Question

pls solve this question fast​

Answer 1

Join QP and BP,

In triangle RBQ and RBP ,

QR = PR =8cm ( given )

[Perpendicular from the centre of a circle to a chord bisects it]

So here perpendicular from the centre of a circle is OB , and it bisects chord QP , so

angle BRQ = angle BRP = 90°

BR = BR ( COMMON )

triangle RBQ is congruent to triangle RBP ( RHS CONGRUENCY RULE )

You can also proof that BRQ = ORQ

BR= RP = 4 cm ( by CPCT )

OB = BR + RP

= 4+4 = 8cm

radius = 8cm

diameter = 2*radius

diameter = 2*8 = 16 cm

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