consider a right triangle. with a & b as sides and c as hypotenuse.
then a^2+b^2 =c^2
now area of this triangle can be given as
Area 1 = 1/2 a * b
also
area = sqrt s(s-a)(s-b)(s-c)
s= (a+b+c)/2
equating we get,
s(s-a)(s-b)(s-c) = a^2 b^2 /4
substitute for s,
(a+b+c) (a-b+c)(c+b-a)(a+b-c) / 16 = a^2 b^2 /4
i.e.,
[(a+b)^2 - c^2].............../16= a^2 b^2/4
this term on the left becomes zero.
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this implies either a=0 or b=0.
what is wrong here? please help?
The left hand side is definitely not zero. I think you missed some signs or left out some terms in the expansion:
(a+b+c)(a-b+c)(c+b-a)(a+b-c) =
[a^2 - (b+c)^2][(b^2 - c^2 - a^2 + 2ac)] =
(a^2 - b^2 - 2bc - c^2)(b^2 - c^2 - a^2 + 2ac) =
2a^2b^2 - a^4 + 2a^3c - b^4 - 2ab^2c - 2b^3c + 2bc^3 + 2a^2bc - 4abc^2 - 2ac^3 + c^4
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Dec 29, 2008 <news:9c5963e2-9645-4bbf-ac07-e266c45a97c9@a12g2000pro.googlegroups.com> in alt. algebra.help:. > I know by the triangle inequality that:
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What exactly are you trying to do here by equating Heron's formula and the "regular" formula for the area of a triangle, though?
hey...only a^2 + b^2 - c^2 is zero....in a right angled triangle...
(a+b)^2 - c^2 is not equal to zero..i think so thats the mistake......
Where did you get this formula:
area = sqrt s(s-a)(s-b)(s-c) ?
.
I do see an error in your reasoning here:
[(a+b)^2 - c^2] you are saying =0
(a+b)^2 /= a^2 + b^2
(a+b)^2 = a^2 + 2ab + b^2
I think dere is a mistake in ur calculation.....
the left term of eq. [(a+b)^2 - c^2].............../16= a^2 b^2/4
doesnt become zero on solving....
infact it comes out to be
-a^4-b^4+c^4+2a^2b^2+2a^2c^2+2^b^2c^2
try one more time.
it looks like there is a mistake in computing this line:
(a+b+c) (a-b+c)(c+b-a)(a+b-c) / 16 (hint: these should lead to a^4...)
try again. good luck
brother, area = sqrt s(s-a)(s-b)(s-c)?????
i think:
area sqr = s(s-a)(s-b)(s-c)
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TRIANGLE HELP!!!!!!!!?