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17thKVS-JMO2014-Solutions5

17thKVS-JMO2014-Solutions5

Probability Introduction :

This Page contains the solution  of 17thKVS-JMO2014-Solutions5  in detail. See  the image for the question . A, B, C, D are four points on the circle. Join AC and BD such that those two lines are perpendicular to each other . We should know that perpendicular drawn from center to any chord , bisects the chord. that is ON is perpendicular to the chord AC so AN = NC , similarly we should know that in a quadrilateral if two opposites angles each = 90 then that quadrilateral is rectangle .  In a rectangle opposite sides are equal . So here in this figure OMEN is rectangle .

 5Q:  A, B, C, D are four points on a circle with radius R such that AC is Perpendicular to BD and meets BD at E. Prove that EA² + EB² + EC² + ED² = 4R²

 Solution :

 

     Given a circle with center O and radius R .   A, B, C, D are points on circle such that AC is perpendicular to BD and they intersect at E .   Let us construct ON , OM perpendicular from center O to AC and BD chords respectively . Join OA, OB, OE .

OMEN forms a rectangle . so MO=EN , ME=ON . We know that N, M bisects chords AC, BD respectively .

  AN=NC  , BM=MD , OA=OB=OC=OD=R (radius)

  OA² = AN² + ON²   i.e., R² = AN² + ON² --------(1)

 OB² = BM² + OM²  i.e., R² = BM² + EN² ---------(2)

 EA = AN - EN so EA² = (AN - EN )²   =  AN² + EN² - 2 AN * EN -----------(3)

 EB = BM - EM  , EB =BM - ON so EB² = BM² + ON² - 2 BM*ON ------------(4)

 EC = EN + NC = EN + AN so       EC² = EN² + AN² + 2 AN*EN ------------(5)

 ED = EM + MD = ON + BM  so    ED² = ON² + BM² + 2 BM*ON -----------(6)

 (3) + (4) + (5) + (6) we get ,

 EA² + EB² + EC² + ED² = 2 AN² + 2 ON² + 2 BM² + 2 EN²

                                   = 2 ( AN² + ON² ) +2 ( BM² + EN² )

                                   = 2 R² + 2 R²

                                  = 4R²

  Therefore  EA² + EB² + EC² + ED² = 4R² .

 Hence proved.

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