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.
$\endgroup$ 22 Answers
$\begingroup$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
$$3\sqrt{3}r^2$$
$\endgroup$ $\begingroup$Using the circle in SS_C4's answer, We know that triangles BOC and ABC are similar, so both are 30-60-90 triangles.
Applying the properties of 30-60-90 triangles $(r,\,{\sqrt 3}r,\,2r)$: We have, $OC=r,\,BC= {\sqrt 3}r,\,\,\text{and }\,AC={\sqrt 3}({\sqrt 3}r)=3r$
Therefore, the area is $\frac{1}{2}bh=\frac{1}{2}(2BC)(AC)=\frac{1}{2}(2{\sqrt 3}r)(3r)=3{\sqrt 3}r^2$
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