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Ex 6.5
Ex 6.5, 2 Important Deleted for CBSE Board 2023 Exams
Ex 6.5, 3 Important Deleted for CBSE Board 2023 Exams
Ex 6.5, 4 Deleted for CBSE Board 2023 Exams
Ex 6.5, 5 Deleted for CBSE Board 2023 Exams
Ex 6.5, 6 Deleted for CBSE Board 2023 Exams
Ex 6.5, 7 Important Deleted for CBSE Board 2023 Exams
Ex 6.5, 8 Important Deleted for CBSE Board 2023 Exams
Ex 6.5, 9 Deleted for CBSE Board 2023 Exams
Ex 6.5, 10 Deleted for CBSE Board 2023 Exams
Ex 6.5, 11 Important Deleted for CBSE Board 2023 Exams
Ex 6.5, 12 Important Deleted for CBSE Board 2023 Exams
Ex 6.5, 13 Deleted for CBSE Board 2023 Exams
Ex 6.5, 14 Deleted for CBSE Board 2023 Exams
Ex 6.5, 15 Important Deleted for CBSE Board 2023 Exams You are here
Ex 6.5, 16 Deleted for CBSE Board 2023 Exams
Ex 6.5, 17 (MCQ) Deleted for CBSE Board 2023 Exams
Last updated at May 14, 2021 by Teachoo
Maths Crash Course - Live lectures + all videos + Real time Doubt solving!
Ex 6.5, 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