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Senior | Next Rank: 100 Posts
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by gmat8000 » Sun Oct 21, 2007 5:14 pm
The perimeter of a certain isosceles right triangle is 16 + 16*sqrt2. What is the length of the hypoteneuse of the triangle?

a) 8
b) 16
c) 4*sqrt2
d) 8*sqrt2
e) 16*sqrt2

The answer is b. Please explain. Thanks![/list]
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by jangojess » Sun Oct 21, 2007 7:29 pm
perimeter = 2x + x rt(2) = 16 + 16 rt(2)
now 2x can be 16 or 16 rt(2)...if 2x=16 then x=8 and x rt(2) = 8 rt(2)..so this cant be possible...if 2x=16 rt(2) then x=8 rt(2) and x rt(2) = 16...bingo!!!!...........So the ans is B
Trying hard!!!
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Thanks

by gmat8000 » Sun Oct 21, 2007 9:19 pm
Would you consider this to be a level 600 or 700 problem? Or worse, even lower?
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by ssy » Mon Oct 22, 2007 5:02 am
The way I would solve it is:

As the triangle is an isoceles, two sides are equal length. Let a be the length of the two equal sides and b be the hypotenuse.

So, 2a+b= 16 + 16*sqrt2

Invoking the rule that the length of any 2 sides of a triangle is greater than the length of the third side, 2a > b.

As 16*sqrt2>16, 2a = 16 sqrt2, b=16.

B=hypotenuse=16.
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by wongee » Mon Oct 22, 2007 6:43 pm
OR:

Longer method but flowed well for me:

x + x + xrt2=16+16rt2

2x+xrt2=16(1+rt2)

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

Now u need to find the hypotenuse which is xrt2, multiplying both side by rt2

xrt2=16rt2 (1+rt2)/2+rt2

= 32 +16rt2/(2+rt2)

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

= 16
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