Ex.12.2 Q13 Areas Related to Circles Solution - NCERT Maths Class 10


A round table cover has six equal designs as shown in figure. If the radius of the cover is \({28 \,\rm{cm},}\) find the cost of making the designs at the rate of

\(\rm{₹}\, 0.35 \)per \(\rm{cm^2.} \)(Use\(\sqrt{3}={ }1.7\) )

 Video Solution
Areas Related To Circles
Ex 12.2 | Question 13

Text Solution

What is known?

A round table cover has \(6\) equal designs as shown in the figure. The radius of the cover \(= 28\, \rm{ cm}\)and the rate of making design is \(\rm{₹\; 0.35}\) per \(\rm{cm^2}\)

What is unknown?

The cost of making the design.


In a circle with radius r and angle at the center with degree measure\(\theta\)

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

(ii) Area of the segment = Area of the sector - Area of the corresponding triangle.

(i) Visually it is clear that the designs are segments of the circle

\(\therefore\;\)Area of the design \(=\) Area of \(6 \)segments of the circle.

(ii) Since the table cover has \(6\) equal design therefore angle of each segment at the center\(\begin{align} = \frac{{{{360}^o}}}{6} = {60^o}\end{align}\)

(iii) Consider segment \({APB.} \)Chord \({AB}\) subtends an angle of \({60^\circ}\) at the centre.

\(\therefore\;\) Area of segment\( {APB} =\)Area of sector \({AOPB}\, -\) Area of \(\Delta {AOB}\)

(iv) To find area of \(\Delta AOB\)

In \(\Delta {AOB}\)

\({OA = OB}\) (radii of the circle)

\(\begin{align}\angle {OAB} = \angle {OBA} \end{align}\) (angles oppositeequal sides of a triangle are equal)

\(\begin{align}\angle {AOB} + \angle {OAB} + \angle {OBA} = {180}^\circ\end{align}\) (Using angle sum property of a triangle)}

\[\begin{align}\angle AOB + 2\angle OAB &= {{180}^\circ }\\2\angle OAB &= {180^\circ } - {60^\circ }\\ \!\!&= 120^\circ \\ \angle OAB &= \frac{{120}}{2} \\&= {60^\circ }\\ &= \angle OBA\end{align}\]

Since all angles of a triangle are of measure \({60^\circ}\)

\(\therefore\;\Delta \,{AOB} \)is an equilateral triangle.

Using area of equilateral triangle \(\begin{align}=\frac{{\sqrt 3 }}{4}{\left( \rm{side} \right)^2}\end{align}\)

Area of\(\begin{align}\Delta AOB = \frac{{\sqrt 3 }}{4}{r^2}\end{align}\)

Area of sector \(\begin{align}{{AOBP = }}\frac{{{{6}}{{{0}}^{{\circ}}}}}{{{{36}}{{{0}}^{{\circ}}}}}\times \pi {{{r}}^{{2}}}\end{align}\)

Area of Segment APB \(\begin{align}{\rm{ = }}\frac{{{{60}^o}}}{{{{360}^o}}}\pi {r^2} - \frac{{\sqrt 3 }}{4}{r^2}\end{align}\)

Area of designs \(\begin{align}{\rm{ = 6}} \times \left( {\frac{{{{60}^o}}}{{{{360}^o}}}\pi {r^2} - \frac{{\sqrt 3 }}{4}{r^2}} \right)\end{align}\)

Since we know cost of making \(1\,\rm{cm}^2 \)of designs we can use unitary method to find cost of designs.


From the figure we observe the designs are made in the segments of a circle.

Since the table cover has \(6\) equal designs

\(\therefore\;\)angle subtended by each chord (bounding the segment) at the center \(\begin{align}= \frac{{{{360}^ \circ }}}{6} = {60^ \circ }\end{align}\)

Consider \(\Delta AOB\)

\(\Delta {OAB}\,=\,\Delta {OBA }\)(\({QOB = OA,}\)angles opposite equal sides in a triangle are equal)

\(\Delta {AOB}\,{+}\,\Delta {OAB}\,{+}\,\,\Delta {OBA}\,{=18}{{{0}}^{{o}}}\) (angle sum of a triangle)

\[\begin{align}{2}\Delta {OAB}\,&{=}\,{18}{{{0}}^{{o}}}-{6}{{{0}}^{{o}}} \\ \Delta {OAB}\,\,&{=}\,\frac{{12}{{{0}}^{{o}}}}{{2}} \\ & {=6}{{{0}}^{{o}}} \end{align}\]

\(\therefore\;\Delta {AOB}\)is an equilateral triangle

Area of \(\Delta AOB\)

\[\begin{align}&= \frac{{\sqrt 3 }}{4}{({\rm{side}})^2}\\ &= \frac{{\sqrt 3 }}{4} \times {(28)^2}\\ &= \sqrt 3 \times 7 \times 28\\& = 196\sqrt 3 \\ &= 196 \times 1.7\\& = 333.2\,\rm{cm}^2\end{align}\]

Area of sector \(OAPB\)

\[\begin{align}&=\,\frac{{{{6}}{{{0}}^{{o}}}}}{{{{36}}{{{0}}^{{o}}}}}{\times}\,{\pi }{{{r}}^{{2}}}\\\,&= \,\frac{{{1}}}{{{6}}}\,{\times }\,\frac{{{{22}}}}{{{7}}}\,{\times}\,{{28}}\,{\times }\,{{28}}\\\,\,& = \,\frac{{{{11} \times 4 \times 28}}}{{{3}}}\\\, &= \,\frac{{{{1232}}}}{{{3}}}\,\rm{{cm}^{{2}}}\end{align}\]

Area of segment \({APB =}\) Area of sector \({OAPB}\, - \)Area of \(\Delta {AOB}\)

\[\begin{align} &= \left( {\frac{{1232}}{3} - 333.2} \right)\,\rm{cm^2}\end{align}\]

Area of designs \(= 6\) Area of segment

\[\begin{align} &= 6 \times\!\!{ }\left( \frac{{1232}}{{3}}{-333}{.2} \right) \\ &= 2464 - 1999.2\,\rm{c}{{{m}}^{{2}}} \\ \\ &= 464{.8}\,\rm{c}{{{m}}^{{2}}} \end{align}\]

Cost of making \(1\, \rm{cm}^2\)of designs \(\rm{= ₹\; 0.35}\)

\(\therefore\;\)Cost of making \(464.8\,\rm {cm}^2\) of designs

\(= \rm{₹}\,0.35 \times 464.8\\ =\rm{₹}\, 162.68\)


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