Consider the following triangle:

There are no right angles in the triangle, so therefore we must use Heron's forumla which first defines a variable s

s=1/2(a+b+c) where a,b,c each represent the length of a side.

then the area of the triangle is equal to

sqrt(s(s-a)(s-b)(s-c))

For the example we see that s=21

and the area is equal to

sqrt(21(21-10)(21-15)(21-17))

=

sqrt(21*11*6*4)=sqrt(5,544)=~74.46 which is the area of the triangle.

How do we prove Herons formula? Consider the triangle:

Here we have taken a triangle with no right angles, and cut a line so that we create two right triangles, and two new lengths at the base: x, and C-x. Now we need to solve for two unknown values x and h. We will solve for x first.

We can use pythagorean theorem to create equations for our new values, we see we have:

x

^{2}+ h

^{2}= B

^{2}

and

(c-x)

^{2}+ h

^{2}= A

^{2}

=

C

^{2}- 2cx + x

^{2}+ h

^{2}= A

^{2}

We know that B= x

^{2}+h

^{2}so we substitute that in to the equation above and get

C

^{2}- 2cx + B

^{2}= A

^{2}

Now we can solve for x!

x= (A

^{2}- B

^{2}- C

^{2}) / - 2c

Now we have x we can plug it back in to the equation

x

^{2}+ h

^{2}= B

^{2}

thus

h = sqrt(B

^{2}- (A

^{2}- B

^{2}- C

^{2}) / - 2c)

^{2})

Now, we know our base is equal to C and our height is equal to h, if we substitute these two values into the formula for the area of a triangle we get

(1/2)*b*h = 1/2*C

^{2}*sqrt(B

^{2}- (A

^{2}- B

^{2}- C

^{2}) / - 2c)

^{2})

That is a complicated formula, so Heron found that you could define a term

s = (1/2)*( A + B + C) and then found the area formula could become a simpler

sqrt(s(s-a)(s-b)(s-c))

If you plug s in to the area formula above you will find the result we saw above:

1/2*C

^{2}*sqrt(B

^{2}- (A

^{2}- B

^{2}- C

^{2}) / - 2c)

^{2})

which is the area of the triangle.

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