AREA RELATED TO CIRCLE

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CIRCLES

AREA AND PERIMETER

THEOREMS

• The perpendicular from the centre of a circle to a chord bisects the chord.
• The line drawn through the

centre of a circle to bisect a chord is perpendicular

to chord

• The angle subtended by an arc double the angle subtended by i on the remaining part of the cir

at the centre is

t at any point cle.

• Angles in the same segment are equal.

• Angle in a semi circle is a right angle.

Parts of a Circle

Radius

diameter

circumference

Major Sector

Major arc

Minor arc

Minor segment

Major segment

Minor Sector

A line drawn at right angles to the radius at the circumference is called the Tangent

What is a sector ?

• We've all had a slice of pie or a piece of pizza. Both are real life examples of a sector of a circle.
• A sector is a wedge of a circle made from two radii.
• Radii is the plural of radius, which is a line segment that starts on the outside of the circle and ends at the center of the circle.
• A radius is like the cut from the crust of the pizza to the middle.

What is a segment ?

• A segment is the section of a circle enclosed by a chord and an arc. Therefore, those halves of the pizza are segments.
• If you eat one half, you would have eaten a semicircle (half of a circle), which is the biggest segment of a circle.
• Since a circle has an infinite number of points on the circumference, there are many possibilities for a chord and, hence, many possibilities for segments.

Area of a sector

Area of the sector = ^θ ×Πr ²

360

Area of a segment

Area of the segment of a circle = area of the corresponding sector - area of then corresponding triangle

Length of an arc of a sector

Length of an arc of a sector of a circle with radius r and angle with degree measure θ =( θ /360)X 2πr

Distance travelled in one revolution

= circumference of the wheel Total distance travelled

=no of revolutions*circumference

=n*c

Distance = speed*time Hence n*c=speed* time

Questions

1. If the diameter of a semi-circular protractor is 14cm, then find its perimeter.

Perimeter = _∏r + d

= ²² × ¹⁴ +14

*

7 2

= 36cm²

2.From each corner of a square of side 4 cm a quadrant of a circle of radius 1 cm is cut and also a circle of diameter 2cm is cut as shown in the figure. Find the area of the remaining portion of the square.

Area of remaining portion of the square

= Area of square – (4*Area of a quadrant + Area of a circle)

7

7

= ⁶⁸ cm²

= 16 − 2× ²²

= 4 × 4 −[4× ⁹⁰ × ( 1)² + ²² × (1)²

360 7

3.In the given figure, a square OABC is inscribed in quadrant OPBQ. If OA=20cm, find the area of the shaded region. (Take π = 3.14)

Using Pythagoras Theorem; BO² = OA² + OC²

= 20² + 20²

BO = 20√2 cm = Radius of circle Area of quadrant = ¼ x πr²

= ¼ x π x (20√2)²

= 628 sq cm

Area of square = Side² = 20² = 400 sq cm

Area of shaded portion = 628 – 400

= 228sq cm

4. A paper is in the form of a rectangular ABCD in which AB = 18 cm and BC = 14 cm. A semicircle with BC as diameter is cut off. Find the area of the remaining portion.

NOTE

• The minute hand takes 60 min (1 hr) for a complete rotation.
• 60 min => 360^o
• 01 min => 360/6 = 6^o

The length of the minute hand of a clock is 5 cm.

Find the area swept by the minute hand during the time period 6:05 am and 6:40 am.

Angle after minute hand movesfor1min = 360^o

There are 35 mins between 6 : 05 am and 6 : 40 am.

∴ total angle = 35× 6 = 210^o Radius = 5cm

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Area of sector = ^θ × Πr ²

360

= ²¹⁰ × ²² × 5 × 5

360 7

₌ 275

6

= 45.83 c m²

CONVERSION OF

UNITS

Conversions of Units

SUMMARY

1. Area of the circle:πr²
2. Circumference of the circle=2πr
3. Length of an arc of a sector of a circle with radius r and angle with degree measure θ is= (θ /360)X 2πr
4. Area of the sector =( θ /360) X πr²
5. Area of the segment of a circle=area of the corresponding sector - area of then corresponding triangle