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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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01 Jul 2018, 20:01
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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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01 Jul 2018, 20:13
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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01 Jul 2018, 21:04
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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02 Jul 2018, 16:52
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

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

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
$$a=\frac{s^2}{2}=\frac{(\frac{h^2}{2})}{2}=(\frac{h^2}{2}*\frac{1}{2})=\frac{h^2}{4}$$
$$a=\frac{h^2}{4}$$
$$4a=h^2$$

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

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02 Jul 2018, 19:54
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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04 Jul 2018, 18:06
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

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

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13 Aug 2019, 23:54
How do we know that the hypothenuse is not the base?
Re: The hypotenuse of an isosceles right triangle has a length of h, and   [#permalink] 13 Aug 2019, 23:54