what is the area of this figure?\nif necessary, round your answer to the nearest hundredth.\nsquare inches

Answer

The figure consists of two rectangles and one quarter-circle.

The top rectangle has dimensions $8\text{ in}$ by $4\text{ in}$. $$A_{1} = 8 \times 4 = 32\text{ in}^{2}$$

The bottom rectangle has dimensions $4\text{ in}$ by $5\text{ in}$. $$A_{2} = 4 \times 5 = 20\text{ in}^{2}$$

The radius is the horizontal segment labeled $6\text{ in}$. $$r = 6\text{ in}$$

The area is one-fourth of a full circle's area. $$A_{3} = \frac{1}{4} \pi r^{2} = \frac{1}{4} \pi (6)^{2} = 9\pi \approx 28.2743\text{ in}^{2}$$

Add the areas of all three components together. $$A_{total} = 32 + 20 + 28.2743 = 80.2743\text{ in}^{2}$$

Round the final result to two decimal places. $$A_{total} \approx 80.27\text{ in}^{2}$$

80.27 square inches