How do you find the area of an equilateral triangle whose area is 4√3 sq.cm? Please show the steps.
Thanks,
Ron
Let "s" be the length of one side of the equilateral triangle. Drop a line from one vertex to the middle of the opposite side. It can be shown that this line is also perpendicular to the opposite side, dividing the equilateral triangle into two right triangles that have hypotenuse of length s and one leg of length s/2.
Taking "x" to be the length of the altitude to the equilateral triangle, which is the other leg of the right triangles, by the Pythagorean theorem, $\displaystyle x^2+ s^2/4= s^2$ so $\displaystyle x^2= 3s^2/4$ and $\displaystyle x= s\sqrt{3}/2$.
Now, the area of the equilateral triangle is "1/2 base times height" or $\displaystyle \frac{1}{2}\left(s\right)\left(\frac{s\sqrt{3}}{2} \right)= \frac{\sqrt{3}}{4}s^2$. Set that equal to $\displaystyle 4\sqrt{3}$ and solve for s.
Then, of course, the perimeter is 3s.
Hey, did anyone READ the problem?
It's an old joke . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . $\displaystyle \begin{array}{c} {\color{red}\uparrow} \\ {\color{red}\text{Here it is!}} \end{array}$How do you find the area of an equilateral triangle whose area is 4√3 sq.cm?
The most direct way of doing this is to use Heron's formula for the area of a triangle:
$\displaystyle
A=\sqrt{s(s-a)(s-b)(s-c)}
$
where $\displaystyle s$ is the semi-perimiter and $\displaystyle a$, $\displaystyle b$ and $\displaystyle c$ are the sides.
For an equilateral triangle we have $\displaystyle a=b=c$ and $\displaystyle s=3a/2$
So:
$\displaystyle A^2=48=s^4 / 3^3$
CB