Answer:
C. [tex] 9\pi [/tex]
Step-by-step explanation:
VU is tangent [tex] \odot W[/tex] at point U and WU is radius.
[tex] \therefore WU \perp VU[/tex]
[tex] \implies m\angle VUW=90\degree[/tex]
[tex] \therefore \triangle VUW[/tex] is right angle triangle.
WZ = WU = x - 1 (Radii of same circle)
WV = x - 1 + 2 = x + 1
By Pythagoras Theorem:
[tex] WU^2 + VU^2 = WV^2 [/tex]
[tex] (x-1)^2 +(2x-4) ^2 = (x+1)^2 [/tex]
[tex]x^2 -2x +1+ 4x^2 -16x + 16=x^2 +2x +1[/tex]
[tex]4x^2 -20x + 16=0[/tex]
[tex]x^2 -5x + 4=0[/tex]
[tex]x^2 -5x + 4=0[/tex]
Equating it with
[tex]ax^2 +bx +c=0[/tex]
We find:
a = 1, b = - 5, c = 4
[tex] b^2 - 4ac =(-5)^2 - 4(1)(4)= 25-16 =9[/tex]
[tex] x=\frac{-b\pm \sqrt{b^2 - 4ac}}{2a} [/tex]
[tex] x=\frac{-(-5)\pm \sqrt{9}}{2\times 1} [/tex]
[tex] x=\frac{5\pm 3}{2} [/tex]
[tex] x=\frac{5+3}{2}, \: x =\frac{5-3}{2} [/tex]
[tex] x=\frac{8}{2}, \: x =\frac{2}{2} [/tex]
[tex] x=4, \: x =1 [/tex]
When x = 4, radius (r) = 4 - 1 = 3
When x = 1, radius (r) = 1 - 1 = 0, which is not possible.
Area of [tex] \odot W [/tex]
[tex] =\pi r^2 [/tex]
[tex] =\pi (3)^2 [/tex]
[tex] =9\pi [/tex]
