Ex.12.3 Q6 Areas Related to Circles Solution - NCERT Maths Class 10

In a circular table cover of radius \(\text{32 cm,}\) a design is formed leaving an equilateral triangle \(ABC\) in the middle as shown in the given figure. Find the area of the design.

Text Solution

 

What is known?

A circular table cover of radius \(\text{= 32 cm,}\) a design is formed leaving an equilateral triangle \(ABC\) in the middle as shown in figure.

What is unknown?

The area of the design.

Reasoning:   

Mark \(O\) as centre of the circle. Join \(BO\) and \(CO. \)

Since ,we know that equal chords of a circle subtends equal angles at the center and all sides of an equilateral triangle are equal.

\(\therefore \;\) Each side of \(\Delta \) will subtend equal angles at centre

\(\begin{align}\therefore \angle BOC = \frac{{{{360}^\circ }}}{3}\,\, = {120^\circ }\end{align}\)

Consider\(\Delta BOC\)

Drop a perpendicular from \(OM\) to \(BC\)

Since we know perpendicular from the center of circle to a chord bisects it

\(\Rightarrow \quad BM = MC\)

\(OB = OC\) (radii) and \(OM = OM \)(common)

\(\therefore \Delta {OBM} \cong \Delta {OCM} \) (by SSS congruency)

\(\Rightarrow \angle BOM\, =\,\angle COM\) (Corresponding Parts of Congruent Triangles are Congrue (CPCTC))

\(\therefore \,2\angle {BOM}=\angle {BOC}={{120}^{\circ }}\)

\[\begin{align}\angle {BOM}&=\frac{{{120}^{\circ }}}{2}={{60}^{\circ }} \\ \frac{{BM}}{{BO}}&=\sin \,{{60}^{\circ }}=\frac{\sqrt{3}}{2} \\{BM}&=\frac{\sqrt{3}}{2}\times {BO}=\frac{\sqrt{3}}{2}\times 32=16\sqrt{3}\\ \quad {BC}&=2 {BM}=32\sqrt{3}\,cm \\ \end{align}\]

Using formula of area of equilateral \(\Delta \)

\(\begin{align} = \frac{{\sqrt 3 }}{4}{({\text{side}})^2}\end{align}\)

We can find area of \(\Delta {ABC}\) since a side \(BC\) of \(\Delta \) is known.

Visually from figure it’s clear

Area of the design \(=\)  Area of circle \(- \)  Area of \(\Delta ABC\)

\[\begin{align}= \pi {r^2} - \frac{{\sqrt 3 }}{4}{({BC})^2}\end{align}\]

This can be solved with ease as both the radius of the circle and \(BC\) are known.

Steps:

Let the center of the circle be \(O.\) Join \(BO\) and \(CO.\)

Since equal chords of a circle subtend equal angles as its centre.

\(\therefore \;\) Sides \(AB, BC\) and \(AC\) of \(\Delta ABC\) will subtend equal angles at the centre of circle

\[\begin{align}\therefore \angle {BOC} = \frac{{{{360}^\circ }}}{3} = {120^\circ }\end{align}\]

Draw \({{OM}} \bot {{BC}}\)

In \(\Delta BOM\ {\rm and} \Delta COM\)

\(BO = CO\) (radii circle)

\(OM = OM\) (common)

\(BM = CM \).... (perpendicular drawn from the center of the circle to a chord bisects it)

\(\Delta {BOM} \cong \Delta {COM}\) (By SSS Congruency)

\(\therefore \;\angle {BOM} = \angle {COM} \dots \dots(2)\)(CPCT)

From figure

\[\begin{align}\angle {BOM} + \angle {COM} &= \angle {BOC}\\2\angle {BOM} &= \angle {BOC}\\ &= {120^ \circ }\left( {{\text{Using}}\,{\text{ }}\left( 2 \right)} \right)\\\angle {BOM} &= \frac{{{{120}^ \circ }}}{2} = {60^ \circ }\end{align}\]

In \(\Delta {BOM}\)

\[\begin{align}\frac{{BM}}{{BO}} &= \sin \,{60^ \circ } = \frac{{\sqrt 3 }}{2}\\\therefore {BM} &= \frac{{\sqrt 3 }}{2}{BO} = \frac{{\sqrt 3 }}{2} \times 32 = 16\sqrt 3 \\{\text{BC}} &= {\text{BM}} + {\text{CM}} = 2{\text{BM}}\qquad({\text{using}}(1))\\\,\,\,\,\,\,\,\, &= 2 \times 16\sqrt 3 \\\,\,\,\,\,\,\,\, &= 32\sqrt 3 \end{align}\]

Radius of circle \((r) =\text{ 32 cm}\)

From figure, we observe

Area of design = Area of circle - Area of \(\Delta {ABC}\)

\[\begin{align} &={\pi {r^2} - \frac{{\sqrt 3 }}{4}{{(BC)}^2}}\\ &={\pi {{(32)}^2} - \frac{{\sqrt 3 }}{4}{{(32\sqrt 3 )}^2}}\\ &= {\pi {{(32)}^2} - \frac{{\sqrt 3 }}{4}{{(32)}^2} \times 3}\\ &={\frac{{22528}}{7} \times 1024 - 32 \times 24\sqrt 3 }\\ &= {\frac{{22528}}{7} - 768\sqrt 3}\end{align}\]

Area of Design \(\begin{align} = \left( {\frac{{22528}}{7} - 768\sqrt 3 } \right)\rm{cm^2}\end{align}\)