Question

What is the area of a triangle that has sides 9 cm, 10 cm, and 13 cm?

Answer 1

`color(blue)(12sqrt(14)cm^2~~44.90` 2 d.p.

Explanation:

There are several ways to find the area of the given triangle. We have the lengths of the sides, but are given no angles. We could find these angles using the cosine rule, but since the given sides are integer values we could use Heron's formula.

Heron's formula is given by:

`"Area"=sqrt(s(s-a)(s-b)(s-c))`

Where `a, b, c` are the triangles sides and `s` is the semi-perimeter:

`s=(a+b+c)/2`

Let: `a=9, b=10, c=13`

`s=(9+10+13)/2=32/2=16`

`"Area"=sqrt(16 * (16-9) * (16-10) * (16-13))`

`\ \ \ \ \ \ \ \ \=sqrt(16 * (7) * (6) * (3))`

`\ \ \ \ \ \ \ \ \=sqrt(2016)=12sqrt(14)cm^2~~44.90` 2 d.p.

Answer 2

`12sqrt{14}`

Explanation:

Heron's Formula is what is usually taught, but Archimedes' Theorem is generally superior.

Given a triangle with sides `a,b and c` and area `S`,

`16S^2 = 4a^2b^2 - (c^2-a^2-b^2)`

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

`=(a+b+c)(-a+b+c)(a-b+c)(a+b-c)`

When we have the lengths the last form is often best; it's pretty much Heron's Theorem but avoids the fraction from the semiperimeter.

`16S^2 = (9+10+13)(-9+10+13)(9-10+13)(9+10-13) = (32)(14)(12)(6)`

`S = \sqrt{2(14)(6)(6)(2)} = 12sqrt{14}`