Question

what is the area of the composite figure if $overline{ab}congoverline{bc}congoverline{cd}congoverline{da}congoverline{dn}$?
$(2pi + 28)\text{ mm}^2$
$(2pi + 32)\text{ mm}^2$
$(2pi + 40)\text{ mm}^2$
$(2pi + 48)\text{ mm}^2$

Explanation:

1. First, identify the component - shapes of the composite figure:

• The composite figure consists of a semi - circle, a rectangle, and a trapezoid.
• Semi - circle:
• The diameter of the semi - circle is \(AB = 2\) mm, so the radius \(r = 1\) mm. The area of a semi - circle is \(A_{semicircle}=\frac{1}{2}\pi r^{2}=\frac{1}{2}\pi(1)^{2}=\frac{\pi}{2}\times2 = \pi\) mm².
• Rectangle:
• The length of the rectangle is \(AB = 2\) mm and the width is \(AD = 2\) mm. The area of the rectangle \(A_{rectangle}=l\times w=2\times4 = 8\) mm².
• Trapezoid:
• The bases of the trapezoid are \(b_1 = 2\) mm and \(b_2 = 8\) mm, and the height \(h = 4\) mm. The area of a trapezoid is \(A_{trapezoid}=\frac{(b_1 + b_2)h}{2}=\frac{(2 + 8)\times4}{2}=20\) mm².

1. Then, find the total area of the composite figure:

• \(A = A_{semicircle}+A_{rectangle}+A_{trapezoid}\).
• \(A=\pi\times2+8 + 20=(2\pi+28)\) mm².

Answer:

\((2\pi + 28)\text{ mm}^2\)