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
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
Updated On: Jul 28, 2025
- 7.8
- 8.5
- 9.1
- 9.3
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The Correct Option is C
Solution and Explanation
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: \( \boxed{BQ = \sqrt{84} \approx 9.17} \)
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