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Geometry Question

Problem Solving — algebra and arithmetic
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by Emeka N. » Sat Apr 16, 2016 5:09 pm
What's the most efficient way of approaching this problem?
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by MartyMurray » Sat Apr 16, 2016 5:41 pm
The ratios of the measurements of an isosceles right triangle can be thought of in two main ways.

Side Side Hypotenuse

..1.....1......√2

√2/2 √2/2...1

Looking at the second way, you can see that each side is 1/2 x length of the hypotenuse x √2.

Compare that relationship to the numbers given, 16√2 and 16.

If you divide 16√2 in half you get 16√2/2, which is 1/2 x 16 x √2.

So 16√2/2 16√2/2 and 16 work as the sides and hypotenuse of an isosceles right triangle.

The perimeter of such a triangle is 16√2/2 + 16√2/2 + 16 = 16√2 + 16.

So the length of the hypotenuse of a right isosceles triangle the perimeter of which measures 16√2 + 16 is 16.

The efficiency of this method comes from knowing the ratios of the sides and hypotenuse of an isosceles right triangle and being able to recognize how to apply them.
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by Brent@GMATPrepNow » Sat Apr 16, 2016 8:49 pm
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

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by DavidG@VeritasPrep » Sun Apr 17, 2016 3:23 am
You can also backsolve:

If the hypotenuse were 16√2, the sides of the triangle would be 16, and the perimeter would be 16√2 + 16 + 16 = 16√2 + 32. That's too big.

If the hypotenuse were 8√2, the sides of the triangle would be 8, and the perimeter would be 8√2 + 8 + 8 = 8√2 + 16. That's too small.

So the answer has to be between 16√2 and 8√2. The only possibility is 16.
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by [email protected] » Sun Apr 17, 2016 10:29 am
Hi Emeka N.,

Here's a larger discussion on this question, including other ways to solve it and some noteworthy aspects about how the prompt is written:

https://www.beatthegmat.com/hypotenuse-o ... 76892.html

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