Area of a triangle - pretty tricky

Since it's a right triangle, we must follow the phythagorean theorem..

Let one side=x, another side=y, and the hypotenuse=z

x^2+y^2=z^2

by rearranging terms, we can see that x^2+y^2=162-z^2

so,

162-z^2=z^2

solving for z, we get z=+/-9, we can ignore the minus and assume that z=9..

we can then plug z into our two equations:

x+y+z=20
x^2+y^2+z^2=162

and find out that xy=20, so the area of the triangle would be xy/2 or 20/2=10
A

yeah, this question was a tricky one indeed. It took me a couple minutes.

If we label the 3 sides of the Triangle a, b, and c so that C is the hypot and a is the height, then this is what we know about the triangle...

a + b + c = 20
a^2 + b^2 + c^2 = 162
also, since is it a right triangle, we know that a^2 = b^2 = c^2

And what we want to solve is A = 1/2 ba = ?

So c^2 + c^2 = 162
2 * c^2 = 162
c^2 = 81 (good sign that it is a perfect square)
c = 9

a + b + 9 = 20
a + b = 11

With all these + signs and ^2, it is hard to see how we will get ba (or ab)...

So, we sqaure both sides of a + b = 11 and we get (a + b)^2 = 11^2
a^2 + 2ab + b^2 = 121 (good sign that we now have ab)
We know that a^2 + b^2 = c^2 and that c^2 = 81 so a^2 + b^2 = 81 so the equation can be rewritten to:
a^2 + b^2 + 2ab = 121
81 + 2ab = 121
2 ab = 40
ab = 20

Therefore the Area = 1/2 ab = 1/2 * 20 = 10

Answer = A

Now, my problem is getting to that in 2 minutes because that took me a lot more than 2 min to solve.

-Carrie

Brent@GMATPrepNow wrote:A right triangle has perimeter 20. If the sum of the squares of the triangle�s three sides is 162, what is the area of the triangle?
(A) 10
(B) 11.5
(C) 12
(D) 13.5
(E) 15

Please note that this is not an official GMAT question; it is my attempt to create difficult (650+ level) GMAT-style questions for this forum.

x+y+z = 20

x^2 + y^2 + z^2 = 162

2z^2 = 162

z= 9

x+y = 11

(x+y)^2 = x^2+y^2 + 2xy

121 = 81 + 2xy

40 = 2xy

xy = 20

area = 1/2 xy = 10