AB is a diameter of a circle of radius 5 cm. Let P and Q be two points on the…
2025
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.
8.5
- B.
9.3
- C.
9.1
- D.
7.8
Show answer & explanation
Correct answer: C
Concept: By Thales' theorem, if AB is a diameter of a circle, then the angle subtended by AB at any other point on the circle is a right angle. So both angle APB and angle AQB are 90°, which turns each unknown chord into the leg of a right triangle whose hypotenuse is the diameter AB — solvable using the Pythagorean theorem.
AB is a diameter of radius 5 cm, so AB = 10 cm. By the angle-in-a-semicircle theorem, angle APB = 90°, and angle AQB = 90°.
In right triangle APB, the Pythagorean theorem gives AP2 + PB2 = AB2. With PB = 6 cm, AP2 + 36 = 100, so AP2 = 64 and AP = 8 cm.
AP is twice AQ, so AQ = AP ÷ 2 = 4 cm.
In right triangle AQB, the Pythagorean theorem gives AQ2 + QB2 = AB2. With AQ = 4 cm, QB2 = 100 − 16 = 84, so QB = √84 ≈ 9.165 cm.
Cross-check: Squaring each offered value against this result shows 9.12 = 82.81 is closer to 84 than 9.32 = 86.49 is, confirming that QB is nearest to 9.1 cm among the given options.