MODUL 3 SMK KUALA KETL : MATHEMATICS SPM TOPIC: CIRCLE, AREA AND PERIMETER
1.
Diagram 1 shows tw two se sector of of ci circl rcle ORQ and ORQ and OPS with OPS with centre O centre O..
R
12 cm
1"#$ O
% cm
P
Q
S
By using
π =
22 7
D&''* 1 , calculate
(a) (a)
the the per perim imet eter er for for the the whole hole diag diagra ram m in in cm, cm,
(b)
area of of th the sh shaded re region in in cm cm 2. [ marks ! marks !
Answer : (a)
(b)
2.
&n diagram 2, ABCD is a rectangle.
21 cm
A
B
14 cm F
D
+&-
E
C
CF is an arc of a circle with center E where E is a point on the line DC with 22 EC = % cm. sing π = , calculate 7 (a) the length, in cm, of arc CF (b)
the area, in cm 2, of the shaded region [ marks !
Answer : (a)
(b)
/.
Diagram / shows two sectors OPQR and OJKL. OPQR and OJKL are three 0uarters of a circle. POL and JOR are straight lines. OP = 21cm and OJ = % cm.
J
P
O
Q L
K
R D&''* / sing π = (a) (b)
22
, calculate 7 the perimeter, in cm, of the whole diagram, the area, in cm 2, of the shaded region. [ marks!
Answer (a)
(b)
.
&n Diagram , JK and PQ are arcs of two circles with centre O. OQRT is a s0uare. K
Q
R
J P
O T 210
D&''*
OT = 1 cm and P is the midpoint of OJ . 22 sing π = , calculate 7 (a) the perimeter, in cm, of the whole diagram, (b) the area, in cm 2 , of the shaded region. [ marks! Answer (a)
(b)
".
Diagram " shows two sectors 3*4 and 56 with the same centre .
M
N
L P
120°
R
O Q
D&''* " 3 = 1 cm. 5 is the midpoint of 3.
[se π =
22 7
!
7alculate (a) (b)
the area of the whole diagram, the perimeter of the whole diagram. [ marks!
Answer (a)
(b)
.
&n Diagram , ABD is an arc of a sector with the centre O and BCD is a 0uadrant. A
OD = OB = 1 cm and ∠ AOB = 45o . 22 sing π = , calculate 7 (a)
the perimeter, in cm, of the whole diagram,
(b)
the area, in cm 2, of the shaded region.
O
B
D
C
[ marks!
D&''* Answer : (a)
(b)
%.
&n Diagram %, the shaded region represents the part of the flat windscreen of a 8an which is being wiped by the windscreen wiper AB. 9he wiper rotates through an angle of 21# o about the centre O. i8en that OA = % cm and AB = 2: cm.
B
′
o
210 A D&''* % O ′
sing π =
22 7
, calculate
the length of arc BB ,
(b)
the ratio of arc lengths , AA : BB
(c)
the area of the shaded region.
′
Answer
(b)
(c)
B
′
(a)
(a)
A
′
[% marks!
:.
Diagram : shows a 0uadrant 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 . i8en OC = OB = BE = 1 cm.
D&''* :
22 , calculate 7 the perimeter, in cm, of the whole diagram,
sing π = (a) (b)
the area, in cm 2, of the shaded region. .
[ marks!
Answer (a)
(b)
Q
;.
R S &n Diagram ;, OPQS is a 0uadrant with the centre O and OSQR is a semicircle with the centre S. 60° T
O
P
D&''* ; i8en that OP = 1 cm. sing < =
22 7
, calculate
(a)
the area, in cm 2, of the shaded region,
(b)
the perimeter, in cm, of the whole diagram. [ marks!
Answer (a)
(b)
1#.
&n diagram 1#, OABC is a sector of a circle with centre O and radius 1 cm.
B
A
60 C
By using
π
=
22 7
O D&''* 1# , calculate
(a)
perimeter, in cm, the shaded area.
(b)
area, in cm2, the shaded area. [% marka!
Answer : (a)
(b)
MODULE 3 - ANSWERS TOPIC: CIRCLE, AREA AND PERIMETER 1
(a)
90 360
90
× 2×
22
× 2×
22
360 57.53
×12 @
7
7
× 12 +
120 360
120 360
22
× 2×
× 2×
7
22 7
×7
1
× 7 + 12 + 5
1 41
(b)
90 360 90
× 22 × 12 2 @ 120 × 22 × 7 2 7
22
×
360 7 122.48
360
× 12 2 +
120 360
1
7
×
22
1
× 7 2 − × 7 × 12
7
2
1 41
2
(a)
2
∠ FEC = 135
135 360 16.5
× 2 × 22 × 7
1
7
41 (b)
L3
=
135 360
×
22 7
×7×7
Shaded area
1
1 = ( 21 × 14) − × 14 × 14 − L3 2 = 138.25
1 41
3
a)
270 360 270 360
×
22
×
22
7
7
× 2 × 21 atau × 2 × 21 >
90 360
90
×
360
×
22 7
22 7
×7× 2
× 7 × 2 > 1 > 1
= 1/: b)
270 360 270 360
1
1 41
× 22 × 21× 21
atau
× 22 × 21× 21
? 2
7
7
= ;2." cm 2
2
90 22 × × 7× 7 × 360 7
1
90 22 × × 7× 7 × 360 7
1 41
4
a)
60
× 2 × 22 × 28
360
60 360
× 2×
113
1
60 360 60 360
22
× 28 + 14 + 14 + 14 + 14 + 28
7
a!a"
3
b)
1
7
11/⋅//
× 22 × 28 × 28 ×
7 22
41
a!a"
60
× 22 × 14 × 14
1
360 7 60 × 22 × 14 ×14 + 1 × 1 360 7
× 28 × 28
7
1
"#
1 41
5
a)
b)
120 360 120 360 /#: 120 360 120
× 22 × 14 × 14 atau
240
×
×
7 22 7
× 14 × 14 >
× 22 × 7 × 7
360 240 22
360
7
7
1
× 7× 7
1 41
× 2× × 2×
22 7 22
360 2 72 3
7
× 14
atau
× 14
>
240
22
× 2× × 7 360 7 240 × 2 × 22 × 7 > % > % 360 7
1 1 41
6
(a)
45 360
×2×
22 7
× 14
1
45 × 2 × 22 × 14 + 14 + 14 + 14 + 14 7 360 2
70
(b)
1
41
3
45 360
×
22 7
× 14 × 14
#r
90 360
× 22 × 14 × 14
1
7
45 × 22 × 14 × 14 + 2 14 × 14 − 360 7
90 360
×
22 7
× 14 × 14
1 161
41
7
(i)
210 360
12:
(ii)
22
× 2×
1
1
@ 12:.//
3
210
× 35
7
41
× 2 × 22 × 7
360
210
7
360
× 2×
22 7
× 35
1
1 " (iii)
41
210 360 210
× 22 × 352 or 7 22
×
360 21"
7
× 35 2 −
210
22
×
360 7 210 22
360
×
7
× 72
1
× 72
1 41
8
(a)
45
× 2 ×
360
22
×
7
1
or
14
2
+ 142 − 14
1
11 > 1 > 1 > 1 > ".%;; ":.:# (2 d. p) (b)
90
×
22
1 41
×7 × 7
45
or
×
360 7 360 1 × 14 × 14 − 90 × 22 × 7 × 7 2 360 7
22 7
>
× 1
A 1
45
×
360
1
22 7
1/."
× 1 ×
1
1 41
(a)
'1 = 90 360
× 22 × 14 × 14 7
and '2 = 60 × 22 × 7 × 7 360 7
'1 '2 12: (b)
1
41
3
51 =
1 1
90 360
×2×
22 7
× 14 or
52 =
180 360
51 > 52 > 1 ":
×2×
22 7
×7
1 1 41
1!
(a)
AB =
14 2
+ 14 2 =
392
= 1;.:#
1
150 360
× 2 × 22 × 14 atau 7
60 360
× 2 × 22 × 14 atau 7
3engCoC AC > 1 > 1 > 1;.:# atau 3engCoC AB > lengCoC B7 > 1 > 1 > 1;.:# :.% (b)
150 360 150 360 770
× 22 × 14 2 atau ×
7
× 14 2 ?
1 2
360
× 2 × 22 × 14 7
1 41
× 14 × 14
1
× 14 × 14 atau
1
2
1
;:
3
158
7 22
1
90
2 3
atau
476 3
atau 1":.%
41
