Let overline{OA} = vec{a},overline{OB} = 10ve | vedclass

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(B) The area of the quadrilateral $OABC$ can be calculated as the sum of the areas of $	riangle OAB$ and $	riangle OBC$.Area of $	riangle OAB = \fr

Vedclass · Accepted Answer

$6$

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(B) The area of the quadrilateral $OABC$ can be calculated as the sum of the areas of $	riangle OAB$ and $	riangle OBC$.Area of $	riangle OAB = \frac{1}{2} |\overline{OA} 	imes \overline{OB}| = \frac{1}{2} |\vec{a} 	imes (10\vec{a} + 2\vec{b})| = \frac{1}{2} |10(\vec{a} 	imes \vec{a}) + 2(\vec{a} 	imes \vec{b})| = \frac{1}{2} |0 + 2(\vec{a} 	imes \vec{b})| = |\vec{a} 	imes \vec{b}|$.Area of $	riangle OBC = \frac{1}{2} |\overline{OB} 	imes \overline{OC}| = \frac{1}{2} |(10\vec{a} + 2\vec{b}) 	imes \vec{b}| = \frac{1}{2} |10(\vec{a} 	imes \vec{b}) + 2(\vec{b} 	imes \vec{b})| = \frac{1}{2} |10(\vec{a} 	imes \vec{b}) + 0| = 5|\vec{a} 	imes \vec{b}|$.Thus,$p = |\vec{a} 	imes \vec{b}| + 5|\vec{a} 	imes \vec{b}| = 6|\vec{a} 	imes \vec{b}|$.The area $q$ of the parallelogram with adjacent sides $OA$ and $OC$ is given by $q = |\vec{a} 	imes \vec{b}|$.Therefore,$p/q = \frac{6|\vec{a} 	imes \vec{b}|}{|\vec{a} 	imes \vec{b}|} = 6$.

Let $\overline{OA} = \vec{a}$,$\overline{OB} = 10\vec{a} + 2\vec{b}$,and $\overline{OC} = \vec{b}$,where $O, A, C$ are non-collinear. Let $p$ be the area of the quadrilateral $OABC$ and $q$ be the area of the parallelogram with adjacent sides $OA$ and $OC$. Then $p/q = \dots$

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