# A right isosceles triangle has an area of 18. What is the length of the longest side of the triangle?

Feb 3, 2017

$\text{The Desired Length"=6sqrt2" units.}$

#### Explanation:

Consider a Right Isosceles $\Delta A B C$ with, $A B = B C$

Clearly, $m \angle B = {90}^{\circ} ,$ so that, $A C$ being the Hypotenuse is

the longest side of $\Delta A B C .$

Now, the Area of $\Delta A B C = \frac{1}{2} \times A B \times B C$

$= \frac{1}{2} A {B}^{2} , \ldots \ldots \ldots \left[\because , A B = A C\right]$.

$\therefore A r e a = 18 \Rightarrow \frac{1}{2} A {B}^{2} = 18 \Rightarrow A {B}^{2} = 36 = B {C}^{2.}$

Using Pythagoras' Theorem, in $\Delta A B C$, we have,

$A {B}^{2} + B {C}^{2} = A {C}^{2} = 36 + 36 = 72$

$\therefore A C = \sqrt{72} = 6 \sqrt{2}$ is the desired lemgth.