Question

If the perimeter of a right isosceles triangle is $\rm 2+ 2 \sqrt{2} \: m$ then find the length of its hypotenuse.​

Answer 1

In right isosceles two sides of triangle are equal except the hypotenuse.

Assume, a = two equal sides of triangle.

and a√2 = hypotenuse.

$\sf \: a + a + a\sqrt{2} = 2 + 2 \sqrt{2}$

$\sf \implies \: 2a + a \sqrt{2} = 2 + 2 \sqrt{2}$

$\sf \implies \: a \sqrt{2} ( \sqrt{2} + 1) = 2 + 2 \sqrt{2}$

$\sf \: a = \dfrac{2 + 2 \sqrt{2} }{ \sqrt{2}( \sqrt{2} + 1)}$

$\sf \implies \: a = \dfrac{2(1 + \sqrt{2} )}{ \sqrt{2} ( \sqrt{2} + 1)}$

$\sf \: a = \dfrac{2 \cancel{( 1 + \sqrt{2} )}}{ \sqrt{2} +( \cancel{\sqrt{2} + 1)} }$

$\sf \: a = \dfrac{2}{ \sqrt{2} }$

$\sf \therefore \: a = \sqrt{2} m$

$\sf \: hypotenuse = a \sqrt{2}$

$\sf \implies hypotenuse \: = \sqrt{2} \sqrt{2}$

$\sf \implies \: hypotenuse \: = 2 \: m$

To check :

Substitute values of each side.

Perimeter of triangle = side + side + hypotenuse.

Perimeter of triangle = √2 + √2 + 2

= 1√2 + 1√2 + 2

= (1+1) √2 + 2

= 2√2 + 2

= 2+ 2√2 m

Therefore , hence the hypotenuse of triangle is 2m.