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Area of a triangle - pretty tricky

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Area of a triangle - pretty tricky

by Brent@GMATPrepNow » Wed Jan 21, 2009 5:19 pm
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.
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by truplayer256 » Wed Jan 21, 2009 6:00 pm
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
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by Brent@GMATPrepNow » Wed Jan 21, 2009 6:09 pm
A it is. Nice solution.
I took a slightly different approach:

Let x and y be lengths of the two legs. We want the area of the triangle, so we want xy/2

First, we can conclude that the hypotenuse has length = sqrt(x^2 + y^2)
So, if perimeter=20, then x+y+ sqrt(x^2 + y^2)=20

If the sum of the squares of its sides=162 then x^2 + y^2 + [sqrt(x^2 + y^2)]^2 = 162
This simplifies to x^2 + y^2 + x^2 + y^2 = 162
Or 2x^2 + 2y^2 = 162
Or x^2 + y^2 = 81

Plug this information into the first (perimeter) equation to get x+y+ sqrt81=20
This tells us that x+y=11

We’re almost done.
If x+y=11, then (x+y)^2 = 11^2
This gives us x^2 + 2xy + y^2 = 121
But we already know that x^2 + y^2 = 81
So, 2xy=40, which means that xy=20 and xy/2=10
The area is 10 (answer choice A)
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by truplayer256 » Wed Jan 21, 2009 6:16 pm
yeah, this question was a tricky one indeed. It took me a couple minutes.
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by gaggleofgirls » Wed Jan 21, 2009 9:39 pm
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.

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by MBA.Aspirant » Fri Jul 15, 2011 5:23 am
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
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