Question

Measurement of a circle (Giving brainliest to correct answer)

Answer 1

Answer:

C. $9\pi$

Step-by-step explanation:

VU is tangent $\odot W$ at point U and WU is radius.

$\therefore WU \perp VU$

$\implies m\angle VUW=90\degree$

$\therefore \triangle VUW$ is right angle triangle.

WZ = WU = x - 1 (Radii of same circle)

WV = x - 1 + 2 = x + 1

By Pythagoras Theorem:

$WU^2 + VU^2 = WV^2$

$(x-1)^2 +(2x-4) ^2 = (x+1)^2$

$x^2 -2x +1+ 4x^2 -16x + 16=x^2 +2x +1$

$4x^2 -20x + 16=0$

$x^2 -5x + 4=0$

$x^2 -5x + 4=0$

Equating it with

$ax^2 +bx +c=0$

We find:

a = 1, b = - 5, c = 4

$b^2 - 4ac =(-5)^2 - 4(1)(4)= 25-16 =9$

$x=\frac{-b\pm \sqrt{b^2 - 4ac}}{2a}$

$x=\frac{-(-5)\pm \sqrt{9}}{2\times 1}$

$x=\frac{5\pm 3}{2}$

$x=\frac{5+3}{2}, \: x =\frac{5-3}{2}$

$x=\frac{8}{2}, \: x =\frac{2}{2}$

$x=4, \: x =1$

When x = 4, radius (r) = 4 - 1 = 3

When x = 1, radius (r) = 1 - 1 = 0, which is not possible.

Area of $\odot W$

$=\pi r^2$

$=\pi (3)^2$

$=9\pi$