two circles with centre at A and B, and of radius 5 cm and 3 cm touch Each Other internally. if the perpendicular bisector of AB meets a bigger circle in P and Q, find the length of PQ.
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Hello Mate!
So, in figure we can see that AC is radius of bigger circle and BC is radius of circle.
As in given question
AC = 5 cm and BC = 3 cm
AC - BC = AB
5 cm - 3 cm = AB
2 cm = AB
Now, PQ is perpendicular bisector of AB so AD = 1 cm.
Now, since ∆ADP is right triangle at D so,
AP² = PD² + AD²
5² - 1² = PD²
√24 = PD
2√6 cm = PD
Now, PQ is chord which intersect AC that passes through centre so AC is perpendicular bisector on PQ.
PD = QD
PD + QD = PQ
2PD = PQ
2(2√6) = PQ
4√6 = PQ.
Hence length of PQ is 4√6.
Have great future ahead!
So, in figure we can see that AC is radius of bigger circle and BC is radius of circle.
As in given question
AC = 5 cm and BC = 3 cm
AC - BC = AB
5 cm - 3 cm = AB
2 cm = AB
Now, PQ is perpendicular bisector of AB so AD = 1 cm.
Now, since ∆ADP is right triangle at D so,
AP² = PD² + AD²
5² - 1² = PD²
√24 = PD
2√6 cm = PD
Now, PQ is chord which intersect AC that passes through centre so AC is perpendicular bisector on PQ.
PD = QD
PD + QD = PQ
2PD = PQ
2(2√6) = PQ
4√6 = PQ.
Hence length of PQ is 4√6.
Have great future ahead!
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