In a right trapezium ABCD,the diagonals are p | vedclass

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(D) Let the vertices be represented by vectors. Since $ABCD$ is a right trapezium with $AB$ perpendicular to $BC$ and $AD$ parallel to $BC$,we can set

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$\sqrt{3}:\sqrt{2}$

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(D) Let the vertices be represented by vectors. Since $ABCD$ is a right trapezium with $AB$ perpendicular to $BC$ and $AD$ parallel to $BC$,we can set $B$ at the origin $(0)$. Let $\vec{BA} = \vec{a}$ and $\vec{BC} = \vec{c}$. Since $AD \parallel BC$ and $AD:BC = 2:3$,we have $\vec{AD} = \frac{2}{3}\vec{c}$. Thus,$\vec{BD} = \vec{BA} + \vec{AD} = \vec{a} + \frac{2}{3}\vec{c}$.The diagonals $AC$ and $BD$ are perpendicular,so $\vec{AC} \cdot \vec{BD} = 0$.$(\vec{c} - \vec{a}) \cdot (\vec{a} + \frac{2}{3}\vec{c}) = 0$.Since $\vec{a} \perp \vec{c}$,$\vec{a} \cdot \vec{c} = 0$. Expanding the dot product:$\frac{2}{3}|\vec{c}|^2 - |\vec{a}|^2 = 0 \Rightarrow |\vec{a}|^2 = \frac{2}{3}|\vec{c}|^2$.Now,the ratio of the lengths of the diagonals is:$\frac{|\vec{AC}|}{|\vec{BD}|} = \frac{|\vec{c} - \vec{a}|}{|\vec{a} + \frac{2}{3}\vec{c}|} = \frac{\sqrt{|\vec{c}|^2 + |\vec{a}|^2}}{\sqrt{|\vec{a}|^2 + \frac{4}{9}|\vec{c}|^2}}$.Substituting $|\vec{a}|^2 = \frac{2}{3}|\vec{c}|^2$:$= \frac{\sqrt{|\vec{c}|^2 + \frac{2}{3}|\vec{c}|^2}}{\sqrt{\frac{2}{3}|\vec{c}|^2 + \frac{4}{9}|\vec{c}|^2}} = \frac{\sqrt{\frac{5}{3}}}{\sqrt{\frac{10}{9}}} = \sqrt{\frac{5}{3} 	imes \frac{9}{10}} = \sqrt{\frac{3}{2}} = \frac{\sqrt{3}}{\sqrt{2}}$.Thus,the ratio is $\sqrt{3}:\sqrt{2}$.

In a right trapezium $ABCD$,the diagonals are perpendicular,and the ratio of the length of bases $AD : BC = 2 : 3$. Then the ratio of the length of diagonals is

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