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MODUL 3MATEMATIK SPM ENRICHMENTTOPIC: CIRCLE, AREA AND PERIMETERTIME: 2 HOURS1.Diagram 1 shows two sector of circle ORQ and OPS with centre O.By using= 722, calculate(a)the perimeter for the whole diagram in cm,(b)area of the shaded region in cm .2[ 6 marks ]Answer :(a)(b)OPQRS1507cm12cmDIAGRAM 1

2.In diagram 2, ABCD is a rectangle.FIGURE 4CF is an arc of a circle with center E where E is a point on the line DC with EC= 7 cm. Using 722, calculate(a)the length, in cm, of arc CF(b)the area, in cm , of the shaded region2[ 6 marks ]Answer :(a)(b) CBADEF14cm21cm F

3.Diagram 3 shows two sectors OPQR and OJKL.OPQR and OJKL are three quarters of a circle.POL and JOR are straight lines. OP = 21cm and OJ= 7 cm.DIAGRAM 3Using 227, calculate(a)the perimeter, in cm, of the whole diagram,(b)the area, in cm , of the shaded region.2[6 marks]Answer:(a)(b) OPQRJLK

4.In Diagram 4, JK and PQ are arcs of two circles with centre O.OQRT is a square.DIAGRAM 4OT = 14 cm and P is the midpoint of OJ.Using 722, calculate(a)the perimeter, in cm, of the whole diagram,(b)the area, in cm , of the shaded region.2[6 marks]Answer:(a)(b) OPJTRQK 210

5.Diagram 5 shows two sectors OLMN and OPQR with the same centre O.OL = 14 cm. P is the midpoint of OL.[Use= 722]Calculate(a)the area of the whole diagram,(b)the perimeter of the whole diagram.[6 marks]Answer:(a)(b)P 120OQRLMNDIAGRAM 5

6.In Diagram 6, ABD is an arc of a sector with the centre O and BCD is aquadrant.OD = OB = 14 cm and45AOB.Using 722, calculate(a)the perimeter, in cm, of the whole diagram,(b)the area, in cm , of the shaded region.2[6 marks]Answer :(a)(b) OABCDDIAGRAM 6

ABO 210oBA7.In Diagram 7, the shaded region represents the part of the flat windscreen of a vanwhich is being wiped by the windscreen wiper AB. The wiper rotates through anangle of 210 about the centre O.oGiven that OA = 7 cm and AB = 28 cm.DIAGRAM 7Using =722, calculate(a)the length of arc BB,(b)the ratio of arc lengths , AA: BB(c)the area of the shaded region.[7 marks]Answer:(a)(b)(c)

8.Diagram 8 shows a quadrant ADO with centre O and a sector BEF with centre B.OBC is a right angled triangle and D is the midpoint of the straight line OC.Given OC = OB = BE = 14 cm.DIAGRAM 8Using= 722, calculate(a)the perimeter, in cm, of the whole diagram,(b)the area, in cm , of the shaded region.2.[6 marks]Answer:(a)(b)

9.In Diagram 9, OPQS is a quadrant with the centre O and OSQR is a semicirclewith the centre S.DIAGRAM 9Given that OP = 14 cm. Using = 722, calculate(a)the area, in cm , of the shaded region,2(b)the perimeter, in cm, of the whole diagram.[6 marks]Answer:(a)(b) OPQRS60T

CBAO6010.In diagram 10, OABC is a sector of a circle with centre O and radius 14 cm.DIAGRAM 10By using 722, calculate(a)perimeter, in cm, the shaded area.(b)area, in cm , the shaded area.2[7 markah]Answer :(a)(b)

MODULE 3 - ANSWERSTOPIC: CIRCLE, AREA AND PERIMETER1(a) 77222360120@12722236090K1 53.575127722236012012722236090K1N1(b) 227722360120@1272236090K1 48.122127217722360120127223609022K1N12(a)135FECK2 5.1677222360135K1N1(b) 777223601353LK1 25.138141421)1421(3LareaShadedK1N13a) 212722360270atau 2772236090K1 212722360270+ 2772236090+14+14K1= 138N1b) 2121722360270atau2 7772236090K1 2121722360270-2 7772236090K1

= 962.5 cm2N14a) 28722236060K1 281414141428722236060K1 31113atau11333N1b) 282872236060atau 141472236060K1 282872236060 141472236060+14 14K1504N15a) 1414722360120atau 77722360240K1 1414722360120+ 77722360240K1308N1b) 147222360120atau 77222360240K1 147222360120+ 77222360240+7+7K1 3272N16(a) 14722236045K1 1414141414722236045K1 3270N1(b) 141472236045or 141472236090K1 14147223609014142141472236045K1161N1

7(i) 357222360210K1128 31@ 128.33N1(ii) 77222360210: 357222360210K11: 5N1(iii) 235722360210or 27722360210K1 22772236021035722360210K12156N18(a) 360452 72214or 14141422K111+14+14+14+5.799K158.80(2 d. p)N1(b) 7772236090or 36045 72214x 14K1 7772236090141421+ 36045 7221414K1136.5N19(a)A =1 141472236090and A =2 7772236060K1A A12K1128 31N1(b)P =1 14722236090or P =2 77222360180K1P + P + 1412K1

58N110(a)AB = 221414= 392= 19.80K1 147222360150atau 14722236060atau 14722236090K1Lengkok AC + 14 + 14 + 19.80 atauLengkok AB + lengkok BC + 14 + 14 + 19.80K184.47N1(b) 214722360150atau 141421K1 214722360150- 141421atauK1 3770 98 32158atau 3476atau 158.67N1