Question

From adjacent figure the semi perimeter of ∆ABC=26cm , find AQ+PC+BR.​

Answer 1

Answer:

52 cm

Step-by-step explanation:

tangents drawn to a circle from an external point are equal .

RP = RQ & PC= PA & AQ = BQ

perimeter of ∆ ABC = 26×2 = 52 cm

in isosceles triangle ABC

,AC = AB = a (Assume) & BC = b

2a + b = 52 cm

here ∆ PAC & ABQ will also be isosceles triangle So PC = AC = a & AC is parallel to BR so BR = AC = a

BC is parallel to AQ So BC = AQ = b

AQ + PC + BR = b + a + a

= 2a + b = 52 cm

Answer 2

Answer:

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