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E-4 Solve any two sub question of following.in figure Z BAC = 90° seg BL, seg СMatemedians then prove that 4(BL+CM) = 5BC320M3.2Nano A |
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Answer» Step-by-step explanation: Given : ∠BAC = 90° seg BL and seg CM are the medians. To PROVE: 4(BL2 + CM2 ) = 5BC2 In ∆BAL, ∠BAL 90° [Given] ∴ BL2 = AB2 + AL2 (i) [Pythagoras theorem] In ∆CAM, ∠CAM = 90° [Given] ∴ CM2 = AC2 + AM2 (ii) [Pythagoras theorem] ∴ BL2 + CM2 = AB2 + AC2 + AL2 + AM2 (iii) [Adding (i) and (ii)] Now, AL = 1/2 AC and AM = 1/2 AB (iv) [seg BL and seg CM are the medians] ∴ BL2 + CM2 = AB2 + AC2 + ( 1/2 AC)2 + (1/2 AB)2 [From (iii) and (iv)] = AB2 + AC2 + AC2/4 + AB2/4 = AB2 + AB2/4 + AC2 + AC2/4 = (5AB2)/4 + 5AC2/4 = BL2 + CM2 = 5/4(AB2 + AC2) ∴ 4(BL2 + CM2 ) = 5(AB2 + AC2 ) (v) In ∆BAC, ∠BAC = 90° [Given] ∴ BC2 = AB2 + AC2 (vi) [Pythagoras theorem] ∴ 4(BL2 + CM2 ) = 5BC2 [From (v) and (vi)] |
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