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Hypotenuse Calculation

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Junior | Next Rank: 30 Posts
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by kindofbluenote » Thu Jun 05, 2008 3:00 am
Perimiter of an isocoles right triangle is 16+16(sqrt)2. What is the hypotenuse?

A. 8
B. 16
C. 4( sqrt)2
D. 8 (sqrt)2
E. 16(sqrt)2

OA is B

This was from the GMAT prep CAT, and it's a great example of a question that I thought looked easy, but has me pulling what little hair I have left out trying to figure out...Any insight will be greatly apreciated.
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by rra » Thu Jun 05, 2008 5:28 am
Perimeter = 16 + 16(root)2
Perimeter Isoceles Right Triangle = 2a + b
where a is the length of perpendicular sides
and b is hypotenous

Since, area is 2a^2 = b^2
And, b = a(root)2

Therefore, 2a + a(root)2 = 16 + 16(root)2
=> a = [16 + 16(root)2]/[2 + (root)2]

Multiplying the equation by [2 - (root)2]

= 16(root)2/2

Therefore, a = 8(root)2

Hence, b = 16

:D
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by reetchatha » Thu Jun 05, 2008 11:04 am
This problem can be tackled easily by substitution. When you are stumped follow this approach start with c and based upon the result of substitution move upwards or downwards.In this case substitute 16 i.e. xsqrt2 = 16 since we know it is 45-45-90 triangle.
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