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

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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

[tex]H = L*\sqrt{2}[/tex]

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

Let's solve for L in terms of e

[tex]H = L*\sqrt{2}\\\\e = L*\sqrt{2}\\\\L*\sqrt{2} = e\\\\L = \frac{e}{\sqrt{2}}\\\\[/tex]

Now let's multiply top and bottom by [tex]\sqrt{2}[/tex] to rationalize the denominator

[tex]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}\\\\[/tex]

If the hypotenuse is e, then the length of each leg is [tex]\frac{e\sqrt{2}}{2}[/tex]

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 [tex]\frac{e\sqrt{2}}{2}[/tex] each.

The area of the triangle is...

[tex]\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}\\\\[/tex]

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.

The required points that are correct are the length of the leg is given as e√2/2 and the area of the triangle is given as e²/4. Options D and E are correct.

Given that,
An isosceles right triangle has a hypotenuse of length e.

In a right-angled triangle, its side, such as hypotenuse, perpendicular, and base is Pythagorean triplets.

Here,
Let the measure of the leg be x,
Apply Pythagoras' theorem,
e² = x ² + x²
e² = 2x²
x = √e²/2
x = e/√2
or
x = e√2/2

Now,?
area of the triangle is given as,
area = 1/2 × e/√2 ×e/√2
area = e² / 4

Thus, the required points that are correct are the length of the leg is given as e√2/2 and the area of the triangle is given as e²/4. Options D and E are correct.

Learn  more about Pythagorean triplets here:

brainly.com/question/22160915

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