Question

o is the centre of circle having radius 5cm triangle ABC is inscribed in the circle such that AB is equal to AC is equal to 6 cm find the length of the side BC

Answer 1

BC = 9.6 cm

Step-by-step explanation:

Given: Two equal chords AB and AC , As AB =  AC  =  6 cm  and Radius of circle  is 5 cm.

As we know, the angle bisector of angle between two equal chords of a circle passes through the center of circle.

Let center of our circle is " O ", So OA bisect ∠ BAC . Join line BC and line OA intersect line BC at " M "

And we also know, Internal angle bisector of angle divide the opposite side in the ratio of the sides containing the angle ( Here these sides are AB and AC ).

So,  

M divide BC in the ratio of 6 : 6   = 1 : 1

Hence, M is mid point of line BC, So MB  =  MC  

And we know when a line from center to chord divide a chord in two equal parts , So joining line is perpendicular to chord .

∠ OMB = ∠ OMC  = 90°

And

As ∠ AMB + ∠ OMB  = 180° ( Linear pair angles ),

So,

∠ AMB = 90°

SO, Δ AMB and Δ OMB are right angles triangle .

Let OM = x , S, AM = 5 - x            [ As OA  = 5 cm ( Radius )]

Now we apply Pythagoras theorem , In Δ  AMB , and we get

$AB^2 = AM^2 + MB^2$

Substitute values, we get

$6^2 = ( 5 - x )^2 + MB^2$

$MB^2 = 36 - ( 25 + x^2 - 10x )$

$MB^2 = 36 - 25 - x^2 + 10x$$MB^2 = 11 - x^2 + 10x$         [1]

Now we apply Pythagoras theorem , In Δ OMB, and we get

$OB^2 = OM^2 + MB^2$

$5^2 = ( x )^2 + MB^2$

$MB^2 = 25 - x^2$                                      [2]

Now from equation 1 and 2, we get

$11 - x^2 + 10x = 25 - x^2$

10x  = 14  

$x=\frac{14}{10}$

x = 1.4, Substitute that value in equation  2, we get

$MB^2 =25 - ( 1.4 )^2$

$MB^2 =25 - 1.96$

$MB^2 =23.04$

$MB=\sqrt{23.04}$

$MB=\sqrt{\frac{2304}{100} }$

$MB=\sqrt{\frac{576}{25} }$

$MB=\sqrt{\frac{24^{2} }{5^2} }$

$MB=\frac{24}{5}$

MB= 4.8 cm

MB = MC = 4.8 cm

And

BC = MB + MC = 4.8 + 4.8 = 9.6 cm.