**Question: **

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 |

**Show Answer**

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

Given that area of parallelogram = 72 sq cm

Area of triangle ABC = $\frac{1}{2}$*area of parallelogram

(1/2)*AB*BC*sinABC = $\frac{1}{2}$*72

sinABC = $\frac{1}{2}$

$\angle$ ABC = 30°

Let us draw a perpendicular CQ from C to AB.

By symmetry we can say that CQ = AP, CP = AQ and QB = PQ,

Therefore, we can say that area of triangle APD = area of triangle CQB

In right angle triangle CQB,

QB = CBcos30° = $16*\frac{\sqrt{3}}{2}$ = $8\sqrt{3}$ cm

CQ = CBsin30° = $16*\frac{1}{2}$ = $8$ cm

Therefore, area of triangle CQB = 1/2*CQ*QB = $1/2*8*8\sqrt{3}$ = $32\sqrt{3}$

Hence, we can say that area of triangle APD = $32\sqrt{3}$.

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