Maine MTA High School (grade 10) Geometry. Practice it free.

$V = \pi r^{2} h = 3 \times 9 \times 2 = 54$.

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

The three angles sum to $180°$: $2(65-x)+(4x+10)=180$, so $140+2x=180$ and $x=20$. Each base angle is $65-20=45°$.

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 diameter length is $\sqrt{(13 - 1)^2 + (13 - (-3))^2} = \sqrt{12^2 + 16^2} = \sqrt{144 + 256} = \sqrt{400} = 20\textsf{.}$ The radius is half the diameter, so $r = \dfrac{20}{2} = 10\textsf{.}$

Circle A has center $(3, -1)$ and radius $3$. Shifting up $4$ moves the center to $(3, 3)$, so Circle B is $(x - 3)^2 + (y - 3)^2 = 9$.

The center is $(-2, 15)$ and the radius is $14\textsf{.}$ A point on the circle must satisfy $-2 - 14 \le a \le -2 + 14\textsf{,}$ so $-16 \le a \le 12\textsf{.}$ The value $-5$ is in this interval.

The smaller denominator is $b^{2} = 9$, so the semi-minor axis has length $b = \sqrt{9} = 3$.