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- 1. CIRCLE Ng Huoy Miin, Trace Gew Yee, “ Prepared by : Pang Kai Yun, Sam Wei Yin, . Liew Poh Ka, Chong Jia Yi
- 2. CIRCLE A circle is a plain figure enclosed by a curved line, every point on which is equidistant from a point within, called the centre.
- 3. DEFINITION Circumference — The circumference of a circle is the perimeter Diameter — The diameter of a circle is longest distance across a circle. Radius — The radius of a circle is the distance from the center of the circle to the outside edge. .
- 4. CIRCUMFERENCE C= T[d C=2’l'[I‘ * Where 1: = 3.142 .
- 5. EXAMPLE (CIRCUMFERENCE) C = TECI C = 2111‘ =3.142x6cm =2x3.142x4cm = 18.85 cm = 25_14 cm .
- 6. AREA OF CIRCLE A= T[I‘2 * Where n = 3.142 .
- 7. EXAMPLE 1 (AREA OF CIRCLE) A = T[I‘2 = 3.142 x 62 = 3.142 X 36 = 113.11 cmz
- 8. EXAMPLE 2 (AREA OF CIRCLE) A= T[I‘2 = 3.142 X 42 = 3.142 X 16 = 50.27 cmz
- 9. ARC A portion of the Circumference of a Circle. arc A
- 10. ARC LENGTH (DEGREE) I“: (36100) 27” * A circle is 360° D
- 11. EXAMPLE 1 (ARC LENGTH)
- 12. RADIAN The angle made by taking the radius and wrapping it along the edge of the Circle.
- 13. FROM DEGREE TO RADIAN 112 1800 Radians = ( ) X Degree FROM RADIAN TO DEGREE 1800 112 Degree = ( ) X Radians
- 14. EXAMPLE (FROM DEGREE TO RADIAN) 1. 300 = ( " )x 300 =3 rad 1800 6 2.1500=( “ )x 1500=5—“rad 1800 6 3. 2700 = (1;‘O0)x 2700 = Bfrad
- 15. EXAMPLE (FROM RADIAN TO DEGREE) 1.§rad= (i)0)X(§)rad= 600 TT 2.2?" rad= (%)0)x(2?ﬂ)rad= 1200 3.? rad: (i)0)x(§Tﬂ)rad= 2250 I C
- 16. ARC LENGTH (RADIAN) la= r0 * Where 0 is radians
- 17. EXAMPLE 2 (ARC LENGTH) la = r 0 = 4.16 cm X 2.5 rad = 10.4 cm
- 18. EXAMPLE 3 (ARC LENGTH) la= r0 la= r0 =10 cmxgrad =25 cmX0.8rad = 7.86 Cm = 20 Cm .
- 19. SECTOR A sector is the part of a circle enclosed by two radii of a circle and their intercepted arc. Minor Sector Major Sector .
- 20. AREA OF SECTCR (DEGREE) By propotion, Area of Sector _ Central Angle Area of Circle _ 3600 A _ n 7Tr2 _ 360° Tl A= <.. ... >
- 21. EXAMPLE 1 (AREA OF SECTOR)
- 22. AREA OF SECTOR (RADIAN) By propotion, Area of Sector _ Central Angle Area of Circle _ Zn A _ 1 E _ 21t 9 2 = — X 71'7" 2a A = lr20 2
- 23. EXAMPLE 2 (AREA OF SECTOR) Area = 1 r26’ 2 . =§<s>2<1.4> 9 =17.5cm2
- 24. SEGMENT The segment of a circle is the region bounded by a chord and the arc subtended by the chord. Chord Minor Segment Major Segment D
- 25. AREA OF SEGMENT E 6 M 4 E . m n S } .1 1 2 g S AI V H _ _ _ m 6 z_. w . .~l. 6 ( (1 a 2 2 _2I_ __ __ m C m Ob .4;
- 26. EXAMPLE (AREA OF SEGMENT) The above diagram shows a sector of a circle, with centre Oand a radius 6 cm. The length of the arc AB is 8 cm. Find (i) 0 A06’ (ii) the area of the shaded segment. Solution: (i) la = 8 cm (ii) the area Ofthe shaded segment Iazre §r2(0-sin0) 8 = r0 0 1 2 . 180 8:69 = 5(6) (1.333-s1n(1.333X G )) 9 = 1-333 radians = §(36)(1.333 — sin 76.380) . B AOB= 1.333 radians = 6.501 cm2
- 27. CHORD Chord of a circle is a line segment whose ends lie on the circle. Chord
- 28. GIVEN THE RADIUS AND CENTRAL ANGLE 0 Chord length = 2r sin (3) D
- 29. EXAMPLE 1 0 Chord length = 2r sin (3) 90 = 2(6) sin (7) = 12 X sin 45 0 ,2 Cm C = 8.49 cm
- 30. GIVEN THE RADIUS AND DISTANCE TO CENTER This is a simple application of Pythagoras‘ Theorem. Chord length = 2/ r2 — d2 D
- 31. EXAMPLE 2 Find the chord of the circle where the radius measurement is about 8 cm that is 6 units from the middle. Solution: Chord length = 2/ T = 27% = 2«/6<I§—36 =2/ E . = 10.58 cm
- 32. SEMICIRCLE
- 33. PERIMETER OF A SEMICIRCLE o Remember that the perimeter is the distance round the outside. A semicircle has two edges. One is half of a circumference and the other is a diameter 0 So, the formula for the perimeter of a semicircle is: Perimeter = nr + 2r Q
- 34. EXAMPLE (PERIMETER) 8cm Perimeter = 1Tr + 2r = (3.142)(§) + 8 = 20.56 cm Q
- 35. AREA OF A SEMICIRCLE OA semicircle is just half of a circle. To find the area of a semicircle we just take half of the area of a circle. 0 So, the formula for the area of a semicircle is: 1 Area = Enrz
- 36. ‘I, _ . n E E)(AIIIlPI_E 2 1 Area = - TE? ‘ 2 4cm 1 2 = X 3.142 X 42 — 25.14 cm2
- 37. SUMMARY Length of Arc: Area of Segment: s = r 0 1 . Length of Chord: A T 5 V2 (9 T Sme ) . 0 I = 2r sin — Area of Triangle: I 2 . Areaofsectorz 3. A T Er Slug 1 A = —r20 2 A” Length Area of Triangle Area of Segment
- 38. THANK YOU I

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