In figure .15.50 , two circles with centres A and B and of radii 5cm and 3cm touch each other internally.If the perpendicular bisector of segment AB meets the bigger circle in P and Q , find the length of PQ.
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Answer:
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.
Explanation:
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Answer:-
It is given that
two circles with centres at A and B, and of radii 5cm and 3cm touch each other internally.
The perpendicular bisector of AB meets the bigger circle in P Join the line AP From the figure we know that PQ intersects the line AB at L So we get AB = (5 – 3) AB = 2cm.
We know that
PQ is the perpendicular bisector of AB So we get AL = ½ × AB Substituting the values AL = ½ × 2 So we get AL = 1 cm.
Consider
△ PLA Using the Pythagoras theorem AP2 = AL2 + PL2
By substituting the values we get 52 = 12 + PL2 So we get PL2 = 52 – 12 On further calculation PL2 = 25 – 1 PL2 = 24 By taking the square root PL = √24 So we get PL = 2 √6 cm We know that PQ = 2 × PL By substituting the values PQ = 2 × 2 √6 PQ = 4 √6 cm
Therefore,
the length of PQ = 4 √6 cm
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