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Question 29 :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

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Given Area (ABCD) = 72 sq cm, CD = 9cms, AD = 16cms

Area of Parallelogram = Base × Height

CD × 16 = 72

CD = \\frac{72}{16}) cms

We extend CD till P such that ∠APD = 90°

So, again by applying the area formula considering CD as base

CD × AP = Area of Parallelogram ABCD

9 × AP = 72

**AP = 8 cms **

We have AP = 8 cms, AD = 16 cms
We observe that the ratio of the sides form a 1 : √3 : 2 triangle
Therefore DP = 8√3 cms
Area (△APD) = 1/2 × AD × DP = 1/2 × 8 × 8√3 sq cms
Area (△APD) = **32 √3 sq cms**

The question is **"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 " **

Choice C is the correct answer.

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