Geometry Practice Test 6

Radius of circle = 6 cm
PO = 10 cm.
PB = 5 cm
Let BC = x cm

ΔOTP is a right angle triangle.
${\left(\mathit{O}\mathit{P}\right)}^2$ = ${\left(\mathit{O}\mathit{T}\right)}^2+{\left(\mathit{P}\mathit{T}\right)}^2$
${\left(10\right)}^2$ = ${\left(6\right)}^2+{\left(\mathit{P}\mathit{T}\right)}^2$
$\sqrt{100-36}$ = PT
PT = 8 cm
We know that
$\mathit{P}{\mathit{T}}^2$ = PB.PC
${\left(8\right)}^2$ = 5 (5 + x)
64 = 25 + 5x
64 - 25 = 5x
39 = 5x
BC = x = 7.8 cm