Question

An isosceles right triangle has a hypotenuse of length e. Select all that are true for the triangle.

Answer 1

2 Answers:

• Choice C) Leg length = $\frac{e\sqrt{2}}{2}$
• Choice E) Area = $\frac{e^2}{4}$

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Explanation:

For any 45-45-90 right triangle, the hypotenuse H is always sqrt(2) times the leg L

In terms of symbols, we can say

$H = L*\sqrt{2}$

We're told the hypotenuse is e, so H = e

Let's solve for L in terms of e

$H = L*\sqrt{2}\\\\e = L*\sqrt{2}\\\\L*\sqrt{2} = e\\\\L = \frac{e}{\sqrt{2}}$

Now let's multiply top and bottom by $\sqrt{2}$ to rationalize the denominator

$L = \frac{e}{\sqrt{2}}\\\\L = \frac{e\sqrt{2}}{\sqrt{2}*\sqrt{2}}\\\\L = \frac{e\sqrt{2}}{\sqrt{2*2}}\\\\L = \frac{e\sqrt{2}}{\sqrt{4}}\\\\L = \frac{e\sqrt{2}}{2}$

If the hypotenuse is e, then the length of each leg is $\frac{e\sqrt{2}}{2}$

This matches with choice C, which is one of the two answers. This allows us to rule out choices A and B, as those expressions are not equivalent to the expression for choice C.

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The two legs of any right triangle form the base and height. The order doesn't matter. The base is always perpendicular to the height. In a right triangle, the two legs are always perpendicular.

The base and height are $\frac{e\sqrt{2}}{2}$ each.

The area of the triangle is...

$\text{Area} = \frac{1}{2}*\text{base}*\text{height}\\\\\text{Area} = \frac{1}{2}*\frac{e\sqrt{2}}{2}*\frac{e\sqrt{2}}{2}\\\\\text{Area} = \frac{e\sqrt{2}*e\sqrt{2}}{2*2*2}\\\\\text{Area} = \frac{e*e\sqrt{2*2}}{2*2*2}\\\\\text{Area} = \frac{e^2\sqrt{4}}{2*2*2}\\\\\text{Area} = \frac{e^2*2}{2*2*2}\\\\\text{Area} = \frac{e^2}{2*2}\\\\\text{Area} = \frac{e^2}{4}$

This matches with choice E, which is the other answer. Choices D and F are eliminated as they are not equivalent to the expression for choice E.