Let us write perimeter and area of circular shaded sector below.

(i) Given: Radius of the circle = 12 cm and angle subtended by the arc = 90°

We know that the length of the arc

⇒ Length of arc AB

Length of AB:

∵ ∠ O = 90°

Using Pythagoras theorem,

OA² +OB² = AB²

⇒ 12² + 12² = AB²

⇒ AB² = 288

⇒ AB=12√2 cm

⇒ AB =16.92 cm

Perimeter of the circular shaded sector = Length of arc AB + length of AB

⇒ Perimeter of the circular shaded sector=18.857 + 16.92 =35.777 cm

Now, Area of the segment AB = area of sector ABO – area of triangle ABO

We know that the area of the minor sector

⇒ Area of ABO

⇒ Area of ABO = 113.14 cm²

In ∆ABO,

∠ O = 90°, AO = BO = 12 cm {radius of the circle}

⇒ Area of ∆ABO = 72 cm²

∴Area of the segment AB = area of sector ABO – area of triangle ABO

⇒Area of the segment AB = 113.14 – 72

⇒ Area of the segment AB = 41.14 cm²

(ii) Given: Radius of the circle = 42 cm and angle subtended by the arc = 60°

We know that the length of the arc

⇒ Length of arc

Length of AC:

In ∆ABC,

∠ B = 60°, AB = BC = 42 cm {radius of the circle}

⇒∠ ABC = ∠ ACB {angles opposite to equal sides are equal}

By the angle sum property of the triangle,

∠ BAC + ∠ ACB + ∠ B = 180°

⇒ 2∠ BAC = 180° - 60°

⇒∠ BAC = 60°

Hence, ∆ ABC is an equilateral triangle.

∴ AC = 42 cm

Perimeter of the circular shaded sector = Length of arc AC + length of AC

⇒ Perimeter of the circular shaded sector = 42 + 44 = 86 cm

Now, Area of the segment AC = area of sector ACB – area of triangle ACB

We know that the area of the minor sector

⇒ Area of ACB =

⇒ Area of ACB = 924 cm²

We know that

Area of a equilateral triangle where a is the side of it.

⇒ Area of ∆ABC = 763.83cm²

∴Area of the segment AC = area of sector ABC – area of triangle ABC

⇒ Area of the segment AC = 924 – 763.83

⇒ Area of the segment AC = 160.17 cm²