Question

PA is tangent to the circle with centre O. If BC = 3cm, AC = 4 cm and $\Delta ACB\sim \Delta PAO,$ then find OA and $\frac{OP}{AP}$.

Answer 1

Consider the problem

$\angle ACB$ is an angle in a semi circle

Therefore, $\angle ACB$$={90}^{\circ }$

Now, in $\Delta ACB$

we use Pythagoras Theorem ${\left(AB\right)}^2={\left(AC\right)}^2+{\left(BC\right)}^2$

$\begin{array}{c}\therefore A{B}^2=16+9\\ AB=\sqrt{25}\\ =5cm\end{array}$

So,

Radius $OA=\frac{1}{2}AB$

$\begin{array}{c}OA=\frac{1}{2}\times 5\\ =\frac{5}{2}cm\end{array}$

Now, since $\Delta ACB\sim \Delta PAO$

$\frac{BA}{CA}=\frac{OP}{AP}$

Therefore, $\frac{OP}{AP}=\frac{5}{4}$