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Ex 10.2, 12 - A triangle ABC is drawn to circumscribe a circle - Ex 10.2

Ex 10.2, 12 - Chapter 10 Class 10 Circles - Part 2
Ex 10.2, 12 - Chapter 10 Class 10 Circles - Part 3
Ex 10.2, 12 - Chapter 10 Class 10 Circles - Part 4
Ex 10.2, 12 - Chapter 10 Class 10 Circles - Part 5
Ex 10.2, 12 - Chapter 10 Class 10 Circles - Part 6
Ex 10.2, 12 - Chapter 10 Class 10 Circles - Part 7

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Ex 10.2,12 A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively (see figure). Find the sides AB and AC. Given: A circle with centre O with OD = radius = 4 cm Let ABC circumscribe the circle Also, BD = 8 cm & CD = 6 cm To find: AB & AC Construction: Join OA, OC& OB Let AC, AB intersect circle at E & F respectively Solution: From theorem 10.2, lengths of tangents drawn from external point are equal Hence, CE = CD = 6 cm BF = BD = 8 cm AE = AF = x Now, our three sides are CD = CD + DB = 6 + 8 = 14 cm AC = AE + EC = (x + 6) cm AB = AF + FC = (x + 8) cm Now, we find Area of ABC using Herons formula Area of triangle = ( ( )( )( )) Here, AC = a = (x + 6) cm AB = b = (x + 8) cm BC = c = 6 + 8 = 14 cm. s = ( + + )/2 = (( + 6) + ( + 8) + 14)/2 = (2 + 28)/2 = (2( + 14))/2 = x + 14 Area ABC= ( ( )( )( )) Putting values = (( +14)( +14 ( +6))( +14 ( +8))( +14 14)) = (( +14)( +14 6)( +14 8)( +14 14)) = (( +14)(8)(6)( )) = (( +14)(48)( )) = (( +14)(48 )) = ( (48 )+14(48 )) = (48 2+672 ) Area ABC = (48 2+672 ) Join OE & OF Here, OD = OF = OE = radius = 4 cm Also, we know that tangent is perpendicular to the radius, So, OD BC , OF AB & OE AC Also, Area of ABC = Area AOC + Area AOB + Area BOC We find Area AOC , Area AOB & Area BOC Now, Area of ABC = Area AOC + Area AOB + Area BOC Putting values (48 2+672 ) = (2x + 12) + (2x + 16) + 28 (48 2+672 ) = 4x + 56 Squaring both sides ( (48 2+672 ))2 = (4x + 56)2 48x2 + 672x = (4x)2 + 562 + 2 4x 56 48x2 + 672x = 16x2 + 3136 + 448x 48x2 16x2 + 672x 448x 3136 = 0 32x2 + 224x 3136 = 0 32(x2 + 7x 98)= 0 x2 + 7x 98 = 0 x2 + 14x 7x 98 = 0 x(x + 14) 7(x + 14) = 0 (x 7) (x + 14) = 0 So, x = 7 & x = 14 Since x cannot be negative, So, x = 7 Now, AC = x + 6 = 7 + 6 = 13 cm & AB = x + 8 = 7 + 8 = 15 cm

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.