Question

O is the centre of a circle of radius 5 cm. T is a point such that OT=13 cm and OT intersects the circle at E. If AB is the tangent to the circle at E, find length of AB.

Answer 1

Step-by-step explanation:

Diagram:- Refer to the attachment

Solution:-

As, Tangent is perpendicular to Radius

⇢ OP ⊥ PT

⇢ ∠OPT = 90°

Applying Pythagoras Theorem in △OPT :-

⇢ OT² = OP² + PT²

⇢ 13² = 5² + PT²

⇢ PT² = 169 - 25

⇢ PT² = 144

⇢ PT = $\sqrt{144}$ = 12cm

Let AP be x cm

Since the length of the tangents are equal.

⇢ AP = AE = x

AT = PT - AP = 12 - x

Since, AB is the tangent to the circle and OE is the radius.

⇢ OE ⊥ AB

⇢ ∠AET = 90°

⇢ ∠OEA = 90°

By applying Pythagoras Theorem in △AET:-

⇢ AT² = AE² + ET²

⇢ (12 - x)² = x² + (13 - 5)²

⇢ 144 - 24x + x² = x² + 8²

⇢ 144 - 24x = 64

⇢ 24x = 80

⇢ x = $\dfrac{80}{24} = \dfrac{10}{3}$

⇢ AE = $\dfrac{10}{3}$ cm

Similarly, BE = $\dfrac{10}{3}$cm

As, AB = AE + BE

⇢ AB = $\dfrac{10}{3} + \dfrac{10}{3} = \dfrac{20}{3}$

$\dashrightarrow \underline{\boxed{\rm \therefore Length \: of \: AB = \dfrac{20}{3} cm}}$