Heron's formula backwards?

Question

Can you solve: Area of a triangle=sqrt(s)(s-a)(s-b)(s-c)
where (s=the semiperimeter of a triangle) and (a, b, and c are the measures of the side of a triangle) for c?

Answer 1

In this case, you can't solve for distinct values of s, a ,b, or c because you have only one equation and 4 variables. You need values on s, a, b, and the area in order to solve for c. Maybe there's extra information in the question (like they're all integers or something)?

Answer 2

Let a,b,c be the sides of a triangle, and let A be the area of the triangle. Heron's formula states that A^2 = s(s-a)(s-b)(s-c), where s = (a+b+c)/2.
Here's one derivation:
Consider the general triangle with edge lengths a,b,c shown here: http://bestshotammo.com/str/heron.JPG

We have a = u+v, b^2 = h^2+u^2, c^2 = h^2+v^2. Subtracting the second from the third gives u^2-v^2 = b^2-c^2. Dividing both sides by a = u+v, we have u-v = (b^2-c^2)/a. Adding u+v = a to both sides and solving for u gives:
u = (a^2+b^2-c^2)/(2a)
Taking h = sqrt(b^2-u^2) we have:
A = ah/2 = (1/2)sqrt{[(ab)^2 - [(a^2+b^2-c^2)/2]^2 }
which is equivalent to Heron's formula. Factoring out 1/4, this gives three different ways of expressing (2ab)^2 - (a^2+b^2-c^2)^2 as a difference of two squares. Equivalently, it gives three different factorizations of 16A^2, each of the form
16A^2 = [(a+b)^2 - c^2] [c^2 - (a-b)^2]
Factoring each of these terms gives the explicitly symmetrical form
16A^2 = (a+b+c)(a+b-c)(c-a+b)(c+a-b)
so if we define s=(a+b+c)/2 we can rewrite the first equation as:
sqrt[s(s-a)(s-b)(s-c)]
which is the area formula as given by Heron.

Answer 3

This is an exercise in basic algebra of symbol manipulation in equations.

Square both sides to get
A^2 = s(s-a)(s-b)(s-c)

Divide by s(s-a)(s-b) to get
s-c = A^2/[s(s-a)(s-b)]

c = s-A^2/[s(s-a)(s-b)]

Answer 4

c+sqrt

Answer 5

no