Question

Can please someone help me? The question is in the picture. Thank you

Answer 1

C.
`9\pi`

Explanation:

We proceed to solve for
`x` by the Pythagorean Theorem:

`(x+1)^(2) = (x-1)^(2) + (2\cdot x - 4)^(2)` (1)

`x^(2) + 2\cdot x + 1 = (x^(2)-2\cdot x + 1) + (4\cdot x^(2)-16\cdot x + 16)`

`x^(2) + 2\cdot x + 1 = 5\cdot x^(2)-18\cdot x + 17`

`4\cdot x^(2) -20\cdot x + 16 = 0`

`(2\cdot x)^(2) -10\cdot (2\cdot x) + 16 = 0`

`(2\cdot x -8)\cdot (2\cdot x -2) = 0`

There are two different solutions:

`x_(1) = 4`,
`x_(2) = 1`

A solution of
`x` is considered realistic when the length of every side of the triangle is a not negative number.

`x_(1) = 4`

`UV = 2\cdot (4) - 4`

`UV = 4`

`WU = 4 - 1`

`WU = 3`

`WV = 4 + 1`

`WV = 5`

`x_(2) = 1`

`UV = 2\cdot (1) - 4`

`UV = -2`

`WU = 1 - 1`

`WU = 0`

`WV = 1 + 1`

`WV = 2`

The only realistic set of solutions is for
`x = 4`, and the radius of the circle is represented by the line segment
`WU`:
`r = 3`

And the area of the circle (
`A`) is calculated from the following formula:

`A = \pi\cdot r^(2)`

`A = \pi\cdot (9)`

`A = 9\pi`

The correct answer is C.