For Acute triangle AB with side lengths a,b,c find a formula for its area.?

Question

Make sure to show the derivation of that formula!

Answer 1

Hi,

Given 3 sides of a triangle, you can find its area by using Hero's (Heron's) Formula, which is:
.........____________
A = √s(s-a)(s-b)(s-c)

where s = (a + b + c)/2. This is the semi-perimeter of the triangle, basically half of the perimeter.

I have included websites discussing the proof of this formula.


I hope that helps!! :-)

Answer 2

If it is a right triangle, you can use the two legs for base and height, otherwise split it up into two right triangles and add up their areas found with this formula:
Area = (1/2) base * height

If it is not right, and you dont want to make it into two right triangles, you can use an area formula.
A = sqrt( s ( s - a ) ( s - b ) ( s - c ) ) where s = (1/2) (a + b + c)

If you know the angles as well (A being the angle opposite side a, and so on)
A = (1/2) b * c * sin A
or
A = (1/2) a^2 (sin B * sin C) / sin A

Answer 3

Hello

We need the semi-perimeter or s =1/2 (a+b+c)

Then we have a formula A = sqrt[s*(s-a)*(s-b)*(s-c)]

Do you need it derived???

Hope This Helps.

Answer 4

area = 1/4 sqrt([a2 + b2 + c2]2 - 2 [a4 + b4 + c4])

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Law of Cosines says
... cos C = (a2 + b2 - c2) / 2 a b

Because sin = sqrt(1 - cos^2), we have
... sin C = sqrt(1 - [(a2 + b2 - c2)/2ab]^2)
... ... = 1/2ab sqrt([2ab]^2 - [a2 + b2 - c2]^2)

The radicand is
... [2ab]^2 - [a2 + b2 - c2]^2 =
... ... = 4a2b2 - (a4 + b4 + c4) - 2a2b2 + 2a2c2 + 2b2c2
... ... = -(a4 + b4 + c4) + (2a2b2 + 2a2c2 + 2b2c2)

It is easy to check that this is equal to
... ... = (a2 + b2 + c2)2 - 2(a4 + b4 + c4)

Now
... area = 1/2 basis * height
... ... = 1/2 a (b sin C)
... ... = ab/2 * 1/2ab sqrt[(a2 + b2 + c2)2 - 2(a4 + b4 + c4)]

which simplifies to the given formula.