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If the hypotenuse of an isosceles right triangle is

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If the hypotenuse of an isosceles right triangle is \(8\sqrt{2}\) , what is the area of the triangle?

(A) 18
(B) 24
(C) 32
(D) 48
(E) 64
[Reveal] Spoiler: OA

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Gnpth wrote:
If the hypotenuse of an isosceles right triangle is \(8\sqrt{2}\) , what is the area of the triangle?

(A) 18
(B) 24
(C) 32
(D) 48
(E) 64

An isosceles right triangle has angle measures of 45-45-90, and sides in ratio \(x: x: x\sqrt{2}\)

The hypotenuse = \(8\sqrt{2}\)
The legs' length is \(8\).*

And area is \(\frac{8*8}{2} = 32\)

ANSWER C

*If it isn't clear, from the ratio of sides, that side length = x = 8, or if you don't recognize that triangle, you can use the Pythagorean theorem to find side length (then area). It is a right triangle, and its sides are equal.

\(x^2 + x^2 = (8\sqrt{2})^{2}\)
\(2x^2 = (64)(2)\)
\(x^2 = 64\)
\(x = 8\)
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If the hypotenuse of an isosceles right triangle is   [#permalink] 22 Sep 2017, 16:13
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