Question 5 - Important questions of Triangle (in Geometry) - Chapter 6 Class 10 Triangles

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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