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

Question

If the perimeter of an isosceles right triangle is  16 + 16 \sqrt{2}  inches long, how long is one of the perpendicular sides?

(A)  4 \sqrt{2} 
(B) 4
(C)  8 \sqrt{2} 
(D) 8
(E)  16 \sqrt{2}

BrentGMATPrepNow · Accepted Answer

The length the sides of an isosceles right triangle look like this: 
 https://i.imgur.com/0UzUwHu.png

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} 
eq 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

unraveled · Answer

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.

AnirudhaS · Answer

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

CEdward · Answer

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

C.

ThatDudeKnows · Answer

\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.

bumpbot · Answer

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

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

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