square ABCD is a trapezium in which overline{ | vedclass

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(B) In trapezium $ABCD$,$\overline{ AD } \| \overline{ BC }$ and $\overline{ AC } \cap \overline{ BD } = \{ P \}$.Since $\overline{ AD } \| \overline{

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$9$

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(B) In trapezium $ABCD$,$\overline{ AD } \| \overline{ BC }$ and $\overline{ AC } \cap \overline{ BD } = \{ P \}$.Since $\overline{ AD } \| \overline{ BC }$,the alternate interior angles are equal,i.e.,$\angle PAD = \angle PCB$ and $\angle PDA = \angle PBC$.Therefore,by $AA$ similarity criterion,$	riangle PAD \sim 	riangle PCB$.Since the triangles are similar,the ratios of their corresponding sides are equal:$\frac{ PA }{ PC } = \frac{ PD }{ PB }$Substituting the given values:$\frac{5}{ PC } = \frac{9}{7.2}$$PC = \frac{5 	imes 7.2}{9} = \frac{36}{9} = 4$.Now,$AC = PA + PC = 5 + 4 = 9$.

$\square ABCD$ is a trapezium in which $\overline{ AD } \| \overline{ BC }, \overline{ AC } \cap \overline{ BD }=\{ P \} .$ If $PD =9, PA =5$ and $PB =7.2,$ then $AC =\ldots \ldots \ldots \ldots$

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