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EMAN · Sun Oct 04, 2009 4:16 pm

The hypotenuse of a right triangle is 10 cm. What is the perimeter, in cm, of the triangle?

(1) The area of the triangle is 25 cm squared
(2) The 2 legs of the triangle are equal in length

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pullabhotla · Mon Oct 05, 2009 4:26 am

given hypotenuse, h = 10 cm.

Let's say sides be a and b.
We have to find out the perimeter = a + b + h.

Statement (1) gives us (1/2) * a * b = 25.
So we cannot solve for a and b. therefore insufficient.

Statement (2) gives us a = b.
As the triangle is right angled. asq2 + bsq2 = hsq2
gives us a = b = sqrt(5). therefore sufficient.

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mbadreams · Mon Oct 05, 2009 8:30 am

Let a be the base, b the height and and h the hypotenuse
Then we know h = 10
Also a^2 + b^2 = 100.
Statement 1 says ½ * a * b = 25  ab = 50
Therefore (a+b)^2 = a^2 + b^2 + 2ab = 100 + 2*50 = 200
Or a + b = √200 = 10√2 . Therefore perimeter = 10 + 10√2 . Hence sufficient.

Statement 2 says both sides are equal. Or a=b = 5√2. Therefore perimeter = 10√2 + 10. Hence sufficient.
Answer D.

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mp2437 · Mon Oct 05, 2009 9:19 am

Agree with choice D as mbadreams, however, he messed up the algebra a bit.

Statement 1: (a+b)^2 = a^2 + b^2 + 2ab = 100 + 2(50) = 200
(a+b)^2 = 200 --> a+b = sqrt(200) = 10*sqrt(2)

Perimiter = a + b + h = 10*sqrt(2) + 10.

Statement 2: 45-45-90 triangle. Each leg is therefore 5*sqrt(2).

Perimeter = 5*sqrt(2) + 5*sqrt(2) + 10 = 10*sqrt(2) + 10

You arrive at choice D.

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mbadreams · Mon Oct 05, 2009 10:16 am

Thanks mp2437... I think I messed up the calculation in both... the choices.. Thanks so much for noting my mistake... I really need to watch out for those...will edit my reply..

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artistocrat · Wed Oct 14, 2009 12:41 pm

(1)SUFF: 2 equations 2 unknowns
(2)SUFF: 1 equation 1 unknown

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okigbo · Fri Oct 16, 2009 10:54 pm

Guys, may be a dumb question but I don't see how a^2+b^2=c^2 is the same as (a+b)^2=c^2. Can someone please explain this to me?

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