Question 3
AB is a diameter of a circle of radius 5 cm. Let P and Q be two points on the circle so that the length of PB is 6 cm, and the length of AP is twice that of AQ. Then the length, in cm, of QB is nearest to
Solution
Since AB is a diameter, AQB and APB will right angles.
In right triangle APB, AP = $$\sqrt{10^2-6^2}=8$$
Now, 2AQ=AP => AQ= 8/2=4
In right triangle AQB, AP = $$\sqrt{10^2-4^2}=9.165$$ =9.1 (Approx)
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