In trapezium ABCD,overline{AB} parallel overl | vedclass

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(C) In trapezium $ABCD$,$\overline{AB} \parallel \overline{CD}$ and $\overline{AC} \cap \overline{BD} = \{P\}$.Since $\overline{AB} \parallel \overlin

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$20$

Vedclass · Answer

(C) In trapezium $ABCD$,$\overline{AB} \parallel \overline{CD}$ and $\overline{AC} \cap \overline{BD} = \{P\}$.Since $\overline{AB} \parallel \overline{CD}$,the alternate interior angles are equal,i.e.,$\angle PAB = \angle PCD$ and $\angle PBA = \angle PDC$.By the $AA$ similarity criterion,$	riangle PAB \sim 	riangle PCD$.Therefore,the ratios of their corresponding sides are equal:$\frac{PA}{PC} = \frac{PB}{PD}$Substituting the given values:$\frac{10}{15} = \frac{PB}{12}$$\frac{2}{3} = \frac{PB}{12}$$PB = \frac{2 	imes 12}{3} = 8$Now,the length of diagonal $BD$ is:$BD = PB + PD = 8 + 12 = 20$.

In trapezium $ABCD$,$\overline{AB} \parallel \overline{CD}$ and $\overline{AC} \cap \overline{BD} = \{P\}$. If $PA = 10$,$PC = 15$ and $PD = 12$,then $BD = \ldots$

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