How to Find Area of a Triangle from $3$ sides

Question

How would I find the area of a triangle, given $3$ side lengths $3$, $4$, and $6$?

Would there be any way to do this?

Answer 1

Given 3 sides $A$, $B$, and $C$, you could use the Heron's Formula to obtain the lengths of the three sides.

However, I like this approach more, since you can easily understand it (although it is LONGER).

Follow this image to see what I mean.

$AB=3$

$AC=4$

$BC=6$

Define side $AD$ (which is an altitude) as $x$, and side $BD$ as $y$.

Therefore, by the pythagorean theorem,

$$y^2+x^2=3^2$$$$(6-y)^2+x^2=4^2$$

Once you solve these two equations, you obtain $x$, which is the height of the triangle from the base $BC$. Now I think you know what to do.

Answer 2

Yes. The formula for finding the area of a triangle is $A=\cfrac{bh}{2}$ where $b$ is the length of the base and $h$ is the height. If it is a right triangle, the height is easy. If not, you have to do some other geometry to find the height. Once you have the height from a particular base plug it into the formula. If it is not right comment on this and I will help you more!

Answer 3

Heron's formula gives that the area $A$ of a triangle of sides $a,b,c$ and semiperimeter $s = \frac 12(a+b+c)$ is $A = \sqrt{s(s-a)(s-b)(s-c)}$. Apply that.

Answer 4

Heron's formula would be the most direct $A = \sqrt {s(s-a)(s-b)(s-c)}$

$s = \frac{a+b+c}{2}$

$A = \sqrt {\frac {13}{2}\frac {7}{2}\frac{5}{2}\frac {1}{2}} = \frac {\sqrt {455}}{4}$