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 semi - circle but inside the rectangle? a(w)=w² + 5w−\frac{1}{2}pi w² a(w)=w²−\frac{1}{2}pi w² a(w)=w² + 5w−\frac{1}{8}pi w² a(w)=w² + 5w+\frac{1}{8}pi w²

Explanation:

1. First, recall the area - formulas:

• The area of a rectangle with length \(l\) and width \(w\) is \(A_{rect}=l\times w\). In the given rectangle, if the width is \(w\) and the length is \(w + 5\), then \(A_{rect}=w(w + 5)=w^{2}+5w\).
• The area of a full - circle with radius \(r\) is \(A_{circle}=\pi r^{2}\). Since the diameter of the semi - circle is \(w\) (the width of the rectangle), the radius of the semi - circle \(r=\frac{w}{2}\), and the area of the semi - circle is \(A_{semicircle}=\frac{1}{2}\pi r^{2}=\frac{1}{2}\pi(\frac{w}{2})^{2}=\frac{1}{8}\pi w^{2}\).

1. Then, find the area of the region outside the semi - circle but inside the rectangle:

• The area \(A(w)\) of the region outside the semi - circle but inside the rectangle is the area of the rectangle minus the area of the semi - circle.
• \(A(w)=w^{2}+5w-\frac{1}{8}\pi w^{2}\).

Answer:

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