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in an equilateral triangle ABC.D is a point on side BC such that BD=1/3BC.prove that 9AD2=7AB2

Posted by Harry Potter 7 years, 8 months ago

CBSE > Class 10 > Mathematics

7 answers

Nitesh Sharma 7 years, 8 months ago

Given :- An equilateral triangle ABC and D be a point on BC such that BD= 1/3 BC. To prove :- 9AD^2 = 7AB^2 Construction :- Draw AE Perpendicular to BC and Join AD. Proof :- Triangle ABC is an equilateral triangle and AE Perpendicular to BC Therefore BE = EC Thus, we have BD = 1/3 BC and DC = 2/3 BC and BE = EC = 1/3BC In Triangle AEB AE^2 + BE^2 = AB^2 [Using Pythagoras Theorem] AE^2 = AB^2 - BE^2 AD^2 - DE^2 = AB^2 - BE^2 ['.' In Triangle AED, AD^2 = AE^2 + DE^2] AD^2 = AB^2 - BE^2 + DE^2 AD^2 = AB^2 - (1/2 BC)^2 + ( BE - BD )^2 AD^2 = AN^2 - 1/4 BC ^2 + (1/2 BC - 1/3 BC )^2 AD^2 = AB^2 - 1/4 BC^2 + ( BC/6 )^2 AD^2 = AB^2 - BC^2 (1/4 - 1/36 ) AD^2 = AB^2 - BC^2 (8/36) 9AD^2 = 9AB^2 - 2BC^2 9AD^2 = 9AB^2 - 2AB^2 [ '.' AB = BC ] 9AD^2 = 7 AB^2

Harry Potter 7 years, 8 months ago

i know the shortest trick. i was just knowing how many here know the short ans??????

Anand Kumar Thakur 7 years, 8 months ago

This is too long

Shikhar Yadav 7 years, 8 months ago

No

Harry Potter 7 years, 8 months ago

in RD it is too long.don't you have any short sol.

Shikhar Yadav 7 years, 8 months ago

Given in Rd Sharma

Amir Sheikh 7 years, 8 months ago

Hi siya

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