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 at the centre is double the angle subtended by it at any point on the remaining part of the circle.
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 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 /360X 2 π r
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. *
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)
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; BO 2 = OA 2 + OC 2 = 20 2 + 20 2 BO = 20√2 cm = Radius of circle Area of quadrant = ¼ x π r 2 = ¼ x π x (20√2) 2 = 628 sq cm Area of square = Side 2 = 20 2 = 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.
CONVERSION OF UNITS
Conversions of Units 1 cm 2 = 10 mm x 10 mm =100 mm 2 1 m 2 = 100 cm x 100 cm = 10 000 cm 2 1 m 2 = 1000 mm x 1000 mm = 10 00 000 mm 2
1.Area of the circle: π r 2 2.Circumference of the circle=2 π r 3.Length of an arc of a sector of a circle with radius r and angle with degree measure θ is= θ /360X 2 π r 4.Area of the sector = θ /360 X π r 2 5.Area of the segment of a circle=area of the corresponding sector - area of then corresponding triangle SUMMARY