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A isosceles right triangle has area of 12.5 then the length of the hyp

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A isosceles right triangle has area of 12.5 then the length of the hyp  [#permalink]

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New post 23 Aug 2018, 03:48
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C
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Re: A isosceles right triangle has area of 12.5 then the length of the hyp  [#permalink]

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New post 23 Aug 2018, 04:03
Bunuel wrote:
A isosceles right triangle has area of 12.5 then the length of the hypotenuse is


A. \(12.5\sqrt{2}\)

B. \(5\sqrt{3}\)

C. \(5\sqrt{2}\)

D. 5

E. \(\sqrt{2}\)


An isosceles right angled triangle is a 45-90-45 triangle, its sides are in the ratio x:\sqrt{x}:1

Given Area=12.5
Or, \(\frac{1}{2}*x*x=12.5\)
Or, \(x^2=25\)
Or, x=5

So, hypotenuse=\sqrt{2}x=[m]5\sqrt{2}

Ans. (c)
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Re: A isosceles right triangle has area of 12.5 then the length of the hyp  [#permalink]

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New post 23 Aug 2018, 07:39
Bunuel wrote:
A isosceles right triangle has area of 12.5 then the length of the hypotenuse is


A. \(12.5\sqrt{2}\)

B. \(5\sqrt{3}\)

C. \(5\sqrt{2}\)

D. 5

E. \(\sqrt{2}\)


\(\frac{1}{2}a^2 = \frac{25}{2}\) ( As \(12.5 = \frac{25}{2}\) )

So, we can safely conclude that \(a = 5\)

Now, Hypotenuse of the isosceles right triangle = \(5\sqrt{2}\) , Answer must be (C)
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A isosceles right triangle has area of 12.5 then the length of the hyp  [#permalink]

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New post 24 Aug 2018, 10:42
Bunuel wrote:
A isosceles right triangle has area of 12.5 then the length of the hypotenuse is


A. \(12.5\sqrt{2}\)

B. \(5\sqrt{3}\)

C. \(5\sqrt{2}\)

D. 5

E. \(\sqrt{2}\)

(1) Find side length from area
Area of right isosceles triangle*: \(\frac{s^2}{2}=12.5\)
\(s^2=25\)
\(s=5\)

(2) Hypotenuse length? Pythagorean theorem
\(s^2 +s^2 = h^2\)
\(25+25=h^2\)
\(h^2=50\)
\(\sqrt{h^2}=\sqrt{(25*2)}\)
\(h=5\sqrt{2}\)

Answer C

Faster: an isosceles right triangle is the special 45-45-90 triangle

The sides opposite those angles are in the ratio \(x: x: x\sqrt{2}\).
Find \(s\), which is opposite the 45° angle and hence corresponds with \(x\) -- then multiply by \(\sqrt{2}\)

From above, \(s=5\)
Hypotenuse, opposite the 90° angle, corresponds with \(x\sqrt{2}\)
\(x=5\)
Hypotenuse = \(5\sqrt{2}\)

Answer C


*The perpendicular legs of ANY right triangle are its base and height. In an isosceles right triangle, legs (non-hypotenuse sides) are equal.
Thus area of a right isosceles triangle is
\(A=\frac{b*h}{2}=\frac{s*s}{2}=\frac{s^2}{2}\)

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Re: A isosceles right triangle has area of 12.5 then the length of the hyp  [#permalink]

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New post 26 Aug 2018, 19:18
Bunuel wrote:
A isosceles right triangle has area of 12.5 then the length of the hypotenuse is


A. \(12.5\sqrt{2}\)

B. \(5\sqrt{3}\)

C. \(5\sqrt{2}\)

D. 5

E. \(\sqrt{2}\)


We can let n = the leg of the triangle and create the equation:

n^2/2 = 12.5

n^2 = 25

n = 5

So the hypotenuse is 5√2.

Answer: C
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Re: A isosceles right triangle has area of 12.5 then the length of the hyp   [#permalink] 26 Aug 2018, 19:18
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