Question

How to proof heron's formula using pythagorean theorem?

Answer 1

Let the sides of a triangle have lengths a,b and c. Introduce the semiperimeter p = (a + b + c)/2 and the area S. Then Heron's formula asserts that

S2 = p(p - a)(p - b)(p - c)

W. Dunham analyzes the original Heron's proof in his Journey through Genius.

For the right triangle with hypotenuse c, we have S = ab/2. We'll modify the right hand side of the formula by noting that

p - a = (- a + b + c)/2,
p - b = (a - b + c)/2,
p - c = (a + b - c)/2.

It takes a little algebra to show that

16S2= (a + b + c)(- a + b + c)(a - b + c)(a + b - c) = 2a2b2 + 2a2c2 + 2b2c2 - (a4 + b4 + c4)

For the right triangle, 16S2 = 4a2b2. So we have

4a2b2= 2a2b2 + 2a2c2 + 2b2c2 - (a4 + b4 + c4)

Taking all terms to the left side and grouping them yields

(a4 + 2a2b2 + b4) - 2a2c2 - 2b2c2 + c4 = 0

With a little more effort

(a2 + b2)2 - 2c2(a2 + b2) + c4 = 0

And finally

[(a2 + b2) - c2]2 = 0

Answer 2

Step-by-step explanation:

Pythagorean theorem, the well-known geometric theorem that the sum of the squares on the legs of a right triangle is equal to the square on the hypotenuse (the side opposite the right angle)—or, in familiar algebraic notation, a2 + b2 = c2.