# How do you find the area of an isosceles right angle triangle with a perimeter of 40 units?

Apr 20, 2017

$A r e a = 400 \left(3 - 2 \sqrt{2}\right)$

#### Explanation:

An isosceles triangle has two of its sides the same length.

This triangle is also a right angled triangle so Pythagoras's Theorem holds.

(1) find the hypotenuse

call the equal lengths  ""x

then we have by Pythagoras

${x}^{2} + {x}^{2} = {c}^{2}$

$\implies c = x \sqrt{2}$

so the sides are:$\text{ } x , x , x \sqrt{2}$

(2) Find $x$

the perimeter is 40

$\therefore x + x + x \sqrt{2} = 40$

$2 x + x \sqrt{2} = 40$

$x \left(2 + \sqrt{2}\right) = 40$

$x = \frac{40}{2 + \sqrt{2}} = 20 \left(2 - \sqrt{2}\right) \text{ on rationalising}$

(3) Find the area

$A r e a = \frac{1}{2} \times x \times x$

$A r e a = \frac{1}{2} {\left(20 \left(2 - \sqrt{2}\right)\right)}^{2}$

$A r e a = \frac{1}{2} \times 400 \times \left(4 - 4 \sqrt{2} + 2\right)$

$A r e a = 200 \times \left(6 - 4 \sqrt{2}\right)$

take out common factors

$A r e a = 400 \left(3 - 2 \sqrt{2}\right)$

$= 68.63$

Apr 20, 2017

$A = 68.63$ square units.

#### Explanation:

Let the two equal sides of the triangle be $x$.

The hypotenuse is $40 - 2 x$ because the perimeter is $40$ units.

By Pythagoras: The square on the hypotenuse is equal to the sum of the squares on the other two sides.

${\left(40 - 2 x\right)}^{2} = {x}^{2} + {x}^{2}$

$1600 - 160 x + 4 {x}^{2} = 2 {x}^{2}$

$2 {x}^{2} - 160 x + 1600 = 0 \text{ } \div 2$

${x}^{2} - 80 x + 800 = 0 \text{ } \leftarrow$ solve for $x$

$x = \frac{- \left(- 80\right) \pm \sqrt{6400 - 4 \left(1\right) \left(800\right)}}{2}$

$x = \frac{80 + \sqrt{3200}}{2} = 68.28 \text{ } \leftarrow$ reject as too big

$x = \frac{80 - \sqrt{3200}}{2} = 11.7157$

The equal sides of the right-angled isosceles triangle are also the base and height. Use $x$ to find the area.

$A = \frac{b \times h}{2}$

$A = \frac{11.7157 \times 11.7157}{2} = 68.63$ square units