Look at the picture in the attachment.
The area of the figure is equal to the sum of the areas A₁ and A₂.
[tex]A_1=9 \ mm \times 4.8 \ mm=43.2 \ mm^2 \\
A_2=\frac{9 \ mm \times 7.7 \ mm}{2}=\frac{69.3 \ mm^2}{2}=34.65 \ mm^2 \\ \\
A=43.2 \ mm^2 + 34.65 \ mm^2=77.85 \ mm^2 \approx 77.9 \ mm^2[/tex]
The answer is A.
