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# Isosceles right triangle with perimeter 16+16(2^0.5). What

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Isosceles right triangle with perimeter 16+16(2^0.5). What [#permalink]

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19 Oct 2006, 23:30
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Isosceles right triangle with perimeter 16+16(2^0.5). What is the length of the hypothenuse?

I know this is a special triangle and therefore the sides are in a ratio of 1:1:2^0.5 but I am having a hard time setting this up. The official answer is 16...how do you get there?

Thak you very much,

Primeminister

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Re: Geometry question - explanation requested [#permalink]

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20 Oct 2006, 00:15
primeminister wrote:
Isosceles right triangle with perimeter 16+16(2^0.5). What is the length of the hypothenuse?

I know this is a special triangle and therefore the sides are in a ratio of 1:1:2^0.5 but I am having a hard time setting this up. The official answer is 16...how do you get there?

Thak you very much,

Primeminister

You must remember that the sum of two sides of the triangle cannot be larger than the length of one of the sides....

Therefore, if the perimeter of isoscles triangle is 16+16(2^0.5)... hypotenuse must be equal 16, because the other two sides are equal in legth and the sum of those two sides must be larger than the remaining side....

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20 Oct 2006, 00:23
Thanks SimaQ...i follow the triangle logic of what you said...not sure how that makes the hypothenuse 16 just by looking at the 16+16(2^0.5)??? Perhaps you can set it up in an equation so I can see it.

I can do the following:
x+x+x(2^0.5)=16+16(2^0.5) Solving for x is very easy if you have a calculator and it gets the right answer. A bit tougher doing it just with a pencil.

Any more thoughts?

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20 Oct 2006, 00:47
I meant the length of one side cannot be larger than the sum of the other two sides of the triangle....

Sorry for that mistake.... Now you should see what i REALLY meant

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20 Oct 2006, 00:52
Perimeter: 16+16(2^0.5).....
From stem we know that the two sides are eqaul....
From the rule above we know that those two sides cannot be equall to 8+8=16, because the remaining side, that is hypotenuse, would be equal to more than 16, that is 16(2^0.5)....

Therefore the sum of the other two sides must be equal to 16(2^0.5)
And the remaining side, hypotenuse, 16....

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20 Oct 2006, 12:39
if x is side hypo should be x*2^0.5

x + x + x*2^0.5 = 16 + 16*2^0.5
x(2 + 2^0.5) = 16( 1 + 2^0.5)
x(2 ^0.5 * 2^0.5 +2^0.5)=16 (1 + 2^0.5)
x*2^0.5 (2^0.5 + 1)=16( 1 + 2^0.5)

if u cancel out (1+2^0.5) from both sides

x*2^0.5=16

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20 Oct 2006, 22:00
Code:
if x is side hypo should be x*2^0.5

x + x + x*2^0.5 = 16 + 16*2^0.5
x(2 + 2^0.5) = 16( 1 + 2^0.5)
x(2 ^0.5 * 2^0.5 +2^0.5)=16 (1 + 2^0.5)
x*2^0.5 (2^0.5 + 1)=16( 1 + 2^0.5)

if u cancel out (1+2^0.5) from both sides

x*2^0.5=16

Very good explaination

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21 Oct 2006, 04:08
Thanks for the explanation damager was really lost on this one..
Guys this from Gmat Prep!!

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23 Oct 2006, 08:29
I approached it in the following way and reached a different value....

given,
2x + x*2^0.5 = 16 + 16^0.5

then,
x(2 + 2^0.5) = 8 (2 + 2^0.5)

therefore,
x = 8

and hypotneous is 8^0.5 which is not 16.

what went wrong in my approach. Damager's method works but what if I started with the above approach .... ???

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23 Oct 2006, 08:43
Isosceles right triangle with perimeter 16+16(2^0.5). What is the length of the hypothenuse?

since a + b + c = 16 + 16*sqrt(2)
but isoscles triangle so a = b

2a + c = 16 + 16 * sqrt(2) --- A

Also, its a right angle triangle

c^2 = 2a^2 (c^2 = a^2 + b^2 ; b = a)

a = c/(sqrt(2)) --- B

plugging in value of B in A

2c/sqrt(2) + c = 16 + 16 * sqrt(2)

solving, c = 16

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23 Oct 2006, 09:07
Just multiply sqrt(2) to your special triangle:
1:1:sqrt(2)
You get
sqrt(2):sqrt(2):2
which gives you perimeter 2+2sqrt(2)
Now you know the ratio is 2:2+2aqrt(2) or 16:16+16sqrt(2).
_________________

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keep on seeking, and you will find;
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24 Oct 2006, 16:31
Damager wrote:
if x is side hypo should be x*2^0.5

x + x + x*2^0.5 = 16 + 16*2^0.5
x(2 + 2^0.5) = 16( 1 + 2^0.5)
x(2 ^0.5 * 2^0.5 +2^0.5)=16 (1 + 2^0.5)
x*2^0.5 (2^0.5 + 1)=16( 1 + 2^0.5)

if u cancel out (1+2^0.5) from both sides

x*2^0.5=16

can someone explain how do one get the bold part?

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2= SqRoot 2 * SqRoot 2 [#permalink]

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24 Oct 2006, 16:43
Bold part explanation

2 = SqRoot 2 * SqRoot 2

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2= SqRoot 2 * SqRoot 2   [#permalink] 24 Oct 2006, 16:43
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