In an exercise, I have to answer the perimeter of a equilateral triangle knowing that its area is $$\sqrt{3}$$
How can I achieve it? I tried inventing equations, but all dead ends. Please explain.
[Math] Perimeter of equilateral triangle from its area
areatriangles
Related Solutions
Since the ratio of sides are in $13:14:15$, let the three sides be $13k$, $14k$ and $15k$ in ascending order.
Now the 3 sides add up to a perimeter length of $84$ m, so
$$\begin{align*} 13k + 14k + 15k =& 84 \text{ m}\\ k =& 2 \text{ m} \end{align*}$$
And so the three sides are $26$ m, $28$ m and $30$ m respectively. Now use Heron's formula to find the area.
(Hint: you should confirm that $s$ is half of perimeter)
Treat the side as the chord of a circle in which it subtends the given angle - the extended sine rule gives the radius. Then use the area of the triangle to calculate the height and you can identify the correct point on the circle.
Let the side you are given be $a$, and take this as the base of the triangle, and the angle opposite be $A$.
The extended version of the sine rule tells us that $\cfrac a{\sin A}=2R$ where $R$ is the circumradius of the triangle, so you can find $R$.
Having found $R$ you take the perpendicular bisector of your side, and locate one of the two points (above/below) which is distance $R$ from the extremities of the side (vertices $B, C$) and construct the circle.
With side $a$ as the base you calculate the height from area $=\cfrac {ah}2$.
Then take a line $l$ parallel to side $a$ and distance $h$ from it (on the correct side for the angle you want - if you get the wrong side the angle will be $180^{\circ}-A$). If the triangle is possible this will cut the circle in two points (you will see a symmetry about the perpendicular bisector) unless $l$ happens to be a tangent, when you get a single point. One of these points can be taken as vertex $A$.
I have not advised a formula, but a method. Following the method and doing the algebra will give you a formula if you need one.
Best Answer
If an equilateral triangle has side length $d$, the length of every heigth is given by $\frac{\sqrt{3}}{2}d$ by the pythagorean theorem, hence the area is $\frac{\sqrt{3}}{4}d^2$ and the perimeter is $3d$. Hence, if the area equals $\sqrt{3}$, the perimeter equals $6$.