Question

in a rhombus of side 2cm the sum of squares of its diagonals is

Answer 1

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Given :-  A rhombus ABCD with diagonals AC and BD intersecting at point O.

To proove : -  AB2 + BC2 + CD2 + AD2 = AC2 + BD2

Proof :-  Since we know that diagonals of rhombus intersect each other at 900  . 

Therefore,  ang. AOB = ang. BOC = ang. COD = ang. AOD = 900

By Pythagores Theorem,

in triangles AOB, BOC, COD and AOD, we will get :-

AB2 = AO2 + BO2 ---------(1) 

BC2 = BO2 + CO2 --------(2)

CD2 = CO2 + DO2 -------(3)

AD2 = AO2 + DO2 ---------(4)

On adding (1),(2),(3) and (4),

AB2 + BC2 + CD2 + AD2 = 2 (AO2 + BO2 + CO2+DO2)

  = 2 ( AC2 / 2 + BD2 /2)  (DIAGONALS BISECT EACH OTHER. AO = CO = AC / 2  BO = DO = BD / 2 )

  = AC2  + BD2

Answer 2

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