how can pythagoras theorem be proved using hero's formula?

Answer 1

we have A^2 = s(s-a)(s-b)(s-c)

s= (a+b+c)/2
multiply by 16

16A^2 = 2s(2s-2a)(2s-2b)(2s-2c)
= (a+b+c)(b+c-a)(a+c-b)(a+b-c)
let b and c be base

A = bc/2
A^2 = b^2c^/4

so we get
4b^2c^2= ((a+b+c)(b+c-a))(a+(b-c))(a-(b-c)

= ((b+c)^2 - a^2)(a^2 - (b-c)^2)
or ((b+c)^2 - a^2)((b-c)^2-a^2) = - 4b^2c^2

or (b^2+c^2-a^2-2bc)(b^+c^2-a^2+2bc) = -4 b^2 c^2

let b^2+c^2-a^2 = x

(x+2bc)(x-2bc) = x^2-4bcx - 4b^2c^2 = - 4b^c^2
or x^2 = 4bc x

x =0 or 4bc
x = 0 means a^2= b^2+c^2 same as pythogoras theorem

x = 4bc means a^2 < (b-c)^2 so not possible

QED

Answer 2

Heron's Formula: A = sqrt(s(s-a)(s-b)(s-c))
Right Triangle: A = ab/2

The two are equal, taking their square...
a²b²/4 = s(s-a)(s-b)(s-c)

After expanding the RHS you eventually get...
a²b²/4 = (2a²b² + 2a²c² + 2b²c² - a^4 - b^4 - c^4) / 16
4a²b² = 2a²b² + 2a²c² + 2b²c² - a^4 - b^4 - c^4

Move all terms not containing c to the left... LHS is a square:
a^4 + b^4 + 2a²b² = 2a²c² + 2b²c² - c^4
(a²+b²)² = 2c²(a²+b²) - c^4

Now move everything else to the left...
c^4 - 2c²(a²+b²) + (a²+b²)² = 0

This has the form of a square (x² - 2xy + y² = (x-y)², where x = c² and
y = a²+b²). So...

(c² - (a²+b²))² = 0
c² - (a²+b²) = 0
a² + b² = c². QED.

Answer 3

Let ABC and DEF be two congruent right triangles such that B lies on DE and A, F, C, E are collinear. BC = EF = a, AC = DF = b, AB = DE = c. Obviously, AB DE. Compute the area of ΔADE in two different ways.

Area(ΔADE) = AB·DE/2 = c2/2 and also Area(ΔADE) = DF·AE/2 = b·AE/2. AE = AC + CE = b + CE. CE can be found from similar triangles BCE and DFE: CE = BC·FE/DF = a·a/b. Putting things together we obtain

c2/2 = b(b + a2/b)/2

Answer 4

This is not a formal geometric proof but an algebraic derivation.
In an algebraic derivation, beginnings and ending points are arbitrary in the sequence of steps in that they are truly logically/algebraically equivalent so long as all steps are allowed (e.g., no dividing by zero). This is another way of saying that once algebraic expressions are proven equivalent, it does not matter which one you start with.

Therefore, take this derivation of Hero from Pythagoras, and follow it algebraically from bottom to top for your "proof".

Answer 5

i dont know...
i cant see the relation...

Answer 6

I DONT KNOW BUDDY............