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
150° O
7 cm
P
Q
S
22 By using π = , calculate 7
DIAGRAM 1
(a)
the perimeter for the whole diagram in cm,
(b)
area of the shaded region in cm2. [ 6 marks ]
Answer : (a)
(b)
2.
In diagram 2, ABCD is a rectangle.
21 cm
A
B
14 cm F 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 = 7 cm. Using π =
22 , calculate 7
(a)
the length, in cm, of arc CF
(b)
the area, in cm2, of the shaded region [ 6 marks ]
Answer : (a)
(b)
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 P
O
Q L
K
R DIAGRAM 3 Using π = (a) (b)
22 , calculate 7
the perimeter, in cm, of the whole diagram, the area, in cm2, of the shaded region. [6 marks]
Answer: (a)
(b)
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 π = (a)
(b)
22 , calculate 7
the perimeter, in cm, of the whole diagram, the area, in cm2 , of the shaded region. [6 marks]
Answer: (a)
(b)
5.
Diagram 5 shows two sectors OLMN and OPQR with the same centre O.
M
L P
120°
N R
O Q5 DIAGRAM OL = 14 cm. P is the midpoint of OL. [Use π =
22 ] 7
Calculate
(a) (b)
the area of the whole diagram, the perimeter of the whole diagram. [6 marks]
Answer: (a)
(b)
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 . Using π =
22 , calculate 7
(a)
the perimeter, in cm, of the whole diagram,
(b)
the area, in cm2, of the shaded region.
O
B
D
C
[6 marks]
DIAGRAM 6 Answer : (a)
(b)
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′ 210o
22 Using π = , calculate 7
DIAGRAM 7 A′
(a)
the length of arc BB′ ,
(b)
the ratio of arc lengths , AA′ : BB′
(c)
the area of the shaded region.
Answer: (a)
(b)
(c)
O
A
B
[7 marks]
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 π = (a)
the perimeter, in cm, of the whole diagram,
(b)
the area, in cm2, of the shaded region. .
Answer: (a)
(b)
22 , calculate 7
[6 marks]
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 P
O DIAGRAM 9 Given that OP = 14 cm. Using π =
22 , 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)
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 π =
22 , calculate 7
(a)
perimeter, in cm, the shaded area.
(b)
area, in cm2, the shaded area. [7 markah]
Answer : (a)
(b)
MODULE 3 - ANSWERS TOPIC: CIRCLE, AREA AND PERIMETER 1 (a)
90 22 120 22 × 2 × ×12 @ × 2× ×7 360 7 360 7 90 22 120 22 × 2 × × 12 + × 2 × × 7 + 12 + 5 360 7 360 7 57.53
K1 K1 N1
(b)
90 22 120 22 2 × × 12 2 @ × ×7 360 7 360 7
K1
90 22 120 22 1 × × 12 2 + × × 7 2 − × 7 × 12 360 7 360 7 2 122.48
K1 N1
2 (a)
∠FEC = 135
K2
135 22 × 2× ×7 360 7 16.5
K1 N1
(b)
135 22 × ×7 ×7 360 7
K1
1 Shaded area = (21 × 14) − × 14 × 14 − L3 2 = 138.25
K1
L3 =
N1 3 a)
b)
270 22 90 22 × × 2 × 21 atau × ×7×2 360 7 360 7
K1
270 22 90 22 × × 2 × 21 + × × 7 × 2 + 14 + 14 360 7 360 7
K1
= 138
N1
270 22 × × 21 × 21 atau 360 7
2×
90 22 × ×7×7 360 7
K1
270 22 90 22 × × 21 × 21 - 2 × × ×7×7 360 7 360 7
K1
= 962.5 cm2
N1
60 22 × 2 × × 28 360 7
K1
4 a)
60 22 × 2 × × 28 + 14 + 14 + 14 + 14 + 28 360 7 1 113 atau 113⋅33 3 b)
60 22 60 22 × × 28 × 28 atau × ×14 ×14 360 7 360 7 60 22 60 22 × × 28 × 28 − × ×14 ×14 + 14 × 14 360 7 360 7 504
K1 N1
K1 K1 N1
5 a)
120 22 240 22 × × 14 × 14 atau × ×7×7 360 7 360 7 120 22 240 22 × × 14 × 14 + × ×7×7 360 7 360 7 308
b)
K1 K1 N1
120 22 240 22 × 2 × × 14 atau × 2× × 7 360 7 360 7 120 22 240 22 × 2 × × 14 + × 2× × 7 + 7 + 7 360 7 360 7 2 72 3
K1 K1 N1
6 (a)
45 22 ×2× × 14 360 7
K1
22 45 ×2× × 14 + 14 + 14 + 14 + 14 7 360
K1
70
(b)
2 3
45 22 × × 14 × 14 360 7
N1 or
90 22 × × 14 × 14 360 7
K1
90 22 45 22 × × 14 × 14 + 2 14 × 14 − × × 14 × 14 360 7 360 7
K1
161
N1
7 (i)
(ii)
(iii)
210 22 × 2 × × 35 360 7 1 128 @ 128.33 3
K1 N1
210 22 210 22 × 2× ×7 : × 2 × × 35 360 7 360 7
K1
1: 5
N1
210 22 210 22 × × 35 2 or × × 72 360 7 360 7 210 22 210 22 × × 35 2 − × × 72 360 7 360 7 2156
K1 K1 N1
8 (a)
45 22 ×2× × 14 360 7
or
14 2 + 14 2 − 14
11 + 14 + 14 + 14 + 5.799 58.80 (2 d. p) (b)
45 22 × × 14 x 14 360 7 45 22 1 90 22 × 14 × 14 − × ×7 × 7 + × × 14 × 14 2 360 7 360 7
90 22 × ×7 × 7 360 7
or
K1 K1 N1 K1 K1
136.5
N1
90 22 60 22 × × 14 × 14 and A2 = × ×7×7 360 7 360 7 A1 – A2 1 128 3
K1 K1
9 (a)
A1 =
N1
(b)
P1 =
90 22 180 22 × 2 × × 14 or P2 = ×2× ×7 360 7 360 7
K1
P1 + P2 + 14 58
K1 N1
AB =
K1
10 (a)
14 2 + 14 2 = 392 = 19.80 150 22 60 22 90 22 × 2 × × 14 atau × 2 × × 14 atau × 2 × × 14 360 7 360 7 360 7
Lengkok AC + 14 + 14 + 19.80 atau Lengkok AB + lengkok BC + 14 + 14 + 19.80 84.47 (b)
150 22 1 × × 14 2 atau × 14 × 14 360 7 2 150 22 1 × × 14 2 - × 14 × 14 atau 360 7 2 770 – 98 3 2 476 158 atau atau 158.67 3 3
K1 K1 N1 K1 K1
N1