Question

If the perimeter of a right-angled isosceles triangle is $(\sqrt{2}+1)cm$, then what is the length of the hypotenuse?

• A. $\frac{1}{\sqrt{2}}$ cm
• B. $\sqrt{2}$ cm
• C. $1$ cm
• D. $2$ cm

Answer 1

The correct option is C $\left(1cm\right)$

Let the equal sides of given right-angled isosceles triangle be $a$ and hypotenuse be $h$.
Then from Pythagoras theorem, we have $h^2=a^2+a^2$
$\Rightarrow h^2=2a^2$
$\Rightarrow h=\sqrt{2a^2}$
$\therefore h=2\sqrt{a}$
Given, perimeter is $\sqrt{2+1}cm$
$a+a+a\sqrt{2}=\sqrt{2+1}$

$\Rightarrow 2a+a\sqrt{2}=\sqrt{2}+1$

$\Rightarrow a\sqrt{2}(\sqrt{2}+1)=\sqrt{2}+1$

$\Rightarrow$ $a=\frac{(\sqrt{2}+1)}{\sqrt{2}(\sqrt{2}+1)}$

$\Rightarrow a=\frac{1}{\sqrt{2}}$

$\therefore$ Hypotenuse $=a\sqrt{2}=\frac{1}{\sqrt{2}}\times \sqrt{2}cm=1cm$