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

1. $18\sqrt 3$
2. $24\sqrt 3$
3. $32\sqrt 3$
4. $12\sqrt 3$

Given that, in a parallelogram $\text{ABCD}$ of area $72$ sq cm, the sides $\text{CD}$ and $\text{AD}$ have lengths $9$ cm and $16$ cm, respectively.

Let $‘h\text{’ cm}$ be the height.

We can draw the parallelogram,

We know that, the area of the parallelogram $= \text{Base} \times \text{Height}$

Thus, $9 \times h = 72$

$\Rightarrow \boxed{h = 8\;\text{cm}}$

In $\triangle \text{APD}$, apply the Pythagoras’ theorem,

$\text{(AD)}^{2} = \text{(AP)}^{2} + \text{(DP)}^{2}$

$\Rightarrow 8^{2} + \text{(DP)}^{2} = (16)^{2}$

$\Rightarrow \text{(DP)}^{2} = 256 – 64$

$\Rightarrow \text{(DP)}^{2} = 192$

$\Rightarrow \text{DP} = \sqrt{192}$

$\Rightarrow \text{DP} = \sqrt{64 \times 3}$

$\Rightarrow \text{DP} = 8 \sqrt{3} \; \text{cm}$

$\therefore$ The area of triangle $\text{APD}= \frac{1}{2} \times \text{Base} \times \text{Height}$

$\quad = \frac{1}{2} \times 8 \sqrt{3} \times 8$

$\quad = 32 \sqrt{3} \; \text{cm}^{2}$

Correct Answer $: \text{C}$

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