Triangle problem

[monteleone82]

Please help me with the problem below!!!

The perimeter of a certain isosceles right triangle is 16 + (16 * square root of 2). What is the length of the hypotenuse of the triangle?

Answer choices:

A. 8
B. 16
C. 4*square root of 2
D. 8*square root of 2
E. 16*square root of 2

[albatross86]

We have here an isosceles right triangle. There are a two things about this kind of triangle that you should absolutely know by-heart for the GMAT:

1. Such a triangle is a 45-45-90 degrees triangle.
2. Such a triangle has its sides in the ratio 1:1:√2

Here we are going to make use of the second property.

Let the sides be x, x and x√2.

Thus the perimeter = x + x + x√2 = x*(2 + √2) = 16 + 16√2 ( as given in the question stem )

Thus x = 16*(1 + √2) / (2 + √2)

Multiply and divide by (2 - √2) to remove the radical sign from the denominator.

x = 16*(1 + √2)*(2 - √2) / (2 + √2)*(2 - √2)

x = 16*(2 - √2 + 2√2 - 2) / (4 - 2)

x = 16*√2 / 2

=> x = 8√2

Thus the hypotenuse = x√2 = 8√2 * √2 = 8 * 2 = 16.... Ans: Pick B

NOTE: Another way to do this quickly is to realize that this perimeter of 16 + 16√2 needs to be in the form x + x + x*√2

Thus you can express it as 8√2 + 8√2 + (8√2) * √2 ... can you see how we did that?

[Rich@VeritasPrep]

Hey there,

When dealing with isosceles right triangles, most people know the x : x : x*sqrt(2) ratio.

However, there's also the x*sqrt(2) : x*sqrt(2) : 2x ratio.

In other words, if you have an isosceles right triangle, you can half the hypotenuse and multiply the result by sqrt(2) to get the leg length.

Examples include: sqrt(2) : sqrt(2) : 2 or 3*sqrt(2) : 3*sqrt(2) : 6.

It turns out that just such a ratio is necessary in this problem.

[monteleone82]

Rich,

Can you please elaborate? I am still a little confused with your response.

Thanks!

[Rich@VeritasPrep]

Sure, I'll use some concrete examples:

So for example, if we have an isosceles right triangle (i.e. a 45-45-90 triangle), and the two legs each have a length of 5, then we know that the hypotenuse must have a length of 5√2.

Whatever the length of the leg is, we just multiply it by √2 to get the length of the hypotenuse.

But there's nothing that says the length of the leg has to be an integer. For example, it could be 2√2. But the principle is still the same: multiply the length of the leg by √2 to get the hypotenuse. In this case, that would be 2√2 * √2 = 2*2 = 4.

So, if the length of the leg is 2√2, then the hypotenuse has length 4.
If the length of the leg were 3√2, then the hypotenuse would be 6.

In general, if the leg has a side length involving a √2 term, then drop the √2 term and double the remaining number.

In the case of this problem, the perimeter is 16 + 16√2. It looks as if 16√2 should be the hypotenuse, but if it were, then each leg would have length 16, and the perimeter would be 32 + 16√2.

Instead, 16√2 is actually the sum of the two leg lengths. This triangle has a leg length of 8√2 and a hypotenuse of 16.

So the big takeaway is that on the GMAT, the hypotenuse of a 45-45-90 triangle will not necessarily have a √2 term. That √2 might show up in the leg length instead.

So while you should definitely know the 1:1:√2 side-length ratio, you should also be prepared for √2 : √2: 2, as is the case in this problem.

Make sense?

[monteleone82]

Thanks Rich!!!

That makes perfect sense. I will be sure to look out for this type of problem on the exam. Thanks again!!

[rsvaishu]

Please help me with the problem below!!!

The perimeter of a certain isosceles right triangle is 16 + (16 * square root of 2). What is the length of the hypotenuse of the triangle?

Isosceles right triangle is 1:1:square root of 2

Perimeter of triangle = 16+16 square root of 2. ie., x+x+x sq of 2 = 16 +16 sq of 2

therefore x(1+1+ sq of 2) = 16(1+ sq of 2)

x = 16(1 + sq of 2) / (2 + 2 sq of 2)

Therefore x = 8 sq of 2

so x : x : x sq of 2 => 8 sq of 2 : 8 sq of 2 : 8 * 2( sq of 2 * sq of 2)

Thus hypotenuse is 16 ( Choice B)