$AB$ is a diameter of a circle of radius $5 \mathrm{~cm}$. Let $P$ and $Q$ be two points on the circle so that the length of $PB$ is $6 \mathrm{~cm}$, and the length of $AP$ is twice that of $AQ$ . Then the length, in cm , of $QB$ is nearest to

We have a circle with radius 5 cm, so the diameter AB = 10 cm.

Points P and Q lie on the circle, with PB = 6 cm and AP = 2×AQ.

---

Since AB is a diameter and P, Q are points on the circle, we can use Thales' theorem: Any angle inscribed in a semicircle is a right angle.

This means: $\angle APB = \angle AQB = 90°$

Right triangles allow us to use the Pythagorean theorem.

---

In right triangle APB:

AB = 10 cm (diameter)

PB = 6 cm (given)

$\angle APB = 90°$ (angle in semicircle)

Using Pythagorean theorem: $AP^2 + PB^2 = AB^2$

$AP^2 + 6^2 = 10^2$

$AP^2 + 36 = 100$

$AP^2 = 64$

$AP = 8$ cm

---

We're told that AP = 2×AQ

So: $AQ = \tfrac{AP}{2} = \tfrac{8}{2} = 4$ cm

---

In right triangle AQB:

AB = 10 cm (diameter)

AQ = 4 cm (calculated above)

$\angle AQB = 90°$ (angle in semicircle)

Using Pythagorean theorem: $AQ^2 + QB^2 = AB^2$

$4^2 + QB^2 = 10^2$

$16 + QB^2 = 100$

$QB^2 = 84$

$QB = \sqrt{84}$

---

$84 = 4 \times 21$

So $\sqrt{84} = \sqrt{4 \times 21} = 2\sqrt{21}$

$\sqrt{21} \approx 4.58$

Therefore: $QB = 2\sqrt{21} \approx 2 \times 4.58 = 9.16$ cm

---

When we see a diameter in a circle problem, we immediately think "angles in semicircle = 90°" - this unlocks the power of Pythagorean theorem!

The length of QB is nearest to 9.1 cm.