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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 $⊥$ to AB

    Then in $\triangle\text{ABC}$
    $\angle\text{BAC}$ is acute
    $∴$ BC2 = AC2 + AB2 - 2AF : AB ...(1)
    and In $\triangle\text{BDA}$
    $\angle\text{DBA}$ is acute
    $∴$ 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)] ($∴$ EF = DC)
    = BC2 + AD2 - 2AB [(-CD)]
    = AD2 + BC2 + 2AB × CD

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