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?
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$\begingroup$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.
$\endgroup$ $\begingroup$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.
$\endgroup$ $\begingroup$Heron's formula would be the most direct $A = \sqrt {s(s-a)(s-b)(s-c)}$
$s = \frac{a+b+c}{2}$
$A = \sqrt {\frac {13}{2}\frac {7}{2}\frac{5}{2}\frac {1}{2}} = \frac {\sqrt {455}}{4}$
$\endgroup$ $\begingroup$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!
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