Find the area of the shaded region in fig.,

The area of the square $\mathrm{ABCD}=(14)^{2} \mathrm{~cm}^{2}=196 \mathrm{~cm}^{2}$

$(\because$ side of the square $14 \mathrm{~cm})$

The sum of the areas of the semicircles APD and BPC

= 2 × {area of semicircle APD}

$(\because$ the areas of the two semicircles are equal)

$=2 \times\left\{\frac{1}{2} \pi r^{2}\right\}=\pi \times\left(\frac{\text { AD }}{2}\right)^{2}=\pi \times\left(\frac{14}{2}\right)^{2}$

$(\because \mathrm{AD}$ is diameter of the semicircle APD)

$=\frac{22}{7} \times 49 \mathrm{~cm}^{2}=154 \mathrm{~cm}^{2}$

The area of the shaded region

= The area of the square ABCD – The sum of the areas of the semicircles APD and BPC.

$=196 \mathrm{~cm}^{2}-154 \mathrm{~cm}^{2}=42 \mathrm{~cm}^{2}$