# Two Circles Inscribed in Triangles in a Semi-Circle +-- {.image} [[TwoCirclesInscribedinTrianglesinaSemiCircle.png:pic]] > A circle of area $8$ is inscribed in an equilateral triangle. What's the area of the other circle? =-- ## Solution by [[Lengths in an Equilateral Triangle]] +-- {.image} [[TwoCirclesinTrianglesinaSemiCircleLabelled.png:pic]] =-- In the above diagram, let $r$ be the radius of the semi-circle, $p$ the radius of the purple circle, and $q$ the radius of the orange circle. The line $O D$ [[bisects]] $A C$, so angle $C \hat{O} D$ is half of angle $C \hat{O} A$. As the [[angles at a point on a line]] add up to $180^\circ$ and the [[interior angle]] of an [[equilateral triangle]] is $60^\circ$, angle $C \hat{O} A$ is $120^\circ$ so angle $C \hat{O} D$ is $60^\circ$. Then $C O E$ is half an [[equilateral triangle]], so $O E$ is half of $O C$, which is $r$, and then the radius of the smaller circle is $q = \frac{r}{4}$. From the relationships between the [[lengths in an equilateral triangle]], the radius of the circle in the equilateral triangle is $\frac{1}{2\sqrt{3}}$ of the side length, which is $r$. Hence $p = \frac{1}{2\sqrt{3}} r$, and so $r = 2 \sqrt{3} p$. Then $q = \frac{2 \sqrt{3}}{4} p$ so $ \pi q^2 = \frac{3}{4} \pi p^2$. Since $\pi p^2 = 8$, the area of the smaller circle is $6$.