Answers

2014-03-07T12:51:29+05:30
Let us consider rhombus ABCD whose diagonals are AC AND BD whose intersection point is O
we know that diagonals of rhombus are perpendicular to each other .
now in ΔABC,BO IS PERPENDICULAR to AC
THEREFORE AREA of ΔABC = \frac{1}{2} ×BO×AC
similarly in ΔADC
area of ΔADC= \frac{1}{2} ×DO×AC
AREA OF RHOMBUS=area of ΔABC+area of ΔADC
                               = \frac{1}{2} ×BO×AC+ \frac{1}{2} ×DO×AC
                               = \frac{1}{2} ×AC[BO+DO]
                               = \frac{1}{2} ×AC×BD
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