Module 3 circle area and perimeter

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<p>1. MODUL 3 MATEMATIK SPM ENRICHMENT TOPIC: CIRCLE, AREA AND PERIMETER TIME: 2 HOURS 1. Diagram 1 shows two sector of circle ORQ and OPS with centre O. R 12 cm 150 O 7 cm QP S 22 7 , calculate DIAGRAM 1 By using = (1) the perimeter for the whole diagram in cm, (2) area of the shaded region in cm2 . [ 6 marks ] Answer : (a) (b) 2 . In diagram 2, ABCD is a rectangle. 21 cm B A 2. 14 cm F CD FIGURE 4 E CF is an arc of a circle with center E where E is a point on the line DC with E C = 7 cm. Using = 22 , calculate 7 (1) the length, in cm, of arc CF (2) the area, in cm2 , of the shaded region [ 6 marks ] Answer : (a) (b) 3. 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. J Q P L O K R DIAGRAM 3 Using = 22 7 , calculate (1) the perimeter, in cm, of the whole diagram, (2) the area, in cm2 , of the shaded region. [6 marks] Answer: (a) 4. (b) 5. 4. In Diagram 4, JK and PQ are arcs of two circles with centre O. OQRT is a square. K Q R J P O T 210 DIAGRAM 4 OT = 14 cm and P is the midpoint of OJ. Using = 22 7 , calculate (1) the perimeter, in cm, of the whole diagram, (2) the area, in cm2 , of the shaded region. [6 marks] Answer: (a) 6. (b) 7. 5. Diagram 5 shows two sectors OLMN and OPQR with the same centre O. M L N 120 P R O Q DIAGRAM 5 OL = 14 cm. P is the midpoint of OL. [Use = 22 7 ] Calculate (1) the area of the whole diagram, (2) the perimeter of the whole diagram. [6 marks] Answer: (a) 8. (b) 9. 6. In Diagram 6, ABD is an arc of a sector with the centre O and BCD is a quadrant. A OD = OB = 14 cm and AOB = 45 . Using = 22 7 , calculate (1) the perimeter, in cm, of the whole diagram, (2) the area, in cm2 , of the shaded region. [6 marks ] Answer : (a) O D DIAGRAM 6 B C 10. (b) 11. 7. In Diagram 7, the shaded region represents the part of the flat windscreen of a van which is being wiped by the windscreen wiper AB. The wiper rotates through an angle of 210o about the centre O. Given that OA = 7 cm and AB = 28 cm. B A 210 o O A DIAGRAM 7 Using = 22 7 , calculate (1) the length of arc BB , (2) the ratio of arc lengths , AA : BB (3) the area of the shaded region. Answer: (a) B [ 7 m a r k s ] ( b ) ( c ) 12. 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 8 Using = 22 7 , calculate (a) the perimeter, in cm, of the whole diagram, (b) the area, in cm2 , of the shaded region. . [6 marks] Answer: (a) 13. (b) 14. 9. In Diagram 9, OPQS is a quadrant with the centre O and OSQR is a semicircle with the centre S. Q R S 60 T O P DIAGRAM 9 Given that OP = 14 cm. Using = 22 7 , calculate (a) the area, in cm2 , of the shaded region, (b) the perimeter, in cm, of the whole diagram. [6 marks] Answer: (a) 15. (b) 16. 10. In diagram 10, OABC is a sector of a circle with centre O and radius 14 cm. B A 60 C O DIAGRAM 10 By using = 22 , calculate 7 (1) perimeter, in cm, the shaded area. (2) area, in cm2 , the shaded area. [7 markah] Answer : (a) (b) 17. MODULE 3 - ANSWERS TOPIC: CIRCLE, AREA AND PERIMETER 1 9 0 2 2 12 @ 120 2 22 7 (a) 2 36 0 7 3 6 0 7 9 0 2 2 1 2 0 22 2 12 + 2 7 +12 + 5 36 0 7 3 6 0 7 57.5 3 (b) 9 0 22 122 @ 1 2 0 22 72 3 6 0 7 3 6 0 7 9 0 2 2 12 2 + 120 22 72 1 7 12 3 6 0 7 360 7 2 122. 48 2 FEC = 135(a) 135 2 2 2 7 360 7 16.5 (b) L = 1 3 2 2 7 3 5 736 0 7 Shaded area =(2114) = 138 3 2 7 0 2 2 90 2 21a) at au 363 6 0 7 2 7 0 2 2 2 21 + 9 0 3 6 0 7 36 0 = 138 b) 2 7 0 2 2 21 21 atau3 6 0 7 2 7 0 2 2 21 21 2 3 6 0 7 18. K1 K1 N1 K1 K1 N1 K2 K1 N1 K1 K1 N1 K1 K1 N1 K1 K1 = 962.5 cm2 4 a) 6 0 2 2 2 2836 0 7 6 0 2 2 2 28 +14 +1 3 6 0 7 113 1 3 atau 1133 3 b) 6 0 2 2 28 28 atau 36 0 7 3 0 6 0 2 2 28 28 6 0 2 36 0 7 36 0 5 0 4 5 1 2 0 2 2 2 a) 14 14 ata u 36 0 7 3 0 1 2 0 2 2 14 14 + 240 36 0 7 36 0 3 0 8 b) 1 2 0 2 2 2 1 4 ata u 36 0 7 3 6 0 1 2 0 2 2 2 1 4 + 2 4 0 19. 36 0 7 3 6 0 7 6 72 2 3 4 5 2 2 (a) 2 1 436 0 7 4 5 2 2 2 14 + 14 + 14 + 14 + 143 6 0 7 70 2 3 (b) 4 5 2 2 o r 9 0 2 2 14 14 36 0 7 14 1436 0 7 4 5 2 2 9 0 2 2 14 14 + 2 14 14 14 14 3 6 0 7 3 6 0 7 N1 K1 K1 N1 K1 K1 N1 K1 K1 N1 K1 K1 N1 K1 K1 N1 K1 K1 1</p>