Geometry Practice Test 2

Length of AB = 24 cm
In ΔOPA,
By Pythagoras theorem,
${\left(\mathit{O}\mathit{A}\right)}^2$ = ${\left(\mathit{O}\mathit{P}\right)}^2+{\left(\mathit{A}\mathit{P}\right)}^2$
${\left(15\right)}^2$ = ${\left(\mathit{O}\mathit{P}\right)}^2+{\left(12\right)}^2$
255 + 144 = $\mathit{O}{\mathit{P}}^2$

OP = $\sqrt{81}$ = 9 cm
Also in ΔO’PA,
${\left(\mathit{O}\prime \mathit{A}\right)}^2$ = ${\left(\mathit{O}\prime \mathit{P}\right)}^2+{\left(\mathit{A}\mathit{P}\right)}^2$
${\left(13\right)}^2$ = ${\left(\mathit{O}\prime \mathit{P}\right)}^2+{\left(12\right)}^2$
169 −144 = ${\left(\mathit{O}\prime \mathit{P}\right)}^2$
$\sqrt{25}$ = O'P
O′P = 5 cm.
OP + O′P = 9 + 5 = 14 cm