In trapezium \(ABCD\) with \(AB\parallel DC\), points \(P\) and \(Q\) lie on \(AD\) and \(BC\) respectively, with \(PQ\parallel DC\). If \(PD=18\,\text{cm}\), \(BQ=35\,\text{cm}\) and \(QC=15\,\text{cm}\), find \(AD\).
60 cm
Step 1: Write down the given information.
Step 2: On side \(BC\), find the ratio of \(BQ\) to \(QC\):
\[ \dfrac{BQ}{QC} = \dfrac{35}{15} = \dfrac{7}{3} \]
Step 3: Because \(PQ \parallel DC\), the trapezium is divided into similar smaller trapeziums/triangles. This means the sides are divided in the same ratio. So, \[ \dfrac{AP}{PD} = \dfrac{BQ}{QC} = \dfrac{7}{3} \]
Step 4: Use this proportion to find \(AP\):
\[ \dfrac{AP}{18} = \dfrac{7}{3} \quad \Rightarrow \quad AP = \dfrac{7}{3} \times 18 \]
\[ AP = 42\,\text{cm} \]
Step 5: Now find the full side \(AD\):
\[ AD = AP + PD = 42 + 18 = 60\,\text{cm} \]
Final Answer: The length of \(AD\) is 60 cm.