This figure is made up of a quadrilateral and a semicircle. What is the area of this figure? Use 3.14 for pi. Round your answer to the nearest tenth. A.) 22.8 units² B.) 40.4 units² C.) 45.5 units² D.) 74.5 units²

Answer:The correct option is B. The area of the figure is 40.4 units².Step-by-step explanation:The line AB divides the figure in two parts one is a rectangle and another is semicircle.The distance formula is[tex]d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]The length of AB is[tex]AB=\sqrt{(2+3)^2+(4-2)^2}=\sqrt{25+4}=\sqrt{29}[/tex]The length of AD is[tex]AD=\sqrt{(-1+3)^2+(-3-2)^2}=\sqrt{4+25}=\sqrt{29}[/tex]Since AB=AD, therefore ABCD is a square. The area of the of square is[tex]A_1=a\times a=\sqrt{29}\times \sqrt{29} =29[/tex]The area of square is 29 units².The area of a semicircle is[tex]A_2=\frac{\pi}{2}r^2[/tex]Since AB is the diameter of the semicircle, therefore the radius of the semicircle is[tex]r=\frac{d}{2}=\frac{\sqrt{29}}{2}[/tex]The area of the semicircle is[tex]A_2=\frac{3.14}{2}(\frac{\sqrt{29}}{2})^2=40.3825[/tex]The area of the figure is[tex]A=A_1+A_2=29+11.2825=40.3825\approx 40.4[/tex]Therefore the area of the figure is 40.4 units². Option B is correct.