Question

In the given figure, ABC is a triangle and PQ is a straight line meeting AB in P and AC in Q. If AP = 1 cm, PB = 3 cm, AQ = 1.5 cm, QC = 4.5 cm, prove that are of ∆APQ is $\frac{1}{16}$ of the area of ∆ABC.

Answer 1

We have:
$$\begin{gathered} \frac{AP}{AB}=\frac{1}{1+3}=\frac{1}{4}\mathrm{and}\frac{AQ}{AC}=\frac{1.5}{1.5+4.5}=\frac{1.5}{6}=\frac{1}{4} \\ \Rightarrow \frac{AP}{AB}=\frac{AQ}{AC} \end{gathered}$$

Also, $\angle A=\angle A$
By SAS similarity , we can conclude that ∆APQ$~$∆ABC.

$$\begin{gathered} \frac{\mathrm{ar}(∆APQ)}{\mathrm{ar}(∆ABC)}=\frac{A{P}^2}{A{B}^2}=\frac{1^2}{4^2}=\frac{1}{16} \\ \Rightarrow \frac{\mathrm{ar}(∆APQ)}{\mathrm{ar}(∆ABC)}=\frac{1}{16} \\ \Rightarrow \mathrm{ar}(∆APQ)=\frac{1}{16}\times \mathrm{ar}(∆ABC) \end{gathered}$$

Hence proved.