ABCD is a quadrilateral with overline{AB}=bar | vedclass

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(B) 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 $	riang

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$\frac{5}{2}$

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(B) 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 (2\bar{a} + 3\bar{b})| = \frac{1}{2} |2(\bar{a} 	imes \bar{a}) + 3(\bar{a} 	imes \bar{b})| = \frac{1}{2} |0 + 3(\bar{a} 	imes \bar{b})| = \frac{3}{2} |\bar{a} 	imes \bar{b}|$.Area of $	riangle ADC = \frac{1}{2} |\overline{AD} 	imes \overline{AC}| = \frac{1}{2} |\bar{b} 	imes (2\bar{a} + 3\bar{b})| = \frac{1}{2} |2(\bar{b} 	imes \bar{a}) + 3(\bar{b} 	imes \bar{b})| = \frac{1}{2} |-2(\bar{a} 	imes \bar{b}) + 0| = |\bar{a} 	imes \bar{b}|$.Total area of quadrilateral $ABCD = \frac{3}{2} |\bar{a} 	imes \bar{b}| + |\bar{a} 	imes \bar{b}| = \frac{5}{2} |\bar{a} 	imes \bar{b}|$.The area of the parallelogram with adjacent sides $AB$ and $AD$ is $|\bar{a} 	imes \bar{b}|$.Given that the area of the quadrilateral is $\alpha$ times the area of the parallelogram,we have $\frac{5}{2} |\bar{a} 	imes \bar{b}| = \alpha |\bar{a} 	imes \bar{b}|$.Thus,$\alpha = \frac{5}{2}$.

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

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