In a parallelogram $ABCD$ of area $72 \mathrm{sq} \mathrm{cm}$, the sides $CD$ and $AD$ have lengths $9 \mathrm{~cm}$ and $16 \mathrm{~cm}$, respectively. Let P be a point on $C D$ such that $A P$ is perpendicular to $C D$. Then the area, in sq cm, of triangle $A P D$ is

We have a parallelogram ABCD with:

Area = 72 sq cm

Side CD = 9 cm

Side AD = 16 cm

Point P is on CD such that AP ⊥ CD

When AP is perpendicular to CD, AP becomes the height of the parallelogram with CD as the base.

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For any parallelogram: Area = Base × Height

Using CD as base and AP as height:

Area of parallelogram ABCD = CD × AP

72 = 9 × AP

AP = 72 ÷ 9 = 8 cm

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Now we have a right triangle APD where:

AD = 16 cm (hypotenuse)

AP = 8 cm (one leg, perpendicular to CD)

PD = ? (other leg, along CD)

Since AP ⊥ CD creates a right angle at P, we can use the Pythagorean theorem:

$AD^2 = AP^2 + PD^2$

$16^2 = 8^2 + PD^2$

$256 = 64 + PD^2$

$PD^2 = 192$

$PD = \sqrt{192} = \sqrt{64 \times 3} = 8\sqrt{3}$ cm

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For triangle APD:

Base = PD = $8\sqrt{3}$ cm

Height = AP = 8 cm

Area of triangle APD = $\tfrac{1}{2} \times \text{base} \times \text{height}$

Area = $\tfrac{1}{2} \times 8\sqrt{3} \times 8$

Area = $\tfrac{1}{2} \times 64\sqrt{3}$

Area = $32\sqrt{3}$ sq cm

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When a point creates a perpendicular from a vertex to the opposite side in a parallelogram, it gives us the height needed for area calculations. This height can then be used with the Pythagorean theorem to find other measurements in the resulting right triangle.

Answer: $32\sqrt{3}$ sq cm