Mensuration Practice Test 7

In the figure PS = 8 cm and QR = 12 cm
OT = OS (radius of the sphere)
ΔPSR is similar to ΔPOT
⇒ $\frac{\mathit{P}\mathit{S}}{\mathit{P}\mathit{T}}$ = $\frac{\mathit{S}\mathit{R}}{\mathit{O}\mathit{T}}$
$\frac{8}{\mathit{P}\mathit{T}}$ = $\frac{6}{\mathit{r}}$ [PT = $\sqrt{\mathit{P}{\mathit{O}}^2-\mathit{O}{\mathit{T}}^2}$]
= $\frac{8}{\sqrt{\mathit{P}{\mathit{O}}^2-\mathit{O}{\mathit{T}}^2}}$ = $\frac{6}{\mathit{r}}$
= $\frac{8}{9}$ = $\frac{\sqrt{\mathit{P}{\mathit{O}}^2-\mathit{O}{\mathit{T}}^2}}{\mathit{r}}$
= ${\left(\frac{8}{6}\right)}^2$ = $\frac{\mathit{P}{\mathit{O}}^2-{\mathit{r}}^2}{{\mathit{r}}^2}$
= ${\left(\frac{8}{6}\right)}^2$ = $\left(\frac{\mathit{P}\mathit{O}}{\mathit{r}}\right)-1$
= $\frac{64+36}{36}$ = ${\left(\frac{\mathit{P}\mathit{O}}{\mathit{r}}\right)}^2$
= $\sqrt{\frac{100}{36}}$ = $\frac{\mathit{P}\mathit{O}}{\mathit{r}}$
= $\frac{10}{6}$ = $\frac{\mathit{P}\mathit{S}-\mathit{O}\mathit{S}}{\mathit{r}}$
= $\frac{10}{6}$ = $\frac{8-\mathit{r}}{\mathit{r}}$
= $\frac{10}{6}$ = $\frac{8}{\mathit{r}}-1$ = $\frac{16}{6}$ = $\frac{8}{\mathit{r}}$
2r = 6
r = 3 cm
Volume of the sphere = $\frac{4}{3}\mathit{\pi }{\mathit{r}}^3$
= $\frac{4}{3}\times \frac{22}{7}\times {\left(3\right)}^2$
≈ 113 $\mathit{c}{\mathit{m}}^3$