Geometry

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Geometry

by rnaah » Fri Sep 30, 2011 2:37 am
The perimeter of a certain isosceles right triangle is 16+16√2. What is the length of the hypotenuse of the triangle?
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by Anurag@Gurome » Fri Sep 30, 2011 4:30 am
rnaah wrote:The perimeter of a certain isosceles right triangle is 16+16√2. What is the length of the hypotenuse of the triangle?

Let the sides of triangle be x, x, and sqrt2*x.
2x + sqrt2*x = 16+16*sqrt2.
x(2 + sqrt2) = 8*sqrt2(2 + sqrt2).
x = 8*sqrt2.
Hypotenuse is x* sqrt2 = 8*sqrt2*sqrt2 = 16.
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by GMATGuruNY » Fri Sep 30, 2011 9:06 am
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
In an isosceles right triangle, the sides are proportioned s : s : s√2.

We can plug in the answer choices, which represent the value of s√2.
Answer choices C, D and E clearly won't work:

C: If s√2 = 4√2, then s=4 and p = 4+4+4√2 = 8 + 4√2.
D: If s√2 = 8√2, then s=8 and p = 8+8+8√2 = 16 + 8√2.
E: If s√2 = 16√2, then s=16 and p = 16+16+16√2 = 32 + 16√2.

Answer choice B: s√2 = 16.
s = 16/√2 = (16√2)*(√2/√2) = (16√2)/2 = 8√2.
p = 8√2 + 8√2 + 16 = 16√2 + 16. Success!

The correct answer is B.
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by phoenix111 » Fri Sep 30, 2011 9:48 pm
rnaah wrote:The perimeter of a certain isosceles right triangle is 16+16√2. What is the length of the hypotenuse of the triangle?
Let lenght of hypt. be X:

Perimeter = X + Xcos45 + Xcos45 = X + X/√2 + X/√2 = X + X√2

Now X + X√2 = 16 + 16√2, so X = 16

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by fueledGMAT » Tue Dec 13, 2011 8:07 pm
GMATGuruNY 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
In an isosceles right triangle, the sides are proportioned s : s : s√2.

We can plug in the answer choices, which represent the value of s√2.
Answer choices C, D and E clearly won't work:

C: If s√2 = 4√2, then s=4 and p = 4+4+4√2 = 8 + 4√2.
D: If s√2 = 8√2, then s=8 and p = 8+8+8√2 = 16 + 8√2.
E: If s√2 = 16√2, then s=16 and p = 16+16+16√2 = 32 + 16√2.

Answer choice B: s√2 = 16.
s = 16/√2 = (16√2)*(√2/√2) = (16√2)/2 = 8√2.
p = 8√2 + 8√2 + 16 = 16√2 + 16. Success!

The correct answer is B.
I am not sure I understand how you derived s on this problem for b. Can you please go into a little more detail on how you got 16/sqt2? I thought S was 16
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by GMATGuruNY » Tue Dec 13, 2011 8:58 pm
fueledGMAT wrote:
GMATGuruNY 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
In an isosceles right triangle, the sides are proportioned s : s : s√2.

We can plug in the answer choices, which represent the value of s√2.
Answer choices C, D and E clearly won't work:

C: If s√2 = 4√2, then s=4 and p = 4+4+4√2 = 8 + 4√2.
D: If s√2 = 8√2, then s=8 and p = 8+8+8√2 = 16 + 8√2.
E: If s√2 = 16√2, then s=16 and p = 16+16+16√2 = 32 + 16√2.

Answer choice B: s√2 = 16.
s = 16/√2 = (16√2)*(√2/√2) = (16√2)/2 = 8√2.
p = 8√2 + 8√2 + 16 = 16√2 + 16. Success!

The correct answer is B.
I am not sure I understand how you derived s on this problem for b. Can you please go into a little more detail on how you got 16/sqt2? I thought S was 16
The sides of a 45-45-90 triangle are proportioned s : s : s√2.
The hypotenuse = s√2.
Answer choice B states that the hypotenuse = 16.
Thus:
s√2 = 16.
s = 16/√2.
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I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

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