4. In an equilateral triangle with side $a$, a circle is inscribed. From the vertex of the triangle, a second circle with radius $0.5 a$ is drawn. Find the area enclosed between these circles.
Solution. Let $ABC$ be the given equilateral triangle with side length $a$. We will make the necessary constructions (see the figure). The required area is the difference between the area of a $60^{\circ}$ sector of the second circle and the difference of one-third of the area of the triangle and the area of the inscribed circle. The area of the sector is $S_{c}=\frac{1}{6} \pi(0.5 a)^{2}=\frac{\pi a^{2}}{24}$, the area of the triangle is $S_{\Delta}=\frac{\sqrt{3} a^{2}}{4}$, and the area of the inscribed circle is $S_{o}=\frac{\pi a^{2}}{12}$. Therefore, the required area is $S=S_{c}-\frac{1}{3}\left(S_{\Delta}-S_{o}\right)=\frac{\pi a^{2}}{24}-\frac{1}{3}\left(\frac{\sqrt{3} a^{2}}{4}-\frac{\pi a^{2}}{12}\right)=\frac{a^{2}}{72}(5 \pi-6 \sqrt{3})$.

Answer: $\frac{a^{2}}{72}(5 \pi-6 \sqrt{3})$ sq. units.

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