Question

In the figure, AC=10cm, PC= 15cm, PQ=12cm, find PB?

• A. AB
• B. BC
• C. CA
• D. AS

Answer 1

The correct option is C CA

In $△ABCand△PQC$
$\angle PQC=\angle ABC={90}^{\circ }$
$\angle C=\angle C$ (common angle)
Therefore, $△ABC\sim △PQC$ by AA similarity
$\frac{AC}{PC}=\frac{AB}{PQ}\Rightarrow \frac{10}{15}=\frac{AB}{12}\Rightarrow AB=8cm$
In $△ABC$ , by using Pythagoras theorem
$A{B}^2+B{C}^2=A{C}^2\Rightarrow BC=6cm$
Now, PB = PC –BC = 15-6 = 9cm