• Areas Related to Circles : Exercise 12.3 (Mathematics NCERT Class 10th)

Unless stated otherwise, Use\pi = {{22} \over 7}
Q.1     Find the area of the shaded region in figure, if PQ = 24 cm, PR = 7 cm and O is the centre of the circle.

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Sol.      Since ROQ is a diameter, therefore, \angle RPQ = {90^o}
             In rt \angle d\,\Delta \,PRQ R{Q^2} = R{P^2} + P{Q^2}
              \Rightarrow R{Q^2} = {7^2} + {24^2} = 49 + 576 = 625
              \Rightarrow RQ = \sqrt {625} = 25\,cm
             Therefore, Radius r = {1 \over 2}RQ = {{25} \over 2}cm
             Area of the semi circle  = {1 \over 2}\pi {r^2} = {1 \over 2} \times {{22} \over 7} \times {{25} \over 2} \times {{25} \over 2}c{m^2}
              = {{6875} \over {28}}c{m^2}
             and area of \Delta \,RPQ = {1 \over 2} \times RP \times PQ
              = \left( {{1 \over 2} \times 7 \times 24} \right)c{m^2} = 84\,c{m^2}
             Area of the shaded region
             = Area of the semi circle – Area (\Delta RPQ)
              = \left( {{{6875} \over {28}} - 84} \right)c{m^2} = \left( {{{6875 - 2352} \over {28}}} \right)c{m^2}
              = {{4523} \over {28}}c{m^2}


Q.2      Find the area of the shaded region in figure, if radii of the two concentric circles with centre O are 7 cm and 14 cm respectively and \angle AOC = {40^o}

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Sol.     Area of the shaded region
            = Area of sector AOC – Area of sector OBD
             = \left( {{{40} \over {360}} \times \pi \times {{14}^2} - {{40} \over {360}} \times \pi \times {7^2}} \right)c{m^2}
             = {1 \over 9} \times {{22} \over 7}\left( {196 - 49} \right)c{m^2}
             = \left( {{{22} \over {63}} \times 147} \right)c{m^2} = {{154} \over 3}c{m^2}


Q.3      Find the area of the shaded region in figure, if ABCD is a square of side 14 cm and APD and BPC are semicircles.

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Sol.       Area of the square ABCD = {\left( {14} \right)^2}c{m^2} = 196\,c{m^2}
             Diameter of the semicircles = AD or BC = 14 cm
             Therefore, Radius of each semicircle = 7 cm
             Area of the semicircular regions  = 2 \times {1 \over 2}\pi {r^2} = \pi {r^2}
              = \left( {{{22} \over 7} \times 49} \right)c{m^2} = 154\,c{m^2}
             Therefore, Area of the shaded portion
             = Area of the square ABCD – Area of the semicircular regions
             = \left( {196 - 154} \right)c{m^2} = 42\,c{m^2}


Q.4      Find the area of the shaded region in figure, where a circular arc of radius 6 cm has been drawn with vertex O of an equilateral triangle OAB of side 12 cm as centre.

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Sol.       Area of the circular portion
             = Area of the circle – Area of the sector
             = \pi {r^2} - {{60} \over {360}}\pi {r^2} = \pi {r^2}\left( {1 - {1 \over 6}} \right)
             = {5 \over 6}\pi {r^2}, where r = 6
             = \left( {{5 \over 6} \times {{22} \over 7} \times 36} \right)c{m^2} = {{660} \over 7}c{m^2}
            Area of the equilateral \Delta \,OAB
             = {{\sqrt 3 } \over 4}{\left( {side} \right)^2} = \left( {{{\sqrt 3 } \over 4} \times 144} \right)c{m^2}
             = 36\sqrt 3 \,c{m^2}
            Therefore, Area of the shaded region
             = \left( {{{660} \over 7} + 36\sqrt 3 } \right)c{m^2}


Q.5      From each corner of a square of side 4 cm a quadrant of a circle of radius 1 cm is cut and also a circle of diameter 2 cm is cut as shown in figure. Find the area of the remaining portion of the square.

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Sol.           The area of the whole square ABCD  = {4^2}c{m^2} = 16\,c{m^2}
                  The sum of the area of the four quadrants at the four corners of the square
                  = The area of a circle of radius 1 cm
                   = {{22} \over 7} \times {1^2}c{m^2} = {{22} \over 7}c{m^2}
                  The area of the circle of diameter 2 cm,
                   i.e., radius 1\,c{m^2}
                    = {{22} \over 7} \times {1^2}c{m^2} = {{22} \over 7}c{m^2}
                   Therefore, Area of the remaining portion
                   = The area of the square ABCD
                   – The sum of the area of 4 quadrants at the four corners of the square
                   – The area of the circle of diameter 2 cm
                    = \left( {16 - {{22} \over 7} - {{22} \over 7}} \right)c{m^2} = \left( {{{112 - 22 - 22} \over 7}} \right)c{m^2}
                    = {{68} \over 7}c{m^2}


Q.6      In a circular table cover of radius 32 cm, a design is formed leaving an equilateral triangle ABC in the middle as shown in figure. Find the area of the design (shaded region).

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Sol.         Let ABC be an equilateral triangle and let O be the circumcentre of the circumcircle of radius 32 cm.
               Area of the circle  = \pi {r^2}
                = \left( {{{22} \over 7} \times 32 \times 32} \right)c{m^2}
                = {{22528} \over 7}c{m^2}
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                Area of \Delta ABC = 3 × Area of \Delta BOC
                 = 3 \times {1 \over 2} \times OB \times OC \times \sin BOC
                \left( {{3 \over 2} \times 32 \times 32 \times \sin \,{{120}^o}} \right)
                 = \left( {3 \times 16 \times 32 \times {{\sqrt 3 } \over 2}} \right)
                 = 768\sqrt 3 \,c{m^2}
                Therefore, Area of the design (i.e., shaded region)
                = Area of the circle – Area of \Delta ABC
                 = \left( {{{22528} \over 7} - 768\sqrt 3 } \right)c{m^2}


Q.7       In figure , ABCD is a square of side 14 cm. With centres A, B, C and D, four circles are drawn such that each circle touch externally two of the remaining three circles. Find the area of the shaded region.

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Sol.           The area of the whole square ABCD = {14^2}c{m^2} = 196\,c{m^2}
                  The sum of the area of the four quadrants at the four corners of the square
                  = The area of a circle of radius 7\,c{m^2}
                   = \pi {\left( 7 \right)^2}c{m^2} = \left( {{{22} \over 7} \times 49} \right)c{m^2} = 154\,c{m^2}
                  Area of the shaded portion = The area of the square ABCD
                  – The sum of the area of four quadrants at the four corners of the square
                   = \left( {196 - 154} \right)c{m^2} = 42\,c{m^2}


Q.8       Figure depicts a racking track whose left and right ends are semicircular.

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The distance between the two inner parallel line segment is 60 m and they are each 106 m long. If the track is 10 m wide, find
              (i) The distance around the track along its inner edge.
              (ii) The area of the track.
Sol.           We have, OB = O'C = 30 m
                  and AB = CD = 10 m
                  OA = O'D = (30 + 10) m = 40 m
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               (i) The distance around the track around along its inner edge.
                = BC + EH + 2 × circumference of the semicrircle of radius OB = 30 m
                = \left( {106 + 106 + 2 \times {1 \over 2} \times 2\pi \left( {30} \right)} \right)m
                = \left( {212 + 2 \times {{22} \over 7} \times 30} \right)m
                = \left( {212 + {{1320} \over 7}} \right)m = \left( {{{1484 + 1320} \over 7}} \right)m = {{2804} \over 7}m

               (ii) Area of the track
               = Area of the shaded region
               = Area of rectangle ABCD + Area of rectangle EFGH
               + 2 (Area of the semicircle of radius 40 m
              – Area of the semicircle with radius 30 m)
               = \left[ {\left( {10 \times 106} \right) + \left( {10 + 106} \right) + 2\left\{ {{1 \over 2} \times {{22} \over 7} \times {{\left( {40} \right)}^2} - {1 \over 2} \times {{22} \over 7} \times {{\left( {30} \right)}^2}} \right\}} \right]{m^2} 

               = \left[ {1060 + 1060 + {{22} \over 7}\left( {{{40}^2} - {{30}^2}} \right)} \right]{m^2}
               = \left[ {2120 + {{22} \over 7}\left( {40 + 30} \right)\left( {40 - 30} \right)} \right]{m^2}
               = \left( {2120 + {{22} \over 7} \times 70 \times 10} \right){m^2}
               = \left( {2120 + 2200} \right){m^2}
               = 4320\,{m^2}


Q.9      In figure AB and CD are two diameters of a circle (with centre O) perpendicular to each other and OD is the diameter of the smaller circle. If OA = 7 cm, find the area of the shaded region.

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Sol.       Area of the sector  = \left( {{{90} \over {360}} \times {{22} \over 7} \times 7 \times 7} \right)c{m^2}
              = {{77} \over 2}c{m^2}
             Area of \Delta OCB  = {1 \over 2} \times OC \times OB
              = {1 \over 2} \times 7 \times 7\,c{m^2}
              = {{49} \over 2}c{m^2}
             Therefore, The area of the segment BPC
              = \left( {{{77} \over 2} - {{49} \over 2}} \right)c{m^2} = {{28} \over 2}c{m^2} = 14\,c{m^2}
             Similarly the area of the segment AQC = 14\,c{m^2}
             Also, the area of the circle with DO as diameter
              = \left( {{{22} \over 7} \times {7 \over 2} \times {7 \over 2}} \right)c{m^2} = {{77} \over 2}c{m^2}
             Hence, the total area of the shaded region
              = \left( {14 + 14 + {{77} \over 2}} \right)c{m^2}
              = \left( {{{28 + 28 + 77} \over 2}} \right)c{m^2} = {{133} \over 2}c{m^2}
              = 66.5\,\,c{m^2}


Q.10      The area of an equilateral triangle ABC is 17320.5 \,c{m^2}. With each vertex of the triangle as centre, 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)

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Sol.       Let each side of the triangle be a cm. Then,
             Area = 17320.5\,c{m^2}
            \Rightarrow {{\sqrt 3 } \over 4}{a^2} = 17320.5 \left[ {\sin ce\,Area = {{\sqrt 3 } \over 4}{{\left( {side} \right)}^2}} \right]
            \Rightarrow {a^2} = {{17320.5 \times 4} \over {\sqrt 3 }} = {{17320.5 \times 4} \over {1.73205}} = 40000
               \Rightarrow a = 200
              Thus, radius of each circle is 100 cm.
              Now, required area
              = Area of \Delta ABC – 3 ×(Area of a sector of angle 60º in a circle of 100 cm)
               = \left[ {17320.5 - 3\left( {{{60} \over {360}} \times 3.14 \times 100 \times 100} \right)} \right]c{m^2}
               = \left( {17320.5 - 15700} \right)c{m^2} = 1620.5\,c{m^2}


Q.11      On a square hand kerchief, nine circular designs each of radius 7 cm are made (see figure). Find the area of the remaining portion of the handkerchief. 

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Sol.      Side of the square ABCD = AB
             = 3 × diameter of circular design
             = 3 × (2 × 7) cm = 42 cm
             Therefore, Area of the square ABCD
              = \left( {42 \times 42} \right)c{m^2} = 1764\,c{m^2}
             Area of one circular design
              = \pi {r^2} = \left( {{{22} \over 7} \times 7 \times 7} \right)c{m^2} = 154\,c{m^2}
             Therefore, Area of 9 such designs
              = \left( {9 \times 154} \right)c{m^2} = 1386\,c{m^2}
             Therefore, Area of the remaining portion of the handkerchief
             = Area of the square ABCD – Area of 9 circular designs
              = \left( {1764 - 1386} \right)c{m^2} = 378\,c{m^2}


Q.12      In figure OACB is a quadrant of a circle with centre O and radius 3.5 cm. If OD = 2 cm , find the area of the
               (i) quadrant OACB,                                            (ii) Shaded region,

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Sol.            (i) Area of quadrant  = {1 \over 4}\pi {r^2}
                    = {1 \over 4} \times {{22} \over 7} \times {\left( {3.5} \right)^2}c{m^2}
                    = {1 \over 4} \times {{22} \over 7} \times {7 \over 2} \times {7 \over 2}c{m^2}
                    = {{77} \over 8}c{m^2}

                   (ii) Area of \Delta \,AOD = {1 \over 2}Base\, \times \,Height
                    = {1 \over 2}\left( {OB \times OD} \right)
                    = {1 \over 2}\left( {3.5 \times 2} \right)c{m^2} = {7 \over 2}c{m^2}
                   Hence, area of the shaded region
                   = Area of quadrant – Area of \Delta AOD
                    = \left( {{{77} \over 8} - {7 \over 2}} \right)c{m^2} = \left( {{{77 - 28} \over 8}} \right)c{m^2}
                    = {{49} \over 8}c{m^2}


Q.13       In figure, a square OABC is inscribed in a quadrant OPBQ . If OA = 20 cm, find the area of the shaded region. (Use \pi = 3.14

 

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Sol.            Radius of the quadrant = OB = \sqrt {O{A^2} + A{B^2}}
                   = \sqrt {{{20}^2} + {{20}^2}} \,cm
                   = 20\sqrt {1 + 1} \,cm
                   = 20\sqrt 2 \,cm
                  Therefore, Area of quadrant OPBQ
                   = {1 \over 4}\pi {r^2} = {1 \over 4} \times 3.14 \times {\left( {20\sqrt 2 } \right)^2}c{m^2}
                   = \left( {{1 \over 4} \times 3.14 \times 800} \right)c{m^2} = 628\,c{m^2}
                  Area of the square OABC
                   = {\left( {20} \right)^2}c{m^2} = 400\,c{m^2}
                  Hence, area of the shaded region
                  = Area of quadrant – Area of square OABC
                   = \left( {628 - 400} \right)c{m^2} = 228\,c{m^2}


Q.14      AB and CD are respectively across of two concentric circles of radii 21 cm and 7 cm and centre O (see figure). If \angle AOB = {30^o}, find the area of the shaded region.

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Sol.      Let {A_1}\,and\,{A_2} be the areas of sectors OAB and OCD respectively. Then,
             {A_1} = Area of a sector of angle 30º in a circle of radius 21 cm.
              = \left( {{{30} \over {360}} \times {{22} \over 7} \times 21 \times 21} \right)c{m^2}\left[ {U\sin g\,A = {\theta \over {360}} \times \pi {r^2}} \right]
              = \left( {{{231} \over 2}} \right)c{m^2}
             {A_2} = Area of a sector of angle 30º in a circle of radius 7 cm
              = \left( {{{30} \over {360}} \times {{22} \over 7} \times 7 \times 7} \right)c{m^2}
              = {{77} \over 6}c{m^2}
             Area of the shaded region
              = {A_1} - {A_2} = \left( {{{231} \over 2} - {{77} \over 6}} \right)c{m^2}
              = {{693 - 77} \over 6}c{m^2}
              = {{616} \over 6}c{m^2} = {{308} \over 3}c{m^2}


Q.15      In figure , ABC is a quadrant of a circle of radius 14 cm and a semicircle is drawn with BC as diameter. Find the area of the shaded region.

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Sol.      Let 14 cm be the radius of the quadrant with A as the centre.
            Then the area of the quadrant ABMC

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             = {1 \over 4}\pi {r^2}
             = \left( {{1 \over 4} \times {{22} \over 7} \times 196} \right)c{m^2}
             = 154\,c{m^2}
            Area of \Delta BAC
             = {1 \over 2} \times AC \times AB
             = \left( {{1 \over 2} \times 14 \times 14} \right)c{m^2} = 98\,c{m^2}
            Therefore, Area of the segment of the circle, BMC
            = Area of the quadrant ABMC – Area of \Delta BAC
            = (154 – 98) c{m^2} = 56\,c{m^2}
             Now, since AC = AB = 14 cm and \angle BAC = {90^o}
             Therefore, By Pyrthagoras Theorem,
             BC = \sqrt {A{C^2} + A{B^2}}
              = \sqrt {{{14}^2} + {{14}^2}}
              = 14\sqrt 2
             Therefore, Radius of the semicircle BNC = 7\sqrt 2 \,cm
             Area of the semicircle BNC = {\pi \over 2}{\left( {7\sqrt 2 } \right)^2}c{m^2}
              = \left( {{1 \over 2} \times {{22} \over 7} \times 98} \right)c{m^2} = 154\,c{m^2}
             Hence, the area of the region between two arcs BMC and BNC
             = The area of the shaded region
             = The area of semicircle BNC – The area of the segment of the circle BMC
              = \left( {154 - 56} \right)c{m^2} = 98\,c{m^2}


Q.16      Calculate the area of the the designed region in figure common between the two quadrants of circle of radius 8 cm each. 

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Sol.       Here, 8 cm is the radius of the quadrants ABMD and BNDC.
              Sum of their areas
               = 2 \times {1 \over 4}\pi {r^2} = {1 \over 2}\pi {r^2}
               = \left( {{1 \over 2} \times {{22} \over 7} \times 64} \right)c{m^2}
               = {{704} \over 7}c{m^2}

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               Area of the square ABCD
                = \left( {8 \times 8} \right)c{m^2}
                = 64\,c{m^2}
               Area of the designed region
               = Area of the shaded region
               = Sum of the area of quadrants
               – Area of the square ABCD
                = \left( {{{704} \over 7} - 64} \right)c{m^2} = \left( {{{704 - 448} \over 7}} \right)c{m^2}
                = {{256} \over 7}c{m^2}


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