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Ex 6.6, 6 (Optional) - Prove that sum of squares of diagonals of

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

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Ex 6.6, 6 Prove that the sum of the squares of the diagonals of parallelogram is equal to the sum of the squares of its sides. Given: ABCD is a parallelogram To Prove: Sum of squares of diagonals = Sum of squares of its sides AC2 + BD2 = AB2 + BC + CD2 + DA2 Construction: Draw AX CD and BY DC extended to Y. Proof: In Right AXC Applying Pythagoras theorem, AC2 = AX2 + CX2 In Right BYD Applying Pythagoras theorem, BD2 = BY2 + DY2 In BYC, By Pythagoras Theorem BC2 = BY2 + CY2 BD2 = BC2 + CD2 + 2CD.CY From equation (1) AC2 = AX2 + CX2 Putting CX = CD DX AC2 = AX2 + (CD DX)2 AC2 = AX2 + CD2 + DX2 2CD.DX AC2 = (AX2 + DX2) + CD2 2CD.DX In AXD, By Pythagoras Theorem AD2 = AX2 + DX2 In parallelogram ABCD, Opposite sides of parallelogram are equal CD = AB AC2 = AD2 + CD2 2CD.DX AC2 = AD2 + AB2 2CD.DX Hence, the equations are BD2 = BC2 + CD2 + 2CD.CY (3) AC2 = AD2 + AB2 2CD.DX (4) Adding equations (3) and (4) BD2 + AC2 = BC2 + CD2 + 2CD.CY + AD2 + AB2 2CD.DX BD2 + AC2 = BC2 + CD2 + AD2 + AB2 + 2CD(CY DX) Now, we need to prove CY = DX In AXD and BYC, AXD = BYC AD = BC AX = BY AXD BYC Thus, DX = CY Putting CY = DX in equation (5) BD2 + AC2 = BC2 + CD2 + AD2 + AB2 + 2CD (CY DX) BD2 + AC2 = BC2 + CD2 + AD2 + AB2 + 2CD (DX DX) BD2 + AC2 = BC2 + CD2 + AD2 + AB2 + 2CD 0 BD2 + AC2 = BC2 + CD2 + AD2 + AB2 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 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.