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Tricky Right Triangle

Problem Solving — algebra and arithmetic (GMAT Focus Edition)
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Tricky Right Triangle

by skalevar » Sun Oct 10, 2010 8:04 am
If a right triangle has area 28 and hypotenuse 12, what is its perimeter?

A. 20
B. 24
C. 28
D. 32
E. 36

This seems not too difficult, but I was stumped when working through the algebra. Can someone help?

Thanks,

OA is C
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by shovan85 » Sun Oct 10, 2010 8:31 am
Let b be base h be height

So (1/2) * b * h = 28 (Area)

Hypo be H then H = 12

b^2 + h ^ 2 = 12^2 = 144

and bh = 56 (from area)
Thus (b+h)^2 = b^2 + h ^ 2+ 2bh = 144 + 2*56 = 256

Then b+h = 16 and we know H = 12
So Perimeter = b+h+H = 28.
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by skalevar » Sun Oct 10, 2010 8:39 am
Thanks.

So that I'm clear - how do you know to use that (b+h)^2 trick? it doesn't seem like an obvious method unless you already know the answer.
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by shovan85 » Sun Oct 10, 2010 8:45 am
skalevar wrote:Thanks.

So that I'm clear - how do you know to use that (b+h)^2 trick? it doesn't seem like an obvious method unless you already know the answer.
When ever you see a right angle triangle the first thing strikes to your brain is b^2 + h^2 = H^2.

Is n't it?

What else do u have? Area. Formula is bh/2.

You have two options either solve 2 variables 2 eqns... possible. Or have a look properly before u jump into solving. I was about to solve the two equations but suddenly (b+h)^2 formula clicked. But I had no idea that I can solve it so fast by using (b+h)^2 formula :)
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by uwhusky » Sun Oct 10, 2010 9:01 am
skalevar wrote:Thanks.

So that I'm clear - how do you know to use that (b+h)^2 trick? it doesn't seem like an obvious method unless you already know the answer.
Practice. Knowing how to approach this question is knowing how to approach a SC question.

If you have never seen an absolute phrase modifier, you probably won't be able to recognize the correct structure of one. But after you have seen one, then absolute phrase modifiers are much easier to spot and solve.
Yep.
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by [email protected] » Thu Oct 14, 2010 5:21 pm
Can yo please demonstrate how to solve this question using two equation?

Thanks,

Arshi
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by Tani » Fri Oct 15, 2010 2:48 pm
pythagorean theorem: a^2 + b^2 = 144

area of a triangle: 1/2 (ab) = 28

therefore ab = 56

therefore a^2 + 2ab + b^2 = 144 + 2(56) = 256

factor the left side (a+b)^2 = 256

square root of both sides (a+b) = 16

perimeter = 12 + 16 = 28

You don't have to know the lengths of the two sides, just their total.
Tani Wolff
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by shovan85 » Sat Oct 16, 2010 12:22 pm
[email protected] wrote:Can yo please demonstrate how to solve this question using two equation?

Thanks,

Arshi
It is possible but quite impossible to solve in GMAT. It will eat most of your time so better not try ;)
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by Tani » Sat Oct 16, 2010 12:32 pm
always assume there is a shortcut. The use of area and perimeter in one problem suggested that somehow you could combine the pythagorean theorem with the area of a triangle. The result took less than 2 minutes.
Tani Wolff
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