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# MODULE 3-Circle Area and Perimeter

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## MODULE 3-Circle Area and PerimeterDocument Transcript

• PPR Maths nbk MODUL 3 SKIM TUISYEN FELDA (STF) 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 7 cm 150° O P Q S DIAGRAM 1 22 By using π = , calculate 7 (a) the perimeter for the whole diagram in cm, (b) area of the shaded region in cm2. [ 6 marks ] Answer : (a) (b)
• PPR Maths nbk 2. In diagram 2, ABCD is a rectangle. A 21 cm B 14 cm F D FIGURE 4 E C CF is an arc of a circle with center E where E is a point on the line DC with EC 22 = 7 cm. Using π = , calculate 7 (a) the length, in cm, of arc CF (b) the area, in cm2, of the shaded region [ 6 marks ] Answer : (a) (b)
• PPR Maths nbk 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 22 Using π = , calculate 7 (a) the perimeter, in cm, of the whole diagram, (b) the area, in cm2, of the shaded region. [6 marks] Answer: (a) (b)
• PPR Maths nbk 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. 22 Using π = , calculate 7 (a) the perimeter, in cm, of the whole diagram, (b) the area, in cm2 , of the shaded region. [6 marks] Answer: (a) (b)
• PPR Maths nbk 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. 22 [Use π = ] 7 Calculate (a) the area of the whole diagram, (b) the perimeter of the whole diagram. [6 marks] Answer: (a) (b)
• PPR Maths nbk 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 = 45o . 22 Using π = , calculate O B 7 (a) the perimeter, in cm, of the whole diagram, (b) the area, in cm2, of the shaded region. [6 marks] D C DIAGRAM 6 Answer : (a) (b)
• PPR Maths nbk 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′ 210o O A B DIAGRAM 7 22 Using π = , calculate 7 (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)
• PPR Maths nbk 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 22 Using π = , calculate 7 (a) the perimeter, in cm, of the whole diagram, (b) the area, in cm2, of the shaded region. . [6 marks] Answer: (a) (b)
• PPR Maths nbk 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 22 Given that OP = 14 cm. Using π = , calculate 7 (a) the area, in cm2, of the shaded region, (b) the perimeter, in cm, of the whole diagram. [6 marks] Answer: (a) (b)
• PPR Maths nbk 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 22 By using π = , calculate 7 (a) perimeter, in cm, the shaded area. (b) area, in cm2, the shaded area. [7 markah] Answer : (a) (b)
• PPR Maths nbk MODULE 3 - ANSWERS TOPIC: CIRCLE, AREA AND PERIMETER 1 90 22 120 22 (a) × 2 × × 12 @ × 2× × 7 K1 360 7 360 7 90 22 120 22 × 2 × × 12 + × 2 × × 7 + 12 + 5 360 7 360 7 K1 57.53 N1 90 22 120 22 2 (b) × × 12 2 @ × ×7 K1 360 7 360 7 90 22 120 22 1 × × 12 2 + × × 7 2 − × 7 × 12 360 7 360 7 2 K1 122.48 N1 2 (a) ∠FEC = 135o K2 135 22 × 2× ×7 360 7 K1 16.5 N1 135 22 (b) L3 = × ×7×7 K1 360 7 ⎛1 ⎞ Shaded area = (21 × 14) − ⎜ × 14 × 14 ⎟ − L3 ⎝2 ⎠ K1 = 138.25 N1 3 270 22 90 22 a) × × 2 × 21 atau × ×7×2 K1 360 7 360 7 270 22 90 22 × × 2 × 21 + × × 7 × 2 + 14 + 14 K1 360 7 360 7 = 138 N1 270 22 90 22 b) × × 21 × 21 atau 2× × ×7×7 K1 360 7 360 7 270 22 90 22 × × 21 × 21 - 2 × × ×7×7 K1 360 7 360 7
• PPR Maths nbk 2 = 962.5 cm N1 4 60 22 a) × 2 × × 28 K1 360 7 60 22 × 2 × × 28 + 14 + 14 + 14 + 14 + 28 K1 360 7 1 113 atau 113⋅33 N1 3 60 22 60 22 b) × × 28 × 28 atau × ×14 ×14 K1 360 7 360 7 60 22 60 22 × × 28 × 28 − × ×14 ×14 + 14 × 14 K1 360 7 360 7 504 N1 5 120 22 240 22 a) × × 14 × 14 atau × ×7×7 K1 360 7 360 7 120 22 240 22 × × 14 × 14 + × ×7×7 K1 360 7 360 7 308 N1 120 22 240 22 b) × 2 × × 14 atau × 2× ×7 K1 360 7 360 7 120 22 240 22 × 2 × × 14 + × 2× ×7 + 7 + 7 K1 360 7 360 7 2 72 N1 3 6 (a) 45 22 K1 ×2× × 14 360 7 ⎛ 45 22 ⎞ K1 ⎜ ×2× × 14 ⎟ + 14 + 14 + 14 + 14 ⎝ 360 7 ⎠ 2 N1 70 3 (b) 45 22 or 90 × 22 × 14 × 14 K1 × × 14 × 14 360 7 360 7 ⎛ 45 22 ⎞ ⎛ 90 22 ⎞ K1 ⎜ × × 14 × 14 ⎟ + 2 ⎜14 × 14 − × × 14 × 14 ⎟ ⎝ 360 7 ⎠ ⎝ 360 7 ⎠ 161 N1
• PPR Maths nbk 7 210 22 (i) × 2 × × 35 K1 360 7 1 128 @ 128.33 N1 3 210 22 210 22 (ii) × 2× ×7 : × 2 × × 35 K1 360 7 360 7 1: 5 N1 210 22 210 22 2 (iii) × × 35 2 or × ×7 K1 360 7 360 7 210 22 210 22 2 × × 35 2 − × ×7 K1 360 7 360 7 2156 N1 8 45 22 (a) ×2× × 14 or 14 2 + 14 2 − 14 K1 360 7 11 + 14 + 14 + 14 + 5.799 K1 58.80 (2 d. p) N1 90 22 45 22 (b) × ×7 × 7 or × × 14 x 14 K1 360 7 360 7 1 90 22 45 22 × 14 × 14 − × ×7 × 7 + × × 14 × 14 K1 2 360 7 360 7 136.5 N1 9 90 22 60 22 (a) A1 = × × 14 × 14 and A2 = × ×7×7 360 7 360 7 K1 A1 – A2 K1 1 128 N1 3 90 22 180 22 (b) P1 = × 2 × × 14 or P2 = ×2× ×7 K1 360 7 360 7 P1 + P2 + 14 K1
• PPR Maths nbk 58 N1 10 (a) AB = 14 2 + 14 2 = 392 = 19.80 K1 150 22 60 22 90 22 × 2 × × 14 atau × 2 × × 14 atau × 2 × × 14 K1 360 7 360 7 360 7 Lengkok AC + 14 + 14 + 19.80 atau Lengkok AB + lengkok BC + 14 + 14 + 19.80 K1 84.47 N1 150 22 1 (b) × × 14 2 atau × 14 × 14 K1 360 7 2 150 22 1 × × 14 2 - × 14 × 14 atau K1 360 7 2 770 – 98 3 2 476 158 atau atau 158.67 N1 3 3