Question

If the diagonal of a square is ${}^{\prime }{a}^{\prime }$ units, What is the diagonal of the square, whose area is double that of the first square?

Answer 1

Given

Diagonal of first square ${=}^{\prime }{a}^{\prime }$ unit

Let side of first square be $x$ unit

then, diagonal = $\sqrt{2}\left(side\right)$

$\therefore a=2\sqrt{x}$

$\Rightarrow a^2=4x$ [On squaring both sides]

$\Rightarrow x=\frac{a^2}{4}$

$\therefore$ side of first square $=\frac{a^2}{4}$

Area of First square $=(side{)}^2$

$=(\frac{a^2}{4}{)}^2$

$=\frac{a^4}{16}$

Area of the Second saquare $=2\times$ Area of First square $=2\times \frac{a^4}{16}=\frac{a^4}{8}$ [Given]

$\therefore (Side{)}^2=\frac{a^4}{8}$

$\Rightarrow side=\sqrt{\frac{a^2}{8}}$

$\Rightarrow side=\frac{a^2}{2\sqrt{2}}$

$\therefore$ diagonal of second square (whose area is twice the first sqaure) $=\sqrt{2}\times \left(\frac{a^2}{2\sqrt{2}}\right)=\frac{a^2}{2}$