## MATH 7200 : Foundations of Geometry
I

## University of Georgia, Fall
2000

## Dr. McCrory, Instructor

### A Trigonometric Proof of Heron's
Formula

### by Shannon Umberger

#### Note: This proof was adapted from a proof on Dr. Jim
Wilson's website (Department of Mathematics Education, College
of Education, University of Georgia).

Given a triangle with sides of lengths a, b, and c. Trigonometry
gives an area formula that states A = (1/ 2)absinC, where C is
the angle opposite side c.

Using the Law of Cosines, c^{2} = a^{2} + b^{2}
- 2abcosC. Solving for cosC gives cosC = (a^{2} + b^{2}
- c^{2})/ 2ab. Rearrange the Pythagorean identity (sinC)^{2}
+ (cosC)^{2} = 1 to state sinC = [1 - (cosC)^{2}]^{1/2}.
By substitution, sinC = [1 - ((a^{2} + b^{2} -
c^{2})/ 2ab)^{2}]^{1/2}.

Factoring gives sinC = [(1 + (a^{2} + b^{2}
- c^{2})/ 2ab)*(1 - (a^{2} + b^{2} - c^{2})/
2ab)]^{1/2}. It follows that sinC = [((2ab + a^{2}
+ b^{2} - c^{2})/ 2ab)*((2ab - a^{2} -
b^{2} + c^{2})/ 2ab)]^{1/2}. Factoring
out the denominator gives sinC = (1/ 2ab)*[(2ab + a^{2}
+ b^{2} - c^{2})(2ab - a^{2} - b^{2}
+ c^{2})]^{1/2}.

After some regrouping, sinC = (1/ 2ab)*[((a^{2} + 2ab
+ b^{2}) - c^{2})(c^{2} - (a^{2}
- 2ab + b^{2}))]^{1/2}. Factoring gives sinC =
(1/ 2ab)*[((a + b)^{2} - c^{2})(c^{2}
- (a - b)^{2})]^{1/2}. After factoring once more,
sinC = (1/ 2ab)*[((a + b) + c)((a + b) - c)(c + (a - b))(c - (a
- b))]^{1/2}. With some rearranging, sinC = (1/ 2ab)*[(a
+ b + c)(a + b - c)(a - b + c)(-a + b + c)]^{1/2}.

Since the semiperimeter of the above triangle is defined as
s = (a + b + c)/ 2, it follows that 2s = a + b + c, and so 2s
- a = b + c. Adding -a to both sides of the equation gives 2s
- 2a = -a + b + c. Therefore, 2(s - a) = -a + b + c. Similar algebra
steps give 2(s - b) = (a - b + c) and 2(s - c) = (a + b - c).

If follows by substitution that sinC = (1/ 2ab)*[2s*[2(s -
c)]*[2(s - b)]*[2(s - a)]]^{1/2}. With some rearranging,
sinC = (1/ 2ab)*[16s(s - a)(s - b)(s - c)]^{1/2} = (1/
2ab)*4*[s(s - a)(s - b)(s - c)]^{1/2} = (2/ ab)*[s(s -
a)(s - b)(s - c)]^{1/2}.

Now, substitute the value of sinC into the area formula A =
(1/ 2)absinC. So, A = (1/ 2)*ab*(2/ ab)*[s(s - a)(s - b)(s - c)]^{1/2}.
Thus, A = [s(s - a)(s - b)(s - c)]^{1/2}, which was to
be proved. QED.

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