The diagonals AC and BD of a rhombus ABCD int | vedclass

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(D) Since $BD=2 \sqrt{2}$,we have $OB=OD=\sqrt{2}$.Given that $(3,4)$ is the midpoint of $AC$,we have $C=(5,6)$.Thus,$OA=OC=\sqrt{(3-1)^2+(4-2)^2}=\sq

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$-3\alpha + 12$

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(D) Since $BD=2 \sqrt{2}$,we have $OB=OD=\sqrt{2}$.Given that $(3,4)$ is the midpoint of $AC$,we have $C=(5,6)$.Thus,$OA=OC=\sqrt{(3-1)^2+(4-2)^2}=\sqrt{4+4}=2 \sqrt{2}$.In $	riangle AOB$,$OA^2+OB^2=AB^2$,so $AB^2=(2\sqrt{2})^2+(\sqrt{2})^2=8+2=10$,which means $AB=\sqrt{10}$.Since $O(3,4)$ is the midpoint of $BD$,we have $\alpha+\gamma=6$ and $\beta+\delta=8$.Also,$OB^2=(\alpha-3)^2+(\beta-4)^2=2$ and $OD^2=(\gamma-3)^2+(\delta-4)^2=2$.Since $ABCD$ is a rhombus,$AB=BC=CD=DA=\sqrt{10}$.Using the distance formula for $CD^2=10$,we get $(\gamma-5)^2+(\delta-6)^2=10$.Solving the system of equations for $\alpha, \beta, \gamma, \delta$ with the condition $\alpha Finally,$\beta+\gamma-\delta=(\beta-\delta)+\gamma=(-2\alpha+6)+(6-\alpha)=-3\alpha+12$.

The diagonals $AC$ and $BD$ of a rhombus $ABCD$ intersect at the point $(3,4)$. If $BD=2 \sqrt{2}$,$A=(1,2)$,$B=(\alpha, \beta)$,$D=(\gamma, \delta)$ and $\alpha < \delta < \gamma < \beta$,then $\beta+\gamma-\delta=$

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