MO MAP Geometry Circles. Practice it free.

Sector area $= \dfrac{60}{360} \times \pi r^2 = \dfrac{60}{360} \times 3 \times 16 = 8$.

The area of the whole circle is $\pi r^{2} = \pi \cdot 12^{2} = 144\pi\textsf{.}$ A $60^{\circ}$ sector is $\dfrac{60}{360} = \dfrac{1}{6}$ of the circle, so the area is $\dfrac{1}{6} \cdot 144\pi = 24\pi\text{ m}^{2}\textsf{.}$

The circumference is $2\pi r = 2\pi \cdot 18 = 36\pi\textsf{.}$ A $30^{\circ}$ arc is $\dfrac{30}{360} = \dfrac{1}{12}$ of the circle, so the arc length is $\dfrac{1}{12} \cdot 36\pi = 3\pi\textsf{.}$

This is a right triangle ($5$-$12$-$13$). Area $= \dfrac{1}{2} \times 5 \times 12 = 30$.

The area of the whole circle is $\pi r^{2} = \pi \cdot 6^{2} = 36\pi\textsf{.}$ A $90^{\circ}$ sector is $\dfrac{90}{360} = \dfrac{1}{4}$ of the circle, so the area is $\dfrac{1}{4} \cdot 36\pi = 9\pi\text{ m}^{2}\textsf{.}$

The circumference of the full circle is $2\pi r = 2\pi \cdot 10 = 20\pi\textsf{.}$ The curved edge is half of that, so $\dfrac{1}{2} \cdot 20\pi = 10\pi\textsf{.}$

Area $= \dfrac{1}{2} \times 10 \times 12 = 60$.

The area of the whole circle is $\pi r^{2} = \pi \cdot 4^{2} = 16\pi\textsf{.}$ A $60^{\circ}$ sector is $\dfrac{60}{360} = \dfrac{1}{6}$ of the circle, so the area is $\dfrac{1}{6} \cdot 16\pi = \dfrac{8\pi}{3}\text{ in}^{2}\textsf{.}$