Pythagoras Theoram - Proving
Question 2 Deleted for CBSE Board 2024 Exams
Question 3 Deleted for CBSE Board 2024 Exams
Question 4 Important Deleted for CBSE Board 2024 Exams
Question 5 Deleted for CBSE Board 2024 Exams You are here
Question 6 Important Deleted for CBSE Board 2024 Exams
Question 7 Deleted for CBSE Board 2024 Exams
Question 8 Deleted for CBSE Board 2024 Exams
Question 9 Important Deleted for CBSE Board 2024 Exams
Question 10 Important Deleted for CBSE Board 2024 Exams
Question 11 Important Deleted for CBSE Board 2024 Exams
Last updated at April 16, 2024 by Teachoo
Question 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