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GMAT Prep Question

by okigbo » Fri Sep 11, 2009 6:41 pm
The perimeter of a certain isosceles right triangle is 16+16sq.rt2. What is the length of the hypotenuse of the triangle?

a. 8
b. 16
c. 4sq.rt2
d. 8sq.rt2
e. 16sq.rt2

I know that the equation is x+x+x(sq.rt.2)=16+16sq.rt2

But the math gets too complicated and I cant see an end in sight at the 90 sec mark. Can someone shed some light on how to tackle this problem?

Thanks.
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by DanaJ » Sat Sep 12, 2009 12:17 am
You have correctly written the equation.

2x + x*(sqrt2) = 16 + 16*(sqrt2)

x[2 + (sqrt2)] = 16[1 + (sqrt2)] --- divide both sides by 2 + (sqrt2) to get that x will be the following fraction:

16[1 + (sqrt2)]
---------------
2 + (sqrt2)

Now, what we do when we have a square root at the denominator is we multiply both the denominator and the numerator with the denominator's "opposite". In this case, 2 + (sqrt2) will have 2 - (sqrt2) as its opposite.

So the above fraction will be equal to:

16[1 + (sqrt2)][2 - (sqrt2)]
----------------------------
[2 + (sqrt2)][2 - (sqrt2)]

You may be wondering why we would do such a thing. Well, fact is we eliminate the square root from the denominator:
[2 + (sqrt2)][2 - (sqrt2)] = 2^2 - (sqrt2)^2 = 2
This happens because of the very important formula of (a + b)(a - b) = a^2 - b^2.

You can also do the calculations of the numerator:
16[1 + (sqrt2)][2 - (sqrt2)] = 16[2 - (sqrt2) + 2(sqrt2) - 2] = 16*(sqrt2)


So in the very end your fraction will be:

16(sqrt2)
--------
2

Which will of course be 8(sqrt2).
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