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Q.

In given figure, two circles with centers A and B of radii 20 cm and 8 cm 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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a

4 93  cm  

b

4 94  cm   

c

4 92  cm  

d

4 91  cm  

answer is A.

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Detailed Solution

Given that two circles with centers A and B of radii 20 cm and 8 cm touch each other internally.
We have to find the length of PQ.
The Pythagoras theorem relates the sides of a triangle and is expressed as the square of the hypotenuse side is equal to the sum of squares of the other two sides.
The required figure geometry is shown below,
Question ImageIn the figure, Since PQ is perpendicular bisector on AB. So,
AB =ACBC AB =208 AB =12cm   The triangle ADP is the right triangle at D. Thus,
A P 2 =P D 2 +A D 2 20 2 =P D 2 + 6 2 40036 =P D 2 PD =2 91  cm   Since PQ is a chord which intersects AC that passes through the center so AC is perpendicular bisector on PQ. Thus, PD = QD Therefore,
PQ =PD+QD PQ =2PD PQ =2(2 91 ) PQ =4 91  cm  
The length of PQ is 4 91  cm  .
Therefore, the correct option is 1.
 
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