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Triangle Question

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by brood1989 » Sat Apr 16, 2011 10:58 am
I apologize if someone has already posted this question, I tried to search. The question is from
the GMAT Practice Test and states that the perimeter of an equilateral triangle is 16 + 16xsquare root of 2. What is the length of the hypotenuse. I'm a little thrown of since I know that x + x + x(times square root of 2) would be the perimeter. Suggestions on how to solve this?
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by maihuna » Sat Apr 16, 2011 11:13 am
can u type the whole q, what is hypot.. here...
brood1989 wrote:I apologize if someone has already posted this question, I tried to search. The question is from
the GMAT Practice Test and states that the perimeter of an equilateral triangle is 16 + 16xsquare root of 2. What is the length of the hypotenuse. I'm a little thrown of since I know that x + x + x(times square root of 2) would be the perimeter. Suggestions on how to solve this?
Charged up again to beat the beast :)
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by brood1989 » Sat Apr 16, 2011 11:28 am
Sorry I forgot to put a question mark. The question
is asking for the hypotenuse, given that the perimeter
of an iso right triangle is 16 + 16*square root of 2.
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by arunpanda22 » Sat Apr 16, 2011 7:22 pm
my solution is as follows
isosceles rt triangle has two equal sides call it 'a' then hypotenuse a sq root 2
perimeter = 2a+a sq root 2
16+16 root 2= 2a+ a root2
16(1+root2)= a root2(1+root2)
a root2=16
a=8 root2
so hypotnuese is a*root2=8 root2*root2=16
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by brood1989 » Sat Apr 16, 2011 7:38 pm
Little lost on this step: 16(1+root2)= a root2(1+root2)
I see how you factor out the 16, but I'm not seeing
how you factor the a root 2 from the equation 2a + a root 2.
Isn't the only common factor here an a?
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by manpsingh87 » Sat Apr 16, 2011 8:54 pm
brood1989 wrote:Little lost on this step: 16(1+root2)= a root2(1+root2)
I see how you factor out the 16, but I'm not seeing
how you factor the a root 2 from the equation 2a + a root 2.
Isn't the only common factor here an a?
16(1+sqrt(2)=(2+sqrt(2))a;
a=16(1+sqrt(2))/(2+sqrt(2));
now multiply numerator and denominator with 2-sqrt(2);
a=16(1+sqrt(2))*(2-sqrt(2)/(2+sqrt(2))*(2-sqrt(2));
denominator is of the form (a+b)(a-b)=a^2-b^2;
hence we have 16(1+sqrt(2))(2-sqrt(2)/4-2,
now upon solving this we have a=8sqrt(2); and hypotenuse = asqrt(2),=8sqrt(2)*sqrt(2)=16

i hope it helps..!!!
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by arunpanda22 » Sat Apr 16, 2011 10:07 pm
brood1989 wrote:Little lost on this step: 16(1+root2)= a root2(1+root2)
I see how you factor out the 16, but I'm not seeing
how you factor the a root 2 from the equation 2a + a root 2.
Isn't the only common factor here an a?
buddy root2*root2=2
so if u split 2 then u get (root2)^2
hence equation becomes (root2)^2*a+root2*a
take out root2*a as a factor
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by Geva@EconomistGMAT » Sat Apr 16, 2011 10:32 pm
brood1989 wrote:I apologize if someone has already posted this question, I tried to search. The question is from
the GMAT Practice Test and states that the perimeter of an equilateral triangle is 16 + 16xsquare root of 2. What is the length of the hypotenuse. I'm a little thrown of since I know that x + x + x(times square root of 2) would be the perimeter. Suggestions on how to solve this?
The concept needed for this problem is the common ratio of x:x:x√2 between the two legs and the hypotenuse of any right isosceles triangle. This means that if the legs of the triangle are x, then the hypotenuse is x times √2; the flips side of this ratio is that you can move from the hypotenuse back to the leg by dividing by √2.

the question is best solved by reverse Plugging in - plug in the answers back into the question (Which is why we always post a question with its answer choices). Assume that each of the answer choices is the hypotenuse, use the properties of right isosceles above to find the two legs, and see if the resulting perimeter is equal to the question's 16+16√2

The answer choices here went something like 8, 16, 4√2, 8√2, 16√2. The last three are the easiest to plug in, but also the easiest to eliminate: if the hypotenuse is 16√2, then each of the legs is simply 16 (divide by √2), but then the perimeter is 16+16+16√2 = 32+16√2 - too big.
C and D will also not work (try it).
B 16: Divide 16 by √2 to get the leg: 16/√2 is equal to 8√2 (multiply both top and bottom by √2, to get rid of the root in the bottom, then reduce)
So the perimeter of the shape is leg+leg+hypotenuse = 8√2+8√2+16 = 16+16√2.
Last edited by Geva@EconomistGMAT on Sun Apr 17, 2011 7:28 am, edited 1 time in total.
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by Geva@EconomistGMAT » Sat Apr 16, 2011 10:38 pm
Before anyone further asks: A generic method for getting rid of roots in the denominator of a fraction (such as 16/√2).

1) Multiply the fraction by the same root as the denominator in both top and bottom. You can do this manipulation - it's the equivalent of multiplying the fraction by √2/√2 = 1, so the fraction does not change its value in the process. The current version of the fraction is (16/√2) * (√2/√2) = 16√2 / √2√2.

2) Combine the roots in the denominator into a nice integer: √2√2 is simply 2. Current version of the fraction is 16√2 / 2.

3) Reduce like terms in top and bottom: 16/2 is 8, leaving 8√2.

As an exercise, Try it for 15/√3. Remember to multiply and divide by √3 this time - the goal is to ge rid of the root in the denominator by multiplying the root by itself.
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by brood1989 » Sun Apr 17, 2011 7:26 am
Thank you, I apologize for not posting the answer choices.
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