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

A
9.3
B
7.8
C
9.1
D
8.5

Question Explanation

Text Explanation


Since AB is a diameter, AQB and APB will right angles.

In right triangle APB, AP = 10262=8\sqrt{10^2-6^2}=8

Now, 2AQ=AP => AQ= 8/2=4

In right triangle AQB, AP = 10242=9.165\sqrt{10^2-4^2}=9.165 =9.1 (Approx)

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