# Ex.12.3 Q3 Areas Related to Circles Solution - NCERT Maths Class 10

## Question

Find the area of the shaded region in the given figure, if \(ABCD\) is a square of side \(\text{14 cm}\) and \(APD\) and \(BPC\) are semicircles.

## Text Solution

**What is known?**

\(ABCD\) is a square of side (*\(l\)*) \(\text{14 cm.}\) \(APD\) and \(BPC\) are semicircles.

**What is unknown?**

Area of the shaded region.

**Reasoning:**

(i) From figure it is clear that diameter of both semicircles \(=\) side of square \(=\text{14 cm}\)

\(\therefore \;\) Radius of each semicircle \(\begin{align} (r) = \frac{{14}}{2} = 7{\text{cm}}\end{align}\)

(ii) Visually, it is clear

Area of shaded region = Area of square \(ABCD\, -\) (Area of semicircle \(APD\, +\) Area of semicircle \(BPC\))

\[\begin{align} &= {\left( \rm{side} \right)^2} - \left( {\frac{{\pi {r^2}}}{2} + \frac{{\pi {r^2}}}{2}} \right)\\ &= {\left( \rm{side} \right)^2} - \pi {r^2}\end{align}\]

**Steps:**

Since semicircles \(APD\) and \(BPC\) are drawn using sides \(AD\) and \(BC\) respectively as diameter.

\(\therefore \;\)Diameter of each semicircle \(=\,\text{14 cm}\)

Radius of each semicircle \(\begin{align}(r) = \frac{{14}}{2} = 7{\text{cm}}\end{align}\)

Area of shaded region \(=\) Area of square \(ABCD \,- \) (Area of semicircle \(APD \,+ \)Area of semicircle \(BPC\))

\[\begin{align}&{ = {{({\text{side}})}^2} - \left( {\frac{1}{2}\pi {r^2} + \frac{1}{2}\pi {r^2}} \right)}\\&{ = {{14}^2} - \pi \times {{(7)}^2}}\\ &{ = 196 - \frac{{22}}{7} \times 7 \times 7}\\&{ = 196 - 154}\\&{ = 42\,{\text{c}}{{\text{m}}^2}}\end{align}\]