The following figure shows a circle with center O and a diameter AB.

(i) Name the angle APB.

(ii) State the measure of angle APB.

(iii) If AP = 12 cm and OA = 10 cm; find the lengths of AB and BP.

(iv) If 4AP = 3PB = 12 cm; find the radius of the circle.

(i) AB is diameter of the circle.

Hence, angle APB is the angle of semi-circle.

(ii) As, the angle subtended by a diameter at any point on the circumference of a circle is a right angle (90°).

Hence, angle APB = 90°.

(iii) Given,

AP = 12 cm and OA = 10 cm

⇒ OA = OB (Radii of the circle)

⇒ OB = 10 cm

From figure,

⇒ AB = OA + OB = 10 cm + 10 cm = 20 cm

Since, angle APB = 90°.

Using pythagoras theorem,

⇒ AP² + PB² = AB²

⇒ 12² + PB² = 20²

⇒ 144 + PB² = 400

⇒ PB² = 400 - 144

⇒ PB² = 256

⇒ PB = $\sqrt{256}$

⇒ PB = 16 cm.

Hence, AB = 20 cm and PB = 16 cm.

(iv) Given,

⇒ 4AP = 3PB = 12 cm

⇒ AP = $\dfrac{12}{4}$ = 3 cm

⇒ PB = $\dfrac{12}{3}$ = 4 cm

Since, angle APB = 90°.

Using pythagoras theorem,

⇒ AP² + PB² = AB²

⇒ 3² + 4² = AB²

⇒ 9 + 16 = AB²

⇒ 25 = AB²

⇒ AB = $\sqrt{25}$

⇒ AB = 5 cm

Now, OA = OB (Radii of the circle)

From figure,

⇒ AB = OA + OB

⇒ AB = 2 x OA

⇒ 5 = 2 x OA

⇒ OA = $\dfrac{5}{2}$ = 2.5 cm.

Hence, the radius of the circle = 2.5 cm.