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.

  • 2
    $\begingroup$ Hint: draw radii that are perpendicular to each side. That will cut up the triangle into 3 smaller triangles whose areas are relatively easy to deduce. $\endgroup$ – Oiler Feb 18 '16 at 6:39
  • 1
    $\begingroup$ Closely related: math.stackexchange.com/questions/172601/… $\endgroup$ – David K Feb 18 '16 at 6:53

Make a construction like so enter image description here

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



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 $\frac{1}{2}bh=\frac{1}{2}(2BC)(AC)=\frac{1}{2}(2{\sqrt 3}r)(3r)=3{\sqrt 3}r^2$


protected by Community Oct 21 at 3:27

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