A 45° - 45° - 90° right triangle has hypotenuse of length

[swerve]

A 45° - 45° - 90° right triangle has hypotenuse of length h. What is the area of the triangle in terms of h?

$$A.\ \frac{h}{\sqrt{2}}$$
$$B.\ \frac{h}{2}$$
$$C.\ \frac{h}{4}$$
$$D.\ h^2$$
$$E.\ \frac{h^2}{4}$$

The OA is E.

Please, can anyone explain this PS question for me? I tried to solve it but I can't get the correct answer. I need your help. Thanks.

[GMATGuruNY]

A 45° - 45° - 90° right triangle has hypotenuse of length h. What is the area of the triangle in terms of h?

$$A.\ \frac{h}{\sqrt{2}}$$
$$B.\ \frac{h}{2}$$
$$C.\ \frac{h}{4}$$
$$D.\ h^2$$
$$E.\ \frac{h^2}{4}$$

The sides of a 45-45-90 triangle are in the following ratio:
s : s : s√2.

Let s=1, implying that h = s√2 = (1)(√2) = √2.
In a 45-45-90 triangle, the legs constitute the base and height.
Since each leg has a length of 1, we get:
Area = (1/2)bh = (1/2)(1)(1) = 1/2.

The question stem asks for the value of the area: 1/2.
Now plug h=√2 into the answers to see which yields the target value of 1/2.
Only E works:
h²/4 = (√2)²/4 = 2/4 = 1/2.

The correct answer is E.

[Scott@TargetTestPrep]

A 45° - 45° - 90° right triangle has hypotenuse of length h. What is the area of the triangle in terms of h?

$$A.\ \frac{h}{\sqrt{2}}$$
$$B.\ \frac{h}{2}$$
$$C.\ \frac{h}{4}$$
$$D.\ h^2$$
$$E.\ \frac{h^2}{4}$$

If we let the side of the triangle = q, then:

h = q√2

h/√2 = q

So the area of the triangle is:

q^2 x 1/2 = (h/√2)^2 x 1/2 = (h^2)/2 x 1/2 = (h^2)/4.

Answer: E