Question

a circle with Centre P has chord AB is equals to 8 cm distance PQ is equals to 3 cm find the radius of the circle​

Answer 1

(note : que can be solved if PQ perpendicular to AB)

Solution :

We know,

Perpendicular from centre bisects the chord .

So,

AQ = BQ = 8/2 cm = 4 cm.

Now,

On joining P to A we get rt angle triangle PAQ. (<Q = 90 deg)

So, by Pythagoras theorem :

AP^2 = AQ^2 + PQ^2

AP^2 = 4^2 + 3^2

AP^2 = 16 + 9

AP^2 = 25

AP = 5 cm.

Hence, radius of circle is 5 cm

Answer 2

Answer:

Join ap

Now, from triangle apq

$ap = \sqrt{aq {}^{2} + pq {}^{2} } \\ = \sqrt{4 {}^{2} + 3 {}^{2} } \\ = \sqrt{16 + 9} \\ = \sqrt{25} \\ = 5$

Ans: Radius of the circle is 5 cm