Question

Find the area of a square one of whose diagonals is $5cm$

Answer 1

Let ${}^{\prime }{a}^{\prime }$ be the side of the square and $d$ be the diagonal.

By Pythagoras theorem, we have

$d^2=a^2+a^2$

$d=\sqrt{2a^2}=\sqrt{2}a$

Given: Diagonal $\left(d\right)=5$

Diagonal of square is given by the formula, $\sqrt{2}a=d$

$\Rightarrow \sqrt{2}a=5$

$\Rightarrow a=\frac{5}{\sqrt{2}}$.

Then, area of the square $=(side{)}^2=a^2=(\frac{5}{\sqrt{2}}{)}^2=\frac{25}{2}c{m}^2$