10 Math Mensuration Circle

Finish Line & Beyond MENSURATION: CIRCLE Important Formulae: 1. Circumference of a circle = 2 Π r. 2. Area of a circle = Π r² 3. Length of an arc of a sector of a circle with radius r and angle with degree measure θ is

θ × 2 Πr 360

4. Area of a sector of a circle with radius r and angle with degrees measure θ is 2

θ × Πr² 360

5. Area of segment of a circle = Area of the corresponding sector – Area of the corresponding triangle.

Sector Segment

Arc

Example: The cost of fencing a circular field at the rate of Rs 24 per metre is Rs 5280. The field is to be ploughed at the rate of Rs 0.50 per m². Find the cost of ploughing the field (Take π =

22 ) 7

Solution : Length of the fence (in metres) = =

TotalCost Rate

5280 = 220 metres which is the circumference of the circle. 24

Now, Circumference= 2Πr Or, 220 = 2Πr

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Finish Line & Beyond Or,

2 r = 220 ×

Or,

r = 35

7 = 70 22

22 × 35 ²= 3850 m² 7 So, Cost of ploughing = 3850 × 0.50 = 1925 Rupees Hence Area =

Example: Find the area of the sector of a circle with radius 4 cm and of angle 30°. Also, find the area of the corresponding major sector (Use π = 3.14). Solution: Area of the sector = =

θ × Πr² 360

30 22 × × 16 360 7

= 4.19 cm² Area of Major Sector = Area of Circle- Area of Minor Sector = 50.2-4.19 =46.01 cm² Example: Find the area of the segment AYB, if radius of the circle is 21 cm and H AOB = 120°. Solution: Y A

P B O

Area of the segment AYB = Area of sector OAYB – Area of Δ OAB Area of Sector =

120 22 × 21²= 462 cm² × 360 7

The perpendicular on AB from O will be angular bisector of So, ∠ AOP = 60° In ∆AOP cos 60 = Or,

∠ O

OP 21

1 OP = 2 21

Or, OP = 11.5

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Finish Line & Beyond Now, tan 60 =

BP OP

2BP 21 21 3 Or, BP = 2 Or,

3=

So, Area of ∆AOP = 21

1 3 × 21 ÷ 4 (Area = × base × height ) 2

= 190.95 So Area of the segment = 462-190.95 = 271.04 cm²

Areas of Combinations of Plane Figures Example: Two circular flower beds have been shown on two sides of a square lawn ABCD of side 56 m. If the centre of each circular flower bed is the point of intersection O of the diagonals of the square lawn, find the sum of the areas of the lawn and the flower beds.

Solution: As the angle of one sector is 90°, so two sectors will be having an area equal to the semicircle. Rest of the area left in the lawn will be half the area of square. Hence the required answer = Area of Semicircle + Half Area of Square =

1 22 × 28 2 × 28 2 + 56²) ( 2 7

Radius of circle will be half of diagonal which is 56

2

= 4032 m²

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Finish Line & Beyond Example: Find the area of the shaded region in where ABCD is a square of side 14 cm.

Solution: Area of Shaded Region = Area of Square- 4 × Area of Circle Area of square ABCD = 14 × 14 = 196 cm² Diameter of each circle =14 ÷ 2 = 7 cm So, Radius of each circle = 7/2 cm. Area of 4 Circles =

22 7 7 × × × 4 = 154 cm² 7 2 2

So, Area of Shaded region = 196-154 = 42 cm²

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