Question

prove that area of a rhombus is equal to 1/2 ×sum of diagonals

Answer 1

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