Two circles with centres A and B of radii 5 cm and 3 cm respectively touch each other internally. If the perpendicular bisector of AB meets the bigger circle at P and Q, then the value of PQ is
A. √6 cm
B. 2√6 cm
C. 3√6 cm
D. 4√6 cm
Answer: Option D
Solution (By Examveda Team)
According to question
AP = 5 cm, DB = 3 cm
AB = 5 - 3 = 2 cm
AO = AB ÷ 2 = 1 cm
PQ is ⊥ bisector
∴ AO = 1, PO = OQ
In right angle ΔPOA
(AP)2 = (OA)2 + (OP)2
(5)2 = (1)2 + (OP)2
(OP)2 = 25 - 1
(OP)2 = 24
(OP) = 2√6 cm
∴ PQ = 2 × OP
PQ = 2 × 2√6
PQ = 4√6 cm
This Question Belongs to Arithmetic Ability >> Geometry
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