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
Correct Answer: 3
Area of the parallelogram ABCD = (base)(height) = (CD)(AP) = 72 sq.cm.
(CD)(AP) = 72 9(AP) = 72 => AP = 8
$D P=\sqrt{A D^{2}-A P^{2}}=\sqrt{16^{2}-8^{2}}=8 \sqrt{3}$
Area of triangle $A P D=\frac{1}{2}(A P)(P D)=32 \sqrt{3}$
Area of the parallelogram ABCD = (base)(height) = (CD)(AP) = 72 sq.cm.
(CD)(AP) = 72 9(AP) = 72 => AP = 8
$D P=\sqrt{A D^{2}-A P^{2}}=\sqrt{16^{2}-8^{2}}=8 \sqrt{3}$
Area of triangle $A P D=\frac{1}{2}(A P)(P D)=32 \sqrt{3}$
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