\left(\frac{1}{2}\right)\left(d1\right)\cdot\left(d2\right)\ =\ 12$

d1*d2 = 24.

The length of the side of a rhombus is given by $\frac{\sqrt{\ d1^2+d2^2}}{2}$. This is because the two diagonals and a side from a right-angled triangle with sides d1/2, d2/2 and the side length.

$\frac{\sqrt{\ d1^2+d2^2}}{2}=\ 5$

Hence $\sqrt{\ d1^2+d2^2}\ =\ 10$

$d1^2+d2^2\ =\ 100$

Using d1*d2 = 24, 2*d1*d2 = 48.

$d1^2+d2^2\ +2\cdot d1\cdot d2=\ 100+48\ =\ 148$

$d1^2+d2^2\ -2\cdot d1\cdot d2=\ 100-48\ =\ 52$

$d1+d2\ =\ \sqrt{\ 148}$  (1)

d1-d2 = $\sqrt{52}$   (2)

(1) + (2)= 2*(d1) = 2*($\sqrt{\ 37}+\sqrt{\ 13}$)

d1 = $\sqrt{\ 37}+\sqrt{\ 13}$

or 

In a rhombus the area of a Rhombus is given by :

The diagonals perpendicularly bisect each other. Considering the length of the diagonal to be 2a, 2b.

The area of a Rhombus is : $\left(\frac{1}{2}\right)\cdot\left(2a\right)\cdot\left(2b\right)\ =\ 12$

ab =6.

The length of each side is : $\sqrt{\ a^2+b^2}$ = 5, $a^2+b^{2\ }=\ 25,\$

$\left(a+b\right)^2=\ 37,\ \left(a+b\right)\ =\ \sqrt{\ 37}$

($\left(a-b\right)^2=\ 13,\ a-b\ =\ \sqrt{\ 13}$

$2a\ =\ \left(\sqrt{\ 37}+\sqrt{\ 13}\right)$,  $2b\ =\ \left(\sqrt{\ 37}-\sqrt{\ 13}\right)$.

2a is longer diagonal which is equal to $\ \left(\sqrt{\ 37}+\sqrt{\ 13}\right)$