Relation between Perimeter and hypotenuse of triangle

amirhakimi wrote: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

Answer is B

An important point here 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 left side to get : x√2(√2 + 1) = 16 + 16√2
Now factor 16 from 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

amirhakimi wrote: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

Answer is B

Another option is to test the answer choices.

We'll use the fact that, for any isosceles right triangle, the sides have length x, x, and x√2 for some positive value of x. The hypotenuse is x√2

So, if x is the length of one leg of the right triangle, the perimeter = x + x + x√2

Start with answer choice D.
If the hypotenuse is 8√2, then x = 8 [since the hypotenuse = x√2]
So, the perimeter = 8 + 8 + 8√2 = 16 + 8√2
NO GOOD! We're told that the perimeter is 16 + 16√2, so we need the triangle to be LARGER THAN 8√2.
This means we can ELIMINATE A, C and D

IMPORTANT: At this point, we can test either B or E. It doesn't matter which one we test. If B works, we're done. If B doesn't work, then we'll automatically choose E.

Let's try answer choice B.
If the hypotenuse is 16, then we know that x√2 = 16
So, x = 16/√2 = 8√2
So, in this instance, the perimeter is 8√2 + 8√2 + 16 = 16√2 + 16
PERFECT! We're told that the perimeter is 16 + 16√2

Answer: B

Cheers,
Brent