Question

O is the centre of a circle with radius 5 centimetre if T is a point such that OT equals to 13 cm find the length of AB.​

Answer 1

Answer:

Step-by-step explanation:

Clearly ∠OPT=90  

o

 

Applying Pythagoras in △OPT, we have

OT  

2

=OP  

2

+PT  

2

 

⇒13  

2

=5  

2

+PT  

2

 

⇒PT  

2

=169−25=144

⇒PT=12 cm

Since lengths of tangents drawn from a point to a circle are equal. Therefore,

AP=AE=x(say)

⇒AT=PT−AP=(12−x)cm

Since AB is the tangent to the circleE. Therefore, OE⊥AB

⇒∠OEA=90  

o

 

⇒∠AET=90  

o

 

⇒AT  

2

=AE  

2

+ET  

2

 [Applying Pythagoras Theorem in △AET]

⇒(12−x)  

2

=x  

2

+(13−5)  

2

 

⇒144−24x+x  

2

=x  

2

+64

⇒24x=80

⇒x=  

3

10

​  

cm