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GMATPrep Test 2 Q8

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Abhijit K Master | Next Rank: 500 Posts Default Avatar
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GMATPrep Test 2 Q8

Post Sun Feb 15, 2015 8:03 am
The perimeter of a certain isosceles right triangle is 16+16root2. What is the length of the hypotenuse of the triangle?
A.8
B.16
C.4root2
D.8root2
E.16root2

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GMAT/MBA Expert

Post Sun Feb 15, 2015 8:10 am
Quote:
The perimeter of a certain isosceles right triangle is 16 + 16√2. What is the length of the hypotenuse?
a. 8
b. 16
c. 4√2
d. 8√2
e. 16√2
The sides of an isosceles right triangle are in the following ratio: s : s : s√2.
So if s=side and h=hypotenuse, then h = s√2 and s = h/√2.

We can plug in the answers, which represent the length of the hypotenuse.

Answer choice C: h = 4√2
s = (4√2)/√2 = 4.
p = 4 + 4 + 4√2 = 8 + 4√2.
Eliminate C. The perimeter needs to be quite a bit larger.

Answer choice B: h = 16
s = 16/√2 = (16*√2)/(√2*√2) = (16√2)/2 = 8√2.
p = 8√2 + 8√2 + 16 = 16 + 16√2. Success!

The correct answer is B.

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GMAT/MBA Expert

Post Sun Feb 15, 2015 8:25 am
Quote:
The Perimeter of a certain isosceles right triangle is 16 + 16√2. What is the length of the hypotenuse of the triangle?

A) 8
B) 16
C) 4√2
D) 8√2
E) 16√2
An IMPORTANT point to remember is that, in any isosceles right triangle, the sides have length x, x, and x√2 for some positive value of x.

Note: x√2 is the length of the hypotenuse, so our goal is to find the value of x√2

From here, we can see that the perimeter will be x + x + x√2

In the question, the perimeter is 16 + 16√2, so we can create the following equation:
x + x + x√2 = 16 + 16√2,
Simplify: 2x + x√2 = 16 + 16√2
IMPORTANT: Factor x√2 from the left side to get : x√2(√2 + 1) = 16 + 16√2
Now factor 16 from the right side to get: x√2(√2 + 1) = 16(1 + √2)
Divide both sides by (1 + √2) to get: x√2 = 16

Answer = B

Cheers,
Brent

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Matt@VeritasPrep GMAT Instructor
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Post Sun Feb 15, 2015 6:17 pm
We'll start with x + x + x√2 = 16 + 16√2.

We know that ONE of these numbers (16 or 16√2) is the hypotenuse, and the other is the sum of the two legs. If 2x = 16, then x = 8, and x√2 = 8√2. That doesn't match our numbers, so it can't work.

Hence the OTHER number, 16√2, must be the sum of the two legs. This gives us legs of 8√2, and a hypotenuse of 8√2(√2), or 16. Success!

If we wanted to do this algebraically, we could. Let x = one of the legs, and x√2 = the hypotenuse.

2x + √2x = 16 + 16√2
x * (2 + √2) = 16 + 16√2
x = (16 + 16√2)/(2 + √2)
x = ((16 + 16√2) * (2 - √2)) / ((2+√2)(2-√2))
x = (16√2)/2
x = 8√2

So our legs are each 8√2, and our hypotenuse is 16.

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