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45-45-90 triangle - radicals question

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45-45-90 triangle - radicals question

by damrkstr » Sun Nov 02, 2008 5:13 pm
The perimeter of a certain isosceles right triangle is 16 + 16√2.
What is the length of the hypotenuse of the triangle?

I tried solving it setting x + x + x√2 = 16 + 16√2 but am not solving it correctly. Could someone help me out with the best way to solve the problem?

The answer was supposed to be 16

Thanks for your help!
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by dmateer25 » Sun Nov 02, 2008 5:37 pm
The equation you set up will work just fine. I think the best way to solve this using your equation is to just estimate √2 as 1.4.

x + x + 1.4x = 16 + 16(1.4)

3.4x = 38.4

x = ~11.3


The hypotenuse would be 1.4 (11.3) = ~15.82

It would be pretty clear that the answer would be 16.

Also you could do the following:

Assume that the hypotenuse is x and the sides of the triangle are x√2/2.

(x√2/2)+(x√2/2)+x=16 + 16√2
2x√2/2 + x = 16 + 16√2
x√2 + x = 16 + 16√2

x =16
x√2=16√2
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Re: 45-45-90 triangle - radicals question

by jimmiejaz » Wed Nov 05, 2008 4:24 am
damrkstr wrote:The perimeter of a certain isosceles right triangle is 16 + 16√2.
What is the length of the hypotenuse of the triangle?

I tried solving it setting x + x + x√2 = 16 + 16√2 but am not solving it correctly. Could someone help me out with the best way to solve the problem?

The answer was supposed to be 16

Thanks for your help!
taking ahead your approach...
we can see
x + x + x√2 = 16 + 16√2
pick numbers...
first take x=8, it gives
8+8+8√2 which is not equal to 16+16√2. hence xis not eual to 8.
now pick x=8√2
8√2+8√2+16 = 16+16√2

hence the side of the triangle is 8√2 and the hypotenuse is 16.
U arrived at the equation correctly but then had to just pick numbers.
hope it helps...
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Re: 45-45-90 triangle - radicals question

by jimmiejaz » Wed Nov 05, 2008 4:30 am
damrkstr wrote:The perimeter of a certain isosceles right triangle is 16 + 16√2.
What is the length of the hypotenuse of the triangle?

I tried solving it setting x + x + x√2 = 16 + 16√2 but am not solving it correctly. Could someone help me out with the best way to solve the problem?

The answer was supposed to be 16

Thanks for your help!
one more simple and elegant approach.....
x+x+x√2= 16 + 16√2
2x+x√2=16(1+√2)
x√2(√2+1)=16(1+√2)
which gives x√2 = 16 or x = 8√2
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by stickyfingers48 » Sat Nov 08, 2008 3:01 am
2x²=y² and 2x+y=16+16√2

y=16[/list]
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Re: 45-45-90 triangle - radicals question

by cartera » Sun Nov 09, 2008 6:02 am
jimmiejaz wrote:
damrkstr wrote:The perimeter of a certain isosceles right triangle is 16 + 16√2.
What is the length of the hypotenuse of the triangle?

I tried solving it setting x + x + x√2 = 16 + 16√2 but am not solving it correctly. Could someone help me out with the best way to solve the problem?

The answer was supposed to be 16

Thanks for your help!
one more simple and elegant approach.....
x+x+x√2= 16 + 16√2
2x+x√2=16(1+√2)
x√2(√2+1)=16(1+√2)
which gives x√2 = 16 or x = 8√2
x = 8√2 is not 16 and also, how did you turn 2x+x√2 into x√2(√2+1)?
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Re: 45-45-90 triangle - radicals question

by Ossa » Sun Nov 09, 2008 5:04 pm
cartera wrote:
jimmiejaz wrote:
damrkstr wrote:The perimeter of a certain isosceles right triangle is 16 + 16√2.
What is the length of the hypotenuse of the triangle?

I tried solving it setting x + x + x√2 = 16 + 16√2 but am not solving it correctly. Could someone help me out with the best way to solve the problem?

The answer was supposed to be 16

Thanks for your help!
one more simple and elegant approach.....
x+x+x√2= 16 + 16√2
2x+x√2=16(1+√2)
x√2(√2+1)=16(1+√2)
which gives x√2 = 16 or x = 8√2
x = 8√2 is not 16 and also, how did you turn 2x+x√2 into x√2(√2+1)?
Hi,
Let me answer the second part of your question:
Remembering that √2*√2=2, then 2x=(√2*√2)x
Accordingly 2x+x√2=(√2*√2)x+x√2=x√2(√2+1)
Ossa Elhadary, PhD, CISA, PMP
Math Specialist
Test Prep New York
maximize your score, minimize your stress

www.testprepny.com
[email protected]
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