Question

In the figure, AC = 10cm, PC = 15cm, PQ = 12 cm, then PB = __ cm .

Answer 1

$In\Delta ABCand\Delta PQC,\angle PQC=\angle ABC={90}^{\circ }\angle C=\angle C\text{(common angle)}\therefore \Delta ABC\sim \Delta PQC\text{by AA similarity}\frac{AC}{PC}=\frac{AB}{PQ}=\frac{BC}{QC}\Rightarrow \frac{10}{15}=\frac{AB}{12}=\frac{BC}{QC}\Rightarrow AB=8cmIn\Delta ABC,A{B}^2+B{C}^2=A{C}^2\Rightarrow {10}^2=8^2+B{C}^2\Rightarrow BC=6cmPB=PC-BC=15-6=9cm\therefore PB=9cm.$