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jolli1 [7]
3 years ago
12

Measurement of a circle (Giving brainliest to correct answer)

Mathematics
1 answer:
xxMikexx [17]3 years ago
4 0

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

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