In the adjoining figure, $P Q\|X Y\| B C, A P=2 \mathrm{~cm}$, $P X=1.5 \mathrm{~cm}$ and $B X=4 \ma

Question

In the adjoining figure, $P Q\|X Y\| B C, A P=2 \mathrm{~cm}$, $P X=1.5 \mathrm{~cm}$ and $B X=4 \mathrm{~cm}$. If $Q Y=0.75 \mathrm{~cm}$, then $A Q+C Y=$\

Professor PC Thomas Classes · Answer

(c): Given, $P Q\|X Y\| B C, A P=2 \mathrm{~cm}, P X=1.5 \mathrm{~cm}$, $B X=4 \mathrm{~cm}$ and $Q Y=0.75 \mathrm{~cm}$.\ Here, $\frac{P X}{Q Y}=\frac{1.5}{0.75}=\frac{2}{1} \Rightarrow Q Y=\frac{P X}{2}$\ Similarly, $A Q=\frac{A P}{2}=1 \mathrm{~cm}$ and $C Y=\frac{B X}{2}=\frac{4}{2}=2 \mathrm{~cm}$\ So, $A Q+C Y=1+2=3 \mathrm{~cm}$\