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
Triangle Problem Help
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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.
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.
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.
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.
Last edited by mbadreams on Mon Oct 05, 2009 9:17 am, edited 1 time in total.
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.
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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