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The perimeter of a certain isosceles right triangle is 16 +

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The perimeter of a certain isosceles right triangle is 16 + [#permalink] New post 12 Sep 2005, 18:50
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The perimeter of a certain isosceles right triangle is 16 + 16 x (root2). What is the length of the hypotenuse of the triangle?



a) 8

b) 16

c) 4(root2)

d) 8(root2)

e) 16(root2)


The correct answer is b - 16. Does anyone know how to solve this problem? Thanks
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 [#permalink] New post 12 Sep 2005, 19:15
Since the triangle is isoceles and right, then we know that the two legs are equal in length and we also know that the ratio of the sides is 1:1:(rt2).

This means that, if we let x=the length of one of the legs, we know that:

x + x + (rt2)x = 16+16(rt2)
(2+(rt2))x = 16(1+rt2)
x = 16(1+rt2)/(2+rt2)
x = 16/(rt2)

Then just multiply by (rt2) to get the length of the hypotenuse, which is 16.
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 [#permalink] New post 12 Sep 2005, 19:20
By the way, since when did I become a "senior manager" around here. I just joined!
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 [#permalink] New post 13 Sep 2005, 05:52
hey coffeeloverfreak how did you determine (1+sqrt2)/(2+sqrt2) to equal 1/sqrt2? my simple math has gone to crap lately
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 [#permalink] New post 13 Sep 2005, 06:45
(1+rt2)/(2+rt2)
= (rt2)(1+rt2)/ (rt2)(2+rt2) - multiplying by rt2/rt2 is like multiplying by 1.
= rt2+2/2rt2+2
= 1(rt2+2)/rt2(2+rt2) - then cancel the like term (rt2+2)
=1/rt2
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 [#permalink] New post 13 Sep 2005, 06:52
Yep. it's 16.

P= x+x+x*sqrt2

2x+x*sqrt2 = 16+16*sqrt2
x*(2+sqrt2) = 8*sqrt2(sqrt2+2)
x=8*sqrt2
H=x*sqrt2=8*sqrt2*sqrt2=16
  [#permalink] 13 Sep 2005, 06:52
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