Question

In a circle of radius 5 cm,AB and AC are two chords such that AB = AC = 6 cm.The perpendicular distance of A from BC is​

Answer 1

Answer: ANSWER

We know that, if AB and AC are two equal chords of a circle, then the centre of the circle lies on the bisector of  

∠BAC.

Here, AB=AC=6cm. So, the bisecor of ∠BAC passes through the centre O i.e. OA is the bisector of ∠BAC.

Since, the internal bisector of an angle divides the opposite sides in the ratio of the sides containing the  

angle. Therefore, M divides BC in the ratio 1:1, i.e. M is the mid-point of BC.

⇒OM⊥BC.  

In the right triangle ABM, we have  

⇒AB  

2

=AM  

2

+BM  

2

 

⇒36=AM  

2

+BM  

2

 

⇒BM  

2

=36−AM  

2

 ...(i)

In the right triangle OBM, we have

⇒OB  

2

=OM  

2

+BM  

2

 

⇒25=(OA−AM)  

2

+BM  

2

 

⇒BM  

2

=25−(OA  

A

​  

M)  

2

 

⇒25−(5−AM)  

2

  .....(ii)

From equations (i) and (ii), we get

36−AM  

2

=25−(5−AM)  

2

 

⇒11−AM  

2

+(5−AM)  

2

=0

⇒11−AM  

2

+25−10AM+AM  

2

=0

⇒10AM=36

⇒AM=3.6

Putting AM=3.6 in equation (i), we get

BM  

2

=36−(3.6)  

2

=36−12.96

⇒BM=  

36−12.96

​  

=  

23.04

​  

=4.8cm

Hence, BC=2BM=2×4.8cm=9.6cm.