Correct. The diameter of the semicircle is 4 cm. This is equal to the length of one of the bases of the trapezoid. The other base is $\displaystyle{3}+{3}={6}\usepackage{cmbright}\Huge\$ cm. The height of the trapezoid is $\displaystyle{9}-{5}={4}\usepackage{cmbright}\Huge\$ cm. Now find the area of each part:

Area of the semicircle: $\displaystyle{A}=\frac{{\pi{{r}}^{{2}}}}{{2}}=\frac{{\pi{{\left({2}\right)}}^{{2}}}}{{2}}=\frac{{\pi\cdot{4}}}{{2}}\approx\frac{{{3.14}\cdot{4}}}{{2}}={6.28}\usepackage{cmbright}\Huge\$

Area of the trapezoid: $\displaystyle{A}=\frac{{{b}_{{1}}+{b}_{{2}}}}{{2}}\cdot{h}=\frac{{{4}+{6}}}{{2}}\cdot{4}=\frac{{10}}{{2}}\cdot{4}={5}\cdot{4}={20}\usepackage{cmbright}\Huge\$

Area of the rectangle: $\displaystyle{A}={l}\cdot{w}={5}\cdot{3}={15}\usepackage{cmbright}\Huge\$

The total area is the sum of these parts: $\displaystyle{6.28}+{20}+{15}={41.28}\usepackage{cmbright}\Huge\$

The approximate area of the composite figure is 41.28 square centimeters.