Question

$ ext{ABCD}$ is a trapezium in which $AB \parallel DC$ and $P$ and $Q$ are points on $AD$ and $BC$ respectively, such that $PQ \parallel DC$. If $PD = 18	ext{ cm}$, $BQ = 3.5	ext{ cm}$ and $QC = 15	ext{ cm}$, then find the length of $AD$.

Accepted Answer

Since $AB \parallel PQ \parallel DC$, by the Thales theorem extended to transversals: $$\frac{AP}{PD} = \frac{BQ}{QC}$$ $$\frac{AP}{18} = \frac{3.5}{15} \implies AP = 18 	imes \frac{3.5}{15} = 4.2	ext{ cm}$$ Therefore, $AD = AP + PD = 4.2 + 18 = 22.2	ext{ cm}$.

Parallel lines in trapezium and segment lengths

Explanation

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