Question

The figure below shows part of a stained-glass window depicting the rising sun. Which function can be used to find the area of the region outside the semicircle but inside the rectangle?

Answer 1

Answer:

$A(w) = w^2 + 5w - \frac{1}{8}\pi w^2$

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)

$A(w) = (l*w) - (\frac{1}{2} \pi r^2)$

Where,

lenght of rectangle (l) = w + 5

width of rectangle (w) = w

r = ½*w = $\frac{w}{2}$

Plug in the values:

$A(w) = ((w + 5)*w) - (\frac{1}{2} \pi (\frac{w}{2})^2)$

$A(w) = ((w + 5)*w) - (\frac{1}{2} \pi (\frac{w}{2})^2)$

Simplify

$A(w) = (w^2 + 5w) - (\frac{1}{2} \pi (\frac{w^2}{4})$

$A(w) = w^2 + 5w - \frac{1}{2}*\pi*\frac{w^2}{4}* \pi$

$A(w) = w^2 + 5w - \frac{1*\pi*w^2}{2*4}$

$A(w) = w^2 + 5w - \frac{1*\pi w^2}{8}$

$A(w) = w^2 + 5w - \frac{1}{8}\pi w^2$