Question

If an isosceles right triangle has area $8c{m}^2$ then the length of its hypotenuse is
(a) $\sqrt{32}cm$
(b) $\sqrt{48}cm$
(c) $\sqrt{24}cm$
(d) $\sqrt{4}cm$

Answer 1

Given: Area of an isosceles right triangle is $8c{m}^2$ and also in an isosceles right triangle, base$=$height=perpendicular
Area of an isosceles right triangle$8=\frac{1}{2}\times \text{Base}\times \text{Height}$
$\Rightarrow 8\times 2=\text{Base}\times \text{Base}$ (Since Base$=$Height )
$\Rightarrow {\text{base}}^2=16$
$\Rightarrow \text{base}=\pm \sqrt{16}$
$\therefore \text{base}=\pm 4$
Since, length can't be negative, thus, base$=$Height$=4$
In a right angled triangle, using Pythagoras theorem,
${\text{(Hypotenuse)}}^2={\text{(Base)}}^2+{\text{(Perpendicular)}}^2$
$\Rightarrow {\text{(Hypotenuse)}}^2=4^2+4^2$
$\Rightarrow {\text{(Hypotenuse)}}^2=16+16$
$\Rightarrow {\text{(Hypotenuse)}}^2=32$
$\therefore \text{(Hypotenuse)}=\sqrt{32}$
Hence, the correct option is (a).