Questions: Jeff's fish tank has a base that is a trapezoid, as shown. The bases of the trapezoid are 24 and 28 inches, and the height is 9 inches. The tank contains 24.3 gallons of water. To the nearest inch, how deep is Jeff's tank? (A gallon is 231 cubic inches.) The tank is inches in height. (Round to the nearest inch as needed.)

Solution Steps

The area $A$ of a trapezoid is given by the formula: $$A = \frac{1}{2} \times (b_1 + b_2) \times h$$ where $b_1$ and $b_2$ are the lengths of the bases, and $h$ is the height.

Given: $$b_1 = 28 \text{ inches}$$ $$b_2 = 24 \text{ inches}$$ $$h = 9 \text{ inches}$$

Substitute the values into the formula: $$A = \frac{1}{2} \times (28 + 24) \times 9$$ $$A = \frac{1}{2} \times 52 \times 9$$ $$A = 26 \times 9$$ $$A = 234 \text{ square inches}$$

Given that 1 gallon is 231 cubic inches, convert 24.3 gallons to cubic inches: $$\text{Volume} = 24.3 \text{ gallons} \times 231 \text{ cubic inches/gallon}$$ $$\text{Volume} = 5613.3 \text{ cubic inches}$$

The volume $V$ of the tank is given by the product of the area of the base and the depth $d$: $$V = A \times d$$

Rearrange to solve for $d$: $$d = \frac{V}{A}$$

Substitute the values: $$d = \frac{5613.3 \text{ cubic inches}}{234 \text{ square inches}}$$ $$d \approx 24 \text{ inches}$$

Final Answer