# What is the area of a 45-45-90 triangle, with a hypotenuse of 8mm in length?

Nov 14, 2015

$4 m {m}^{2}$

#### Explanation:

The formula for calculating the area of a triangle is $\frac{1}{2} b a s e \cdot h e i g h t$. Thanks to the fact that this is a 45-45-90 triangle the base of the triangle and the height of the triangle are equal. So we simply need to find the values of the two sides and plug them into the formula.

We have the length of the hypotenuse, so we can use the pythagorean theorem to calculate the length of the two sides.

(we know the area is going to be measured in $m {m}^{2}$ so we'll leave units out of the equations for now)

${a}^{2} + {b}^{2} = {8}^{2}$

$a = b$

We can simplify here, because we know the two remaining sides are equal. So we're just going to solve for

${a}^{4} = 16$
${a}^{2} = 8$
$a = \sqrt{8}$

Both non-hypotenuse sides of the triangle are $\sqrt{8 m m}$ long. Now we can use the triangle area formula so solve.

$a r e a = \frac{1}{2} b a s e \cdot h e i g h t = \frac{1}{2} \cdot \sqrt{8} \cdot \sqrt{8} = \frac{1}{2} \cdot 8 = 4 m {m}^{2}$