Geometry Practice Test 1

Chord AB of the outer circle intersects inner circle at point C and R.
OP = 3 cm

Radius of the outer circle = ${\mathit{r}}_1$ = 13 cm
Radius of the inner circle = 7 cm
In right triangle OPB,
${\left(\mathit{O}\mathit{P}\right)}^2+{\left(\mathit{P}\mathit{B}\right)}^2$ = ${\left(\mathit{O}\mathit{B}\right)}^2$
${\left(3\right)}^2+{\left(\mathit{P}\mathit{B}\right)}^2$ = ${\left(13\right)}^2$
$\mathit{P}{\mathit{B}}^2$ = 169 − 9 = 160
PB = $\sqrt{160}$ = $4\sqrt{10}$ cm
Also ΔOPC = ${\left(\mathit{O}\mathit{C}\right)}^2$ = ${\left(\mathit{O}\mathit{P}\right)}^2+{\left(\mathit{P}\mathit{C}\right)}^2$
${\left(7\right)}^2$ = ${\left(3\right)}^2+{\left(\mathit{P}\mathit{C}\right)}^2$
49 - 9 = $\mathit{P}{\mathit{C}}^2$
PC = $\sqrt{40}$ = $2\sqrt{10}$ cm
AP = PB = $4\sqrt{10}$ cm
CP = PD = $2\sqrt{10}$ cm
AC = AP − CP
AC = $4\sqrt{10}-2\sqrt{10}$
AC = $2\sqrt{10}$ cm