Question : In the given figure, AC and DE are perpendicular to the tangent CB. AB passes through the centre O of the circle, whose radius is 20 cm. If AC is 36 cm, what is the length (in cm) of DE?
Option 1: 4
Option 2: 6
Option 3: 2
Option 4: 8
Correct Answer: 4
Solution : Given: AC $\perp$ CB DE $\perp$ CB OP $\perp$ CB ( radius $\perp$ tangent) AC || DE || OP AO = OD = OP= radius = 20 cm AC = 36 cm Let BD = $x$ cm AB = $x + 20 + 20$ = $x$+40 OB = $x + 20$ Since $\angle$ A = $\angle$ O = $\angle$ D and $\angle$ P = $\angle$ C = $\angle$ E, $\triangle$ OBP ~ $\triangle $ABC ~ $\triangle $DBE $\frac{OB}{OP}$ = $\frac{AB}{AC}$ ⇒ $\frac{x+20}{20}$ = $\frac{x+40}{36}$ ⇒ $36x+720$ = $20x+800$ ⇒ $16x$ = 80 ⇒ BD = $x$ = 5 ⇒ $\frac{AB}{AC}$ = $\frac{DB}{DE}$ ⇒ $\frac{5+40}{36}$ = $\frac{5}{DE}$ ⇒ DE = $\frac{36}{45}×5$ = 4 cm Hence, the correct answer is 4.
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