The area of a circle inside a triangle in relation to the area of the triangle

Question

So, I already calculated this and got π/√3 as the result. I'm just not sure whether or not that's correct.

The question was: a circle is drawn inside an equilateral triangle. What's the area of the circle in relation to the area of the triangle?

Thanks in advance!

Answer 1

Let $1$ be a length-side of the triangle.

Thus, the need relation it's $$\frac{\pi\left(\frac{1}{2\sqrt3}\right)^2}{\frac{\sqrt3}{4}}=\frac{\pi}{3\sqrt3}.$$

Answer 2

I will solve it in another approach, letting the radius of the circle be $1$. Let the triangle be $ABC$, with $D$ on $BC$, such that $BD=DC$. $E$ is on $AC$ and $F$ is on $AB$, with the same conditions as above. Let $O$ be the centre of the circle inscribed within the equilateral triangle.

So $\angle OBD = 30°, \angle BOD =60°, \angle ODB =90°$

In $\triangle ODB$, $tan30° = \frac{OD}{BD}$, so we have $\frac{1}{\sqrt3}= \frac{1}{BD}$. Now, let each side of the equilateral triangle be $x$.

We have $\frac{1}{\sqrt3}=\frac{1}{\frac{x}{2}}$. This means $x=2\sqrt3$.

Now, the area of the triangle is $\frac{\sqrt3}{4}x^2=3\sqrt3$.

Answer 3

A bit of geometry:

Given equilateral $\triangle ABC$, side length $a$, and inscribed circle radius $r$.

Let $h$ be the height on $AB$.

(heights = medians =angle bisectors = perp. bisectors )

Pythagoras: $h = (a/2)√3.$

1) Area of $\triangle ABC: $

$A_t= (1/2)ah = (1/4)a^2√3.$

2) Angle bisectors intersect at $M$.

Since $\triangle ABC$ is equilateral :

$M$ is the intersection of the medians which divide $h$ in the ratio $2:1$.

Hence $r=(1/3)h$.

Area of inscribed circle:

$A_c= πr^2= π( 1/3)^2h^2=π (a/2)^2 /3$

Ratio : $A_c/A_t = π√3/9.$

How about the circumscribed circle? Where is the centre of this circle, what is the radius?