Answer:
C. [tex]9\pi[/tex]
Step-by-step explanation:
We proceed to solve for [tex]x[/tex] by the Pythagorean Theorem:
[tex](x+1)^{2} = (x-1)^{2} + (2\cdot x - 4)^{2}[/tex] (1)
[tex]x^{2} + 2\cdot x + 1 = (x^{2}-2\cdot x + 1) + (4\cdot x^{2}-16\cdot x + 16)[/tex]
[tex]x^{2} + 2\cdot x + 1 = 5\cdot x^{2}-18\cdot x + 17[/tex]
[tex]4\cdot x^{2} -20\cdot x + 16 = 0[/tex]
[tex](2\cdot x)^{2} -10\cdot (2\cdot x) + 16 = 0[/tex]
[tex](2\cdot x -8)\cdot (2\cdot x -2) = 0[/tex]
There are two different solutions:
[tex]x_{1} = 4[/tex], [tex]x_{2} = 1[/tex]
A solution of [tex]x[/tex] is considered realistic when the length of every side of the triangle is a not negative number.
[tex]x_{1} = 4[/tex]
[tex]UV = 2\cdot (4) - 4[/tex]
[tex]UV = 4[/tex]
[tex]WU = 4 - 1[/tex]
[tex]WU = 3[/tex]
[tex]WV = 4 + 1[/tex]
[tex]WV = 5[/tex]
[tex]x_{2} = 1[/tex]
[tex]UV = 2\cdot (1) - 4[/tex]
[tex]UV = -2[/tex]
[tex]WU = 1 - 1[/tex]
[tex]WU = 0[/tex]
[tex]WV = 1 + 1[/tex]
[tex]WV = 2[/tex]
The only realistic set of solutions is for [tex]x = 4[/tex], and the radius of the circle is represented by the line segment [tex]WU[/tex]: [tex]r = 3[/tex]
And the area of the circle ([tex]A[/tex]) is calculated from the following formula:
[tex]A = \pi\cdot r^{2}[/tex]
[tex]A = \pi\cdot (9)[/tex]
[tex]A = 9\pi[/tex]
The correct answer is C.
