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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