# 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

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