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Last updated at May 29, 2018 by Teachoo
Ex 12.3, 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). Area of design = Area of circle β area of triangle ABC For circle Radius = r = 32 cm Area of circle = ππ2 = 22/7Γ322 = 22/7Γ32Γ32 = 22528/7 cm2 Area Ξ ABC Now for equilateral triangle ABC, We find side first Let centre of circle be O. Join OA, OB and OC For Ξ AOB, Ξ BOC & Ξ AOC OA = OB = OC OB = OC = OA AB = BC = AC Hence, AOB β BOC β AOC β΄ β "AOB"=β "BOC"=β AOC But, sum of angles around a point = 360Β° β β AOB + β BOC + β AOC = 360Β° β BOC + β BOC + β BOC = 360 3β BOC = 360Β° β BOC = (360Β°)/3 β BOC = 120Β° Also, Since Ξ AOB, Ξ AOC & Ξ BOC are congruent Their areas would also be equal. Hence, Area Ξ AOB = Area Ξ AOC = Area Ξ BOC So, Area of β π΅ππΆ = 1/3 (Area of ABC) Now, we find area Ξ BOC. Finding area of Ξ BOC Area Ξ BOC = 1/2 Γ Base Γ Height We draw OM β₯ BC β΄ β OMB = β OMC = 90Β° In Ξ OMB & Ξ OMC β OMB = β OMC OB = OC OM = OM β΄ Ξ OMB β Ξ OMC β β BOM = β COM β΄ β BOM = β COM = 1/2 β BOC β β BOM = β COM = 1/2 Γ 120Β° = 60Β° Also, since Ξ OMB β Ξ OMC β΄ BM = CM β BM = CM = 1/2 BC From (1) BM = 1/2BC 2BM = BC BC = 2BM Putting value of BM BC = 2 Γ 16β3 BC = 32β3 Now, Area of Ξ BOC = 1/2 Γ Base Γ Height = 1/2 Γ BC Γ OM = 1/2 Γ 32β3 Γ 16 = 16β3 Γ 16 = 256β3 We know that Area of β BOC" = " 1/3 "Area of" β ABC 256 β3 = 1/3 Area of ABC 256 β3Γ3 = Area of ABC 768 β3 = Area of ABC Area of β ABC=768β3 cm2 Now , Area of design = Area of circle β area of triangle ABC = 22528/7β768β3 cm2