Diagonals AC and BD of a parallelogram ABCD intersect at O. Given that AB = 12 cm and perpendicular distance between AB and DC is 6 cm. Calculate the area of the triangle AOD.

Let ABCD be a parallelogram with AC and BD the diagonals intersecting at O.

From figure,

AB = 12 cm and DM = 6 cm.

By formula,

Area of parallelogram ABCD = base × height = AB × DM

= 12 × 6

= 72 cm².

Since, diagonals of parallelogram intersect each other so O is the mid-point of BD.

∴ AO is the median of the △ABD.

Since, median divides the triangle into two triangles of equal area,

∴ Area of △AOD = $\dfrac{1}{2}$ × Area of △ABD ……(1)

Since, diagonal of a parallelogram divides it into two triangles of equal area.

∴ Area of △ABD = $\dfrac{1}{2}$ × Area of || gm ABCD.

Substituting above value of △ABD in equation 1 we get,

Area of △AOD = $\dfrac{1}{2} \times \dfrac{1}{2}$ Area of || gm ABCD

= $\dfrac{1}{4} \times 72$ = 18 cm².

Hence, area of △AOD = 18 cm².