# Ex.12.3 Q4 Areas Related to Circles Solution - NCERT Maths Class 10

## Question

Find the area of the shaded region in the given figure, where a circular arc of radius \(\text{6 cm}\) has been drawn with vertex \(O\) of an equilateral triangle \(OAB\) of side \(\text{12 cm}\) as centre.

## Text Solution

**What is known?**

A circular arc of radius \(\text{= 6 cm}\) is drawn with vertex \(O\) of an equilateral \(\Delta {OAB}\) of side \(\text{12 cm}\) as centre.

**What is unknown?**

Area of the shaded region.

**Reasoning:**

From the figure it is clear that the shaded region has an overlap region of area of a sector of angle \({60^ \circ }\) (since each angle of an equilateral \(\Delta \) is of measure \({60^ \circ }\) area of a circle with radius \(\text{6 cm}\) and area of triangle \(OAB\) with side \(\text{12 cm.}\)

\(\therefore\;\)Area of shaded region \(=\) Area of circle with radius \(\text{6 cm}\; +\) Area of \(\Delta {OAB}\) - Area of sector of angle \({60^ \circ }\)

\[\begin{align} = \pi {(6)^2} + \frac{{\sqrt {3{{(12)}^2}} }}{4} - \frac{{{{60}^\circ}}}{{{{360}^\circ}}} \times \pi {(6)^2}\end{align}\]

Using formula, Area of an equilateral \(\begin{align} \Delta = \frac{{\sqrt 3 }}{4}{({\mathop{\text sid}\nolimits} e)^2}\end{align}\)

Area of the sector of angle \(\begin{align} \theta = \frac{\theta }{{{{360}^\circ }}} \times \pi {r^2} \end{align}\)

Where \(r \) is the radius of the circle

**Steps:**

Radius of circle \((r) =\text{ 6 cm}\)

Side of equilateral \(\Delta {OAB(s)} = 12\,{\text{cm}}\)

We know each interior angle of equilateral \(\Delta = {60^\circ }\)

Since they overlap, part of area of a sector \(OCD\) is in area of the circle and triangle.

\(\therefore \;\) Area of shaded region = Area of circle \(+\) Area of \(\Delta {OAB}\,- \)Area of sector \(OCD\) of angle \(60^\circ\)

\[\begin{align} &= \pi {\left( 6 \right)^2} + \frac{{\sqrt 3 }}{4}{\left( {12} \right)^2} - \frac{{60}}{{360}} \times \pi {\left( 6 \right)^2}\\ &= 36\pi + 36\sqrt 3 - \frac{{36}}{6}\pi \\ &= 36\left( {\pi + \sqrt 3 - \frac{\pi }{6}} \right)\\ &= 36\left( {\sqrt 3 + \frac{{5\pi }}{6}} \right)\\ &= 36\sqrt 3 + 30\pi \\ &= 36\sqrt 3 + 30 \times \frac{{22}}{7}\\ &= 36\sqrt 3 + \frac{{660}}{7}\,{\text{c}}{{\text{m}}^2}\end{align}\]