Arc $AC$ is a quarter-circle with center $B$. The shaded region $ABC$ is "rolled" along a straight board $PQ$ until it reaches its original orientation for the first time with point $B$ landing at point $B^{\prime}$. If $BC = \frac{2}{\pi}$ cm, what is the length of the path that point $B$ travels? Express your answer in the simplest form.

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You have 2 sides that are 2 cms and you have 6 sides that are 1/2 cm and you have 4 sides that are 1 cm. the smallest sides are 1/2 the medium sized sides are 1 and the smallest are 1/2. so to find the perimeter you have to add them all up together so your answer will be 11 cm's

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x^2 + 81

we can factor this using difference of square binomials.

rewrite the equation:

x^2 - (-81)

factor:

(a^2 - b^2) = (√a^2 + √b^2)(√a^2 - √b^2)

(x^2 - (-81) = (x + √-81)(x - √-81)

we can remove the - from under the radical by multiplying it by i.

(x^2 - (-81) = (x + i√81)(x - i√81)

(x^2 - (-81) = (x + 9i)(x - 9i)

(x + 9i)(x - 9i)

we can check our answer by using foil:

x^2 - 9ix + 9ix - 81i^2

x^2 - 81^2

i^2 = -1

x^2 + 81