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

Referring to the given figure :
∠APB and ∠AQB both equal 90º (as angles in a semicircle are right angles).
Let AQ be denoted as x, making AP equal to 2x.
In right triangle ΔAPB:
Applying the Pythagorean theorem, AP² = AB² - BP².
This simplifies to 8² = 10² - 6², leading to AP = 8 and consequently, x = 4.
Similarly, in right triangle ΔAQB:
Using the Pythagorean theorem :
BQ² = AB² - AQ².
⇒ BQ² = 10² - 4² = 84
⇒ BQ = √84 ≈ 9.1
This results in BQ = √84, approximately 9.1.