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

- 18√3
- 24√3
- 32√3
- 12√3

### Answer

Area of the parallelogram ABCD = base × height

CD × AP = 72

9 × AP = 72

AP = 8

In right triangle, APD

DP^{2} = AD^{2} − AP^{2}

DP^{2} = 16^{2} − 8^{2}

DP = 8√3

Area of triangle APD = 1/2 × AP × PD

= 32√3

**The correct option is C.**