Answer:
[tex] A(w) = w^2 + 5w - \frac{1}{8}\pi w^2 [/tex]
Step-by-step explanation:
A = the area of the region outside the semicircle but inside the rectangle
w = the width of the rectangle or diameter of the semicircle
Since "A" is determined by "w", therefore, "A" is a function of "w" = A(w).
A(w) = (area of rectangle) - (area of semicircle)
[tex] A(w) = (l*w) - (\frac{1}{2} \pi r^2) [/tex]
Where,
lenght of rectangle (l) = w + 5
width of rectangle (w) = w
r = ½*w = [tex] \frac{w}{2} [/tex]
Plug in the values:
[tex] A(w) = ((w + 5)*w) - (\frac{1}{2} \pi (\frac{w}{2})^2) [/tex]
[tex] A(w) = ((w + 5)*w) - (\frac{1}{2} \pi (\frac{w}{2})^2) [/tex]
Simplify
[tex] A(w) = (w^2 + 5w) - (\frac{1}{2} \pi (\frac{w^2}{4}) [/tex]
[tex] A(w) = w^2 + 5w - \frac{1}{2}*\pi*\frac{w^2}{4}* \pi [/tex]
[tex] A(w) = w^2 + 5w - \frac{1*\pi*w^2}{2*4} [/tex]
[tex] A(w) = w^2 + 5w - \frac{1*\pi w^2}{8} [/tex]
[tex] A(w) = w^2 + 5w - \frac{1}{8}\pi w^2 [/tex]
