Class 10 Mathematics

# Area of Circle

PrevNext## Exercise 12.3 (NCERT Book) Part 1

**Solution:** Here, OR is the hypotenuse of ∆PQR, because lines from two ends of diameter always make a right angle when they meet at circumference.

So, OR^{2} = PQ^{2} + PR^{2}

= 24^{2} + 7^{2}

= 576 + 49 = 625

Or, OR = 25 cm

Or radius = 12.5 cm

Area of triangle = ½ x base x height

= ½ x 24 x 7 = 84 sq cm

Area of semicircle = ½ x πr^{2}

= ½ x π x 12.5^{2} = 245.3125 sq cm

So, area of shaded region = 245.3125 – 84 = 161.3125 sq cm

**Solution:** Area of shaded region

= Area of Bigger Sector – Area of Smaller Sector

If R and r are the two radii, then area of shaded region

**Solution:** Area of Square = Side^{2} = 14^{2} = 196 sq cm

Area of two semicircles = πr^{2}

= π x 7^{2} = 154 sq cm

Area of shaded region = 196 – 154 = 42 sq cm

**Solution:** Area of sector outside triangle

Area of triangle

Area of shaded region = 94.28 + 62.352 = 156.632 sq cm

**Solution:** Area of square = Side^{2} = 4^{2} = 16 sq cm

Area of cut portion = Area of two circles = 1 circle + 4quadrants

= 2 x π x 1^{2} = 6.28 sq cm

So, area of remaining portion of square = 16 – 6.28 = 22.28 sq cm

**Solution:** In ∆OCB; OB = 32 cm, ∠OBC = 30^{o}

Hence, area of equilateral triangle

Area of circle = πr^{2} = π x 32^{2} = 3215.36 sq cm

Area of shaded region = 3215.36 – 1330.176 = 1885.184 sq cm

**Solution:** Area of square = Side^{2} = 14^{2} = 196 sq cm

Area of four quadrants = Area of 1 circle

= πr^{2} = π x 7^{2} = 154 sq cm

Area of shaded portion = 196 – 154 = 42 sq cm

**Solution:** Distance = Length of straight track + circumference of inner loop

Circumference = 2πr = 3.14 x 60 = 188.4

Hence, distance = 188.4 + 106 + 106 = 400.4 m

Area of straight portion = 2 x length x width

= 2 x 106 x 10 = 2120 sq m

Area of circular portion = π(R^{2} – r^{2})

Where, R = radius of outer circle and r = radius of inner circle

Area = π(40^{2} – 30^{2}) = 2200 sq m

So, Total area of track = 2200 + 2120 = 4320 sq m

**Solution:** Area of smaller circle = πr^{2} = π x (3.5)^{2} = 38.5 sq cm

Area of bigger circle = 4 x 38.5 (because radius is double that of smaller circle)

Hence, area of bigger semicircle = 2 x 38.5 = 77 sq cm

Area of ∆ABC = ½ x AB x OC

= ½ x 14 x 7 = 49 sq cm

Area of shaded portion in semicircle = 77 – 49 = 28 sq cm

Total area of shaded portion = 28 + 38.5 = 66.5 sq cm

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Exercise 1

Exercise 2 (Part 1)

Exercise 2 (Part 2)

Exercise 3 (Part 2)