# Ex.12.3 Q10 Areas Related to Circles Solution - NCERT Maths Class 10

## Question

The area of an equilateral triangle \(ABC\) is \(17320.5\, \rm{cm}^2. \) With each vertex of the triangle as center, a circle is drawn with radius equal to half the length of the side of the triangle (see Figure). Find the area of the shaded region.

(Use \(\pi = 3.14\) and \(\sqrt 3 = 1.73205\))

## Text Solution

**What is known?**

(i) Area of an equilateral triangle \(\Delta {ABC} = 17320.5\,{\text{c}}{{\text{m}}^2}\)

(ii) with each vertex of a \(\Delta \) as center a circle is drawn with radius

\[\begin{align} = \frac{1}{2}({\text{ length of side of }}\Delta {ABC})\end{align}\]

(Use \(\pi = 3.14\) and \(\sqrt 3 = 1.73205\))

**What is unknown?**

Area of the shaded region.

**Reasoning:**

(i) Since area of triangle is given , we can find side of \(\Delta {ABC}\) using the formula of area of equilateral

\[\begin{align}\Delta = \frac{{\sqrt 3 }}{4}{({\text{side}})^2}\\{({\text{side}})^2} = \frac{{{\text{area}} \times 4}}{{\sqrt 3 }}\end{align}\]

(ii) Since radius\(\begin{align} {(r)} = \frac{1}{2} \end{align}\)( length of side of \(\Delta\) ) we can find \(\rm{}r\)

Also, all angles of an equilateral triangle are equal.

\(\therefore \;\) Angle subtended by each sector \(\begin{align}(\theta ) = \frac{{{{180}^0}}}{3} = {60^\circ }\end{align}\)

Using the formula

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

We can find Area of each sector \( \begin{align} = \frac{{60}}{{360}} \times \pi {r^2}\, = \frac{\pi }{6}{r^2}\end{align}\)

All sectors are equal as they have same radius \({r}\) and \(\begin{align}{ \theta = 6}{{\text{0}}^{{o}}}\end{align}\)

(iii) Visually from the figure it is clear that:

Area of the shaded region \(=\) Area of \(\Delta {ABC} - 3\; \times\) Area of each sector

\[\begin{align}&= 17320.5 - 3 \times \frac{{\pi {r^2}}}{6}\\ &= 17320.5 - \frac{{\pi {r^2}}}{2}\end{align}\]

which can be easily solved using \(\pi = 3.14\) (as given) and we already know \(\rm{}r\)

**Steps:**

(i) Area of equilateral \(\Delta = 17320.5\,{\text{c}}{{\text{m}}^2}\)

\[\begin{align}\frac{{\sqrt 3 }}{4}{({\text{side}})^2} &= 17320.5\,\,{\text{c}}{{\text{m}}^2}\\{\left( {{\text{side}}} \right)^2} &= \frac{{17320.5 \times 4}}{{\sqrt 3 }}\\&= { \frac{{17320.5 \times 4}}{{1.73205}}}\\&{\rm side = \sqrt {10000 \times 4} }\\&= { 100 \times 2}\\& ={ 200\,\,{\text{cm}}}\end{align}\]

\(\text{radius of each sector(r)}\)\[\begin{align}&= \frac{1}{2}\left( \rm{side} \right) \\&= \frac{1}{2}\left( {200} \right)\,{\text{cm}}\\ &= 100\,{\text{cm}}\end{align}\]

All interior angles of an equilateral \(\Delta \) are of measure \({60^ \circ }\)

And all \(3\) sectors are made using these interior angles.

\(\therefore\;\) Angles subtended at the center by each sector \((\theta) = {60^ \circ }\)

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

\(\text{Area of 3 sectors}\)\[\begin{align} &= 3 \times \frac{{{{60}^ \circ }}}{{{{360}^ \circ }}} \times \pi {r^2}\\&= 3 \times \frac{1}{6} \times 3.14 \times {\left( {100\rm{}\,cm} \right)^2}\\&= 15700\rm{}\,cm^2

\end{align}\]

Area of shaded region \(=\) Area of Quadrant \(\rm{}\Delta{ABC}\,-\) Area of 3 sector

\[\begin{align}&= 17320.5c{m^2} -15700c{m^2}\\&= 1620.5c{m^2}\end{align}\]