Module 3-circle Area And Perimeter
This document was uploaded by user and they confirmed that they have the permission to share it. If you are author or own the copyright of this book, please report to us by using this DMCA report form. Report DMCA
Overview
Download & View Module 3-circle Area And Perimeter as PDF for free.
More details
- Words: 1,538
- Pages: 14
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
P
Q
S
22 By using = , 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
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
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 (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 (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
N
L P
120
R
O Q DIAGRAM 5 OL = 14 cm. P is the midpoint of OL. [Use =
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 45 . Using
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
A
210o O
A
B
DIAGRAM 7 Using π =
22 , calculate 7
(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)
[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 =
22 , calculate 7
(a)
the perimeter, in cm, of the whole diagram,
(b)
the area, in cm2, of the shaded region. .
Answer: (a)
(b)
[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
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 135 135 22 2 7 360 7 16.5
K2 K1 N1
(b)
135 22 L3 77 360 7
K1
1 Shaded area (21 14) 14 14 L3 2 138.25
K1 N1
3 a)
b)
270 22 90 22 2 21 atau 72 360 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 77 360 7
270 22 90 22 21 21 - 2 77 360 7 360 7
K1
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 11333 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 77 360 7 360 7 120 22 240 22 14 14 + 77 360 7 360 7 308
b)
6 (a)
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
45 22 2 14 360 7
K1
22 45 2 14 14 14 14 14 7 360
K1
70
(b)
K1
2 3
45 22 14 14 360 7
N1 or 90 22 14 14 360
K1
7
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 2 35 2 or 7 360 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 14 14 7 7 + 2 360 7 360 7
90 22 7 7 360 7
or
136.5
K1 K1 N1 K1 K1 N1
9 (a)
(b)
90 22 60 22 14 14 and A2 = 77 360 7 360 7 A1 – A2 1 128 3 90 22 180 22 P1 = 2 14 or P2 = 2 7 360 7 360 7
A1 =
P1 + P2 + 14
K1 K1 N1 K1 K1
58
N1
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
K1
10 (a)
AB =
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
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 = , 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
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
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 (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 (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
N
L P
120
R
O Q DIAGRAM 5 OL = 14 cm. P is the midpoint of OL. [Use =
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 45 . Using
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
A
210o O
A
B
DIAGRAM 7 Using π =
22 , calculate 7
(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)
[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 =
22 , calculate 7
(a)
the perimeter, in cm, of the whole diagram,
(b)
the area, in cm2, of the shaded region. .
Answer: (a)
(b)
[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
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 135 135 22 2 7 360 7 16.5
K2 K1 N1
(b)
135 22 L3 77 360 7
K1
1 Shaded area (21 14) 14 14 L3 2 138.25
K1 N1
3 a)
b)
270 22 90 22 2 21 atau 72 360 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 77 360 7
270 22 90 22 21 21 - 2 77 360 7 360 7
K1
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 11333 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 77 360 7 360 7 120 22 240 22 14 14 + 77 360 7 360 7 308
b)
6 (a)
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
45 22 2 14 360 7
K1
22 45 2 14 14 14 14 14 7 360
K1
70
(b)
K1
2 3
45 22 14 14 360 7
N1 or 90 22 14 14 360
K1
7
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 2 35 2 or 7 360 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 14 14 7 7 + 2 360 7 360 7
90 22 7 7 360 7
or
136.5
K1 K1 N1 K1 K1 N1
9 (a)
(b)
90 22 60 22 14 14 and A2 = 77 360 7 360 7 A1 – A2 1 128 3 90 22 180 22 P1 = 2 14 or P2 = 2 7 360 7 360 7
A1 =
P1 + P2 + 14
K1 K1 N1 K1 K1
58
N1
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
K1
10 (a)
AB =
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
Related Documents
Area And Perimeter Dominoes
August 2019 40
Module 3-circle Area And Perimeter
May 2020 1
Module 3-circle Area And Perimeter
October 2019 7
Area And Perimeter
April 2020 9
Area And Perimeter
April 2020 17