In a parallelogram ABCD of area 72 sq cm, the sides CD and AD have lengths 9 cm and 16 cm, respectively. Let P be a point on CD such that AP is perpendicular to CD. Then the area, in sq cm, of triangle APD is

1. 18√3
2. 24√3
3. 32√3
4. 12√3

Answer

Area of the parallelogram ABCD = base × height

CD × AP = 72

9 × AP = 72

AP = 8

In right triangle, APD

DP² = AD² − AP²

DP² = 16² − 8²

DP = 8√3

Area of triangle APD = 1/2 × AP × PD

= 32√3

The correct option is C.