SSC CGL Model Paper 19

PS = 10 cm [diameter]
PR = 9 cm
PR ∥ OM ∥ ST [perpendicular to the same line]
[and OM is a radius, radius makes 90° with the tangent]
ΔPRQ ∼ ΔOMQ ∼ ΔSTQ
[∠Q is common & one angle = 90°]
$\frac{\mathrm{PR}}{\mathrm{PQ}}=\frac{\mathrm{OM}}{\mathrm{OQ}}=\frac{\mathrm{ST}}{\mathrm{SQ}}$ [property of similar triangle]
OM = 5 cm [radius]
$\frac{9}{10+\mathit{x}}=\frac{5}{5+\mathit{x}}$ [x = length of SQ]
45 + 9x = 50 + 5x
x = 5/4 cm
$\frac{\mathrm{OM}}{\mathrm{ST}}=\frac{\mathrm{OQ}}{\mathrm{SQ}}$
$\mathrm{OQ}=5+\frac{5}{4}+\frac{25}{4}\mathrm{cm}$
$\frac{5}{\mathrm{ST}}=\frac{\frac{25}{4}}{\frac{5}{4}}\Rightarrow \mathrm{ST}=1\mathrm{cm}$