Ex 6.6, 5 (Optional) - AD is a median of triangle ABC,  AM BC. Prove

Ex 6.6, 5 (Optional) - Chapter 6 Class 10 Triangles - Part 2
Ex 6.6, 5 (Optional) - Chapter 6 Class 10 Triangles - Part 3 Ex 6.6, 5 (Optional) - Chapter 6 Class 10 Triangles - Part 4 Ex 6.6, 5 (Optional) - Chapter 6 Class 10 Triangles - Part 5 Ex 6.6, 5 (Optional) - Chapter 6 Class 10 Triangles - Part 6 Ex 6.6, 5 (Optional) - Chapter 6 Class 10 Triangles - Part 7 Ex 6.6, 5 (Optional) - Chapter 6 Class 10 Triangles - Part 8

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Question 5 In Fig. 6.60, AD is a median of a triangle ABC and AM BC. Prove that : AC2 = AB2 + BC2 2 BC . DM + ( /2)^2. Given: ABC is a triangle AD is a Median of ABC BD = CD = 1/2 BC Also, AM BC To Prove: (i) AC2 = AB2 + BC2 2 BC . DM + ( /2)^2. Proof : In Right AMD Applying Pythagoras Theorem, AD2 = AM2 + DM2 In Right AMC Applying Pythagoras Theorem, AC2 = AM2 + CM2 Thus, our equations are AD2 = AM2 + DM2 (2) AC2 = AM2 + CM2 (3) Subtracting (3) (2) AC2 AD2 = AM2 + CM2 (AM2 + DM2) AC2 AD2 = AM2 + CM2 AM2 DM2 AC2 AD2 = CM2 DM2 AC2 = AD2 + CM2 DM2 Putting CM = CD + DM AC2 = AD2 + (CD + DM)2 DM2 AC2 = AD2 + CD2 + DM2 + 2CD.DM DM2 AC2 = AD2 + CD2 + 2CD.DM Putting CD = / BC Form equation (1) AC2 = AD2 + ( / )^2+ 2( /2) (DM) AC2 = AD2 + BC.DM + ( /2)^2 Hence Proved. Question 5 In Fig. 6.59, ABC is a triangle in which ABC < 90 and AD BC.Prove that (ii) AB2 = AD2 BC.DM + [ /2]^2. In Right AMB Applying Pythagoras Theorem, AB2 = AM2 + BM2 In Right AMD Applying Pythagoras Theorem, AD2 = AM2 + DM2 Thus, our equations are AB2 = AM2 + BM2 (4) AD2 = AM2 + DM2 (5) Subtracting (4) (5) AB2 AD2 = AM2 + BM2 (AM2 + DM2) AB2 AD2 = AM2 + BM2 AM2 DM2 AB2 AD2 = BM2 DM2 AB2 = AD2 + BM2 DM2 Putting BM = BD DM AB2 = AD2 + (BD DM)2 DM2 AB2 = AD2 DM2 + (BD DM)2 AB2 = AD2 DM2 + BD2 + DM2 2BD.DM AB2 = AD2 + BD2 2BD.DM Putting BD = 1/2 BC from equation (1) AB2 = AD2 + ( / )^2 2 ( / ).DM AB2 = AD2 BC.DM + ( /2)^2 Hence Proved Question 5 In Fig. 6.59, ABC is a triangle in which ABC < 90 and AD BC. Prove that (iii) AC2 = AB2 = 2 AD2 + 1/2 BC2 . Results of (i) and (ii) part are AC2 = AD2 + BC.DM + ( /2)^2 (6) AB2 = AD2 BC.DM + ( /2)^2 (7) Adding equations (6) & (7) AC2 + AB2 = AD2 + BC.DM + ( /2)^2 + AD2 BC.DM + ( /2)^2 AC2 + AB2 = 2AD2 + ( /2)^2 + ( /2)^2 AC2 + AB2 = 2AD2 + ( ^2)/4 + ( ^2)/4 AC2 + AB2 = 2AD2 + 2(( ^2)/4) AC2 + AB2 = 2AD2 + 1/2 ^2 Hence Proved

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.