Perimeter of isosceles right triangle

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shibal: Master | Next Rank: 500 Posts; Posts: 345; Joined: Wed Mar 18, 2009 6:53 pm; Location: Sao Paulo-Brazil; Thanked: 12 times; GMAT Score:660

by shibal » Thu Jun 25, 2009 6:39 pm

perimeter of isosceles right triangle is 16 +16√2, what is hypotenuse?

oa 16

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Source: Beat The GMAT — Problem Solving | Thu Jun 25, 2009 6:39 pm

ssmiles08: Master | Next Rank: 500 Posts; Posts: 472; Joined: Sun Mar 29, 2009 6:54 pm; Thanked: 56 times

by ssmiles08 » Thu Jun 25, 2009 7:37 pm

its a 45-45-90 triangle. this perimeter works when one of the two sides is 8*sqrt(2)

hypotenuse = 16

8*sqrt(2) + 8*sqrt(2) + 16 = 16 + 16sqrt(2)

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shanmugam.d: Junior | Next Rank: 30 Posts; Posts: 29; Joined: Thu Apr 30, 2009 1:32 pm; Thanked: 2 times

by shanmugam.d » Thu Jun 25, 2009 11:39 pm

as ssmiles08 said yeah 45-45-90 rule applies here:
to put it another way: consider side of traingle as x
the perimeter of a traingle (45-45-90) is 2x + x√2 = 16 +16√2

so either 2x = 16 or 2x = 16√2. If 2x = 16 then x=8 -> this is not matching with the value of hypotenuse (i.e 16√2 not matching with 8√2)
Hence 2x = 16√2 -> x =8√2

so hypotenuse = 8√2 * √2 = 16

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nitya34: Master | Next Rank: 500 Posts; Posts: 179; Joined: Tue Jan 08, 2008 7:23 pm; Thanked: 11 times; GMAT Score:590

by nitya34 » Fri Jun 26, 2009 12:53 am

P=x+x+x(root 2)=2x+x(root2)=x(2+root 2)

Now given P is 8(root 2)(root 2 +2)

hence x=8(root 2)

hence Hypo is x(root 2)=16

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