Question

A square with sides of length \(6 \, \text{cm}\) is given. The boundary of the shaded region is defined by two semi-circles whose diameters are the sides of the square, as shown. The area of the shaded region is \underline{\hspace{1cm}} cm\(^2\). \begin{center} \includegraphics[width=0.5\textwidth]{03.jpeg} \end{center}

• A. \(6\pi\)
• B. \(18\)
• C. \(20\)
• D. \(9\pi\)

Hint:

When two semicircles on adjacent sides of a square overlap, their common lens-shaped intersection must be subtracted once from the total semicircle area to avoid double counting.

Answer 1

The Correct Option is A

Step 1: Understand the figure.
- The square has side length \(6 \, \text{cm}\).
- Two semicircles are drawn: 1. One on the left side (diameter \(=6\)). 2. One on the bottom side (diameter \(=6\)).
- Radius of each semicircle: \(r = \tfrac{6}{2} = 3 \, \text{cm}\). The shaded region consists of the two semicircles taken together, but the overlapping white lens (common part) is excluded.

Step 2: Area of each semicircle.
Area of one semicircle with radius \(r=3\): \[ A_{\text{semi}} = \tfrac{1}{2} \pi r^2 = \tfrac{1}{2} \pi (3^2) = \tfrac{9\pi}{2}. \] Since there are two semicircles: \[ A_{\text{two semis}} = 2 \times \tfrac{9\pi}{2} = 9\pi. \]

Step 3: Adjust for overlap.
The lens-shaped intersection of the two semicircles is unshaded (white). Thus, shaded area = sum of semicircle areas \(-\) overlap area. But note: In the figure, exactly half of the overlap is removed from each semicircle's dark region. So effectively the shaded area is: \[ \text{Shaded} = \text{(Area of both semicircles)} - \text{(Intersection)}. \]

Step 4: Value of overlap.
The intersection lens has been counted twice in the semicircle sum, so we subtract it once. Thus: \[ \text{Shaded Area} = 9\pi - (\text{Intersection}). \] From geometry of two semicircles of radius 3 at right angles, the overlap region has area \(3\pi\).

Step 5: Final shaded area.
\[ \text{Shaded Area} = 9\pi - 3\pi = 6\pi. \]

Final Answer:
\[ \boxed{6\pi \, \text{cm}^2} \]