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Ex 6.5, 15 - In equilateral triangle ABC, D is a point on BC - Pythagoras Theoram - Proving

Ex 6.5, 15 - Chapter 6 Class 10 Triangles - Part 2
Ex 6.5, 15 - Chapter 6 Class 10 Triangles - Part 3 Ex 6.5, 15 - Chapter 6 Class 10 Triangles - Part 4 Ex 6.5, 15 - Chapter 6 Class 10 Triangles - Part 5

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Question 15 In an equilateral triangle ABC, D is a point on side BC such that BD = 1/3BC. Prove that 9AD2 = 7 AB2 Given: Equilateral triangle ABC D is a point an BC Such that BD = 1/3 BC To prove: 9 AD2 = 7 AB2 Construction: Lets draw AE BC Proof: All sides of equilateral triangle is equal, AB = BC = AC Let AB = BC = AC = x Given BD = 1/3BC BD = /3 In AE = AE AB = AC = Hence by RHS congruency BE = EC So, BE = EC = 1/2BC BE = EC = /2 So, BE = /2 BD + DE = /2 /3+ = /2 DE = /2 /3 DE = (3 2 )/(2 3) DE = /6 Using Pythagoras theorem (Hypotenuse)2 = (Height)2 + (Base)2 Similarly In right AD2 = AE2 + DE2 AD2 = 3 2/4+( /6 )^2 AD2 = 3 2/4+ 2/36 AD2 = ((3 2) 9 + 2)/36 AD2 = (27 2 + 2)/36 AD2 = (28 2)/36 AD2 = 7 2/9 9AD2 = 7x2 9 AD2 = 7 AB2 Hence proved

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Davneet Singh

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.