Draw Projections of Objects and Determine True Dimensions

TO DRAW PROJECTIONS OF ANY OBJECT,
ONE MUST HAVE FOLLOWING INFORMATION
A) OBJECT
{ WITH IT’S DESCRIPTION, WELL DEFINED.}
B) OBSERVER
{ ALWAYS OBSERVING PERPENDICULAR TO RESP. REF.PLANE}.
C) LOCATION OF OBJECT,
{ MEANS IT’S POSITION WITH REFFERENCE TO H.P. & V.P.}
TERMS ‘ABOVE’ & ‘BELOW’ WITH RESPECTIVE TO H.P.
AND TERMS ‘INFRONT’ & ‘BEHIND’ WITH RESPECTIVE TO V.P
FORM 4 QUADRANTS.
OBJECTS CAN BE PLACED IN ANY ONE OF THESE 4 QUADRANTS.
IT IS INTERESTING TO LEARN THE EFFECT ON THE POSITIONS OF VIEWS ( FV, TV )
OF THE OBJECT WITH RESP. TO X-Y LINE, WHEN PLACED IN DIFFERENT QUADRANTS.
ORTHOGRAPHIC PROJECTIONS
OF POINTS, LINES, PLANES, AND SOLIDS.
STUDY ILLUSTRATIONS GIVEN ON HEXT PAGES AND NOTE THE RESULTS.TO MAKE IT EASY
HERE A POINT A IS TAKEN AS AN OBJECT. BECAUSE IT’S ALL VIEWS ARE JUST POINTS.

NOTATIONS
FOLLOWING NOTATIONS SHOULD BE FOLLOWED WHILE NAMEING
DIFFERENT VIEWS IN ORTHOGRAPHIC PROJECTIONS.
IT’S FRONT VIEW a’ a’ b’
SAME SYSTEM OF NOTATIONS SHOULD BE FOLLOWED
INCASE NUMBERS, LIKE 1, 2, 3 – ARE USED.
OBJECT POINT A LINE AB
IT’S TOP VIEW a a b
IT’S SIDE VIEW a” a” b”

X
Y
1ST
Quad.2nd
Quad.
3rd
Quad. 4th
Quad.
X Y
VP
HP
Observer
THIS QUADRANT PATTERN,
IF OBSERVED ALONG X-Y LINE ( IN RED ARROW DIRECTION)
WILL EXACTLY APPEAR AS SHOWN ON RIGHT SIDE AND HENCE,
IT IS FURTHER USED TO UNDERSTAND ILLUSTRATION PROPERLLY.

HP
VP
a’
a
A
POINT A IN
1ST
QUADRANT
OBSERVER
VP
HP
POINT A IN
2ND
QUADRANT
OBSERVER
a’
a
A
OBSERVER
a
a’
POINT A IN
3RD
QUADRANT
HP
VP
A
OBSERVER
a
a’
POINT A IN
4TH
QUADRANT
HP
VP
A
Point A is
Placed In
different
quadrants
and it’s Fv & Tv
are brought in
same plane for
Observer to see
clearly.
Fv is visible as
it is a view on
VP. But as Tv is
is a view on Hp,
it is rotated
downward 900
,
In clockwise
direction.The
In front part of
Hp comes below
xy line and the
part behind Vp
comes above.
Observe and
note the
process.

FV & TV of a point always lie in the same vertical line
FV of a point ‘P’ is represented by p’. It shows position of the point
with respect to HP only. While drawing FV, XY line represents the
HP and the plane of paper represents the VP.
If the point P lies above HP, p’ lies above the XY line.
If the point P lies in the HP, p’ lies on the XY line.
If the point P lies below the HP, p’ lies below the XY line.
TV of a point ‘P’ is represented by p. It shows position of the point
with respect to VP only. While drawing TV, XY line represents the
VP and the plane of paper represents the HP.
If the point P lies in front of VP, p lies below the XY line.
If the point P lies in the VP, p lies on the XY line.
If the point P lies behind the VP, p lies above the XY line.
Basic concepts for drawing projection of point
FV does not give any information about position of the point w.r.t. VP
TV does not give any information about position of the point w.r.t. HP

A
a
a’
A
a
a’
A
a
a’
X
Y
X
Y
X
YFor Fv
For Tv
For Fv
For Tv
For Tv
For Fv
POINT A ABOVE HP
& INFRONT OF VP
POINT A IN HP
& INFRONT OF VP
POINT A ABOVE HP
& IN VP
PROJECTIONS OF A POINT IN FIRST QUADRANT.
PICTORIAL
PRESENTATION
PICTORIAL
PRESENTATION
ORTHOGRAPHIC PRESENTATIONS
OF ALL ABOVE CASES.
X Y
a
a’
VP
HP
X Y
a’
VP
HP
a X Y
a
VP
HP
a’
Fv above xy,
Tv below xy.
Fv above xy,
Tv on xy.
Fv on xy,
Tv below xy.

PROJECTIONS OF A POINT IN SECOND QUADRANT.
POINT A ABOVE HP
& BEHIND VP
X
Y
A
For Fv
For Tv

SIMPLE CASES OF THE LINE
1. A VERTICAL LINE ( LINE PERPENDICULAR TO HP & // TO VP)
2. LINE PARALLEL TO BOTH HP & VP.
3. LINE INCLINED TO HP & PARALLEL TO VP.
4. LINE INCLINED TO VP & PARALLEL TO HP.
5. LINE INCLINED TO BOTH HP & VP.
STUDY ILLUSTRATIONS GIVEN ON NEXT PAGE
SHOWING CLEARLY THE NATURE OF FV & TV
OF LINES LISTED ABOVE AND NOTE RESULTS.
PROJECTIONS OF STRAIGHT LINES.
INFORMATION REGARDING A LINE means
IT’S LENGTH,
POSITION OF IT’S ENDS WITH HP & VP
IT’S INCLINATIONS WITH HP & VP WILL BE GIVEN.
AIM:- TO DRAW IT’S PROJECTIONS - MEANS FV & TV.

X
Y
V.P.
X
Y
V.P. b’
a’
b
a
F.V.
T.V.
a b
a’
b’
B
A
TV
FV
A
B
X Y
H.P.
V.P.
a’
b’
a b
Fv
Tv
X Y
H.P.
V.P.
a b
a’ b’Fv
Tv
For Fv
For Tv
For Tv
For Fv
Note:
Fv is a vertical line
Showing True Length
&
Tv is a point.
Note:
Fv & Tv both are
// to xy
&
both show T. L.
1.
2.
A Line
perpendicular
to Hp
&
// to Vp
A Line
// to Hp
&
// to Vp
Orthographic Pattern
Orthographic Pattern
(Pictorial Presentation)
(Pictorial Presentation)

A Line inclined to Hp
and
parallel to Vp
(Pictorial presentation)
X
Y
V.P.
A
B
b’
a’
b
a
θ
θ
F.V.
T.V.
A Line inclined to Vp
and
parallel to Hp
Ø
V.P.
a b
a’
b’
BA
Ø
F.V.
T.V.
X Y
H.P.
V.P.
F.V.
T.V.
a b
a’
b’
θ
X Y
H.P.
V.P.
Øa
b
a’ b’
Tv
Fv
Tv inclined to xy
Fv parallel to xy.
3.
4.
Fv inclined to xy
Tv parallel to xy.
Orthographic Projections

X
Y
V.P.
For Fv
a’
b’
a b
B
A
α
β
For Tv
F.V.
T.V.
X
Y
V.P.
a’
b’
a b
α
β
F.V.
T.V.
For Fv
For Tv
B
A
X Y
α
β
H.P.
V.P.
a
b
FV
TV
a’
b’
A Line inclined to both
Hp and Vp
5.
Note These Facts:-
Both Fv & Tv are inclined to xy.
(No view is parallel to xy)
Both Fv & Tv are reduced
lengths.
(No view shows True Length)
Fv is seen on Vp clearly.
To see Tv clearly, HP is
rotated 900
downwards,
Hence it comes below xy.
On removal of object
i.e. Line AB
Fv as a image on Vp.
Tv as a image on Hp,

X Y
H.P.
V.P.
X Y
β
H.P.
V.P.
a
b
TV
a’
b’
FV
TV
b1
b1’
TL
X Y
α
β
H.P.
V.P.
a
b
FV
TV
a’
b’
Here TV (ab) is not // to XY line
Hence it’s corresponding FV
a’ b’ is not showing
True Length &
True Inclination with Hp.
In this sketch, TV is rotated
and made // to XY line.
Hence it’s corresponding
FV a’ b1’ Is showing
True Length
&
True Inclination with Hp.
Note the procedure
When Fv & Tv known,
How to find True Length.
(Views are rotated to determine
True Length & it’s inclinations
with Hp & Vp).
Note the procedure
When True Length is known,
How to locate FV & TV.
(Component a’b2’ of TL is drawn
which is further rotated
to determine FV)
a
a’
b’
b
β
b1
’
θ
α
TL
b2
Ø
TL
Fv
Tv
Means Fv & Tv of Line AB
are shown below,
with their apparent Inclinations
α & β
Here a’b1’ is component
of TL ab1 gives length of FV.
Hence it is brought Up to
Locus of a’ and further rotated
to get point b’. a’ b’ will be Fv.
Similarly drawing component
of other TL(a’b1‘) TV can be drawn.
θ
b1
b2’

a’
X Y
a
20
20
50
55
b’
75
b1
b1’
b2’
b2
θ α
Φ β
TL=83
TL=83
b
θ: True inclination of the
line with HP = 26º
α : Inclination of FV of the
line with HP/XY
Ø: True inclination of the
line with VP = 41º
β : Inclination of TV of the
line with VP/XY
List of questions of type I : 10.13,10.7,5,11,12, 10.28,18,23,26,10.18,10.23
h’
h
v
v’
A: 20mm ↑HP,20 mm →VP
B: 55mm ↑HP,75 mm →VP
End A of a line AB is 20mm above HP & 20mm in front of VP while its end B is 55mm above HP and 75mm
in front of VP. The distance between end projectors of the line is 50mm. Draw projections of the line and
find its true length and true inclination with the principal planes. Also mark its traces.
Given,
a0 b0 = 50 mm
a0 b0
To draw FV & TV,
To find TL, θ ,Ø, HT & VT
Type-I:Given projections (FV & TV) of the line. To find True length & true inclination of the line
with HP (θ) and with VP(Φ).
h’v’a’b’ are always collinear.
h v a b are always collinear.

Problems for finding TL, θ,Ø, HT & VT
Problem1: End A of a line AB is in the HP and 12 mm in front of VP, while its end B
is 45mm above HP and 75mm in front of VP. The distance between their end
projectors is 30 mm. Draw its projections and find its true length and true inclinations
with the principle planes. Also mark its traces.
Problem 2: End A of a line AB is 25mm below the HP and 20 mm behind VP, while
its end B is 45 mm below HP and 65mm behind VP. The distance between their end
Problem 4: End A of a line AB is 25mm below the HP and 20 mm behind VP, while
its end B is 40mm below HP and 65mm behind VP. The distance between their end
Problem 3: End A of a line AB is 70 mm above HP and 20 mm in front of VP, while
its end B is 20mm above HP and 50mm in front of VP. The distance between their
end projectors is 60 mm. Draw its projections and find its true length and true
inclinations with the principle planes. Also mark its traces.

X Y
25
a’
25
70
b1’
a
b2
70
30°
45° b1
b
b’
b2’
h’
h
v
v’
List of questions of type II :
10.8,10.11,10.12,10.14,10.17,
10.10,1,4,6,8,
Line inclined to both HP & VP
Type –II Given (i) T.L., θ and Ø,
(ii) T.L., F.V., T.V.
to draw projections, find α, β,H.T. and V.T.
PROBLEM
A line AB, 70mm long, has its end A 25 mm
above HP and 25mm in front of VP. It is
inclined at 30° to HP and 45°to VP. Draw its
projections and mark its traces.
Given,
A: 25mm ↑HP,25 mm →VP
TL=70mm,
θ=30º,
Ø=45º
To draw/find,
FV, TV, HT & VT

Trapezoidal Method
This method is used to determine true length, true inclination, θ and Ø when
projections of the line (FV & TV) are already given.
Let a’b’ and ab are FV & TV of a line.
X Y
a’
a
b’
b
For solving the problem by trapezoidal
method draw perpendiculars at a and b
Measure distance of a’ from the XY line
and mark it on the perpendicular drawn
from a.
Similarly measure the distance of b’
from the XY line and mark it on the
perpendicular drawn from b.
Mark the points as A1and B1 respectively
and join them. This will give True length of
the line.
Extend this line to intersect the top view.
The point of intersection gives HT and the
angle between TL and TV will be θ.
A1
B1
θ
h
Now draw perpendiculars at a’ and b’
Measure distance of a from the XY line and
mark it on the perpendicular drawn from a’.
Similarly measure the distance of b from
the XY line and mark it on the
perpendicular drawn from b’.
Mark the points as A2and B2 respectively
and join them. This will give True length of
the line.
A2
B2
v’
Extend this line to intersect the front view.
The point of intersection gives VT and the
angle between TL and FV will be Ø.
Ø
Note: If the distances of the points are measured on
the opposite sides of XY line, then perpendiculars
are drawn on opposite sides and, points are also
marked on opposite sides.

Compulsory problems in projection of Straight Lines
Type I 10.13, 5, 10.17,10.18, 10.23
Type II 10.11,10.12,10.14, 6, 8,10.10
Type III 10.19

Q10.11 The top view of a 75mm long line AB measures 65mm,while its front view measures
50mm. Its one end A is in HP and12mm in front of VP. Draw the projections of AB and
determine its inclination with HP and VP
X Y
a’
12
a
b1
Hint: Draw ab1=65mm // to XY.
Because when TV is // to XY, FV
gives TL.
75
b1’b’
5065
65
b
75
b2
Given,
TL=75mm,TV=65mm,FV=50mm
A is in HP & 12mm→
VP
To draw FV &TV of the line AB
To find θ & Ø
30º
48º
Ans. θ=30º
Ans. Ø=48º

Q 5. The end A of a line AB is in the HP and 25 mm behind the VP. The end B is in the VP
and 50 mm above the HP. The distance between the end projectors is 75 mm. draw the
projections of AB and determine its true length, traces and inclination with two planes.
Given,
A is in HP & 25mm ←
V.P.
B is in VP & 50 ↑ V.P.
a0b0=75mm
To find,
True Length, θ,Ø, H.T. and V.T.
X Y
75
a’
50
b’
25
a
b
b1’
b1
93 Ans. TL= a’b1’=93 mm
33º
θ=33º
b2’
b215º
93
Ø=15º
h’
h
v
v’

Q10.12 A line AB, 65mm long has its end A 20mm above H.P. and 25mm in front of VP. The
end B is 40mm above H.P. and 65mm in front of V.P. Draw the projections of AB and show its
inclination with H.P. and V.P.
X Y
20
a’
25
a
4065
Given,
TL=65mm
A is 20mm ↑ HP & 25mm →
V.P.
B is 40mm ↑ & 65mm →
V.P.
To draw FV &TV of the line
AB
To find θ & Ø
Hint1:Mark a’ 20mm above
H.P & a 25mm below XY
Hint2:Draw locus of b’ 40mm
above XY & locus of b 65 mm
below XY
65
b1’
b1
b
b’
b2’
b2
65
18º
38º
Ans. θ=18º
Ans. Ø=38º

Q10.13:The projectors of the ends of a line AB are 5cm apart. The end A is 2cm above the
H.P and 3cm in front of V.P. The end B is1cm below H.P. and 4cm behind the V.P. Determine
the true length and traces of AB, and its inclination with the two planes
X Y50
20
a’
30
a
40
b
10
b’
b1
91
Ans. θ=20º
20º
b2’
b2
50º
Ans. Ø=50º
h’
h
v
v’
Given,
a0b0=50mm
A is 20mm ↑ HP & 30mm →
V.P.
B is 10mm ↓ & 40mm ←
V.P.
To find,
b1’

Q10.13:The projectors of the ends of a line AB are 5cm apart. The end A is 2cm above the
H.P and 3cm in front of V.P. The end B is1cm below H.P. and 4cm behind the V.P. Determine
the true length and traces of AB, and its inclination with the two planes
X Y50
20
a’
30
a
40
b
10
b’
91
Ans. θ=20º
Ans. Ø=50º
h
v’
Given,
A0B0=50mm
A is 20mm ↑ HP & 30mm →
V.P.
B is 10mm ↓ & 40mm ←
V.P.
To find,
A1
B1
θ
TL
TL=91
A2
B2
TL
Ø

Q10.18: The ends of a line PQ are on the same projector. The end P is 30 mm below the
HP and 12 mm behind the VP. The Q is 55 mm above the HP and 45mm in front of the
VP. Determine the true length and traces of PQ and its inclination with two planes.
Given,
p0q0= 0mm
P is 30mm ↓ HP &
12mm ←
V.P. Q is
55mm ↑ & 45mm →
V.P.
To find,
X Y
30
p’
12
p
55
q’
45
q
For solving the problem by trapezoidal method,
first draw perpendiculars at p & q but on opposite
sides as p’ & q’ are on the opposite sides of the xy
line.
P1
Q1
h
θ
θ= 34º
T.L. T.L.=102
mm
P2
Q2
T.L.
v’
Ø
ø= 56º

Q10.14:A line AB, 90mm long, is inclined at 45º to the H.P. and its top view makes an angle of
60º with the V.P. The end A is in the H.P. and 12mm in front of V.P. Draw its front view and
find its true inclination with the V.P.
YX
a’
a
12
90
45º
b1’
b1
60º
b
b’
Given,
T.L.=90mm, θ=45º, β=60º
A is in the H.P. & 12mm→V.P.
To find/draw,
F.V.,T.V. & Ø
90
b2
38º
Ans. Ø = 38º

Q 4:A line AB, 75mm long, is in the second quadrant with the end A in the H.P. and the end B
in the V.P. The line is inclined at 30º to the H.P. and at 45º to the V.P. Draw the projections of
AB and determine its traces.
X Y
Given,
A: in H.P.,
TL=75mm,
θ=30º,
Ø=45º
B: in V.P.,
To find H.T. and V.T.
To draw F.V. and T.V.
a’
30º
75
b1’
45º
75
a
75
45º b2
X Y
a’
a
b1’
θ
TL
ø
TL
b2
b2’
b’
b2’
b’
b
b
h’
h
v
v’

10.17:A line AB, 90mm long, is inclined at 30 to the HP. Its end A is 12mm above the HP and
20mm in front of the VP. Its FV measures 65mm. Draw the TV of AB and determine its
inclination with the VP
X Y
12
a’
20
a
90
30°
b1’
65
b’
b1
b
90
b2
44°
Ans: Ø = 44º
Given,
T.L.=90mm
θ = 30º
A is 12mm ↑H.P. &20mm→V.P.
F.V.=65mm
To find/draw,
T.V. & Ø

Q10.23:Two lines AB & AC make an angle of 120 between them in their FV & TV. AB is
parallel to both the HP & VP. Determine the real angle between AB & AC.
X Y
c’
b’
ab
a’120°
c
120°
c1
c1’
c2
c2’
112°
C
Ans. 112º
Given,
AB parallel to both HP and VP,
a’c’ & ac inclined at 120 with
a’b’ and ab respectively
To find,
True angle between AB & AC
Assume any arbitrary length of
AB & AC. Also assume any
distances of A & B from HP &
VP

Q10.28: The distance between the end projectors of a line AB is 70 mm and the projectors through traces
are 110 mm apart. The end A of the line is 10 mm above the H.P. If the top view and the front view of
the line make 30° and 60° with XY line respectively. Draw the projections of the line and determine (i)
the traces, (ii) the angles with the H.P. and the V.P. and (iii) the true length of the line.
Assume that the line is in the first quadrant.
X Y
Given,
a0b0= 70
mm,
h’v= 110 mm,
A: 10 mm
↑ H.P.,
α =60º,
β=30ºTo find,
True Length,
θ,
Ø,
H.T. and
V.T.
a0
70
b0
10
a’ 60º
b’
h’
110
v
30º
a
h
b
v’
Ans: HT is 63 mm in front of
V.P.
63
VT is 190 mm above the H.P.
190
b1
b1’
56º
θ = 56º
T.L.= 146 mm
b2’
b2
16º
ø = 16º
T.L.
T.L.

Q18: The projectors drawn from H.T. and V.T. of a straight line AB are 80 mm apart while
those drawn from its ends are 50 mm apart. The H.T. of the line 35 mm in front of V.P. ,the
V.T. is 55 mm above the H.P and the end A is 10 mm above the H.P. Draw the projections of
AB and determine its length and inclinations with the reference planes.
Given,
a0b0= 50
mm,
h’v= 80 mm,
A: 10 mm ↑
H.P.,
To find,
True Length,
θ,
Ø
X Yh’
80
v
35
h
55
v’
a’
a
a0
10
50
b’
b
b0
b1
b1’
T.L.
Ans. : T.L.= 65 mm
32º
θ = 32º
b2’
b2
T.L.
ø = 20º20º

Q23: The front view of a line makes an angle of 30º with the xy. The H.T. of the line is 45 mm
in front of the V.P., while its V.T. is 30 mm below the H.P. One end of the line is 10 mm above
the H.P. and the other end is 100 mm in front of the V.P.
Draw the projections of the line and determine (i) its true length, and (ii) its inclinations with
the H.P. and the V.P.
X Y
Given,
α =30º,
H.T.: 45 mm→V.P.,
V.T. : 30 mm ↓H.P.
A: 10 mm ↑ H.P.,
B: 100 mm→V.P.
To find,
True Length,
θ,
Ø
10
a’
h’
45
h
30
v’
v
100
a
b’
b
b1’
b1
T.L.
Ans. : T.L.= 66mm
23º
θ = 23º
b2’
b2
T.L.
37º
ø = 37º
30º

Q26: The projectors of the ends of a line PQ are 90 mm apart. P is 20 mm above the H.P.
while Q is 45 mm behind the V.P. The H.T. and the V.T. of the line coincide with each other
on xy, between the two end projectors and 35 mm away from the projector of the end P.
Draw the projections of PQ and determine its true length and inclinations with the two
planes.
X Yp0
90
To find,
True Length,
θ,
Ø
Given,
p0q0= 90,
h’v= 0 mm,
P: 20 mm ↑ H.P.,
Q: 45 mm←V.P.,
p0h’=35 mm
q0
20 p’
45
q
35
h’
v
h
v’
q’
p
q1
q1’
T.L.
Ans. : T.L.= 127 mm
24º
θ = 24º
q2
T.L.
35º
ø = 35º
q2’

Q10.19 A line AB, inclined at 40º to the V.P. has its end 50mm and 20mm above the
H.P. the length of its front view is 65mm and its V.T. is 10mm above the H.P.
determine .the true length of AB its inclination with the H.P. and its H.T.
Given,
Ø = 40º, A is 20mm↑HP,
B is 50 mm ↑ HP, FV=65mm,
VT is 10mm ↑ HP
To find,
TL, θ & HT
X Y
50
b’
20
a’
10
v’
h’
b2’
v40º
b2
a
b1
b1’
85
21º h
Ans,
TL = 85 mm,
θ = 21º &
HT is 17 mm
behind VP
Note: Here we do not know position of a
and b in top view, but we know v’ and v.
So for time being we consider the line as
v’b’ instead of a’b’. Now if we make
v’b’(FV) parallel to xy, we will get TL of
VB in top view at angle Ø.

Q10.16:The end A of a line AB is 25 mm behind the V.P. and is below the H.P. The end B is 12
mm in front of the VP and is above the HP The distance between the projectors is 65mm. The
line is inclined at 40 to the HP and its HT is 20 mm behind the VP. Draw the projections of the
line and determine its true length and the VT
X Y
Given,
A0B0=65mm
A is 25mm ←V.P.& is ↓H.P.
B is 12mm →V.P. & is above HP
θ = 40º
To find/draw,
F.V., T.V., T.L., v’
25
a
65 12
b
20
h
h’
b1
40º
b1’
b’
a’
v
v’
b2
b2’
Ans. TL= a’b2’=98 mm

Q10. A line AB is in the first quadrant. Its end A and B are 20 mm and 60 mm in front of
the V.P. respectively. The distance between the end projectors is 75 mm. The line is
inclined at 30º to the H.P. and its H.T. is 10 mm above the xy. Draw the projections of
AB and determine its true length and the V.T.
Given,
AB is in first quadrant,
A is 20mm→V.P.,
B is 60 mm → V.P.,
a0b0= 75 mm,
θ = 30º
H.T. is 10mm↑ xy
To draw F.V.
and T.V.
to find,TL, θ &
V.T.
X Y
75
a0
b0
20
a
b
60
10
h
h’
b1
b1’
30º
Note: Here we do not know position
of a’ and b’ in front view, but we
know h’ and h. So for time being we
consider the line as hb instead of ab.
Now if we make hb(TV) parallel to
xy, we will get TL of HB in front
view at angle θ.
b’
a’
Note: As b1’ and b’ lies on
the same locus, similarly
a1’ and a’ will also lie on
the same locus. And a1’b1’
will be true length of AB
a1’
T.L.
Ans. TL= a1’b1’=98mm
v
v’
VT is 13 mm above the H.P.

Q9. The front view of a line AB measures 65 mm and makes an angle of 45º with xy. A is
in the H.P. and V.T. of the line is 15 mm below the H.P. The line is inclined at 30º to the
V.P. draw the projections of AB and find its true length and inclination with the H.P. Also
locate its H.T.
Given,
FV=65mm,
α = 45º,
A is inHP,
VT is 15 mm ↓HP
Ø = 30º,
To find,
TL, θ & HT
X Y45º
a’
b’
15
v’
v
b2’
30º
b2
b
a
b1
b1’
T.L.
38ºh’
h
12
Ans: TL=75mm
HT is 12 mm in front of VP
θ = 38º
65
of a and b in top view, but we know
v’ and v. So for time being we
consider the line as v’b’ instead of
a’b’. Now if we make v’b’(FV)
parallel to xy, we will get TL of VB
in top view at angle Ø.

Q 33. The front view of a line AB measures 70 mm and makes an angle of 45º with xy. A
is in the H.P. and V.T. of the line is 15 mm below the H.P. The line is inclined at 30º to
the V.P. draw the projections of AB, and find its true length and inclination with the H.P.
Also locate its H.T.
Given,
FV=70 mm,
α = 45º,
A is in HP,
VT is 15mm ↓HP
Ø = 30º,
To find,
TL, θ & HT
X Y45º
b’
15
v’
a’v
b2’
30º
b2b
a
b1
b1’
T.L.
38º
Ans: TL=81mm
θ = 38º
h’
h
of a and b in top view, but we know
v’ and v. So for time being we
consider the line as v’b’ instead of
a’b’. Now if we make v’b’(FV)
parallel to xy, we will get TL of VB
in top view at angle Ø.

Q34. A line AB measures 100 mm. The projectors through its V.T. and the end A are 40
mm apart. The point A is 30 mm below the H.P. and 20 mm behind the V.P. The V.T. is
10 mm above the H.P. Draw the projections of the line and determine its H.T. and
inclinations with the H.P. and the V.P.
Given,
TL=100 mm,
a0v= 40 mm, A:
30mm↓ H.P., 20mm ← V.P.
VT is 10mm ↑HP
To draw FV &
TV and find,
HT,θ, & ø
X Yv
40
a0
10
v’
30
a’
20
a
a1
a1’
100
b1’
b’
b
42º
θ = 42º
b2
100
ø = 20º
20º
h’
h
5
HT is 5 mm behind the VP

Q9. The front view of a line AB measures 65 mm and makes an angle of 45º with
xy. A is in the H.P. and V.T. of the line is 15 mm below the H.P. The line is inclined
at 30º to the V.P. draw the projections of AB and find its true length and inclination
with the H.P. Also locate its H.T.
Q10. A line AB is in the first quadrant. Its end A and B are 20 mm and 60 mm in
front of the V.P. respectively. The distance between the end projectors is 75 mm.
The line is inclined at 30º to the H.P. and its H.T. is 10 mm above the xy. Draw
the projections of AB and determine its true length and the H.T.
Q33. The front view of a line AB measures 70 mm and makes an angle of 45º with
xy. A is in the H.P. and V.T. of the line is 15 mm below the H.P. The line is inclined
at 30º to the V.P. draw the projections of AB, and find its true length and inclination
with the H.P. Also locate its H.T.
Q34. A line AB measures 100 mm. The projectors through its V.T. and the end A
are 40 mm apart. The point A is 30 mm below the H.P. and 20 mm behind the V.P.
The V.T. is 10 mm above the H.P. Draw the projections of the line and determine
its H.T. and inclinations with the H.P. and the V.P.
Q10.9. A line PQ 75 mm long, has its end P in the V.P. and the end Q in the H.P.
The line is inclined at 30 to the H.P. and at 60 to the V.P. Draw its projections.

Q10.19 A line AB, inclined at 40º to the V.P. has its end 50mm and 20mm above the H.P. the length of its front view is
65mm and its V.T. is 10mm above the H.P. determine .the true length of AB its inclination with the H.P. and its H.T.
Given,
Ø = 40º, A is 20mm↑HP,
B is 50 mm ↑ HP,
FV=65mm, VT is 10mm ↑ HP
To find,
TL, θ & HT
X Y
50
b’
20
a’
10
v’
h’v
65
40º
a
b
h
21º
A1’
B1’
Ans: A1’B1’=TL=85mm
Ans:HT is 17 mm behind VP
Ans:θ = 21º
Step1: For solving the problem by
trapezoidal method, draw a line at 40º(Ø)
from VT’. Then draw perpendiculars from
a’ and b’ on this line.
Step2: Then draw projectors from a’ and b’
and mark the distance of b’B1’ on the
projector of b’ below XY. Similarly mark
the distance a’A1’ on the projector of a’
below XY

Q10.25 The straight line AB is inclined at 30º to H.P., while its top view at 45º to XY line. The
end A is 25 mm in front of the V.P. and it is below the H.P. The end B is 75 mm behind the
V.P. and it is above the H.P. Draw the projections of the line when its V.T. is 40 mm below the
H.P. Find the true length of the portion of the straight line which is in the second quadrant
and locate its H.T..
Given,
θ=30º ,
β = 45º,
A is 25mm
→V.P. & ↓ H.P. B is
75mm ← V.P. & ↑ H.P
V.T is 40 mm ↓ H.P.
To draw/find,
F.V. and T.V,
TL of portion in second quadrant
H.T.
X Y
25
a β=45º
75
b
v
v’
40
b1
θ=30º
b1’b’
a’
h’
h
HBis the portion of the line in
II quadrant as its FV and TV
are above XY line
b2
b2’
42
Ans: HT= 49 mm behind the V.P.
TL of HB=42mm

Q10.26 The front view of a line PQ makes an angle of 30º with XY line (ground line). The H.T.
of the line is 45 mm behind the V.P. while its V.T. is 30 mm above H.P. The end P of the line is
10 mm below H.P. and the end Q is in the first quadrant. The line is 150 mm long. Draw the
projections of the line and determine the true length of the portion of the line in second
quadrant. Also find the angle of the line with H.P. and V.P.
Given, α
= 30º, H.T.:
45 mm←VP, in the HP
V.T.: 30 mm↑HP. In the VP
P:10 mm ↓HP.
Q; in first quadrant.
TL=150 mm
To draw/find,
F.V. and T.V,
TL of portion in second quadrant
θ. ø
X Y
h’
45
h
10
p’
30
v’
v
α=30º
p v1
v1’
150
q1’
q’
q
θ=24º
HV is the portion of the line in
II quadrant as its FV and TV
are above XY line
Ans: θ=24º
v2’
v2
75
TL of HV=75 mm
ø=37º
ø=37º

Q10.27 The end P of a line PQ 130 mm long, is 55 mm in front of the VP. The H.T. of the line
is 40 mm in front of the V.P. and the V.T. is 50 mm above the HP. The distance between H.T.
and V.T. is 110 mm. Draw the projections of the line PQ and determine its angles with the
H.P. and the V.P.
Given,
T.L.= 130 mm,
P:55 mm →V.P.,
H.T.: 40
mm→V.P. , V.T.:
50 mm↑H.P.,
h’v = 110 mm
To draw/find,
F.V. and T.V,θ. ø
X Y
110
h’
40
h
v
50
v’
55
p
p’
v1
v1’
130
q1’
q’
q
θ=23º
q2
ø=19º
130

Q6. The top view of a 75mm long line CD measures 50 mm. C is 50 mm in front of the VP &
15mm below the HP. D is 15 mm in front of the VP & is above the HP. Draw the FV of CD &
find its inclinations with the HP and the VP. Show also its traces.
X Y
15
50
Given,
TL = 75 mm, TV =50 mm,
C is 15mm ↓ HP & 50 mm → VP,
D is 15 mm → VP
To draw,
FV & to find θ & Ø
To mark HT & VT
c’
c
Locus of D
Hint 1: Cut an arc of 50 mm
from c on locus of D to get
the TV of the line
d
Hint 2: Make TV (cd), // to XY
so that FV will give TL
d1
d1’d’
θ=48º Ans: θ=48º
d2
50
75
Ø=28º
Ans: Ø=28º
75
h’
h
v
v’
d2’

Q10.10 A line PQ 100 mm long is inclined at 30º to the H.P. and at 45º to the V.P. Its
mid point is in the V.P. and 20 mm above the H.P. Draw its projections, if its end P is in
the third quadrant and Q is in the first quadrant.
Given,
TL = 100, θ = 30º,Ø=45º, Mid point M is
20mm↑HP & in the VP
End P in third quadrant &
End Q in first quadrant
To draw,
FV & TV
X Y
20
m’
m
30º
q1’
p1’
50
50
50
q2
p2
50 q2’
q’
p’
q
p
q1
p1
p2’
45º

Problem 3: The front view of a 125 mm long line PQ measures 75 mm while its top view
measures 100 mm. Its end Q and the mid point M are in the first quadrant. M being 20 mm
from both the planes. Draw the projections of line PQ.
X Y
20
20
m’
m
q1’
p1’
p1
q1
q’
p’
50 50
62.5
62.5
37.5
37.5
p
q
Given,
TL = 125mm, FV=75mm, TV=100mm,
Mid point M is 20mm↑HP &20mm in front of the VP
End Q and mid
point M are in first quadrant
To draw,
FV & TV

Q8:A line AB 65 mm long has its end A in the H.P. & 15 mm in front of the V.P. The end B is
in the third quadrant. The line is inclined at 30 to the H.P. and at 60 to the V.P. Draw its
projections.
X Y
15
a
a’
30º
b1’
b2
65
65
60º b1
b
b’
b2’ HP
VP
X Y
15
a”30º
60º
b”
b’
a’
a
b
Given,
TL =65mm,
A: in the HP & 15mm →VP
B: in third quadrant
θ = 30º,
Ø=60º
To draw,
FV & TV

4.8m
4.2m
3.6m
A
B
X Y
4.8m
3.6m
a’
b’
4.2m
a
b
b2’
b2
T.L.
Ans. T.L.= 7.32m
Q11:A room is 4.8 m X 4.2 m X 3.6 m high, determine graphically the distance between a top
corner and the bottom corner diagonally opposite to it.

PROBLEM 12 :- Two oranges on a tree are respectively 1.8 m and 3.0 m above ground, and
those are 1.2 m & 2.1 m from a 0.3 m thick wall, but on the opposite sides of it. The distance
between the oranges, measured along the ground and parallel to the wall is 2.7m. Determine
the real distance between the oranges.
1.2m
1.8m
0.3 m
2.7m
3.0m
2.1m
F.V.
X Y
1.2m
1.8m
2.1m
3.0m
0.3m
A
B
a’
b’
T.V.
a
2.7m
b
b1
b1’
T.L.
Ans. T.L.= 4.65m

Problems for practice
Q1. The top view of a straight line AB is 50 mm long and it makes an angle of 30 with the
xy line. End A is in V.P. and 25 mm from the H.P. End B is 54 mm from the H.P. The line
lies in the first quadrant. Draw projections of the line AB and determine its true length
and true inclinations also mark its traces.

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