3
$\begingroup$

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.

$\endgroup$
  • 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
5
$\begingroup$

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

$$3\sqrt{3}r^2$$

$\endgroup$
1
$\begingroup$

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

$\endgroup$

protected by Community Oct 21 at 3:27

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.