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

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Question

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.

 Video Solution
Areas Related To Circles
Ex 12.3 | Question 6

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