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Quadrilaterals — Maths STD 9 — Question

Gujarat BoardEnglish MediumSTD 9MathsQuadrilaterals1 Mark
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Answer

  1. BC2 + AD2 + 2AB × CD
    Solution:
    Given: ABCD is a trapezium with $\text{AB || CD}$
    Construction: Draw DE and CF $\bot$ to AB.
    Then in $\triangle\text{ABC}$
    $\angle\text{BAC}$ is acute
    $\therefore$ BC2 = AC2 + AB2 - 2AF : AB ...(1)
    and in $\triangle\text{BDA}$
    $\angle\text{DBA}$ is acute
    $\therefore$ AD2 = BD2 + AB2 - 2BE : AB ...(2)
    Adding (1) and (2) we get
    BC2 + AD2 = AC2 + BD2 + 2AB2 - 2AF·AB - 2BE·AB
    ⇒ AC2 + BD2 = BC2 + AD2 - 2AB [AB - AF - BE)
    = BC2 + AD2 - 2AB [AB - (AE + EF) - (BF+ EF)]
    = BC2 + AD2 - 2AB [AB - (AE + EF +BF+ EF)]
    = BC2 + AD2 - 2AB [AB - (AB+ CD)] ($\therefore$ EF = DC)
    = BC2 + AD2 - 2AB [- (CD)]
    = AD2 + BC2 + 2AB × CD

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