,
In DP.OR line AB 11. OR
line AB II.OR, AO = 4.5
PB.
13, B-4 Then find P.A
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Answer:
Step-by-step explanation:
AP = AQ , BP = PR and CR = CQ (tangents from an external point)
Perimeter of ∆ABC = AB + BR + RC + CA
= AB + BP + CQ + CA
= AP + AQ
= 2AP
∆APO is a right-angled triangle. AO2 = AP2 + PO2
132 = AP2 + 52
AP2 = 144
AP = 12
∴ Perimeter of ∆ABC = 24 cm
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