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:

Step1: Calculate the area of the semi - circle

The diameter of the semi - circle is $AB = 4$ mm, so the radius $r = 2$ mm. The area formula for a semi - circle is $A_{semicircle}=\frac{1}{2}\pi r^{2}$. Substituting $r = 2$ mm, we get $A_{semicircle}=\frac{1}{2}\pi(2)^{2}=2\pi$ $mm^{2}$.

Step2: Calculate the area of the square

The side - length of the square $ABCD$ is $4$ mm. The area formula for a square is $A_{square}=s^{2}$, where $s = 4$ mm. So $A_{square}=4\times4 = 16$ $mm^{2}$.

Step3: Calculate the area of the trapezoid

The bases of the trapezoid are $b_1 = 4$ mm and $b_2 = 8$ mm, and the height $h = 4$ mm. The area formula for a trapezoid is $A_{trapezoid}=\frac{(b_1 + b_2)h}{2}$. Substituting the values, we have $A_{trapezoid}=\frac{(4 + 8)\times4}{2}=\frac{12\times4}{2}=24$ $mm^{2}$.

Step4: Calculate the total area of the composite figure

The total area $A = A_{semicircle}+A_{square}+A_{trapezoid}$. Substituting the values of the areas of the sub - figures, we get $A=2\pi+16 + 24=(2\pi + 40)$ $mm^{2}$.

Answer:

$(2\pi + 40)$ $mm^{2}$