# In a certain isosceles right triangle, the altitude to the hypotenuse has length 4sqrt(2) . What is the area of the triangle?

Jun 26, 2018

$32$

#### Explanation:

Any isosceles right triangle is half a square, cut by its diagonal.

So, the altitude to the hypotenuse is half the diagonal of the square (which also means that the altitude to the hypotenuse is half the hypotenuse, by the way).

This means that the diagonal of the square is $8 \sqrt{2}$.

In any square, you have

$d = l \sqrt{2}$

where $d$ is the diagonal and $l$ is the side of the square.

So, solving for $l$ we have

$l = \frac{d}{\sqrt{2}}$

In this case, this means that the side of the square is $8$ unit long.

This, in turn, means that the area of the square is $64$ units squared.

Since the triangle is half the square, its area will be $32$ units squared.

Alternative strategy

Since the altitude to the hypotenuse is half the hypotenuse, we know that the hypotenuse is $8 \sqrt{2}$. So, we know a side of the triangle and its altitude, so we can use the formula

$A = \setminus \frac{b h}{2} = \setminus \frac{4 \sqrt{2} \cdot 8 \sqrt{2}}{2} = \setminus \frac{32 \cdot 2}{2} = 32$