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What is the area of a rhombus(in cm$$^2$$) whose side is 20 cm and one of the diagonals is 24 cm ?
In a rhombus
diagonals are perpendicular bisectors of each other :
Also In a rhombus
$$side^2\ =\ \left(\frac{Diagonal_1}{2}\ \right)^{^{^2}}+\left(\frac{Diagonal_2}{2}\right)^{^{^2}}$$
Substituting we get
400 = 144 + $$K^2$$
we get K =16
so 2nd diagonal = 2(16) =32
Therefore area = $$\frac{1}{2}\times\ 32\times\ 24=32\times\ 12=384$$
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