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?
the options are:
$a(w)=w^{2}+5w - \frac{1}{2}\pi w^{2}$
$a(w)=w^{2}+5w + \frac{1}{8}\pi w^{2}$
$a(w)=w^{2}-\frac{1}{2}\pi w^{2}$
$a(w)=w^{2}+5w - \frac{1}{8}\pi w^{2}$

Explanation:

Step1: Find Area of Rectangle

The rectangle has length \( w + 5 \) and width \( w \). Area of rectangle is \( \text{length} \times \text{width} = w(w + 5)=w^{2}+5w \).

Step2: Find Area of Semicircle

The diameter of the semicircle is \( w \), so radius \( r=\frac{w}{2} \). Area of a full circle is \( \pi r^{2} \), so area of semicircle is \( \frac{1}{2}\pi r^{2}=\frac{1}{2}\pi(\frac{w}{2})^{2}=\frac{1}{8}\pi w^{2} \)? Wait, no, wait. Wait, diameter is \( w \), so radius is \( \frac{w}{2} \)? Wait, no, maybe I misread. Wait, the semicircle is at the bottom, with base \( w \), so diameter is \( w \), radius \( \frac{w}{2} \)? Wait, no, wait the options have \( \frac{1}{2}\pi w^{2} \) or \( \frac{1}{8}\pi w^{2} \)? Wait, no, let's re - check. Wait, the rectangle's width is \( w \), and the semicircle is inside the rectangle, with diameter equal to \( w \). So area of semicircle is \( \frac{1}{2}\pi(\frac{w}{2})^{2}=\frac{1}{8}\pi w^{2} \)? No, that can't be. Wait, maybe the diameter is \( w \), so radius \( r = \frac{w}{2} \), area of semicircle is \( \frac{1}{2}\pi r^{2}=\frac{1}{2}\pi(\frac{w}{2})^{2}=\frac{\pi w^{2}}{8} \)? But the options have \( \frac{1}{2}\pi w^{2} \) and \( \frac{1}{8}\pi w^{2} \). Wait, maybe I made a mistake. Wait, the region outside the semicircle but inside the rectangle is area of rectangle minus area of semicircle. Wait, rectangle area: \( w(w + 5)=w^{2}+5w \). Semicircle area: the diameter is \( w \), so radius \( \frac{w}{2} \), area of semicircle is \( \frac{1}{2}\pi(\frac{w}{2})^{2}=\frac{\pi w^{2}}{8} \)? No, that's not matching. Wait, maybe the semicircle has radius \( \frac{w}{2} \), but maybe the diameter is \( w \), so area of semicircle is \( \frac{1}{2}\pi(\frac{w}{2})^{2}=\frac{\pi w^{2}}{8} \). Wait, but the last option is \( A(w)=w^{2}+5w-\frac{1}{8}\pi w^{2} \). Wait, let's recalculate:

Wait, rectangle: length \( w + 5 \), width \( w \), area \( w(w + 5)=w^{2}+5w \).

Semicircle: diameter \( w \), so radius \( r=\frac{w}{2} \). Area of semicircle \(=\frac{1}{2}\pi r^{2}=\frac{1}{2}\pi(\frac{w}{2})^{2}=\frac{\pi w^{2}}{8} \).

So the area outside the semicircle and inside the rectangle is area of rectangle minus area of semicircle: \( (w^{2}+5w)-\frac{1}{8}\pi w^{2}=w^{2}+5w - \frac{1}{8}\pi w^{2} \), which is the last option.

Answer:

\( A(w)=w^{2}+5w-\frac{1}{8}\pi w^{2} \) (the last option among the given choices)