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Root

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by umaa » Tue Dec 23, 2008 9:41 pm
My method:

Isosceles triangle: so, Perimeter is, x+x+x sqrroot 2

Perimeter given: 16+16 sqrroot 2

2x+x sqrroot 2 = 16+16 sqrroot 2

x qrroot 2 (sqrroot 2 + 1) = 16 (1 + sqrroot 2)

x sqrroot 2 = 16 = 2*2*2*2

x = 8 sqrroot 2
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by vishubn » Tue Dec 23, 2008 9:50 pm
yaaa...
a:a:root2*a=16+16 root2

2a+a root2= 16+16root2

therefore

2a=16

a=8

root2*a=16 root2

a=16 !!

hyp is 16

pleaes remeber given is the ratio !! and not the actualy values ... so will differ

vishu
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Re: Root

by vishubn » Tue Dec 23, 2008 9:52 pm
umaa wrote:My method:

Isosceles triangle: so, Perimeter is, x+x+x sqrroot 2

Perimeter given: 16+16 sqrroot 2

2x+x sqrroot 2 = 16+16 sqrroot 2

x qrroot 2 (sqrroot 2 + 1) = 16 (1 + sqrroot 2)

x sqrroot 2 = 16 = 2*2*2*2

x = 8 sqrroot 2
also do a quick chekc using the answer back to q~~ i am sure it wont take much time in actual exam tooo

IF i use ur answer .... 16+ 8root2

VIshu
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by umaa » Tue Dec 23, 2008 9:59 pm
How will you get two answers for a?

Even if I plugin my answer, I got the original answer.

2x + x sqrroot 2 = 16 + 16 sqrroot 2

x = 8 sqrroot

2 (8 sqrroot 2) + (8 sqrroot 2) * (sqrroot 2) = 16 + 16 sqrroot 2

16 sqrroot 2 + 8 * 2 = 16 + 16 sqrroot 2

16 + 16 sqrroot 2 = 16 + 16 sqrroot 2.

I'm sorry guys. I don't find anything wrong in it. But I did a mistake. Please clarify me.
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by amitabhprasad » Tue Dec 23, 2008 10:22 pm
umaa
you only mistake was you end up selecting length of equal side and question asked for the hypo.
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Re: Root

by acecoolan » Tue Dec 23, 2008 10:30 pm
umaa wrote:My method:

Isosceles triangle: so, Perimeter is, x+x+x sqrroot 2

Perimeter given: 16+16 sqrroot 2

2x+x sqrroot 2 = 16+16 sqrroot 2

x qrroot 2 (sqrroot 2 + 1) = 16 (1 + sqrroot 2)

x sqrroot 2 = 16 = 2*2*2*2

x = 8 sqrroot 2
Uma - ur method is right albeit a bit long

but please note that what u have eventually arrived at is the side of the triangle and not the hypotenuse i.e.

x = 8 sqrt 2 is the length of the the two equal sides

so 16 is the hypotenuse
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