Question

Please answer me of this question..​

Answer 1

In ∆ BPO and ∆ BQO,

OB = OB (Common sides)

OP =OQ. (Radius )

angle BPO = angle BQO. (both 90°)

then ,

∆ BPO is congruent to∆ BQO. (RHS)

BP = BQ. (CPCT)

BQ = 8 cm

Similarly

∆BQO is congruent to ∆ CQO,

BQ = QC

QC = 8cm

BC = QC + BQ

BC = 8 + 8 = 16cm

Similarly,

∆ CQO and ∆ COR are congruent,

CQ = CR. (CPCT)

CR = 8cm.

In ∆ APO and ∆ ARO,

AO = AO. (Common)

angle P = angle R. (each 90°)

PO = RO. ( Radius)

then ,

∆ APO is congruent to ∆ ARO

AP = AR. (CPCT)

AR = 6 cm

And,

AC = AR + RC = 6 + 8 = 14 cm

AC = 14 cm

hope it will help you...

ONE REQUEST PLZ MARK AS BRAINLEST