Check sibling questions

Ex 6.5, 14 - The perpendicular from A on side BC of a ABC - Pythagoras Theoram - Proving

Ex 6.5, 14 - Chapter 6 Class 10 Triangles - Part 2
Ex 6.5, 14 - Chapter 6 Class 10 Triangles - Part 3


Transcript

Ex 6.5, 14 The perpendicular from A on side BC of a ΔABC intersects BC at D such that DB = 3 CD (see figure). Prove that 2AB2 = 2AC2 + BC2 Given: ΔABC with AD ⊥ BC Also DB = 3 CD To prove: 2AB2 = 2AC2 + BC2 Proof: Let BC = x Using Pythagoras theorem (Hypotenuse)2 = (Height)2 + (Base)2 Now in right ∆ 𝐴𝐷𝐶 AC2 = AD2 + DC2 AC2 = AD2 + (𝑥/4)^2 AC2 = AD2 + 𝑥2/16 Similarly in right ∆ 𝐴𝐷𝐵 AB2 = AD2 + DB2 AB2 = AD2 + ( 3𝑥/4 )2 AB2 = AD2 + 9𝑥2/16 Subtracting (2) from (1) AC2 – AB2 = AD2 + 𝑥2/16 – (𝐴𝐷2+9𝑥2/16) AC2 – AB2 = AD2 + 𝑥2/16−𝐴𝐷2−9𝑥2/16 AC2 – AB2 = AD2 – AD2 + 𝑥2/16 −9𝑥2/16 AC2 – AB2 = (𝑥2 − 9𝑥2)/16 AC2 – AB2 = −8𝑥2/16 AC2 – AB2 = −( 𝑥2)/2 2AC2 – 2AB2 = – x2 Putting BC = x 2AC2 – 2AB2 = – BC2 2AC2 + BC2 = 2AB2 2AB2 = 2AC2 + BC2 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 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.