geometry
What is the area of an equilateral triangle whose inscribed circle has radius $r$? I would like to learn how to deduce the formula.
I deduced the circle outside the triangle, so now I tried to do it with the circle inside the triangle, but I haven't arrived to a solution yet.
$\angle OBC = 30^\circ$, Hence $BM=r\cos 30^\circ = \frac{\sqrt3 r}{2}.$
Task given to you, find the length of $BC$.
Drop a perpendicular from the center of the circle to one of the sides and draw the radius to one of its endpoints. This should give you some nice triangles to play with :)
Best Answer
Make a construction like so
Here, $OC = r$, $BC = \frac{l}{2}$, $AB = l$. Since $ABC \sim BOC$, taking ratios, we get $AC = \frac{l^2}{4r}$.
By the Pythagorean theorem, $AB^2 = AC^2 + BC^2$,
Therefore, $$l = \sqrt{\frac{l^2}{4} + \frac{l^4}{16r^2}}$$
Simplifying, we get $l = r\sqrt{12}$
The area would be $\frac{\sqrt{3}}{4}l^2$, which would be
$$3\sqrt{3}r^2$$