PSAT/NMSQT Geometry and Trigonometry. Practice it free.

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

$180 - 120 = 60°$.

The circumference of a circle is $C = \pi d$. Substitute $d = 20$: $C = \pi(20) = 20\pi\textsf{.}$

Trapezoid area $= \dfrac{1}{2}(b_1 + b_2) \cdot h = \dfrac{1}{2}(7 + 10) \cdot 6 = \dfrac{1}{2}(17) \cdot 6 = 51$.

The area of a rhombus is half the product of the diagonals: $\dfrac{1}{2} \cdot 7 \cdot 8 = 28\text{ ft}^{2}\textsf{.}$

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 diameter length is $\sqrt{(6 - 0)^2 + (8 - 0)^2} = \sqrt{36 + 64} = \sqrt{100} = 10\textsf{.}$ The radius is half the diameter, so $r = \dfrac{10}{2} = 5\textsf{.}$

Circle A has center $(2, -1)$ and radius $4$. Shifting down $3$ moves the center to $(2, -4)$, so Circle B is $(x - 2)^2 + (y + 4)^2 = 16$.