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geometry-Hypotenuse

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Source: — Problem Solving |
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by maihuna » Sat Aug 01, 2009 6:18 am
Cubical this should be a 45 45 90 triangle side being x x x\/2(sq root 2)

so sum is 2x+\/2x = \/2(\/2+1)x = 16(\/2+1)

So x = 8\/2

and hyp = x\/2 = 16
Charged up again to beat the beast :)
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by mohitsharda » Sat Aug 01, 2009 6:23 am
Since its a right angled isoceles so two sides are equal and the third one will be the hypotenuse.
So, let the equal side both be equal to a
The hypotenuse will be a*(root 2)

Perimeter = sum of all sides => a+a+ a(root 2) = 16 + 16 (root 2)
2a+(root 2)a = 16+ 16 (root 2)
=> a = 8(root 2)


Could not find the symbol for square root :P
MS
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by mohitsharda » Sat Aug 01, 2009 6:38 am
Since its a right angled isoceles so two sides are equal and the third one will be the hypotenuse.
So, let the equal side both be equal to a
The hypotenuse will be a*(root 2)

Perimeter = sum of all sides => a+a+ a(root 2) = 16 + 16 (root 2)
2a+(root 2)a = 16+ 16 (root 2)
=> a = 8(root 2)


Could not find the symbol for square root :P
MS
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by vinayakdl » Sat Aug 01, 2009 8:32 am
IMO: 16 is the answer.

8sqrt2 is comes out to be the side. try putting in 8sqrt2 and solving perimeter, it does not match the given condition.

Vinayak
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by ithoughtshewas18 » Sat Aug 01, 2009 9:58 am
I say 16.

16/square root of 2 = 8 square root of 2

So hyp = 16 + 2 sides
2 sides = 2(8 sqrt 2)

16 square root of 2 + 16

como sey llamaaaaa?
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by tom4lax » Tue Aug 04, 2009 4:52 am
OA please?

I got 8V2 for one side, so hyp is 16.
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