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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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23 Aug 2018, 03:48
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74% (00:51) correct 26% (01:20) wrong based on 49 sessions

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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. $$\sqrt{2}$$

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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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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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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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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
$$s^2 +s^2 = h^2$$
$$25+25=h^2$$
$$h^2=50$$
$$\sqrt{h^2}=\sqrt{(25*2)}$$
$$h=5\sqrt{2}$$

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 = $$5\sqrt{2}$$

*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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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.

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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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