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Isosceles Triangle problem

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by kshin78 » Sat Aug 02, 2008 1:15 pm
Hi all, I can't seem to solve this prob...please help.

Perimeter of isosceles triangle is 16+16sqroot2. what is the length of hypotenuse? Please provide how you derived the answer.

I started with 2x+(x sqr2) = 16+16sqr2 from here, i have no way of getting the hypotenuse..

Do i square both sides? is there a faster way to solve this problem?

Thanks!
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by sibbineni » Sun Aug 03, 2008 8:16 am
Perimeter of isosceles triangle is 16+16sqroot2

2x+(x sqr2) = 16+16sqr2

take x sqr2 as common from LHS then we have x sqr2(sqr2+1)
take 16 as common from RHS then we have 16(sqr2+1)

x sqr2(sqr2+1)=16(sqr2+1)

x sqr2=16

x=16/sqr2

therefore hypotenuse =x sqr2
substitute the value of x =16/sqr2 in above we have
==>(16/sqr2)*sqr2
==>16
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by kshin78 » Sun Aug 03, 2008 12:57 pm
thank you so much for the answer.

I see that you factored out sqr2 from the right side and 16 from the left to get the both sides in the similar format.

How did you know to do that? Looks easy when you explained it, but my problem is during the questions, i can't seem to think that one through...is there a tip for these kind of problems?
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by jeffedwards » Tue Dec 27, 2011 9:16 am
If you get lost in the details, just plug in the answers.

For 16:

x(sqr2) = 16
x = 16/(sqr2)

Now plug in 16/(sqr2) for x into your original formula. If the numbers match that's the right choice
2x + x(sqr2) = Original Formula
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by LalaB » Tue Dec 27, 2011 10:14 am
H+2legs=16+16sqroot2

so H is either 16, or 16sqroot2

now remember (45:45:90) x : x : xsqroot2 (where xsqroot2=hypotenuse)

so, if xsqroot2=16sqroot2, then x=16
so, the perimeter is equal to 16+16+16sqroot2=32+16sqroot2

of xsqroot2=16,then x=8sqroot2

then the perimeter will be 8sqroot2+8sqroot2+16=16sqroot2+16
bingo :)
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