Question

find the perimeter and area of each figure. round to the nearest tenth if necessary.
11.
12.

Explanation:

Step1: Calculate the perimeter of figure 11

The perimeter is the sum of all the sides.

$$P = 23+12 + 4+34+41+9$$

$$P=123\ \text{ft}$$

Step2: Calculate the area of figure 11

Divide the figure into a rectangle and a trapezoid.
The area of the rectangle: \(A_1=23\times12 = 276\ \text{ft}^2\)
The bases of the trapezoid: \(b_1 = 12\ \text{ft}\), \(b_2=40\ \text{ft}\), height \(h = 9\ \text{ft}\)
The area of the trapezoid: \(A_2=\frac{(12 + 40)\times9}{2}=234\ \text{ft}^2\)
Total area \(A=A_1+A_2=276+234 = 510\ \text{ft}^2\)

Step3: Calculate the perimeter of figure 12

The perimeter of the semicircle part: \(P_{\text{semicircle}}=\frac{1}{2}\times2\pi r=\pi r\), \(r = 15\ \text{yd}\) (since diameter \(d = 30\ \text{yd}\)), \(P_{\text{semicircle}}=15\pi\approx47.1\ \text{yd}\)
The perimeter of the whole figure: \(P=47.1+17+17+23+23\)

$$P\approx127.1\ \text{yd}$$

Step4: Calculate the area of figure 12

The area of the semicircle: \(A_{\text{semicircle}}=\frac{1}{2}\pi r^2=\frac{1}{2}\pi\times15^2=\frac{225\pi}{2}\approx353.4\ \text{yd}^2\)
The area of the kite (using the formula \(A=\frac{d_1\times d_2}{2}\), \(d_1 = 30\ \text{yd}\), \(d_2=23 - 15=8\ \text{yd}\)): \(A_{\text{kite}}=\frac{30\times8}{2}=120\ \text{yd}^2\)
Total area \(A = 353.4+120=473.4\ \text{yd}^2\)

Answer:

1. \(P = 123\ \text{ft}\), \(A = 510\ \text{ft}^2\)
2. \(P\approx127.1\ \text{yd}\), \(A\approx473.4\ \text{yd}^2\)