CAT 2019 Slot 1 QA Question & Solution
GeometryMedium
Question
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
Options
9.3
7.8
9.1
8.5
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)
