Let ABCD be a quadrilateral with overline{AB} | vedclass

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(A) The area of a quadrilateral $ABCD$ can be calculated as the sum of the areas of two triangles,$	riangle ABC$ and $	riangle ADC$.
Area of $	ria

Vedclass · Accepted Answer

$2.5$

Vedclass · Answer

(A) The area of a quadrilateral $ABCD$ can be calculated as the sum of the areas of two triangles,$	riangle ABC$ and $	riangle ADC$.
Area of $	riangle ABC = \frac{1}{2} |\overline{AB} 	imes \overline{AC}| = \frac{1}{2} |\bar{a} 	imes (3\bar{a} + 2\bar{b})| = \frac{1}{2} |3(\bar{a} 	imes \bar{a}) + 2(\bar{a} 	imes \bar{b})| = \frac{1}{2} |0 + 2(\bar{a} 	imes \bar{b})| = |\bar{a} 	imes \bar{b}|$.
Area of $	riangle ADC = \frac{1}{2} |\overline{AD} 	imes \overline{AC}| = \frac{1}{2} |\bar{b} 	imes (3\bar{a} + 2\bar{b})| = \frac{1}{2} |3(\bar{b} 	imes \bar{a}) + 2(\bar{b} 	imes \bar{b})| = \frac{1}{2} |-3(\bar{a} 	imes \bar{b}) + 0| = \frac{3}{2} |\bar{a} 	imes \bar{b}|$.
Total Area of quadrilateral $ABCD = |\bar{a} 	imes \bar{b}| + \frac{3}{2} |\bar{a} 	imes \bar{b}| = \frac{5}{2} |\bar{a} 	imes \bar{b}|$.
The area of the parallelogram with adjacent sides $\bar{a}$ and $\bar{b}$ is $|\bar{a} 	imes \bar{b}|$.
Given that the area of the quadrilateral is $\alpha$ times the area of the parallelogram,we have $\frac{5}{2} |\bar{a} 	imes \bar{b}| = \alpha |\bar{a} 	imes \bar{b}|$.
Therefore,$\alpha = \frac{5}{2} = 2.5$.

Let $ABCD$ be a quadrilateral with $\overline{AB}=\bar{a}$,$\overline{AD}=\bar{b}$ and $\overline{AC}=3\bar{a}+2\bar{b}$. If its area is $\alpha$ times the area of the parallelogram with $AB$ and $AD$ as adjacent sides,then the value of $\alpha$ is equal to

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