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
![d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}](https://tex.z-dn.net/?f=d%3D%5Csqrt%7B%28x_2-x_1%29%5E2%2B%28y_2-y_1%29%5E2%7D)
The length of AB is
![AB=\sqrt{(2+3)^2+(4-2)^2}=\sqrt{25+4}=\sqrt{29}](https://tex.z-dn.net/?f=AB%3D%5Csqrt%7B%282%2B3%29%5E2%2B%284-2%29%5E2%7D%3D%5Csqrt%7B25%2B4%7D%3D%5Csqrt%7B29%7D)
The length of AD is
![AD=\sqrt{(-1+3)^2+(-3-2)^2}=\sqrt{4+25}=\sqrt{29}](https://tex.z-dn.net/?f=AD%3D%5Csqrt%7B%28-1%2B3%29%5E2%2B%28-3-2%29%5E2%7D%3D%5Csqrt%7B4%2B25%7D%3D%5Csqrt%7B29%7D)
Since AB=AD, therefore ABCD is a square. The area of the of square is
![A_1=a\times a=\sqrt{29}\times \sqrt{29} =29](https://tex.z-dn.net/?f=A_1%3Da%5Ctimes%20a%3D%5Csqrt%7B29%7D%5Ctimes%20%5Csqrt%7B29%7D%20%3D29)
The area of square is 29 units².
The area of a semicircle is
![A_2=\frac{\pi}{2}r^2](https://tex.z-dn.net/?f=A_2%3D%5Cfrac%7B%5Cpi%7D%7B2%7Dr%5E2)
Since AB is the diameter of the semicircle, therefore the radius of the semicircle is
![r=\frac{d}{2}=\frac{\sqrt{29}}{2}](https://tex.z-dn.net/?f=r%3D%5Cfrac%7Bd%7D%7B2%7D%3D%5Cfrac%7B%5Csqrt%7B29%7D%7D%7B2%7D)
The area of the semicircle is
![A_2=\frac{3.14}{2}(\frac{\sqrt{29}}{2})^2=40.3825](https://tex.z-dn.net/?f=A_2%3D%5Cfrac%7B3.14%7D%7B2%7D%28%5Cfrac%7B%5Csqrt%7B29%7D%7D%7B2%7D%29%5E2%3D40.3825)
The area of the figure is
![A=A_1+A_2=29+11.2825=40.3825\approx 40.4](https://tex.z-dn.net/?f=A%3DA_1%2BA_2%3D29%2B11.2825%3D40.3825%5Capprox%2040.4)
Therefore the area of the figure is 40.4 units². Option B is correct.
The answer is C) 14 years.
Plug it in to check:
12500(1+0.0525)^14
12500(2.05) = 25,587.01 (rounded to 25000)
Answer:
B is your answer to this question.
Step-by-step explanation: