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

• A. 7.8
• B. 8.5
• C. 9.1
• D. 9.3

Answer 1

The Correct Option is C

From right triangle \( \triangle APB \):
Using the Pythagorean Theorem:
\[ AP^2 = AB^2 - BP^2 \] Given: \( AB = 10 \), \( BP = 6 \)
\[ AP^2 = 10^2 - 6^2 = 100 - 36 = 64 \Rightarrow AP = \sqrt{64} = 8 \] Since AP = 2x, we get: \[ 2x = 8 \Rightarrow x = 4 \] So, \( AQ = x = 4 \)

Now from right triangle \( \triangle AQB \):
Again using the Pythagorean Theorem: \[ BQ^2 = AB^2 - AQ^2 = 10^2 - 4^2 = 100 - 16 = 84 \] So: \[ BQ = \sqrt{84} \approx 9.17 \]

Answer: \( \boxed{BQ = \sqrt{84} \approx 9.17} \)