If the perimeter of an isosceles right triangle is 16 + 16 2 inches lo

The length the sides of an isosceles right triangle look like this:


So the given information is telling us that: \(x + x + x\sqrt{2} = 16 + 16 \sqrt{2}\)

At this point, it's probably fastest to start testing some answer choices.
Notice that if \(x=8\), we get: \(8 + 8 + 8\sqrt{2} = 16 + 16 \sqrt{2}\)
Simplify to get: \(16 + 8\sqrt{2} = 16 + 16 \sqrt{2}\)

This result tells us two things:
1) \(x=8\) is NOT a solution to the above equation, since \(16 + 8\sqrt{2} \neq 16 + 16 \sqrt{2}\)

2) Since \(16 + 8\sqrt{2} < 16 + 16 \sqrt{2}\), the actual solution to the equation must be GREATER THAN 8.
This means we can eliminate A, B and D

Let's now test \(x = 8\sqrt{2}\)
We get: \(8\sqrt{2} + 8\sqrt{2} + (8\sqrt{2})(\sqrt{2}) = 16 + 16 \sqrt{2}\)
Simplify: \(16\sqrt{2} + 16 = 16 + 16 \sqrt{2}\)

Perfect!

Answer: C

We know that a perpendicular needs to be find out.
\(16 + 16 \sqrt{2}\) can be perimeter of an isosceles triangle only when 16 is the largest side(sum of any two sides will be larger than the third side).
and \(16 \sqrt{2} > 16\)
Hence two sides of same length = 2x = \(16 \sqrt{2}\)
x = \(8 \sqrt{2}\)
Let the largest side = s

perimeter = 2x + s
Taking a perpendicular on s of length h, we have a right triangle with sides x(hypotenuse), \(\frac{s}{2}\) and h.
From here we can either use pythagoras theorem to find out h. OR
As we know, if in a right triangle the hypotenuse is of the form \( \sqrt{2}a\) then other two side are equal of length 'a'.

Hence \(8 \sqrt{2}\) gives h = 8.

Answer D.
_________________

Let the perpendicular sides = x
So, hypothenuse = \(x\sqrt{2}\)
Therefore, \(x+x+x\sqrt{2} = 16+16\sqrt{2}\)
or, \(x=16\frac{(\sqrt{2}+1)}{(\sqrt{2}+2)}\)
or, \(x=16\frac{(\sqrt{2}+1)}{\sqrt{2}(\sqrt{2}+1)}\)
or, \(x = \frac{16}{\sqrt{2}}\)
or,\( x=8\sqrt{2}\)

Answer: C

x + x + √2 = 16 + 16√2
2x + x√2 = 16 + 16√2
x(2 + √2) = 16 + 16√2
x = [16 (1 + √2) / 2 + √2 ] x (2 - √2)/(2-√2)
x = 16(1 + √2) (2 - √2) / 2
x = 16√2 / 2
x = 8√2

C.
_________________

\(\sqrt{2}\) is roughly 1.4.

\(16+16\sqrt{2}\) is roughly 39.

We don't need to know a single thing about right triangles. In fact, we don't even need to know that it is a right triangle in order to eliminate four answer choices. All we need to do is apply some basic possible triangle logic.

A, B, and D are too small to be two sides of any triangle with perimeter 39. What are you going to do, have sides of 6, 6, and 27? You can't make a triangle with that! 4, 4, and 31? Nope. 8, 8, and 23? Nope. A, B, and D are out.

E is more than half the total perimeter, so we can't have two sides of that length, either. E is out.

Answer choice C.

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________