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Right angle Triangle.

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by goyalsau » Sat Oct 09, 2010 3:35 am
HI! Friends.

I have been solving these problems just by plugging in values but at times it becomes too slow, When the number is big or quite difficult to calculate, So

I want to know is there a particular formula to solve these kind of problems.

Please post if you have any suggestions to solve these kind of problems.
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by selango » Sat Oct 09, 2010 3:57 am
Sides of Isosceles triangle are a,a and asqrt(2) with hypotenuse asqrt(2)

Perimeter=a+a+asqrt(2)=2a+sqrt(2)--Eqn1

-->2a+asqrt(2)=16+16sqrt(2)

we need to find value of a to find value of hypotenuse.

Assume 2a=16 or 2a=16sqrt(2)

If 2a=16,a=8

sub a=8 in Eqn1

Perimeter=16+8sqrt(2).Not satisfied


If 2a=16sqrt(2),a=8sqrt(2)

sub a=8sqrt(2) in Eqn1

Perimeter=16sqrt(2)+16.Satisfied

Hypotenuse=asqrt(2)=8sqrt(2)*sqrt(2)=16

Hope this clarify!!!
--Anand--
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by neerajkumar1_1 » Sat Oct 09, 2010 4:35 am
just remember that sides dont always have to be integer...
take side as 8 root(2)
so p = 8 root(2) + 8 root(2) + root(2) 8 root(2) = 16 root(2) + 16...
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