In the figure, $□ A B C D$ is a trapezium. $S i d e A B ∥ s e g P Q ∥ s i d e D C a n d A P = 15, P D = 12, Q C = 14$ , then find $10 B Q$

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Solution

Given $A B ∥ C D ∥ P Q$
thus, $\frac{A P}{P D} = \frac{B Q}{Q C}$ (Equal intercept theorem)
$\frac{15}{12} = \frac{B Q}{14}$
$B Q = \frac{15 \times 14}{12}$
$B Q = 17.5$ cm

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