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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

Question Figure

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)