Xavier Aptitude Test 2022 Paper

Given:
$\mathrm{BD}=6\mathrm{cm}$, $\mathrm{AC}=9\mathrm{cm}$, $\mathrm{DC}=5\mathrm{cm}$, $\mathrm{BP}=8\mathrm{cm}$, and $\mathrm{DP}=5\mathrm{cm}$.
Since $\mathrm{AD}$ is the angular bisector applying the angular bisector theorem we have :
$\frac{\mathrm{A}\mathrm{B}}{\mathrm{B}\mathrm{D}}=\frac{\mathrm{A}\mathrm{C}}{\mathrm{C}\mathrm{D}}$
Hence : Considering $\mathrm{AB}=\mathit{x}\mathrm{cm}$
$\frac{9}{5}=\frac{\mathit{x}}{6}$
$\mathit{x}=10.8\mathrm{cm}$.
Now since $\mathrm{BD}$ is the angular bisector for angle $\mathrm{PBA}$ we have :
Applying the internal angle bisector theorem
$\frac{\mathrm{PB}}{\mathrm{PD}}=\frac{\mathrm{BA}}{\mathrm{AD}}$
Considering $\mathrm{AD}=\mathit{y}\mathrm{cm}$
$\frac{8}{5}=\frac{10.8}{\mathit{y}}$
$\mathit{y}=6.75\mathrm{cm}$
$\mathrm{AP}=\mathrm{AD}+\mathrm{DP}$
$=6.75+5=11.75\mathrm{cm}$