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# Need Help With Math Problem: Triangles

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Intern
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Need Help With Math Problem: Triangles [#permalink]

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18 Nov 2010, 08:33
00:00

Difficulty:

35% (medium)

Question Stats:

50% (01:59) correct 50% (00:24) wrong based on 12 sessions

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The perimeter of a certain isoceles right triangle is 16 + (16 * Square Root (2)). What is the length of the hypotenuse.

8
4 * Square Root (2)
8 * Square Root (2)
16 * Square Root (2)

Thanks!
[Reveal] Spoiler: OA
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Re: Need Help With Math Problem: Triangles [#permalink]

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18 Nov 2010, 09:34
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ANS : B

lets say we have an isoceles triangle ABC with B = 90 degress and AB=BC=X and CA=Y.

now applying rt angle formula sqr Y = sqr X + sqr X = 2(sqr X).......................1
also, perimeter = 2X + Y = 16 + 16(sqrt of 2).................2

solve both the equations for Y.. in the end, we would get the equation something like:

Y + Y(sqrt 2) = 16 + 16(sqrt 2).

From this, its easily seen that Y=16.

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Re: Need Help With Math Problem: Triangles [#permalink]

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18 Nov 2010, 10:51
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Since we are dealing with isosceles triangle, we know that its 2 legs are equal.

The equation for its perimeter = hypotenuse + 2*leg.

and since isosceles right triangle is a 45 45 right triangle,

$$leg = \frac{hypotenuse}{\sqrt{2}}$$

which can be simplified to $$leg = \frac{hypotenuse}{2}\sqrt{2}$$

So the perimeter equation becomes

$$perimeter = hypotenuse + 2\frac{hypotenuse}{2}\sqrt{2}$$

$$perimeter = hypotenuse + hypotenuse\sqrt{2}$$

and since we are given

$$perimeter = 16 + 16\sqrt{2}$$

therefore hypotenuse = 16.
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Re: Need Help With Math Problem: Triangles [#permalink]

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18 Nov 2010, 17:02
Remember the equation for isoscelleous right triangle sides- x:x:x*sqrt(2)

Let one leg= a Thus,
2a+sqrt(2)*a= 16+16*sqrt(2)

2a+sqrt(2)*a= 16*sqrt(2)+16

thus a = 8 *sqrt(2)
And hypotenus= 16

Ans:- B
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Re: Need Help With Math Problem: Triangles [#permalink]

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22 Nov 2010, 14:45
messed up....
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Re: Need Help With Math Problem: Triangles   [#permalink] 22 Nov 2010, 14:45
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# Need Help With Math Problem: Triangles

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