Chapter 6 Angles and Tangents of Circles
MIND TEST 6.1d
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What is the value of angles at the circumference subtended by the diameter?
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Aim: 7R GHWHUPLQH WKH DQJOHV VXEWHQGHG E\ WKH GLDPHWHU Q
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R VXEWHQGHG E\ WKH GLDPHWHU LV ,I PQR LV
Q Is a diameter a chord?
D VHPLFLUFOH WKHQ PQR Discuss.
P
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Example 6 Q R
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\ ± ±
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CHAPTER 6 S
If arc lengths PRQ = 2PS
then, PRQ = POS
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142
Chapter 6 Angles and Tangents of Circles
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O Q
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How do you solve problems involving angles in circles? LEARNING
STANDARD
Example 7
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LQ WKH GLDJUDP 7KH SRLQWV RQ WKH FLUFXPIHUHQFH IRUP DUF PQ ZKLFK LV angles in circles.
RI WKH VDPH OHQJWK DV DUF QR /LQH SQ SDVVHV WKURXJK O 'HWHUPLQH
WKH YDOXH RI P
D QSR Q
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O
SR
Solution:
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Q
MIND TEST 6.1f O P
R
1. 7KH GLDJUDP RQ WKH ULJKW VKRZV D FLUFOH ZLWK FHQWUH O OSU DQG CHAPTER 6
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[ y O y
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143
6.2 Cyclic Quadrilaterals
What do you know about cyclic quadrilaterals? P LEARNING
R STANDARD
$ cyclic quadrilateral LV D TXDGULODWHUDO LQ D FLUFOH
ZKHUH all four vertices RI WKH TXDGULODWHUDO OLH RQ WKH Recognise and describe
circumference of the circle Q cyclic quadrilaterals.
3456 LQ WKH GLDJUDP RQ WKH ULJKW LV D F\FOLF S
TXDGULODWHUDO P DQG R DV ZHOO DV S DQG Q DUH
NQRZQ DV opposite angles LQ WKH F\FOLF TXDGULODWHUDO
Example 8
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L A LL D G LLL K LY S Y
C O P O O V
BO R T
Q
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E 6WDWH WKH RSSRVLWH DQJOHV LQ HDFK F\FOLF TXDGULODWHUDO WKDW \RX KDYH LGHQWL¿HG
Solution:
D L 9HUWH[ D GRHV QRW OLH RQ WKH FLUFXPIHUHQFH KHQFH ABCD LV QRW D F\FOLF TXDGULODWHUDO
LL $OO YHUWLFHV DUH RQ WKH FLUFXPIHUHQFH KHQFH DEFG LV D F\FOLF TXDGULODWHUDO
CHAPTER 6 LLL 9HUWH[ O GRHV QRW OLH RQ WKH FLUFXPIHUHQFH KHQFH KLON LV QRW D F\FOLF TXDGULODWHUDO
LY $OO YHUWLFHV DUH RQ WKH FLUFXPIHUHQFH KHQFH PQRS LV D F\FOLF TXDGULODWHUDO
Y 9HUWH[ O GRHV QRW OLH RQ WKH FLUFXPIHUHQFH KHQFH OTUV LV QRW D F\FOLF TXDGULODWHUDO
E L 1RQH LL D DQG F E DQG * LLL 1RQH
LY P DQG R Q DQG S Y 1RQH
MIND TEST 6.2a
1. )RU HDFK RI WKH IROORZLQJ FLUFOHV O LV WKH FHQWUH RI WKH FLUFOH
L S LL D LLL Q P LY A F
TR Ɣ N B E
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144
Chapter 6 Angles and Tangents of Circles
What are the relationships between angles of a cyclic quadrilateral?
Brainstorming 6 In pairs LEARNING
STANDARD
Aim: 7R GHWHUPLQH WKH UHODWLRQVKLS EHWZHHQ RSSRVLWH LQWHULRU DQJOHV
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Materials: '\QDPLF VRIWZDUH relationships between
angles of cyclic
Steps: quadrilaterals, and hence
use the relationships to
1. 6WDUW ZLWK 1HZ 6NHWFK DQG FOLFN RQ &RPSDVV 7RRO WR GUDZ D FLUFOH determine the values
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quadrilaterals.
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.
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[
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Example 9 K FLASHBACK
L
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180°.
TXDGULODWHUDO KLMN &DOFXODWH WKH YDOXH RI
D [ E y N y Angle of a full rotation is
360°.
[
Solution: M
QU I Z
D 7KH LQWHULRU DQJOHV LKN DQG LMN DUH RSSRVLWH LQ WKH F\FOLF
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y
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[ Calculate the value of
[ + y.
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MIND TEST 6.2b
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146
Chapter 6 Angles and Tangents of Circles
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What is the relationship between the exterior angle with the corresponding opposite interior
angle?
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S
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T
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Example 10 PQ CHAPTER 6
PQ
,Q WKH GLDJUDP RQ WKH ULJKW PQRS LV D F\FOLF TXDGULODWHUDO *LYHQ WKDW P
DQG z DUH H[WHULRU DQJOHV VWDWH WKH RSSRVLWH LQWHULRU DQJOHV FRUUHVSRQGLQJ [R
WR P DQG z
y
Solution: Sz
y LV WKH RSSRVLWH LQWHULRU DQJOH FRUUHVSRQGLQJ WR P
Q LV WKH RSSRVLWH LQWHULRU DQJOH FRUUHVSRQGLQJ WR z
MIND TEST 6.2c H
FG
1. &RS\ DQG FRPSOHWH WKH WDEOH EHORZ EDVHG RQ WKH GLDJUDP RQ WKH ULJKW
E
Exterior angle Corresponding opposite Df
interior angle
ș
2. 'UDZ D FLUFOH DV VKRZQ LQ WKH GLDJUDP /DEHO WKH FRUUHVSRQGLQJ Į
RSSRVLWH LQWHULRU DQJOHV IRU WKH H[WHULRU DQJOH θ DQG α ZLWK V\PEROV
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How do you solve problems involving cyclic LEARNING
quadrilaterals? STANDARD
Example 11 ED C Solve problems involving
E cyclic quadrilaterals.
7KH GLDJUDP RQ WKH ULJKW VKRZV D F\FOLF D
B
TXDGULODWHUDO ABCD DQG D VWUDLJKW OLQH CDE
&DOFXODWH WKH YDOXH RI A
D D
E E
Solution: E E D
D $&% &$% 7KXV E
$&% CAB D
D
D ± ±
D
Example 12 P
7KH GLDJUDP RQ WKH ULJKW VKRZV D F\FOLF TXDGULODWHUDO PQRS DQG Q
D VWUDLJKW OLQH RST &DOFXODWH WKH YDOXH RI PST y
T y
S
Solution:
PQR PSR PST PQ R R
\ \ y
CHAPTER 6
\
y PST
Example 13 PN M
7KH GLDJUDP RQ WKH ULJKW VKRZV D F\FOLF TXDGULODWHUDO KLMN DQG
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D [ y
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Solution: [
L
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WR LW LV DQJOH [
7KXV y $ ± NML
7KXV y $ ± $
[ $ y $
148
Chapter 6 Angles and Tangents of Circles
MIND TEST 6.2d N
$
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149
6.3 Tangents to Circles
What do you understand about the tangents to circles? LEARNING
STANDARD
<RX KDYH OHDUQW WKDW WKH FLUFOH LV D XQLTXH VKDSH DQG KDV PDQ\ VSHFLDO
SURSHUWLHV Recognise and describe
the tangents to circles.
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URDG RQFH ZKHQ LW PDNHV D FRPSOHWH FLUFOH 7KH URDG VHUYHV DV D
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Q SRLQW A DQG SRLQW B RQ WKH FLUFOH 7KXV
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CHAPTER 6 Tangent WR D FLUFOH LV D VWUDLJKW OLQH WKDW WRXFKHV WKH FLUFOH DW RQO\ RQH SRLQW 7KH SRLQW RI
FRQWDFW EHWZHHQ WDQJHQW DQG WKH FLUFOH LV WKH point of tangency
Example 14 P E Q
R GS
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WDQJHQWV WR WKH FLUFOH DQG SRLQWV RI WDQJHQF\" 6WDWH WKH UHDVRQV IRU
\RXU DQVZHU
Solution: T M
UN
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PQ DQG TU UHVSHFWLYHO\
RS LV QRW D WDQJHQW WR WKH FLUFOH EHFDXVH LW SDVVHV WKURXJK WZR SRLQWV RQ WKH FLUFOH +HQFH SRLQW F
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WZR SRLQWV RQ WKH FLUFOH LI H[WHQGHG 7KXV SRLQW M LV QRW D SRLQW RI WDQJHQF\
150
Chapter 6 Angles and Tangents of Circles
MIND TEST 6.3a
1. ,Q WKH GLDJUDPV EHORZ LGHQWLI\ SRLQWV DQG OLQHV ZKLFK DUH
L WDQJHQWV LL SRLQWV RI WDQJHQF\ LLL QRW D WDQJHQW LY QRW D SRLQW RI WDQJHQF\
6WDWH WKH UHDVRQV IRU \RXU DQVZHU
D P Q E DE
A B F
F GB
T
RX Y
H
C
S
What do you know about the value of the angle between tangent and radius at the
point of tangency?
Brainstorming 7 In pairs LEARNING
STANDARD
Aim: 7R PHDVXUH WKH DQJOH EHWZHHQ WDQJHQW DQG UDGLXV RI D FLUFOH
DW WKH SRLQW RI WDQJHQF\ Make and verify
conjectures about the
Materials: '\QDPLF VRIWZDUH angle between tangent
and radius of a circle at
Steps: the point of tangency.
1. 6WDUW ZLWK 1HZ 6NHWFK DQG FOLFN RQ WKH &RPSDVV 7RRO WR GUDZ D
FLUFOH 'LDJUDP
2. &OLFN RQ 6WUDLJKWHGJH 7RRO WR GUDZ CHAPTER 6
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Discussion:
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ABC
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CHAPTER 6 Example 15 A C
[ B
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PHHWV WKH VWUDLJKW OLQH ABC DW SRLQW B RQO\ &DOFXODWH WKH O
YDOXH RI [
Solution:
/LQH ABC LV D WDQJHQW WR WKH FLUFOH DQG LW WRXFKHV WKH FLUFOH DW
SRLQW B 7KXV WKH DQJOH 2%$
AOB [ AOB
AOB ± [ ± AOB
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MIND TEST 6.3b AB C
1. ,Q WKH GLDJUDP RQ WKH ULJKW ABC LV D VWUDLJKW OLQH DQG [
O LV WKH FHQWUH RI WKH FLUFOH *LYHQ WKDW AB OB DQG O
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152
Chapter 6 Angles and Tangents of Circles
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O *LYHQ WKDW OQS LV DQ HTXLODWHUDO WULDQJOH DQG PQR
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FLUFOH DQG PQR LV D WDQJHQW WR WKH FLUFOH *LYHQ
WKDW QT ST DQG QTS FDOFXODWH WKH YDOXH RI T
[ y z
PO
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What are the properties related to two tangents to a circle?
Brainstorming 8 In pairs LEARNING
STANDARD
Aim: 7R GHWHUPLQH WKH SURSHUWLHV UHODWHG WR WZR WDQJHQWV WR D FLUFOH
Make and verify
Materials: 'UDZLQJ SDSHU FRPSDVVHV SURWUDFWRU UXOHU DQG SHQFLO conjectures about the CHAPTER 6
properties related to two
tangents to a circle.
Steps:
1. 'UDZ D FLUFOH RI UDGLXV FP ZLWK FHQWUH O 'UDZ D VWUDLJKW OLQH FP IURP WKH FHQWUH O DQG
ODEHO DV OA 'LDJUDP
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D WKH YDOXH RI [ R
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Chapter 6 Angles and Tangents of Circles
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S O
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N
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Chapter 6 Angles and Tangents of Circles
How do you solve problems involving tangents to circles?
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STANDARD
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Chapter 6 Angles and Tangents of Circles
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Chapter 6 Angles and Tangents of Circles
CONCEPT MAP
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Chapter 6 Angles and Tangents of Circles
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CHAPTER Plans and
7 Elevations
What will you learn?
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Each building in Putrajaya has its own
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Diamond Building is a very beautiful building
with a unique design. The Diamond Building has
received the ASEAN Energy Award for its structure
and design that maximises the use of sunlight. The
Malaysia Green Building Index and Singapore
Green Mark Scheme also awarded platinum ratings
to recognise the building’s design that enable
sustainable recycling of rainwater. The uniqueness
and creativity of the Diamond Building architecture
is distinctive when viewed from various directions.
Have you ever visited the Diamond Building?
168
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Sophia (Ayasofya in Turkish) was converted into a
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as a benchmark when designing other mosques.
This is why most mosques in Turkey are similar
in design.
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ExploringEraChapter7.pdf
WORD B A N K
origin DVDODQ
geometrical shape EHQWXN JHRPHWUL
elevation GRQJDNDQ
solid line JDULV SDGX
dashed line JDULV VHPSDQJ
orthogon RUWRJRQ
plan SHODQ
scale VNDOD
quadrant VXNXDQ
projection XQMXUDQ
169
7.1 Orthogonal Projections
What is a plane and a normal to a plane? LEARNING
STANDARD
You have studied objects in two and three dimensions. Each of these
REMHFWV FRQVLVWV RI ÀDW VXUIDFHV RU FXUYHG VXUIDFHV RU ERWK Draw orthogonal
projections.
curved curved
surface surface
ÀDW VXUIDFH
T
The diagram on the right shows a quarter of a right cylinder with a SR
horizontal base PQRS. Both PSTU and PQRS are planes and QRTU is U
a curved surface. PQ
A plane is the ÀDW VXUIDFH RI DQ REMHFW 7KHUH DUH WKUHH W\SHV RI SODQHV QDPHO\ horizontal
plane YHUWLFDO SODQH and LQFOLQHG SODQH
The diagram on the right shows a right prism with a horizontal plane E
ABCD. ABF and CDE are vertical planes. BCEF and ADEF are
inclined planes. The lines FM and EN are perpendicular to the lines FD N C
AB and CD respectively. The lines FM and EN are also known as the
normal to the plane $%&'.
A MB
CHAPTER 7 A normal to a plane is a straight line that is SHUSHQGLFXODU or that forms a right angle to any
line on the plane.
Example 1 TW
The diagram on the right shows a cube. State the normal to the U 9 R
following planes. P S
(a) 3456 (b) 3678 (c) 567: (d) QRTU
Solution: The order of letters to specify a Q
(a) 83 94 :5 76 normal is important. TS means
(b) 43 56 :7 98 the line TS is perpendicular to
(c) 45 36 87 9: the plane PQRS at point S.
(d) 39 6:
170
Chapter 7 Plans and Elevations
Example 2
The diagram on the right shows a right prism with a rectangular base EH
ABCD. M and N are the midpoints of AB and CD respectively. Given FG
= EH = DN = NC = AM = MB, state the normal to the following planes:
(a) ABCD (b) ADEF F G
N
C
D
Solution: (b) BA, CD, GF, HE MB
(a) FA, GM, HN, ED
A
What do you understand about orthogonal projections?
Object Normal In Diagram 1, PQ is a straight line where point Q lies
P C on the horizontal plane ABCD. PR is a normal line to
Diagram 1 the plane ABCD. The straight line RQ which lies on the
D plane ABCD is an orthogonal projection of the straight
line PQ on the plane ABCD.
QR
AB
Orthogonal projection
Q Object In Diagram 2, the lines PR and QS are the normal to
P the plane ABCD. RS is an orthogonal projection of the
Normal straight line PQ on the plane ABCD.
D Normal
C
SR
AB Diagram 2
Orthogonal projection
Orthogonal projections are images formed on a plane when the projected line from an CHAPTER 7
object is perpendicular to the plane.
,Q 'LDJUDP DQG 'LDJUDP ZH KDYH LGHQWL¿HG WKH RUWKRJRQDO SURMHFWLRQ IRU D OLQH 'LDJUDP DQG
Diagram 4 shows the orthogonal projections of a two-dimensional plane and a three-dimensional
object. Vertical plane EF
AB HG
DC RS
PQ
UT Diagram 4
SR Horizontal plane
'LDJUDP
171
,Q 'LDJUDP PQRS is projected on a vertical plane and in Diagram 4 EFGH is projected on a
horizontal plane.
'LDJUDP 2EMHFW Normal to the plane 2UWKRJRQDO SURMHFWLRQ RQ WKH SODQH
Diagram 3 PQRS 3$ 4% 5& 6' ABCD
Diagram 4 EFGH (5 )6 *7 +8 RSTU
TW
U S R R EH
V
PQ S FG
D
C T Q LK
U DC
AB
Horizontal plane IJ
Diagram 5 PA B
Vertical plane
'LDJUDP
,Q 'LDJUDP D FXERLG LV SURMHFWHG RQ D KRUL]RQWDO SODQH DQG LQ 'LDJUDP D ULJKW SULVP ZLWK WKH
surface BCHGKJ as a uniform cross section is projected on a vertical plane.
'LDJUDP 2EMHFW Normal to the plane 2UWKRJRQDO SURMHFWLRQ RQ WKH SODQH
Diagram 5 Cuboid 3$ 4% 5& 6' ABCD
'LDJUDP Right prism
$3 ,8 /7 '4 )6 (5 PQRSTU
Example 3
CHAPTER 7
Î
Î
Each of the following diagrams shows the projection of an object on a vertical plane or a horizontal
plane. Determine whether the resulting projection is an orthogonal projection.
(a) (b) (c)
Î
Solution:
(a) Yes
(b) Yes
F 1R EHFDXVH WKH OLQHV SURMHFWHG IURP WKH REMHFW WR WKH SODQH LV QRW D QRUPDO
172
Chapter 7 Plans and Elevations
MIND TEST 7.1a
1. Each diagram below shows the object and its projection on a plane. Determine whether the
projection is an orthogonal projection.
(a) (b)
(c) (d)
2. A student looks at the following object from a given viewing direction. Which of the following
combinations shows the correct orthogonal projection?
2EMHFW 2UWKRJRQDO SURMHFWLRQ
(a)
Î
(b) Î
CHAPTER 7
173
+RZ GR \RX GUDZ DQ RUWKRJRQDO SURMHFWLRQ"
You can draw an orthogonal projection of an object on a horizontal plane or a vertical plane using
the following steps.
1. Identify the type of plane and the direction in which the object that should be projected.
2. Draw normal lines from all vertices of the object to the plane. Make sure all the normal lines
are straight and upright so that the length of projected sides and the length of sides of object are
the same.
3. Connect the points of intersection of the normal to the plane to draw the shape of the orthogonal
projection.
4. Redraw the orthogonal projection with actual measurements. Label all vertices and side lengths.
Example 4 Z
The diagram on the right shows a right prism with
rectangular base ABCD on a horizontal plane. $%./*)
is a uniform cross section of the prism. The sides AF and
BK are vertical.
Draw the orthogonal projection of the object on a
(a) horizontal plane as viewed from Z
(b) vertical plane as viewed from X
(c) vertical plane as viewed from Y
Solution:
CHAPTER 7 E 3 cm H
Î F G , 2 cm J
4 cm D 2 cm
Î C
A /K
5 cm Î X
FP B
Î
Y
9LHZLQJ GLUHFWLRQ 2UWKRJRQDO SURMHFWLRQ
(a)
The order of letters is following the
Z viewing direction. Point D is below
point E as viewed from Z.
E 3 cm H
, 2 cm J E/D H, J/C
F G 2 cm
4 cm D
C
A /K
5 cm
FP B 5 cm
Horizontal F/A 3 cm G 1 cm / 2 cm K/B
plane
174
9LHZLQJ GLUHFWLRQ Chapter 7 Plans and Elevations
(b) 2UWKRJRQDO SURMHFWLRQ
Vertical G/F H/E
plane 2 cm
E 3 cm H . / - ,
2 cm
F G , 2 cm J
D 2 cm B/A 5 cm C/D
4 cm C
A /K
5 cm ÎX
FP B
Point A is behind point B as viewed
from X.
(c) Vertical
plane
F/E 3 cm G/H
E 3 cm H 4 cm / , 2 cm K/J
A/D 2 cm
F G , 2 cm J CHAPTER 7
4 cm D 2 cm FP B/C
C
A /K
5 cm
Î FP B Point D is behind point A as viewed
from Y.
Y
175
Example 5 Z
D
The diagram on the right shows a cylindrical object on a horizontal FP
plane. It is given that the diameter of the cylinder is 4 cm and its height
LV FP 4 cm
A
Draw the orthogonal projection of the cylindrical object on a
(a) horizontal plane as viewed from Z Y
(b) vertical plane as viewed from Y
Solution:
CHAPTER 7 C
B
Î
ÎÎ9LHZLQJ GLUHFWLRQ2UWKRJRQDO SURMHFWLRQ
(a) Z
DC
AB A 4 cm B
Horizontal
plane
(b) DC
Vertical
plane
FP
DC
A B
Y
Î A 4 cm B
176
Chapter 7 Plans and Elevations
Brainstorming 1 In groups
Aim: To determine the orthogonal projections of an object.
Materials: '\QDPLF VRIWZDUH GUDZLQJ SDSHU
Steps:
1. Open 9LHZ and select 3D JUDSKLFV
2. Select the shape of pyramid .
3. Basic display is formed (Diagram 1).
4. Drag the cursor to display and select the four points:
D 3RLQW ± RQ WKH UHG OLQH
E 3RLQW ± RQ WKH JUHHQ OLQH
F 3RLQW RQ WKH UHG OLQH
G 3RLQW RQ WKH JUHHQ OLQH DQG FRQQHFW LW WR WKH VWDUWLQJ SRLQW ± DW WKH UHG OLQH
(Diagram 2).
5. The display will show a brownish shape (Diagram 3).
6. 'UDJ WKH FXUVRU XS WR WKH EOXH OLQH 'LDJUDP
7. Select the ' URWDWH LFRQ VHOHFW YLHZ LQ IURQW RI .
8. Place the arrow at the top end of the blue line to see the orthogonal projection on the
horizontal plane (Diagram 5).
Diagram 1 Diagram 2 Diagram 3 Diagram 4 Diagram 5 CHAPTER 7
9. Repeat step 8 on the red line and the green line to see various orthogonal projections on
vertical planes.
10. Draw the resulting orthogonal projections as in steps 8 and 9 in the given table.
11. 6HOHFW D QHZ ¿OH %XLOG RWKHU ' VKDSHV DQG GUDZ RUWKRJRQDO SURMHFWLRQV IURP GL൵HUHQW
perspectives.
177
5HVXOWV RI ¿QGLQJV 2UWKRJRQDO SURMHFWLRQ
Pyramid
The view on the horizontal plane as seen from the blue line
The view on the vertical plane as seen from the red line
The view on the vertical plane as seen from the green line
'LVFXVVLRQ
Discuss the resulting shape of the orthogonal projection as compared to the actual shape of the
object.
)URP %UDLQVWRUPLQJ LW LV IRXQG WKDW: 2UWKRJRQDO SURMHFWLRQ
Pyramid
The view on the horizontal plane as seen from the blue line
The view on the vertical plane as seen from the red line
The view on the vertical plane as seen from the green line
MIND TEST 7.1b
1. Each object below lies on a horizontal plane. Draw orthogonal projections of each object on a
(a) horizontal plane as viewed from Z
CHAPTER 7
Î
Î
Î
(b) vertical plane as viewed from Y
(i) (ii) (iii)
ZZ Z
9C 2 cm
E H 1 cm
, J
5 cm 5 cm F *
N M /K C
K 4 cm / 4 cm 4 cm
AB B
4 cm D
A FP
Î
Î
Î
Y YY
178
Chapter 7 Plans and Elevations
+RZ GR \RX FRPSDUH DQG FRQWUDVW REMHFWV ZLWK WKHLU LEARNING
SURMHFWLRQV" STANDARD
Brainstorming 2 In groups Compare and contrast
between objects and the
Aim: Compare and contrast an object with an orthogonal projection corresponding orthogonal
in terms of length of side and size of angle. projections.
Materials: &DUGERDUG D SHQFLO D SDLU RI VFLVVRUV DGKHVLYH WDSH DQG GUDZLQJ SDSHU
Steps:
1. Draw the following shape according to the size given on a cardboard (Diagram 1).
2. Cut out the shape in Diagram 1 and use adhesive tape to build the shape in Diagram 2.
Z
9
Î
45$ 14 cm
14 cm $ A B CHAPTER 7
45$ 19.8 cm 19.8 cmÎ
Diagram 1
C
Diagram 2
Y
3. Draw an orthogonal projection for the shape that you built on a horizontal plane as viewed
from Z and on a vertical plane as viewed from Y.
4. Produce the orthogonal projections on the horizontal plane and the vertical plane as follows:
Projection from direction Z Projection from direction Y
(horizontal plane) (vertical plane)
9 $ 9
14 cm 19.8 cm
45$ 19.8 cm B
C
C/A 14 cm B
179
5. Measure each of the length of sides and angles of the two orthogonal projections you drawn.
Complete the table below.
3URMHFWLRQ 3URMHFWLRQ
Side 2EMHFW IURP GLUHFWLRQ $QJOH 2EMHFW IURP GLUHFWLRQ
ZZ
AC 14 cm 14 cm 9&% $ 45$
AB 9%&
BC 19.8 cm 19.8 cm BAC $ $
9& 19.8 cm 14 cm CAB
9%
3URMHFWLRQ 3URMHFWLRQ
Side 2EMHFW IURP GLUHFWLRQ $QJOH 2EMHFW IURP GLUHFWLRQ
YY
$9 14 cm 14 cm 9&% $ $
AB 9%& $ 45$
BC 19.8 cm 14 cm &9%
9& $9% 45$ 45$
9% 19.8 cm 19.8 cm
'LVFXVVLRQ
Are all sides and angles of the orthogonal projection of the same size as those of the
object? Discuss.
CHAPTER 7 )URP %UDLQVWRUPLQJ LW LV IRXQG WKDW
(a) For orthogonal projections on a horizontal plane from direction Z WKH OHQJWKV RI $& $%
and BC and the size of BAC $&% and $%& remain unchanged.
(b) For orthogonal projections on a vertical plane from direction Y WKH OHQJWKV RI $9 $% and
9% and the size of $9% and $%9 remain unchanged.
,Q JHQHUDO
The OHQJWK RI VLGHV and VL]H RI DQJOHV of the RUWKRJRQDO SURMHFWLRQV of an object can remain
unchanged or vary according to the YLHZLQJ GLUHFWLRQ.
180
Chapter 7 Plans and Elevations
Example 6 Z
TU
The diagram on the right shows a right prism with a rectangular base
PQRS which lies on a horizontal plane. The plane URQ is a uniform 8 cm
cross section of the object.
SR
(a) Draw to full scale the orthogonal projection of the prism on FP Î X
P 2 cm Q
(i) a horizontal plane as viewed from Z
(ii) a vertical plane as viewed from X
(b) State your conclusion about the length of sides and the size of
angles of the object and its orthogonal projections. Explain your
conclusions.
Î
Solution: Î
(a) (i) (ii) Î(b) (i) The length of sides of TU SR
U/T PQ PS and QR and the right
T/S U/R CHAPTER 7
angle remain unchanged on
FP FP
orthogonal projections as viewed
8 cm from Z. The length of sides TP
and UQ are changed.
P 2 cm Q Q/P FP R/S (ii) The length of sides of TP UQ PS
QR TS and UR as well as the size
of all angles remain unchanged
on the orthogonal projection as
viewed from X.
MIND TEST 7.1c Z
Z 1 cm T 1 cm
E
U
F S
3 cm
D 3 cm 2 cm
A 2 cm Q
R
4 cm 2 cm C ÎP Diagram 2
B
ÎX X
Diagram 1
1. (a) Diagram 1 and Diagram 2 above show two objects placed on a horizontal plane.
Draw a full scale orthogonal projection of both objects on a
(i) horizontal plane as viewed from Z
(ii) vertical plane as viewed from X
(b) State your conclusion about the length of sides and the size of angles of the objects and
their orthogonal projections for Diagram 1 and Diagram 2. Explain your conclusion.
181
7.2 Plans and Elevations
What are plans and elevations? LEARNING
STANDARD
You have learnt that the orthogonal projection of an object or a solid
can be drawn on a horizontal plane and a vertical plane. Draw the plan and
elevations of an object to
The orthogonal projection on a horizontal plane, which is seen scale.
from the top view, is known as a plan. The orthogonal projection on
a vertical plane, which is seen from either the side view or the front TIPS
view, is known as elevations. Orthogonal projection drawings give
accurate information on the design as well as the size of an object. Full scale means the
actual size.
How do you draw a plan and elevations of an object to scale?
The diagram below shows a right prism with a rectangular base ABKJ which lies on a horizontal
plane. ABCDEFGH is a uniform cross section of the prism. The sides AH, FG, ED and BC are
vertical. The plan of the right prism can be drawn as viewed from Z and the elevations of the object
can be drawn as viewed from X and Y. Plan and elevations should be drawn to full scale.
Right prism (object) Plan
As viewed from Z, which is the view from the
CHAPTER 7 Z L top.
M
Î 1 cm
IP
I/J P/O M/N L/K
O 3 cm
4 cm
J N
K
D C
1 cm
H 4 cm Î Y
G
2 cm H/A G/F D/E C/B
F 1 cm E
A 3 cm B Note:
X All sides are drawn with solid lines because
Î they are visible from the top.
182
Chapter 7 Plans and Elevations
Front elevation Side elevation
As viewed from X. As viewed from Y.
D/M 1 cm C/D L/M
1 cm C/L 1 cm
H/I G/P G/H P/I
2 cm F/O 3 cm 1 cm
E/N E/F N/O
A/J 3 cm B/K 1 cm
B/A K/J
Note:
All sides are drawn with solid lines because 4 cm
they can be seen when viewed from X.
Note:
Lines GP, HI, EN and FO are drawn with
dashed lines because the sides are hidden
when viewed from Y.
The drawings of a plan, a front elevation and a side elevation of an object can also be combined on
a piece of paper which is divided into four quadrants. Here are two commonly used methods.
Method 1 Method 2
Second First Second First
quadrant quadrant quadrant quadrant
Front
Side elevation Front Side
elevation elevation elevation
Plan 45°
45° Plan
Third Fourth Third Fourth
quadrant quadrant quadrant quadrant
The position of the front elevation is at the top of the plan. The side elevation is drawn on the left CHAPTER 7
side or the right side of the front elevation, depending on the viewing direction.
In method 1, the side view is from right to left as in Example 7. Thus, the position of this elevation
is on the left side of the front elevation as method 1. In method 2, a side view is from left to right
as in example 8. Thus, the position of this elevation is on the right side of the front elevation as
method 2.
Example 7 E 3 cm J I H
1.5 cm 5 cm ÎY
The diagram on the right shows a right prism with C B
rectangle ABCD that lies on a horizontal plane. D G
ABHGF is a uniform cross section of the prism.
The sides of AF and BH are vertical. Draw to full 4 cm F 5 cm
scale,
(a) the plan of the prism A
(b) the elevation of the prism as viewed from X
(c) the elevation of the prism as viewed from Y
Î
X
183
Solution: Steps:
Side elevation Front elevation 1 The direction of
HI
H/I the side elevation
(direction Y) is from
3.5 cm J/E F/E 3 cm G/J 5 cm right to left, thus the
B/C position of the side
G/F C/D 5 cm I/C elevation is in the
1.5 cm 4 cm A/D J second quadrant.
B/A 45° 2 Draw the plan to full
E/D
scale in the fourth
4 cm quadrant.
F/A 3 cm G 2 cm H/B 3 Project sides of the
Plan
plan with thin solid
CHAPTER 7 Example 8 J OLQHV WR WKH ¿UVW
E quadrant as a guide
The diagram on the right shows a to draw the front
combination of a cuboid and a right prism elevation (direction X).
with rectangle ABCD on a horizontal plane.
ABGHIF is a uniform cross section of the 4 Project the sides
object. BH and FI are vertical. Draw to full
scale, of the plan and the
front elevation to the
(a) the plan of the object second quadrant
to draw the side
(b) the elevation of the object as viewed elevation.
from X
TIPS
(c) the elevation of the object as viewed
from Y Guide for drawing plan
and elevation.
Ƈ Thick solid lines for
visible sides.
Ƈ Dashed lines for
hidden sides.
Ƈ Thin solid lines for
construction lines.
4 cm K
D L H
Y Î 5 cm I 3 cm
G
A C 3 cm
F
B
7 cm
Î
X
184
Chapter 7 Plans and Elevations
Solution: TIPS
Front elevation Side elevation I/H The direction of the side
I/J 4 cm H/K J/K 5 cm elevation (direction Y)
is from left to right, thus
3 cm the position of the side
F/E G/L E/L HOHYDWLRQ LV RQ WKH ¿UVW
quadrant.
3 cm
F/G
A/D 7 cm B/C D/C A/B DISCUSSION CORNER
D J/E K/L/C 45° In the subject Reka
Bentuk dan Teknologi
(RBT), the plan and
elevations of an
object are drawn with
orthographic projection
method. Is this method
the same as the
method you use in this
chapter? Discuss.
A 3 cm I/F 4 cm H/G/B
Plan
MIND TEST 7.2a
1. The diagram below shows a prism with rectangle PQUT on a horizontal plane. PQSR is a
uniform cross section of the prism. Draw to full scale,
(a) the plan of the prism
(b) the elevation of the prism as viewed from X
(c) the elevation of the prism as viewed from Y
Î
W 3 cm V
CHAPTER 7
4 cm TU Î
R S
2 cm
Y
P 1 cm Q
X
185
2. The diagram below shows a block where rectangle ABCD lies on a horizontal plane.
ABVSRONKJGF is a uniform cross section of the block. AF, JG, KN, RS and BV are vertical.
Draw to full scale,
(a) the plan of the object
(b) the elevation of the object as viewed from X
(c) the elevation of the object as viewed from Y
E HM P 1 cm
IL TU
F G DN O S Q Î
JK R 3 cm
V
C
3 cm Y
A 6 cm B
CHAPTER 7 X
Î
3. The diagram below shows a combination of a cuboid and a right prism placed on a horizontal
plane. A semi-cylinder is removed from the cuboid. ADEJKF is a uniform cross section of the
object. AD and FEJ are vertical. Draw to full scale,
(a) the plan of the object
(b) the elevation of the object as viewed from X I
(c) the elevation of the object as viewed from Y
4 cm
C 4 cm H J
2 cm L
G
B
Î 5 cm D E
Y
A F 2 cm KÎ
X
186
Chapter 7 Plans and Elevations
How do you synthesise plan and elevations of an object LEARNING
and sketch the object? STANDARD
The drawings of plan and elevations on four quadrants are connected Synthesise plan and
to each other and can be used to sketch the three-dimensional shape of elevations of an object
an object with ease. and sketch the object.
Example 9 Side elevation Front elevation
The diagram on the right shows the plan, G/F M/H N/E F/E G/H M/N
front elevation and side elevation of a right 1 cm L/I 1 cm
prism with a rectangular base. A cuboid- K/J J/I K/L
shaped block has been removed from the 1 cm 2.5 cm C/D A/D 3 cm 1 cm
prism. Sketch the three-dimensional shape
of the prism. B/A B/C
Solution: 45°
The position of the side elevation is on the
second quadrant. Thus, the view of the side E/D 3 cm N/C
elevation is from the right. H/I 1 cm
F/A G/J 2 cm M/L
Plan 1.5 cm
K/B
Step 1 Step 2
Sketch the three orthogonal projections
given on the planes using the Project the surfaces I, II and III so that
measurements given. Surfaces marked they meet as shown in the diagram below.
I, II and III are surfaces of the cuboid
block. >>
>
Side elevation Front elevation > II
>> IIII
>II>
>
III
CHAPTER 7
I
Scan QR Code or browse
Plan http://yakin-pelajar.com/
Bab%207%video/ to watch
a video about orthographic
projection drawings using
dynamic software.
187
Step 3 Step 4
Sketch the object and label the vertices Complete the sketched object by labelling
with the letters in the orthogonal the length of sides.
projections using the colours as the guide.
Plan
EN
Î
E N Î H 2 cm M 2 cm
H M
=1 cm =
GD I L F GD I 1.5 cm L
G C G= C
F SideÎ
elevation
2 cm J J KK
= 2.5 cm
J J KK
A 3 cm B
AB Front
elevation
Example 10 Front elevation Side elevation
J/I IJ
The diagram on the right shows the plan, front
elevation and side elevation of a combination F/E 60° 4 cm 3.5 cm
of a cuboid and a right prism. Sketch the 1 cm
three-dimensional shape of the object. G/1HEc/mH F/E
A/D B/C D/C A/B
H/C 45° 4.4 cm
E/D I
Solution:
7KH SRVLWLRQ RI VLGH HOHYDWLRQ LV LQ WKH ¿UVW 4.4 cm
quadrant. Thus, the view of side elevation is from
CHAPTER 7 left to right. F/A 2 cm J 4 cm G/B
Plan J
Step 1
Sketch the three orthogonal projections given on I H
the planes using the measurements given. This E 60° C
object contains an angle of 60° on a triangular
surface. Thus, the angle of 60° must be built with
the correct method.
D
Step 2
Connect the vertices to create a combined object. F G
Label the vertices according to the projections. A B
188
Chapter 7 Plans and Elevations
Step 3 I
Draw the combined object and label the vertices 4 cm
and the length of the sides. CHAPTER 7
60° H
E J
D
C 4.4 cm
F 60° G
A 6 cm 1 cm
B
MIND TEST 7.2b Side elevation Front elevation
1. The diagram on the right shows the plan, front F/E 2 cm E/H 5 cm F/G
elevation and side elevation of a combination G/H
of a cuboid and a right prism. Sketch the
three-dimensional shape of the combined 2 cm
object.
C/D J/I D/I C/J
4 cm 4 cm
B/A K/L A/L 3 cm B/K
45° J/K G
H I/L
2 cm
E D/A 3 cm C/B F
1 cm Plan 1 cm
Side elevation Front elevation
2. The diagram on the right shows the plan, L K L/K
front elevation and side elevation of a
combination of a cuboid, a right prism and a 2 cm J/G I/J
semi-cylinder. Sketch the three-dimensional I/H H/G
shape of the combined object.
8 cm
C/B/A 10 cm D/E/F A/F 4 cm B/E 6 cm C/D
45° G/F J/E D
10
cm
H/A I/B 6 cm C
4 cm Plan
189
How do you solve problems involving plans and LEARNING
elevations? STANDARD
Example 11 Solve problems involving
plans and elevations.
The diagram below shows the plan, front elevation and side elevation
of a right prism. Side elevation
Front elevation
N/K/F M/L/E F/E K/L N/M
2 cm I/H G/H J/I 4 cm
J/G
2 cm
B/A 5 cm C/D A/D 7 cm B/C
45q 2 cm
E/D H 3 cm L/I M/C
5 cm
F/A G K/J N/B
Plan
(a) Draw the right prism to full scale.
(b) State the length of FG, in cm, correct to one decimal place.
(c) Originally the prism was a cuboid of size 7 cm × 5 cm × 4 cm. Calculate the volume of the
right prism EFGJKLIH, in cm3, which was removed from the cuboid.
(d) State the ratio of the volume of the right prism that was removed to the volume of the right
prism you drew in question (a).
CHAPTER 7 Solution: (b) FG = 2.8 cm
(a)
(c) The volume of the removed prism
E L 2 cm M = —12 (2 cm)(3 + 5) cm × 5 cm
= 40 cm3
H I 4 cm
C (d) The volume of the projected right prism
F N = the volume of the cuboid – the volume of
4 cm 5 cm the prism EFGJKLIH
K = (7 cm × 5 cm × 4 cm) – 40 cm3
A D = 140 cm3 – 40 cm3
G 3 cm J = 100 cm3
7 cm B Thus, the ratio is
40 : 100
2:5
190
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