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triangles

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Senior | Next Rank: 100 Posts
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by hpgmat » Fri Nov 13, 2009 9:51 pm
perimeter of a certain isolesces right triangle is 16+16 x (square root of 2) . what is the length of hypothesnues.

8
16
4 x squre root of 2
8 x squre root of 2
10 x square root of 2

please show calcs
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Re: triangles

by Abdulla » Sat Nov 14, 2009 12:11 am
hpgmat wrote:perimeter of a certain isolesces right triangle is 16+16 x (square root of 2) . what is the length of hypothesnues.

8
16
4 x squre root of 2
8 x squre root of 2
10 x square root of 2

please show calcs
IMO B

since it's isolesces RIGHT triangle then it's 45:45:90 and its sides x : x : x sqrt2
so, x+x+ x sqrt2 =16+16 sqrt2
x(2+sqrt2)=16+16 sqrt2
x=16+16 sqrt /(2+ sqrt2) now, to get rid of the denominator multiply by (2-sqrt2)/(2-sqrt2)
x= 32+32 sqrt2-32-16sqrt2/ (4-2)
x= 16 sqrt2/2
x= 8 sqrt, now , we know that the hypothesnues = x sqrt2
Therefore, 8 sqrt2 * sqrt2 = 8 *2 = 16.
Abdulla
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