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Consider a parallelogram whose vertices are $\mathrm{A}(1,2), \mathrm{B}(4, \mathrm{y}), \mathrm{C}(\mathrm{x}, 6)$ and $\mathrm{D}(3,5)$ taken in order.
What is the area of the parallelogram?
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What is the area of the parallelogram?
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The correct answer is:
7 square units
Area of parallelogram $=2$ area of $\Delta \mathrm{ADB}$ $\vec{a}=\overline{\mathrm{AB}}=(4-1) \hat{i}+(3-2) \hat{j}$
$\vec{b}=\overrightarrow{\mathrm{AD}}=(3-1) \hat{i}+(5-2) \hat{j}$
$\therefore$ Area of parallelogram $=2\left[\frac{1}{2}|\vec{a} \times \vec{b}|\right]=|\vec{a} \times \vec{b}|$
Now; $\vec{a} \times \vec{b}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & 0 \\ 2 & 3 & 0\end{array}\right|=7 \hat{k}$.
Area $=|\vec{a} \times \vec{b}|=|7 \hat{\mathrm{k}}|=\sqrt{49}=7$ square unit
$\vec{b}=\overrightarrow{\mathrm{AD}}=(3-1) \hat{i}+(5-2) \hat{j}$
$\therefore$ Area of parallelogram $=2\left[\frac{1}{2}|\vec{a} \times \vec{b}|\right]=|\vec{a} \times \vec{b}|$
Now; $\vec{a} \times \vec{b}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & 0 \\ 2 & 3 & 0\end{array}\right|=7 \hat{k}$.
Area $=|\vec{a} \times \vec{b}|=|7 \hat{\mathrm{k}}|=\sqrt{49}=7$ square unit
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