PSAT 8/9 Geometry. Practice it free.

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

$3 \times 2 = 6$.

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

$A = \pi r^2 = 3 \times 9 = 27$.

Set up the proportion $\dfrac{12\text{ kilograms}}{1\text{ cubic meter}} = \dfrac{540\text{ kilograms}}{x\text{ cubic meters}}\textsf{.}$ Cross-multiplying gives $12x = 540\textsf{,}$ so $x = 45\text{ m}^{3}\textsf{.}$

The area ratio is $\dfrac{64}{16} = 4 = k^{2}\textsf{,}$ so the scale factor is $k = 2\textsf{.}$ Then $AB = 5 \cdot 2 = 10\textsf{.}$

Perimeter $P = 2\ell + 2w$, so $30 = 2\ell + 2(5)$. Then $30 = 2\ell + 10$, so $2\ell = 20$ and $\ell = 10\text{ m}\textsf{.}$

$V = \ell w h$ and $\ell = 2w$, so $72 = (2w)(w)(4) = 8w^2$. Then $w^2 = 9$ and $w = 3\text{ cm}\textsf{.}$