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The hypotenuse of an isosceles right triangle has a length of h, and

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The hypotenuse of an isosceles right triangle has a length of h, and  [#permalink]

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New post 01 Jul 2018, 21:01
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The hypotenuse of an isosceles right triangle has a length of h, and the triangle has an area of a. Which of the following must be true?


A) \(a = 4 * h^2\)

B) \(a = 2 * h^2\)

C) \(a = h^2\)

D) \(4a = h^2\)

E) \(2a = h^2\)

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Re: The hypotenuse of an isosceles right triangle has a length of h, and  [#permalink]

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New post 01 Jul 2018, 21:13
Sides will be \(h/\sqrt{2}\). Area a = \((1/2) * (h/\sqrt{2}) * (h/\sqrt{2})\) = \(h^2/4\).

Hence \(4a = h^2\), Option D.

Similar post I found: https://gmatclub.com/forum/in-a-right-i ... fl=similar
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Re: The hypotenuse of an isosceles right triangle has a length of h, and  [#permalink]

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New post 01 Jul 2018, 22:04
Bunuel wrote:
The hypotenuse of an isosceles right triangle has a length of h, and the triangle has an area of a. Which of the following must be true?


A) \(a = 4 * h^2\)

B) \(a = 2 * h^2\)

C) \(a = h^2\)

D) \(4a = h^2\)

E) \(2a = h^2\)


Its a 45-45-90 Triangle. Therefore sides will be in the ratio 1:1:squareroot2
Hence the other 2 sides will be h/squareroot2

Area of Triangle is 1/2*h/squareroot2*h/squareroot2=a
=> 4a=h^2.

Option D
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The hypotenuse of an isosceles right triangle has a length of h, and  [#permalink]

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New post 02 Jul 2018, 17:52
1
Bunuel wrote:
The hypotenuse of an isosceles right triangle has a length of h, and the triangle has an area of a. Which of the following must be true?


A) \(a = 4 * h^2\)

B) \(a = 2 * h^2\)

C) \(a = h^2\)

D) \(4a = h^2\)

E) \(2a = h^2\)

Area of isosceles right triangle, \(a=\frac{s^2}{2}\)
\(2a=s^2\)

Hypotenuse:
\(h^2=2s^2\)
\(\frac{h^2}{2}=s^2\)
Substitute for \(s^2\) in area formula:
\(2a=\frac{h^2}{2}\)

\(4a=h^2\)

OR

\(a=\frac{s^2}{2}=\frac{(\frac{h^2}{2})}{2}=\frac{h^2}{4}\)
\(a=\frac{h^2}{4}\)
\(4a=h^2\)

Answer D

You do not need any of those formulas. Use the Pythagorean theorem, and solve for \(s^2\) (i.e., don't solve for \(s\) - it gets squared in area of triangle)

In an isosceles right triangle, sides are equal.
\(s=s\)
(1) Area
Area of triangle, \(a=\frac{b*h}{2}\)
\(a=\frac{s*s}{2}=\frac{s^2}{2}\)

(2) Find hypotenuse in terms of \(s^2\):
\(s^2+s^2=h^2\)
\(2s^2=h^2\)
\(s^2=\frac{h^2}{2}\)

(3) Substitute and solve
\(a=\frac{s^2}{2}=\frac{(\frac{h^2}{2})}{2}=(\frac{h^2}{2}*\frac{1}{2})=\frac{h^2}{4}\)
\(a=\frac{h^2}{4}\)
\(4a=h^2\)

Answer D
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The hypotenuse of an isosceles right triangle has a length of h, and  [#permalink]

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New post 02 Jul 2018, 20:54
OA: D

as it is \(45^{\circ}-45^{\circ}-90^{\circ}\) triangle , its side will be in ratio \(1:1:\sqrt{2}\)

Hypotenuse = \(h\)
Base = Height =\(\frac{h}{\sqrt{2}}\)

Area,\(a = \frac{1}{2}*\frac{h}{\sqrt{2}}*\frac{h}{\sqrt{2}}\)
\(4a=h^2\)
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Re: The hypotenuse of an isosceles right triangle has a length of h, and  [#permalink]

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New post 04 Jul 2018, 19:06
Bunuel wrote:
The hypotenuse of an isosceles right triangle has a length of h, and the triangle has an area of a. Which of the following must be true?


A) \(a = 4 * h^2\)

B) \(a = 2 * h^2\)

C) \(a = h^2\)

D) \(4a = h^2\)

E) \(2a = h^2\)



An isosceles right triangle has angle measures of 45-45-90. Thus, the two legs are of equal length.

Let x = a leg of the triangle. Using the Pythagorean theorem, we can create the equation:

x^2 + x^2 = h^2

2x^2 = h^2

x^2 = h^2/2

x = h/√2

Recall that in an isosceles right triangle, one leg is the base, and the other leg is the height. Thus, the area of the triangle is ½ x^2, and we have:

½ x^2 = a

½(h/√2)^2 = a

½ (h^2/2) = a

h^2/4 = a

h^2 = 4a

Answer: D
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Re: The hypotenuse of an isosceles right triangle has a length of h, and &nbs [#permalink] 04 Jul 2018, 19:06
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