Question

what is the area of the composite figure if \\(\overline{ab} \cong \overline{bc} \cong \overline{cd} \cong \overline{da} \cong \overline{dn}\\)? \\(\bigcirc\\ (2\pi + 28)\\ \text{mm}^2\\) \\(\bigcirc\\ (2\pi + 32)\\ \text{mm}^2\\) \\(\bigcirc\\ (2\pi + 40)\\ \text{mm}^2\\) \\(\bigcirc\\ (2\pi + 48)\\ \text{mm}^2\\)
(image of composite figure: top is a semicircle with ab as diameter (2 mm marked), middle is a rectangle abcd with right angles, bottom is a trapezoid mdck with base mk = 8 mm, and dn is a segment with right angle at n.)

Explanation:

Step1: Calculate the area of the semicircle

The formula for the area of a semicircle is \(A_{semicircle}=\frac{1}{2}\pi r^{2}\). Given \(AB = 2\) mm (radius \(r = 2\) mm), so \(A_{semicircle}=\frac{1}{2}\pi\times(2)^{2}=2\pi\) \(mm^{2}\).

Step2: Calculate the area of the square

The side - length of the square is \(AB = 2\) mm. The area of a square \(A_{square}=s^{2}\), so \(A_{square}=(2\times2)\times4 = 16\) \(mm^{2}\) (since \(AB = BC=CD = DA = 2\) mm, and the square - like part has an area composed of \(4\) unit - squares of side \(2\) mm).

Step3: Calculate the area of the trapezoid

The formula for the area of a trapezoid is \(A_{trapezoid}=\frac{(a + b)h}{2}\). Here, \(a = 4\) mm (\(CD = 4\) mm as \(CD=DA = 2\) mm and there are two segments), \(b = 8\) mm, \(h = 2\) mm. So \(A_{trapezoid}=\frac{(4 + 8)\times2}{2}=12\) \(mm^{2}\).

Step4: Calculate the total area

The total area \(A=A_{semicircle}+A_{square}+A_{trapezoid}\). Substitute the values: \(A = 2\pi+16 + 12=(2\pi + 28)\) \(mm^{2}\).

Answer:

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