Chapter 6 of 8

Updated 6 August 2026

NCERT Solutions

Orthographic Projections of Simple Machine Blocks — NCERT Solutions

CBSE · Class 11 · Engineering Graphics

NCERT Solutions for Orthographic Projections of Simple Machine Blocks, CBSE Class 11 Engineering Graphics: 59 textbook questions solved step by step.

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Think, Discuss and Write — II and IV Quadrant Observation

1Keep the same object in II quadrant and IV quadrant in the glass box arrangement. Turn/Rotate the H.P. in clockwise direction and open up the box. Then project the views. Discuss your observation with your partner. What do you find out?Show solution

Given: Object placed in II quadrant and IV quadrant separately; H.P. is rotated clockwise to open the box.

Concept: In the standard glass-box method, the H.P. is rotated downward (clockwise when viewed from the right) to bring it into the plane of the V.P.

Observation and Discussion:

When the object is in the II quadrant (above H.P., behind V.P.):

The Front View (projection on V.P.) appears above XY.
The Top View (projection on H.P.) also appears above XY after rotation.
Therefore, both views overlap each other on the same side of XY, making it impossible to distinguish them clearly.

When the object is in the IV quadrant (below H.P., in front of V.P.):

The Front View appears below XY.
The Top View also appears below XY after rotation.
Again, both views overlap each other on the same side of XY.

Conclusion: In both II and IV quadrants, the Front View and Top View fall on the same side of the reference line XY and overlap each other. This makes the drawing ambiguous and unreadable. This is why the II and IV quadrants are NOT used in practice for orthographic projection. Only I quadrant (First Angle Projection) and III quadrant (Third Angle Projection) are used, because in these cases the two views fall on opposite sides of XY and do not overlap.

Assignment 4.1

1(i)Fill in the blank: In ... projection, the ... are perpendicular to the plane of projection.Show solution

In orthographic projection, the projectors (lines of sight) are perpendicular to the plane of projection.

Complete sentence: In orthographic projection, the projectors are perpendicular to the plane of projection.

1(ii)Fill in the blank: In I angle projection, the ... comes between the ... and ... .Show solution

In I angle (First Angle) projection, the object comes between the observer and the plane of projection.

Complete sentence: In I angle projection, the object comes between the observer and the plane of projection.

1(iii)Fill in the blank: In III angle projection, the ... comes between the ... and ... .Show solution

In III angle (Third Angle) projection, the plane of projection comes between the observer and the object.

Complete sentence: In III angle projection, the plane of projection comes between the observer and the object.

2Explain briefly how the reference line represents both the principal planes of projection.Show solution

The reference line XY (also called the ground line or hinge line) is the line of intersection of the two principal planes — the Vertical Plane (V.P.) and the Horizontal Plane (H.P.).

Explanation:

When the H.P. is rotated 90° downward (in First Angle) or upward (in Third Angle) to coincide with the V.P., the line along which they were hinged becomes the reference line XY on the drawing sheet.
Any point above XY in the front view represents a height above H.P. (i.e., distance from H.P. measured on V.P.).
Any point below XY in the top view represents a distance in front of V.P. (i.e., distance from V.P. measured on H.P.).
Thus, XY simultaneously represents:

(a) The edge view of H.P. when looking at the Front View (V.P.).
(b) The edge view of V.P. when looking at the Top View (H.P.).

In this way, the single reference line XY represents both principal planes of projection on the 2D drawing sheet.

3Sketch neatly the symbols used for indicating the method of projection adopted in a drawing.Show solution

Two standard symbols are used to indicate the method of projection:

First Angle Projection Symbol:

It consists of the front view and side view of a frustum of a cone (truncated cone).
The wider (base) end of the frustum is shown on the left and the narrower end on the right in the front view.
The side view (circle with a smaller circle inside, representing the two diameters) is placed to the RIGHT of the front view.
This indicates that the view is placed on the opposite side from the direction of viewing (object between observer and plane).

$\text{[Front view: trapezium wider on left]} \quad \text{[Side view: two concentric circles — placed to the RIGHT]}$

Third Angle Projection Symbol:

The same frustum of a cone is used.
The side view (two concentric circles) is placed to the LEFT of the front view.
This indicates that the view is placed on the same side as the direction of viewing (plane between observer and object).

$\text{[Side view: two concentric circles — placed to the LEFT]} \quad \text{[Front view: trapezium wider on right]}$

Note: These symbols are placed in the title block of the drawing sheet to indicate which projection method has been used.

4Why second and fourth quadrants are not used in practice? (Not to be asked in the exam)Show solution

(Note: This question is marked as not to be asked in the exam, but the explanation is given for understanding.)

Reason:

In the II quadrant, the object is above H.P. and behind V.P. When the H.P. is rotated downward to open the glass box, the Top View (on H.P.) moves upward and falls above XY. The Front View (on V.P.) is also above XY. Thus, both views appear on the same side (above XY) and overlap each other.
In the IV quadrant, the object is below H.P. and in front of V.P. When H.P. is rotated downward, the Top View falls below XY. The Front View is also below XY. Again, both views overlap.
Since the views overlap in both II and IV quadrants, the drawing becomes ambiguous and uninterpretable.
Therefore, II and IV quadrants are not used in practice.

Try These — Projection of Points

(i)State the quadrant: A point whose Top View is 40 mm above XY and Front View is 20 mm below the Top View.Show solution

Given:

Top View is 40 mm above XY.
Front View is 20 mm below the Top View.

Step 1: Determine position of Front View.
Front View = 40 mm above XY − 20 mm = 20 mm above XY.

Step 2: Interpret the positions.

Front View above XY → the point is above H.P.
Top View above XY → the point is behind V.P.

Step 3: Match with quadrant table.

Above H.P. and behind V.P. → Second Quadrant (II).

$\boxed{\text{The point is in the Second Quadrant (II).}}$

(ii)State the quadrant: A point Q whose projections coincide with each other 40 mm below XY.Show solution

Given:

Both Front View and Top View coincide at 40 mm below XY.

Step 1: Interpret Front View below XY.

Front View below XY → the point is below H.P.

Step 2: Interpret Top View below XY.

Top View below XY → the point is in front of V.P.

Step 3: Match with quadrant table.

Below H.P. and in front of V.P. → Fourth Quadrant (IV).

$\boxed{\text{Point Q is in the Fourth Quadrant (IV).}}$

Assignment — Projection of Points

1A point D is 25 mm from H.P. and 30 mm from V.P. Draw its projections considering it in first and third quadrants.Show solution

Given: Point D is 25 mm from H.P. and 30 mm from V.P.

Case 1: First Quadrant (above H.P., in front of V.P.)

Step 1: Draw reference line XY.

Step 2: Mark the Front View (d').

Point is above H.P., so d' is above XY.
Distance from H.P. = 25 mm.
Mark d' at 25 mm above XY.

Step 3: Mark the Top View (d).

Point is in front of V.P., so d is below XY (First Angle).
Distance from V.P. = 30 mm.
Mark d at 30 mm below XY, directly below d'.

Step 4: Draw a vertical projector connecting d' and d perpendicular to XY.

Case 2: Third Quadrant (below H.P., behind V.P.)

Step 1: Draw reference line XY.

Step 2: Mark the Front View (d').

Point is below H.P., so d' is below XY.
Mark d' at 25 mm below XY.

Step 3: Mark the Top View (d).

Point is behind V.P., so d is above XY.
Mark d at 30 mm above XY, directly above d'.

Step 4: Draw a vertical projector connecting d' and d.

Summary Table:
$\begin{array}{|c|c|c|}\hline \text{Quadrant} & \text{Front View (d')} & \text{Top View (d)} \\ \hline \text{I} & 25\text{ mm above XY} & 30\text{ mm below XY} \\ \hline \text{III} & 25\text{ mm below XY} & 30\text{ mm above XY} \\ \hline \end{array}$

2A point P is 15 mm above the H.P. and 20 mm in front of the V.P. Another point Q is 25 mm behind the V.P. and 40 mm below the H.P. Draw the projections.Show solution

Given:

Point P: 15 mm above H.P., 20 mm in front of V.P. → First Quadrant.
Point Q: 25 mm behind V.P., 40 mm below H.P. → Third Quadrant.

Projections of Point P (First Quadrant):

Step 1: Draw reference line XY.

Step 2: Front View p':

Above H.P. → above XY.
Mark p' at 15 mm above XY.

Step 3: Top View p:

In front of V.P. → below XY (First Angle Projection).
Mark p at 20 mm below XY, on the projector through p'.

Projections of Point Q (Third Quadrant):

Step 4: Front View q':

Below H.P. → below XY.
Mark q' at 40 mm below XY.

Step 5: Top View q:

Behind V.P. → above XY.
Mark q at 25 mm above XY, on the projector through q'.

Step 6: Both sets of projections are drawn on the same XY line with vertical projectors connecting each pair.

Result:
$p': 15\text{ mm above XY}, \quad p: 20\text{ mm below XY}$
$q': 40\text{ mm below XY}, \quad q: 25\text{ mm above XY}$

3Draw the projections of the following points on the same XY: B — 20 mm above H.P. and 25 mm in front of V.P.; D — 25 mm below H.P. and 15 mm behind V.P.; E — 15 mm above H.P. and 10 mm behind V.P.; F — 20 mm below H.P. and 25 mm in front of V.P.Show solution

Given: Four points to be projected on the same XY line (First Angle Projection).

Step 1: Draw a horizontal reference line XY.

Step 2: For each point, determine the positions of Front View and Top View using the rule:

Front View above XY → point above H.P.; below XY → point below H.P.
Top View below XY → point in front of V.P.; above XY → point behind V.P. (First Angle).

Point B (I Quadrant — above H.P., in front of V.P.):

Front View b': 20 mm above XY.
Top View b: 25 mm below XY.

Point D (III Quadrant — below H.P., behind V.P.):

Front View d': 25 mm below XY.
Top View d: 15 mm above XY.

Point E (II Quadrant — above H.P., behind V.P.):

Front View e': 15 mm above XY.
Top View e: 10 mm above XY.

Point F (IV Quadrant — below H.P., in front of V.P.):

Front View f': 20 mm below XY.
Top View f: 25 mm below XY.

Step 3: Mark all points at appropriate horizontal positions along XY (spaced apart for clarity), draw vertical projectors for each.

Summary:
$\begin{array}{|c|c|c|c|}\hline \text{Point} & \text{Quadrant} & \text{Front View} & \text{Top View} \\ \hline B & I & 20\text{ mm above XY} & 25\text{ mm below XY} \\ D & III & 25\text{ mm below XY} & 15\text{ mm above XY} \\ E & II & 15\text{ mm above XY} & 10\text{ mm above XY} \\ F & IV & 20\text{ mm below XY} & 25\text{ mm below XY} \\ \hline \end{array}$

Assignment — Projection of Lines

(a)Line CD is in V.P., parallel to H.P., and end C is 30 mm above the H.P. Draw the projections. (Length = 40 mm)Show solution

Given: Line CD, length = 40 mm; lies in V.P.; parallel to H.P.; end C is 30 mm above H.P.

Concept: A line in V.P. and parallel to H.P. will show its true length in the Front View (on V.P.), and the Top View will be a point on XY (since the line is in V.P., its distance from V.P. = 0).

Step 1: Draw reference line XY.

Step 2: Front View (c'd'):

Line is in V.P. and parallel to H.P., so the front view is a horizontal line parallel to XY.
End C is 30 mm above H.P., so c' is 30 mm above XY.
Since the line is parallel to H.P., d' is also 30 mm above XY.
Draw c'd' = 40 mm (true length), horizontal, 30 mm above XY.

Step 3: Top View (cd):

Line lies in V.P. (distance from V.P. = 0), so the top view lies on XY.
Draw cd as a point or a line on XY directly below c'd'.
Since the line is parallel to H.P. and in V.P., the top view is a line of 40 mm on XY.

Result: Front View = horizontal line 40 mm long, 30 mm above XY. Top View = line of 40 mm on XY.

(b)Line EF is parallel to and 25 mm in front of V.P. and is in the H.P. Draw the projections. (Length = 40 mm)Show solution

Given: Line EF, length = 40 mm; parallel to V.P.; 25 mm in front of V.P.; lies in H.P. (height above H.P. = 0).

Concept: A line in H.P. and parallel to V.P. shows its true length in the Top View, and the Front View is a point on XY.

Step 1: Draw reference line XY.

Step 2: Top View (ef):

Line is in H.P. and parallel to V.P., so the top view is a horizontal line parallel to XY.
Line is 25 mm in front of V.P., so ef is 25 mm below XY (First Angle).
Draw ef = 40 mm (true length), horizontal, 25 mm below XY.

Step 3: Front View (e'f'):

Line lies in H.P. (height = 0), so the front view lies on XY.
Draw e'f' as a line of 40 mm on XY, directly above ef.

Result: Top View = horizontal line 40 mm long, 25 mm below XY. Front View = line of 40 mm on XY.

(c)Line GH is in both H.P. and V.P. Draw the projections. (Length = 40 mm)Show solution

Given: Line GH, length = 40 mm; lies in both H.P. and V.P. simultaneously.

Concept: A line lying in both H.P. and V.P. must lie along their line of intersection, which is the XY line itself.

Step 1: Draw reference line XY.

Step 2: Since the line lies on XY:

Front View (g'h') lies on XY (height above H.P. = 0).
Top View (gh) also lies on XY (distance from V.P. = 0).
Both views coincide on XY.

Step 3: Draw gh = g'h' = 40 mm along XY.

Result: Both Front View and Top View are the same line of 40 mm lying on XY.

(d)Line JK is perpendicular to H.P. and 20 mm in front of V.P. The nearest point from H.P. is J, which is 15 mm above H.P. Draw the projections. (Length = 40 mm)Show solution

Given: Line JK, length = 40 mm; perpendicular to H.P.; 20 mm in front of V.P.; J (nearer end) is 15 mm above H.P.; K is the farther end.

Concept: A line perpendicular to H.P. appears as a point in the Top View and as a vertical line (true length) in the Front View.

Step 1: Draw reference line XY.

Step 2: Determine heights:

J is 15 mm above H.P. → j' is 15 mm above XY.
K is the other end: K = J + 40 mm above H.P. = 15 + 40 = 55 mm above H.P. → k' is 55 mm above XY.

Step 3: Front View (j'k'):

Draw a vertical line from j' (15 mm above XY) to k' (55 mm above XY).
Length = 40 mm (true length), vertical.
Horizontal position: 20 mm in front of V.P. does not affect the front view position horizontally (it is a fixed vertical line).

Step 4: Top View (jk):

Line is perpendicular to H.P., so top view is a single point.
The point is 20 mm below XY (20 mm in front of V.P., First Angle).
Mark point j(k) at 20 mm below XY, directly below j'k'.

Result: Front View = vertical line 40 mm long (j' at 15 mm above XY, k' at 55 mm above XY). Top View = a single point 20 mm below XY.

(f)Line UV is perpendicular to the V.P., with the farthest end V from V.P. at 65 mm in front of V.P. and 20 mm above H.P. Draw the projections. (Length = 40 mm)Show solution

Given: Line UV, length = 40 mm; perpendicular to V.P.; farthest end V is 65 mm in front of V.P.; both ends are 20 mm above H.P.

Concept: A line perpendicular to V.P. appears as a point in the Front View and as a horizontal line (true length) in the Top View.

Step 1: Draw reference line XY.

Step 2: Determine distances from V.P.:

V (farthest) is 65 mm in front of V.P.
U (nearer) = 65 − 40 = 25 mm in front of V.P.

Step 3: Front View (u'v'):

Line is perpendicular to V.P., so front view is a single point.
Height = 20 mm above H.P. → u'(v') is 20 mm above XY.
Mark point u'(v') at 20 mm above XY.

Step 4: Top View (uv):

Draw a horizontal line perpendicular to XY (i.e., vertical on the drawing, going away from XY).
u is 25 mm below XY (25 mm in front of V.P.).
v is 65 mm below XY (65 mm in front of V.P.).
Draw uv = 40 mm, perpendicular to XY, from 25 mm to 65 mm below XY.

Result: Front View = single point 20 mm above XY. Top View = horizontal line 40 mm long, perpendicular to XY, from 25 mm to 65 mm below XY.

Additional Assignment — Projection of Lines

*(a)Line AB parallel to H.P. as well as V.P., 25 mm behind V.P. and 30 mm below H.P. (Not to be asked in exam)Show solution

Given: Line AB, length = 40 mm; parallel to both H.P. and V.P.; 25 mm behind V.P.; 30 mm below H.P. → Third Quadrant.

Concept: A line parallel to both planes shows its true length in both Front View and Top View.

Step 1: Draw reference line XY.

Step 2: Front View (a'b'):

30 mm below H.P. → 30 mm below XY.
Parallel to H.P. → horizontal line.
Draw a'b' = 40 mm, horizontal, 30 mm below XY.

Step 3: Top View (ab):

25 mm behind V.P. → 25 mm above XY (Third Quadrant, behind V.P.).
Parallel to V.P. → horizontal line.
Draw ab = 40 mm, horizontal, 25 mm above XY.

Result: Front View = 40 mm horizontal line, 30 mm below XY. Top View = 40 mm horizontal line, 25 mm above XY.

*(b)Line LM is 30 mm behind V.P. and perpendicular to H.P. The nearest point from H.P. is L, which is 10 mm above H.P. (Not to be asked in exam)Show solution

Given: Line LM, length = 40 mm; perpendicular to H.P.; 30 mm behind V.P.; L (nearest to H.P.) is 10 mm above H.P.

Step 1: Draw reference line XY.

Step 2: Heights:

L: 10 mm above H.P. → l' is 10 mm above XY.
M: 10 + 40 = 50 mm above H.P. → m' is 50 mm above XY.

Step 3: Front View (l'm'):

Vertical line from l' (10 mm above XY) to m' (50 mm above XY).
True length = 40 mm.

Step 4: Top View (lm):

Perpendicular to H.P. → single point in top view.
30 mm behind V.P. → 30 mm above XY.
Mark l(m) at 30 mm above XY.

Result: Front View = vertical line 40 mm (10 mm to 50 mm above XY). Top View = single point 30 mm above XY.

*(c)Line NP is 30 mm below H.P. and perpendicular to V.P. The nearest point from V.P. is P, which is 10 mm in front of V.P. (Not to be asked in exam)Show solution

Given: Line NP, length = 40 mm; perpendicular to V.P.; 30 mm below H.P.; P (nearest to V.P.) is 10 mm in front of V.P.

Step 1: Draw reference line XY.

Step 2: Distances from V.P.:

P (nearest): 10 mm in front of V.P.
N (farthest): 10 + 40 = 50 mm in front of V.P.

Step 3: Front View (n'p'):

Perpendicular to V.P. → single point in front view.
30 mm below H.P. → 30 mm below XY.
Mark n'(p') at 30 mm below XY.

Step 4: Top View (np):

Perpendicular to V.P. → horizontal line perpendicular to XY.
P at 10 mm below XY; N at 50 mm below XY.
Draw np = 40 mm, from 10 mm to 50 mm below XY.

Result: Front View = single point 30 mm below XY. Top View = line 40 mm long, from 10 mm to 50 mm below XY.

*(d)Line QR is 10 mm below H.P. and perpendicular to V.P. The farthest point from V.P. is Q, 65 mm behind V.P. (Not to be asked in exam)Show solution

Given: Line QR, length = 40 mm; perpendicular to V.P.; 10 mm below H.P.; Q (farthest from V.P.) is 65 mm behind V.P.

Step 1: Draw reference line XY.

Step 2: Distances from V.P.:

Q (farthest): 65 mm behind V.P. → 65 mm above XY in top view.
R (nearest): 65 − 40 = 25 mm behind V.P. → 25 mm above XY in top view.

Step 3: Front View (q'r'):

Perpendicular to V.P. → single point.
10 mm below H.P. → 10 mm below XY.
Mark q'(r') at 10 mm below XY.

Step 4: Top View (qr):

Line perpendicular to V.P. → horizontal line perpendicular to XY.
Draw qr = 40 mm, from 25 mm to 65 mm above XY.

Result: Front View = single point 10 mm below XY. Top View = line 40 mm long, from 25 mm to 65 mm above XY.

*(e)Line ST is perpendicular to H.P. and behind V.P. The nearest point from H.P. is S, which is 20 mm from V.P. and 15 mm below H.P. (Not to be asked in exam)Show solution

Given: Line ST, length = 40 mm; perpendicular to H.P.; behind V.P.; S (nearest to H.P.) is 20 mm behind V.P. and 15 mm below H.P.

Step 1: Draw reference line XY.

Step 2: Heights:

S: 15 mm below H.P. → s' is 15 mm below XY.
T: 15 + 40 = 55 mm below H.P. → t' is 55 mm below XY.

Step 3: Front View (s't'):

Vertical line from s' (15 mm below XY) to t' (55 mm below XY).
True length = 40 mm.

Step 4: Top View (st):

Perpendicular to H.P. → single point.
20 mm behind V.P. → 20 mm above XY.
Mark s(t) at 20 mm above XY.

Result: Front View = vertical line 40 mm (15 mm to 55 mm below XY). Top View = single point 20 mm above XY.

Assignment — Projection of Plane Surfaces

1A thin pentagonal plate of 35 mm sides is inclined at 30° to the HP and perpendicular to the V.P. One of the edges of the plate is ⊥ to V.P., 20 mm above the H.P. and its one end, which is nearer to the V.P., is 30 mm in front of the V.P. Draw the projections of the plate.Show solution

Given:

Pentagonal plate, side = 35 mm.
Inclined at 30° to H.P., perpendicular to V.P.
One edge ⊥ to V.P., 20 mm above H.P.
Nearer end of that edge is 30 mm in front of V.P.

Concept: Two-step method (plane perpendicular to V.P. and inclined to H.P.).

Step 1 (Initial Position — plane parallel to H.P., perpendicular to V.P.):

Assume the pentagonal plate is lying with its surface parallel to H.P. (i.e., inclined at 0° to H.P.).
Top View: True shape — regular pentagon with 35 mm sides. Draw the pentagon in the top view with one edge perpendicular to XY (i.e., parallel to the projector direction).
Front View: A horizontal line (edge view of the plate) at 20 mm above XY.

Step 2 (Tilt the plate at 30° to H.P.):

Redraw the Front View by tilting the edge-view line at 30° to XY.
The edge that is ⊥ to V.P. and 20 mm above H.P. remains at 20 mm above XY.
The nearer end of this edge is 30 mm in front of V.P. → in the top view, this end is 30 mm below XY.
Rotate the front view (edge line) to 30° with XY, keeping the reference edge fixed.
Project the new front view points down to get the new top view.
The top view will now be a distorted (foreshortened) pentagon.

Result: Front View = inclined line at 30° to XY (showing edge view of pentagon). Top View = foreshortened pentagon (apparent shape).

2Draw the projections of a triangular lamina of 30 mm sides, having one of its sides AB in the VP and with its surface inclined at 60° to the V.P.Show solution

Given:

Equilateral triangular lamina, side = 30 mm.
Side AB lies in V.P.
Surface inclined at 60° to V.P. (perpendicular to H.P. implied, as AB is in V.P.).

Concept: Two-step method (plane perpendicular to H.P., inclined to V.P.).

Step 1 (Initial Position — plane perpendicular to H.P. and V.P., i.e., parallel to V.P.):

Assume the triangular lamina is parallel to V.P. with side AB on V.P.
Front View: True shape — equilateral triangle, 30 mm sides. AB is on XY.
Top View: A horizontal line on XY (edge view, since plane is parallel to V.P.).

Step 2 (Tilt the plane at 60° to V.P.):

Redraw the Top View by tilting the edge-view line at 60° to XY.
Side AB remains on XY (in V.P.).
Project the new top view points up to get the new front view.
The front view will now be a foreshortened triangle.

Result: Top View = inclined line at 60° to XY (edge view). Front View = foreshortened triangle with AB on XY.

3A square plate with 35 mm sides is inclined at 45° to the V.P. and perpendicular to the H.P. Draw the projections of the plate if one of its corners is in the V.P. and the two sides containing that corner are equally inclined to the V.P.Show solution

Given:

Square plate, side = 35 mm.
Inclined at 45° to V.P., perpendicular to H.P.
One corner in V.P.
Two sides containing that corner are equally inclined to V.P.

Concept: Two-step method (plane perpendicular to H.P., inclined to V.P.).

Step 1 (Initial Position — plane parallel to V.P., perpendicular to H.P.):

Assume the square plate is parallel to V.P. with one corner on V.P. (on XY).
The two sides from that corner are equally inclined to V.P. → they make 45° each with V.P. in the final position, but in Step 1, the plate is parallel to V.P.
Front View: True shape — square, 35 mm sides, with one corner on XY.
Top View: A horizontal line on XY (edge view).

Step 2 (Tilt at 45° to V.P.):

Redraw the Top View: tilt the edge-view line at 45° to XY, keeping the corner on XY.
Since the two sides are equally inclined, the plate is symmetrically tilted.
Project the new top view points up to get the new front view.
Front View = foreshortened square (rhombus-like shape).

Result: Top View = line at 45° to XY. Front View = foreshortened square with one corner on XY.

4A hexagonal plate of 30 mm sides is resting on the ground on one of its sides which is parallel to the V.P. and surface of the lamina is inclined at 45° to H.P. Draw its projections.Show solution

Given:

Regular hexagonal plate, side = 30 mm.
Resting on one side on H.P. (ground).
That side is parallel to V.P.
Surface inclined at 45° to H.P.

Concept: Two-step method (plane perpendicular to V.P., inclined to H.P.).

Step 1 (Initial Position — plane parallel to H.P., perpendicular to V.P.):

Assume the hexagonal plate lies flat on H.P. with one side parallel to V.P. (parallel to XY).
Top View: True shape — regular hexagon, 30 mm sides, with one side parallel to XY and on XY.
Front View: A horizontal line on XY (edge view).

Step 2 (Tilt at 45° to H.P.):

Redraw the Front View: tilt the edge-view line at 45° to XY, keeping the resting side on XY.
Project the new front view points down to get the new top view.
Top View = foreshortened hexagon.

Result: Front View = inclined line at 45° to XY (edge view of hexagon). Top View = foreshortened hexagon with one side on XY.

5A rectangular lamina measuring 25 mm × 20 mm is parallel to and 15 mm above H.P. Draw the projections of the lamina when one of its longer edges makes an angle of 30° to V.P.Show solution

Given:

Rectangular lamina, 25 mm × 20 mm.
Parallel to H.P. and 15 mm above H.P.
One longer edge (25 mm) makes 30° to V.P.

Concept: One-step problem (plane parallel to H.P. — no tilt needed for H.P. inclination). The angle to V.P. affects the top view orientation.

Step 1: Draw reference line XY.

Step 2: Front View:

Plane is parallel to H.P. → Front View is a horizontal line (edge view).
15 mm above H.P. → draw a horizontal line 15 mm above XY.
Length of front view = 25 mm (the longer edge, projected).

Step 3: Top View:

Plane is parallel to H.P. → Top View shows true shape.
True shape = rectangle 25 mm × 20 mm.
The longer edge (25 mm) makes 30° to V.P. → draw the rectangle with its longer edge at 30° to XY.
The rectangle is 15 mm below XY (since plane is 15 mm above H.P., in First Angle, top view is below XY; but since plane is parallel to H.P., the top view is at a distance corresponding to the projection).
Centre of rectangle is directly below the front view line.

Result: Front View = horizontal line 25 mm long, 15 mm above XY. Top View = rectangle 25 mm × 20 mm with longer edge at 30° to XY, positioned below XY.

6Draw the projections of a circle of 30 mm diameter, having its plane vertical and inclined at 30° to the V.P. Its centre is 25 mm above the H.P. and 20 mm in front of V.P.Show solution

Given:

Circle, diameter = 30 mm (radius = 15 mm).
Plane is vertical (perpendicular to H.P.) and inclined at 30° to V.P.
Centre: 25 mm above H.P., 20 mm in front of V.P.

Concept: Two-step method (plane perpendicular to H.P., inclined to V.P.).

Step 1 (Initial Position — plane perpendicular to H.P. and parallel to V.P.):

Front View: True shape — circle of 30 mm diameter. Centre at 25 mm above XY.
Top View: A horizontal line (edge view) of length 30 mm, 20 mm below XY (20 mm in front of V.P.).

Step 2 (Tilt at 30° to V.P.):

Redraw the Top View: tilt the edge-view line at 30° to XY, keeping the centre at 20 mm below XY.
Divide the circle in Front View into 12 equal parts (or use key points).
Project each point from the Front View down to the new inclined top view line.
The new Top View is an ellipse (foreshortened circle).
The new Front View is obtained by projecting from the new top view back up — it remains a circle (since the plane is still vertical, the front view shape is an ellipse only if the plane is also inclined to H.P.; here the plane is vertical so the front view is still a circle of 30 mm diameter at 25 mm above XY).

Result: Front View = circle of 30 mm diameter, centre 25 mm above XY. Top View = ellipse with major axis = 30 mm and minor axis = 30 × sin30° = 15 mm, centre 20 mm below XY, inclined at 30° to XY.

Assignment — Projection of Solids (Axis Perpendicular to H.P. or V.P.)

1Project the Front View and Top View of a square prism of 35 mm base edges and 50 mm vertical height, rests on H.P., with two of its vertical rectangular faces parallel to V.P.Show solution

Given:

Square prism: base edge = 35 mm, height = 50 mm.
Resting on H.P. (axis perpendicular to H.P.).
Two vertical rectangular faces parallel to V.P.

Concept: Axis ⊥ H.P. → Start with Top View.

Step 1: Draw reference line XY.

Step 2: Top View:

Square prism resting on H.P. → Top View shows the base square.
Two faces parallel to V.P. → two sides of the square are parallel to XY.
Draw a square of 35 mm × 35 mm with sides parallel and perpendicular to XY.
Centre the square below XY at appropriate position.

Step 3: Front View:

Project up from the top view.
Front View = rectangle, width = 35 mm, height = 50 mm.
Draw the rectangle above XY: 35 mm wide, 50 mm tall.
The two visible vertical edges are the outer edges; the hidden edges (rear face) are shown as dashed lines.

Result: Top View = 35 mm square (sides parallel to XY). Front View = 35 mm × 50 mm rectangle above XY.

2A triangular prism of 40 mm base edges and 60 mm height, standing on its base on H.P. with one of its vertical rectangular faces on the rear, parallel to V.P. Draw its projections.Show solution

Given:

Triangular prism: base edge = 40 mm, height = 60 mm.
Standing on H.P. (axis ⊥ H.P.).
One vertical rectangular face at rear, parallel to V.P.

Step 1: Draw reference line XY.

Step 2: Top View:

Equilateral triangle, side = 40 mm.
One side (the rear face) is parallel to XY and at the back.
The apex of the triangle points toward the observer (toward XY).
Draw the equilateral triangle with one side parallel to XY (at the top of the triangle in top view).

Step 3: Front View:

Project up from top view.
The rear face (parallel to V.P.) appears as a rectangle: 40 mm wide, 60 mm tall.
The apex edge appears as a vertical line in front.
Front View = triangle outline: a rectangle with a vertical line from the apex.
Actually: Front View shows the rectangular face (40 mm × 60 mm) as the outline, with the apex edge as a vertical line inside (visible).

Result: Top View = equilateral triangle (40 mm side) with one side parallel to XY at rear. Front View = rectangle 40 mm × 60 mm with a vertical centre line (apex edge).

3Draw the projections of a cylinder, which rests on H.P. on its base, with 30 mm base diameter and 40 mm long axis.

Given:

Cylinder: base diameter = 30 mm, height (axis) = 40 mm.
Resting on H.P. on its base (axis ⊥ H.P.).

Step 1: Draw reference line XY.

Step 2: Top View:

Circle of diameter 30 mm (true shape of base).
Draw circle of ∅30 mm below XY.

Step 3: Front View:

Rectangle: width = 30 mm (diameter), height = 40 mm.
Draw rectangle 30 mm × 40 mm above XY.
The two vertical lines represent the extreme generators.
Top edge = top circle (shown as a line).

Result: Top View = circle ∅30 mm. Front View = rectangle 30 mm × 40 mm.

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4Project the Front View and Top View of a hemisphere which rests on H.P. with its circular face on top. (φ = 60 mm)

Given:

Hemisphere: diameter = 60 mm (radius = 30 mm).
Resting on H.P. with its flat circular face on top (curved surface on H.P.).

Note: The hemisphere rests on its curved surface on H.P., with the flat circular face facing upward.

Step 1: Draw reference line XY.

Step 2: Top View:

Looking from above, the flat circular face is on top.
Top View = circle of ∅60 mm.
Draw circle of ∅60 mm below XY.

Step 3: Front View:

The curved surface is a semicircle.
Front View = a semicircle of radius 30 mm sitting on XY (flat side up).
The flat face appears as a horizontal line at the top (30 mm above XY).
The curved outline is a semicircle from XY level up to 30 mm height.

Result: Top View = circle ∅60 mm. Front View = semicircle of radius 30 mm with flat face as a horizontal line at top.

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5Project the Front View and Top View of the frustum of a hexagonal pyramid, of 25 mm base edges and 70 mm height, cut at mid-height, parallel to its base.

Given:

Hexagonal pyramid: base edge = 25 mm, height = 70 mm.
Cut at mid-height (35 mm) parallel to base.
Frustum: base edge = 25 mm, top edge = 25/2 = 12.5 mm (by similar triangles), height = 35 mm.

Step 1: Draw reference line XY.

Step 2: Top View:

Two concentric regular hexagons.
Outer hexagon (base): side = 25 mm.
Inner hexagon (top face): side = 12.5 mm.
Both hexagons are concentric and similarly oriented.
Draw with one pair of sides parallel to XY.

Step 3: Front View:

Trapezium shape.
Bottom width = distance across flats of base hexagon (for the visible face).
For a regular hexagon with side 25 mm: across-flats = $25\sqrt{3} \approx 43.3$ mm; across-corners = 50 mm.
If one pair of sides is parallel to V.P., the front view width = 50 mm (across corners) at base and 25 mm at top.
Height = 35 mm.
Draw a trapezium: bottom = 50 mm, top = 25 mm, height = 35 mm, symmetric about centre line.
Show slant edges and hidden lines as appropriate.

Result: Top View = two concentric hexagons (25 mm and 12.5 mm sides). Front View = trapezium, 50 mm base, 25 mm top, 35 mm height.

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Assignment — Projection of Solids (Axis Perpendicular to V.P.)

1Project the Front View and Top View of a hollow cylinder (Pipe) having outer diameter = 50 mm, inner diameter = 40 mm and length = 50 mm, resting on the H.P., with its axis ⊥ to V.P.

Given:

Hollow cylinder (pipe): outer ∅ = 50 mm, inner ∅ = 40 mm, length = 50 mm.
Resting on H.P., axis ⊥ V.P.

Concept: Axis ⊥ V.P. → Start with Front View.

Step 1: Draw reference line XY.

Step 2: Front View:

Axis ⊥ V.P. → Front View shows the circular end face.
Two concentric circles: outer ∅50 mm, inner ∅40 mm.
Centre is at 25 mm above XY (resting on H.P., outer radius = 25 mm).
Draw two concentric circles, outer ∅50 mm and inner ∅40 mm, centre 25 mm above XY.

Step 3: Top View:

Axis ⊥ V.P. → Top View shows the length.
Two rectangles (outer and inner).
Outer rectangle: 50 mm (length) × 50 mm (outer diameter) — but since it rests on H.P., the top view shows the plan.
Outer rectangle: 50 mm long × 50 mm wide.
Inner rectangle (bore): 50 mm long × 40 mm wide, centred.
The cylinder rests on H.P., so the bottom of the outer rectangle is on XY.
Top view is below XY: outer rectangle 50 mm × 50 mm, inner rectangle 50 mm × 40 mm (shown as hidden lines or visible depending on position).

Result: Front View = two concentric circles (∅50 and ∅40), centre 25 mm above XY. Top View = two rectangles (50×50 outer, 50×40 inner) below XY.

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2A triangular pyramid, of 50 mm base and 50 mm axis, is resting on its base corner on the H.P., so that the upper edge of the base is horizontal. The base of the pyramid is on the rear and parallel to V.P. Draw its projections.

Given:

Triangular pyramid (tetrahedron): base edge = 50 mm, axis = 50 mm.
Resting on one base corner on H.P.
Upper edge of base is horizontal.
Base is on the rear and parallel to V.P.

Concept: Base parallel to V.P. → axis ⊥ V.P. → Start with Front View.

Step 1: Draw reference line XY.

Step 2: Front View:

Base is parallel to V.P. → Front View shows true shape of base = equilateral triangle, 50 mm side.
The pyramid rests on one corner on H.P. → one corner of the base triangle is on XY.
The upper edge of the base is horizontal → the base triangle has one vertex at bottom (on XY) and the opposite side horizontal at top.
Draw equilateral triangle: bottom vertex on XY, top side horizontal.
The apex of the pyramid is behind the base (toward V.P. side), so it projects as a point at the centroid of the triangle in front view.
Show the apex as a point (hidden) at the centroid.

Step 3: Top View:

Project down from front view.
The base (parallel to V.P.) appears as a vertical line in top view.
The apex is 50 mm behind V.P. (axis ⊥ V.P., axis = 50 mm).
Top view shows the base as a vertical line on XY and the apex as a point 50 mm above XY.

Result: Front View = equilateral triangle (50 mm) with one vertex on XY, apex shown as hidden point at centroid. Top View = vertical line (base edge view) on XY and apex point 50 mm above XY.

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3Project the Front View and Top View of a pentagonal prism of 30 mm base edges and 60 mm long edges which are ⊥ to V.P., and its rectangular face on top is parallel to H.P.

Given:

Pentagonal prism: base edge = 30 mm, length = 60 mm.
Long edges (axis) ⊥ V.P.
Rectangular face on top is parallel to H.P.

Concept: Axis ⊥ V.P. → Start with Front View.

Step 1: Draw reference line XY.

Step 2: Front View:

Axis ⊥ V.P. → Front View shows the pentagonal end face (true shape).
Regular pentagon, side = 30 mm.
One face on top parallel to H.P. → one side of the pentagon is at the top, horizontal.
The pentagon rests on H.P. → the lowest vertex or edge is on XY.
For a pentagon with top face parallel to H.P.: orient the pentagon with one side at top (horizontal) and the opposite vertex at bottom.
Draw the pentagon with one side horizontal at top and vertex at bottom on XY.

Step 3: Top View:

Shows the length of the prism.
Top View = rectangle: length = 60 mm, width = distance across the top face = 30 mm.
The top face is parallel to H.P. → it appears as a rectangle in top view.
Draw rectangle 60 mm × 30 mm below XY.
Show the other edges as hidden lines.

Result: Front View = regular pentagon (30 mm side) with one side horizontal at top, vertex at bottom on XY. Top View = rectangle 60 mm × 30 mm.

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4The frustum of a square pyramid of 40 mm base edges and 20 mm cut face (top) edges is resting on H.P. on a base edge with its 50 mm long axis horizontal and at right angles to V.P. The cut face is in front. Draw its projections.

Given:

Frustum of square pyramid: base edge = 40 mm, top edge = 20 mm, axis = 50 mm.
Resting on H.P. on a base edge (one edge of base on H.P.).
Axis horizontal and ⊥ V.P.
Cut face (top face, 20 mm square) is in front.

Concept: Axis ⊥ V.P. → Start with Front View.

Step 1: Draw reference line XY.

Step 2: Front View:

Axis ⊥ V.P. → Front View shows the end face.
Cut face (20 mm square) is in front → Front View shows the 20 mm square face.
The frustum rests on a base edge on H.P. → one edge of the base (40 mm) is on XY.
The cut face (20 mm square) is centered above, tilted because the frustum rests on an edge.
Draw the front view: the 20 mm square cut face is visible; the base (40 mm) is behind.
Since the cut face is in front and axis is ⊥ V.P., the front view shows the 20 mm square.

Step 3: Top View:

Shows the length (axis = 50 mm) and the profile.
The frustum rests on one base edge → the base edge is on XY.
Top View = trapezium: bottom = 40 mm (base edge on H.P.), top = 20 mm (cut face), length = 50 mm.
Draw trapezium with appropriate dimensions.

Result: Front View = 20 mm square (cut face). Top View = trapezium (40 mm base, 20 mm top, 50 mm long).

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5A hexagonal prism, of 25 mm base and 60 mm axis, is resting on one of its base edges on the H.P. and its axis is perpendicular to V.P. Project its Front View and Top View.

Given:

Hexagonal prism: base edge = 25 mm, axis = 60 mm.
Resting on one base edge on H.P.
Axis ⊥ V.P.

Concept: Axis ⊥ V.P. → Start with Front View.

Step 1: Draw reference line XY.

Step 2: Front View:

Axis ⊥ V.P. → Front View shows the hexagonal end face (true shape).
Regular hexagon, side = 25 mm.
Resting on one edge on H.P. → one side of the hexagon is on XY.
Draw the hexagon with one side on XY (horizontal, at bottom).

Step 3: Top View:

Shows the length of the prism.
The prism rests on one edge → the bottom edge is on XY.
Top View = rectangle: length = 60 mm, width = distance across the resting edge to the opposite edge.
For a regular hexagon with side 25 mm resting on one edge: the width in top view = across-flats = $25\sqrt{3} \approx 43.3$ mm.
Draw rectangle 60 mm × 43.3 mm below XY.
Show internal edges as hidden lines.

Result: Front View = regular hexagon (25 mm side) with one side on XY. Top View = rectangle 60 mm × 43.3 mm.

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6Project the Front View and Top View of a cylinder, with base diameter = 50 mm and height = 70 mm, resting on H.P., with its axis perpendicular to V.P.

Given:

Cylinder: base ∅ = 50 mm, height = 70 mm.
Resting on H.P., axis ⊥ V.P.

Concept: Axis ⊥ V.P. → Start with Front View.

Step 1: Draw reference line XY.

Step 2: Front View:

Axis ⊥ V.P. → Front View shows the circular end face.
Circle of ∅50 mm.
Cylinder rests on H.P. → centre is 25 mm above XY.
Draw circle ∅50 mm, centre 25 mm above XY.

Step 3: Top View:

Shows the length of the cylinder.
Rectangle: length = 70 mm, width = 50 mm (diameter).
The cylinder rests on H.P. → the bottom of the rectangle is on XY.
Draw rectangle 70 mm × 50 mm below XY.
The two long sides represent the extreme generators.

Result: Front View = circle ∅50 mm, centre 25 mm above XY. Top View = rectangle 70 mm × 50 mm.

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7Draw the Front View and Top View of a cone of base diameter = 30 mm and axis = 65 mm, with its axis perpendicular to V.P., keeping the vertex in front.

Given:

Cone: base ∅ = 30 mm, axis = 65 mm.
Axis ⊥ V.P.
Vertex in front (vertex is nearer to V.P.).

Concept: Axis ⊥ V.P. → Start with Front View.

Step 1: Draw reference line XY.

Step 2: Front View:

Axis ⊥ V.P. → Front View shows the circular base (true shape) since the base is away from observer and vertex is in front.
Wait: vertex is in front → the vertex is closer to V.P. → Front View shows the vertex as a point at the centre and the base circle behind.
Front View = circle ∅30 mm (base) with a point at centre (vertex).
The cone rests on H.P. → the base circle centre is 15 mm above XY.
Draw circle ∅30 mm, centre 15 mm above XY, with a centre point (vertex).

Step 3: Top View:

Shows the length of the cone.
Triangle: base = 30 mm (at the back), apex at front.
The cone rests on H.P. → the base is on H.P. side.
Top View = isosceles triangle: base = 30 mm (at the far end from XY), apex = point (at XY side, 65 mm from base).
Draw the triangle: apex on XY side, base 65 mm away.
Rectangle-like outline: two slant lines from apex to base edges, base line.

Result: Front View = circle ∅30 mm (centre 15 mm above XY) with centre point (vertex). Top View = isosceles triangle, base 30 mm, height 65 mm, apex toward XY.

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8A square prism, base 40 mm side and axis 70 mm long is lying on one of its rectangular faces. Its axis is perpendicular to V.P. Draw its Front View and Top View.

Given:

Square prism: base edge = 40 mm, axis = 70 mm.
Lying on one rectangular face (on H.P.).
Axis ⊥ V.P.

Concept: Axis ⊥ V.P. → Start with Front View.

Step 1: Draw reference line XY.

Step 2: Front View:

Axis ⊥ V.P. → Front View shows the square end face.
Square, side = 40 mm.
Lying on one rectangular face → one side of the square is on XY (on H.P.).
Draw square 40 mm × 40 mm with bottom side on XY.

Step 3: Top View:

Shows the length of the prism.
Rectangle: length = 70 mm, width = 40 mm.
The prism lies on one face on H.P. → the bottom face is on XY.
Draw rectangle 70 mm × 40 mm below XY.

Result: Front View = 40 mm square with bottom side on XY. Top View = rectangle 70 mm × 40 mm.

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9The frustum of a triangular pyramid of 50 mm base edge and 20 mm top edge, rests on H.P. with its base edge on it and the 60 mm long axis parallel to H.P. and at right angles to V.P. The cut face is in front. Project its Front View and Top View.

Given:

Frustum of triangular pyramid: base edge = 50 mm, top edge = 20 mm, axis = 60 mm.
Resting on H.P. with one base edge on H.P.
Axis parallel to H.P. and ⊥ V.P.
Cut face (top face, 20 mm triangle) is in front.

Concept: Axis ⊥ V.P. → Start with Front View.

Step 1: Draw reference line XY.

Step 2: Front View:

Axis ⊥ V.P. → Front View shows the end face.
Cut face (20 mm equilateral triangle) is in front → Front View shows the 20 mm triangle.
The frustum rests on one base edge on H.P. → one edge of the base is on XY.
The cut face triangle (20 mm) is oriented similarly.
Draw the 20 mm equilateral triangle as the front view, with appropriate orientation.

Step 3: Top View:

Shows the length (axis = 60 mm) and the profile.
Top View = trapezium: one end = 50 mm base edge (on H.P.), other end = 20 mm top edge, length = 60 mm.
Draw trapezium with base 50 mm, top 20 mm, length 60 mm.

Result: Front View = equilateral triangle 20 mm side (cut face). Top View = trapezium (50 mm base, 20 mm top, 60 mm long).

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10A right regular pentagonal pyramid of base edge = 25 mm and height = 60 mm, having its axis perpendicular to V.P., with its base parallel to V.P. Draw its projections.

Given:

Pentagonal pyramid: base edge = 25 mm, height = 60 mm.
Axis ⊥ V.P. (base parallel to V.P.).

Concept: Axis ⊥ V.P. → Start with Front View.

Step 1: Draw reference line XY.

Step 2: Front View:

Base parallel to V.P. → Front View shows the true shape of the base = regular pentagon, 25 mm side.
The pyramid rests on H.P. → one vertex or edge of the base is on XY.
Assume the pyramid rests on H.P. with its base on H.P. (axis horizontal, ⊥ V.P.).
The base is vertical (parallel to V.P.) and the axis is horizontal.
The pyramid rests on H.P. → the lowest point of the base pentagon is on XY.
Draw the pentagon with one vertex at the bottom on XY (or one side at bottom, depending on orientation).
The apex is 60 mm behind the base (60 mm from V.P.).
Show the apex as a point at the centroid of the pentagon (hidden, as it is behind).

Step 3: Top View:

Shows the length (axis = 60 mm) and the profile.
Top View = triangle (outline of pyramid from above): base line = width of pentagon, apex point 60 mm from base.
The base pentagon projects as its width in top view.
Draw the outline: two slant lines from apex to the extreme corners of the base, and the base line.

Result: Front View = regular pentagon (25 mm side) with apex shown as hidden point at centroid. Top View = triangular outline with base = pentagon width and height = 60 mm.

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Assignment — Projection of Solids (Axis Parallel to Both V.P. and H.P.)

1A hexagonal prism, base 25 mm side, axis 60 mm long, is lying on the ground on one of its faces with the axis parallel to both V.P. and H.P. Draw its projections.

Given:

Hexagonal prism: base edge = 25 mm, axis = 60 mm.
Lying on one rectangular face on H.P.
Axis parallel to both V.P. and H.P.

Concept: Axis parallel to both planes → axis parallel to XY. Start with Side View or use the standard approach.

Step 1: Draw reference line XY.

Step 2: Front View:

Axis parallel to V.P. → Front View shows the true length of the axis and the profile.
The prism lies on one face → one rectangular face is on H.P.
Front View = rectangle: length = 60 mm (axis), height = distance from H.P. to top face.
For a regular hexagon with side 25 mm lying on one face: height = across-flats = $25\sqrt{3} \approx 43.3$ mm.
Draw rectangle 60 mm × 43.3 mm above XY.
Show the hexagonal end outlines as visible lines on both ends.

Step 3: Top View:

Axis parallel to H.P. → Top View shows the true length.
Top View = rectangle: length = 60 mm, width = 50 mm (across corners of hexagon, since one face is on H.P., the width in top view = side of hexagon × 2 = 50 mm for the face width... actually for a regular hexagon with side 25 mm, the width across the face on H.P. = 25 mm, and the total width across = 50 mm).
Draw rectangle 60 mm × 50 mm below XY.
Show the hexagonal end outlines.

Result: Front View = rectangle 60 mm × 43.3 mm with hexagonal ends. Top View = rectangle 60 mm × 50 mm.

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2A triangular pyramid, base 25 mm side, axis 50 mm long, is resting on the ground on one of its edges of the base. Its axis is parallel to both the planes. Draw its projections.

Given:

Triangular pyramid: base edge = 25 mm, axis = 50 mm.
Resting on one base edge on H.P.
Axis parallel to both V.P. and H.P.

Step 1: Draw reference line XY.

Step 2: Front View:

Axis parallel to V.P. → Front View shows the true length and profile.
The pyramid rests on one base edge → that edge is on XY.
Front View shows the triangular profile: the base edge (25 mm) on XY, the opposite base vertex above, and the apex.
Draw the front view showing the outline of the pyramid.

Step 3: Top View:

Axis parallel to H.P. → Top View shows the true length.
Top View shows the base edge on XY and the projection of the pyramid.
Draw the top view with the base edge on XY and the apex projected below.

Result: Front View = outline of pyramid with base edge on XY, showing triangular profile. Top View = projection showing base edge on XY and apex.

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3A cylinder, base 40 mm diameter, axis 60 mm long, is lying on the ground on its generators with the axis parallel to both V.P. and H.P. Draw its projections.

Given:

Cylinder: base ∅ = 40 mm, axis = 60 mm.
Lying on its generator (on H.P.).
Axis parallel to both V.P. and H.P.

Step 1: Draw reference line XY.

Step 2: Front View:

Axis parallel to V.P. → Front View shows the true length and circular ends.
Rectangle: length = 60 mm, height = 40 mm (diameter).
The cylinder lies on H.P. → bottom of rectangle on XY.
Draw rectangle 60 mm × 40 mm above XY.
Show circles of ∅40 mm at both ends (as ellipses in the front view — but since axis is parallel to V.P., the end circles appear as circles of ∅40 mm).

Step 3: Top View:

Axis parallel to H.P. → Top View shows the true length.
Rectangle: length = 60 mm, width = 40 mm.
Draw rectangle 60 mm × 40 mm below XY.

Result: Front View = rectangle 60 mm × 40 mm with circular ends (∅40 mm). Top View = rectangle 60 mm × 40 mm.

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4The frustum of a hexagonal pyramid of 20 mm base edges and 10 mm cut face top edges is resting on H.P. on a base edge with its 50 mm long axis horizontal and parallel to V.P. The cut top face is in front. Draw its projections.

Given:

Frustum of hexagonal pyramid: base edge = 20 mm, top edge = 10 mm, axis = 50 mm.
Resting on H.P. on one base edge.
Axis horizontal and parallel to V.P.
Cut face (top face, 10 mm hexagon) is in front.

Step 1: Draw reference line XY.

Step 2: Front View:

Axis parallel to V.P. → Front View shows the true shape of the cut face (in front) and the profile.
Cut face is a regular hexagon with 10 mm sides → draw hexagon ∅10 mm side in front view.
The frustum rests on one base edge → one edge of the base is on XY.
The cut face hexagon is at the front.
Front View = hexagon (10 mm side) at the front, with the outline of the frustum behind it.

Step 3: Top View:

Axis parallel to H.P. → Top View shows the length.
Top View = trapezium-like shape: base hexagon width at back, cut face hexagon width at front, length = 50 mm.
Draw the top view showing the two hexagonal outlines and the slant edges.

Result: Front View = 10 mm hexagon (cut face) with frustum outline. Top View = trapezoidal outline showing base (20 mm hexagon) and top (10 mm hexagon) with 50 mm length.

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Assignment — Projection of Solids (Axis Inclined to H.P., Parallel to V.P.)

1A triangular prism with 25 mm edges at its base and the axis 60 mm long is resting on one of the edges of its base with axis parallel to V.P. and inclined at 30° to the H.P. Draw the projections of the prism.

Given:

Triangular prism: base edge = 25 mm, axis = 60 mm.
Resting on one base edge on H.P.
Axis parallel to V.P. and inclined at 30° to H.P.

Concept: Two-step method — axis inclined to H.P. and parallel to V.P.

Step 1 (Axis ⊥ H.P.):

Assume the prism stands with its axis perpendicular to H.P. (vertical), resting on one base edge.
Top View: Equilateral triangle, 25 mm side, with one side on XY (the resting edge).
Front View: Rectangle showing the prism profile — width = 25 mm (base), height = 60 mm.

Step 2 (Tilt axis at 30° to H.P.):

Redraw the Front View by tilting the axis at 30° to XY.
The resting edge remains on XY.
Rotate the front view so that the axis makes 30° with XY.
The new front view shows the prism inclined at 30°.
Project the new front view points down to get the new top view.
The new top view is the foreshortened projection of the prism.

Result: Step 1 Front View = rectangle 25 mm × 60 mm. Step 2 Front View = inclined rectangle at 30° to XY. Top View = foreshortened projection of the triangular prism.

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2A pentagonal pyramid of 25 mm edges of its base and axis 50 mm, has its axis perpendicular to the V.P. and 50 mm above the H.P. Draw the projections of the pyramid if one edge of its base is inclined at 30° to the H.P.

Given:

Pentagonal pyramid: base edge = 25 mm, axis = 50 mm.
Axis ⊥ V.P. and 50 mm above H.P.
One edge of base inclined at 30° to H.P.

Concept: Axis ⊥ V.P. → Start with Front View. The inclination of base edge to H.P. is achieved by rotation.

Step 1: Draw reference line XY.

Step 2: Front View (Initial):

Axis ⊥ V.P. → Front View shows the pentagonal base (true shape).
Regular pentagon, 25 mm side.
Axis is 50 mm above H.P. → centre of pentagon is 50 mm above XY.
One edge of base is inclined at 30° to H.P. → rotate the pentagon so that one edge makes 30° with XY.
Draw the pentagon with one edge at 30° to XY, centre 50 mm above XY.

Step 3: Top View:

Project down from front view.
The apex is 50 mm behind the base (axis = 50 mm, ⊥ V.P.).
Top View shows the outline of the pyramid: pentagon base projected and apex point.
Draw the top view with the pentagon outline and apex.

Result: Front View = regular pentagon (25 mm side) with one edge at 30° to XY, centre 50 mm above XY, with apex shown as hidden point. Top View = projected outline of pyramid.

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3A frustum of square pyramid of 20 mm edges at the top, 40 mm edges at the bottom and 50 mm length of the axis has its side surface (face) inclined at 45° to the H.P. with axis parallel to V.P. Draw the projections of the frustum.

Given:

Frustum of square pyramid: base edge = 40 mm, top edge = 20 mm, axis = 50 mm.
Side face inclined at 45° to H.P.
Axis parallel to V.P.

Concept: Two-step method — axis parallel to V.P., inclined to H.P.

Step 1 (Axis ⊥ H.P.):

Assume the frustum stands with axis ⊥ H.P. (vertical), base on H.P.
Top View: Two concentric squares (40 mm and 20 mm), sides parallel to XY.
Front View: Trapezium — base 40 mm, top 20 mm, height 50 mm.

Step 2 (Tilt so that side face is at 45° to H.P.):

The side face of the frustum is a trapezium.
Tilt the frustum so that one side face makes 45° with H.P.
Redraw the Front View: tilt the trapezium so that the slant face is at 45° to XY.
Project the new front view to get the new top view.

Result: Step 1 Front View = trapezium (40 mm base, 20 mm top, 50 mm height). Step 2 Front View = tilted trapezium with side face at 45° to XY. Top View = foreshortened projection.

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4The frustum of a cone, which is 90 mm base diameter and 30 mm top diameter. Draw the projections of the cone frustum when its axis is parallel to the V.P. and inclined at (i) 30° to H.P. (ii) 60° to the H.P.

Given:

Frustum of cone: base ∅ = 90 mm, top ∅ = 30 mm.
Axis parallel to V.P.
(i) Axis inclined at 30° to H.P. (ii) Axis inclined at 60° to H.P.

Note: The axis length is not given; assume a standard length or proceed with the given diameters. Let axis length = H (to be determined from the original cone geometry or taken as given in the problem — assume it is given or use a reasonable value).

Concept: Two-step method — axis parallel to V.P., inclined to H.P.

Step 1 (Axis ⊥ H.P. — common for both cases):

Assume the frustum stands with axis ⊥ H.P., base on H.P.
Top View: Two concentric circles — outer ∅90 mm (base), inner ∅30 mm (top).
Front View: Trapezium — base 90 mm, top 30 mm, height = axis length.

Step 2(i) — Axis inclined at 30° to H.P.:

Redraw the Front View by tilting the trapezium at 30° to XY.
The base (90 mm) end rests on H.P. → the base end touches XY.
Tilt the front view trapezium so that the axis makes 30° with XY.
Project the new front view points down to get the new top view.
New Top View = foreshortened elliptical outline.

Step 2(ii) — Axis inclined at 60° to H.P.:

Same procedure as above but tilt the front view trapezium at 60° to XY.
The new front view is more steeply inclined.
Project to get the new top view.

Result:

(i) Front View = trapezium inclined at 30° to XY; Top View = foreshortened projection.
(ii) Front View = trapezium inclined at 60° to XY; Top View = more foreshortened projection.

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Assignment — Projection of Solids (Axis Inclined to V.P., Parallel to H.P.)

1Draw the projections of a pentagonal prism having 25 mm edge of its base and the axis 50 mm long when it is resting on its base with an edge of its base inclined at 30° to the V.P.

Given:

Pentagonal prism: base edge = 25 mm, axis = 50 mm.
Resting on its base on H.P. (axis ⊥ H.P.).
One edge of base inclined at 30° to V.P.

Concept: Axis ⊥ H.P. → one-step problem. The inclination of base edge to V.P. affects the top view orientation.

Step 1: Draw reference line XY.

Step 2: Top View:

Axis ⊥ H.P. → Top View shows the true shape of the base = regular pentagon, 25 mm side.
One edge of base inclined at 30° to V.P. → one side of the pentagon makes 30° with XY.
Draw the regular pentagon with one side at 30° to XY.

Step 3: Front View:

Project up from top view.
Front View = outline of the prism as seen from front.
Height = 50 mm (axis).
The front view shows the visible faces and edges.
Draw the front view by projecting the pentagon vertices up to height 50 mm.

Result: Top View = regular pentagon (25 mm side) with one side at 30° to XY. Front View = projected outline of prism, 50 mm tall.

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2A triangular pyramid of 50 mm edges of the base and axis 60 mm long has one of its corners of the base touching V.P. with axis parallel to H.P. and inclined at 45° to the V.P. Draw the projections of the pyramid.

Given:

Triangular pyramid: base edge = 50 mm, axis = 60 mm.
One base corner touching V.P.
Axis parallel to H.P. and inclined at 45° to V.P.

Concept: Two-step method — axis parallel to H.P., inclined to V.P.

Step 1 (Axis ⊥ V.P.):

Assume the pyramid has its axis ⊥ V.P. (base parallel to V.P.).
Front View: True shape of base = equilateral triangle, 50 mm side. One corner on XY (touching V.P.).
Top View: Triangle outline with apex 60 mm from base.

Step 2 (Tilt axis at 45° to V.P.):

Redraw the Top View by tilting the axis at 45° to XY.
The base corner on V.P. remains on XY.
Rotate the top view so that the axis makes 45° with XY.
Project the new top view points up to get the new front view.
New Front View = foreshortened projection of the pyramid.

Result: Step 1 Front View = equilateral triangle (50 mm) with one corner on XY. Step 2 Top View = inclined at 45° to XY. New Front View = foreshortened projection.

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3The frustum of a cone of 40 mm base diameter and 20 mm cut face diameter, rests on H.P., with its axis 50 mm long, parallel to H.P. and inclined to V.P. at 30° towards right. Project the Top View and Front View.

Given:

Frustum of cone: base ∅ = 40 mm, top ∅ = 20 mm, axis = 50 mm.
Resting on H.P. (on its curved surface, axis horizontal).
Axis parallel to H.P. and inclined at 30° to V.P. (towards right).

Concept: Two-step method — axis parallel to H.P., inclined to V.P.

Step 1 (Axis ⊥ V.P.):

Assume axis ⊥ V.P. (base parallel to V.P.).
Front View: Two concentric circles — outer ∅40 mm (base), inner ∅20 mm (top face). Centre at 20 mm above XY (resting on H.P., base radius = 20 mm).
Top View: Rectangle/trapezium showing the length: base ∅40 mm at one end, top ∅20 mm at other end, length = 50 mm.

Step 2 (Tilt axis at 30° to V.P.):

Redraw the Top View by tilting the axis at 30° to XY (towards right).
The base end remains at the same position.
Rotate the top view so that the axis makes 30° with XY.
Project the new top view points up to get the new front view.
New Front View = foreshortened elliptical outline of the frustum.

Result: Step 1 Front View = concentric circles (∅40 and ∅20). Step 2 Top View = inclined at 30° to XY. New Front View = foreshortened projection of frustum.

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4A square duct is in the form of a frustum of a square pyramid. The sides of top and bottom are 90 mm and 60 mm respectively, and the length is 110 mm. It is situated in such a way that its axis is parallel to H.P. and inclined at 60° to V.P. Draw the projections of the duct, assuming the thickness of the duct sheet to be negligible.

Given:

Square duct (frustum of square pyramid): bottom side = 90 mm, top side = 60 mm, length (axis) = 110 mm.
Axis parallel to H.P. and inclined at 60° to V.P.

Concept: Two-step method — axis parallel to H.P., inclined to V.P.

Step 1 (Axis ⊥ V.P.):

Assume axis ⊥ V.P. (base parallel to V.P.).
Front View: Two concentric squares — outer 90 mm × 90 mm (bottom), inner 60 mm × 60 mm (top), both centred. The duct rests on H.P. → bottom of outer square on XY.
Top View: Trapezium — bottom 90 mm, top 60 mm, length 110 mm.

Step 2 (Tilt axis at 60° to V.P.):

Redraw the Top View by tilting the axis at 60° to XY.
Rotate the top view trapezium so that the axis makes 60° with XY.
Project the new top view points up to get the new front view.
New Front View = foreshortened projection of the duct.

Result: Step 1 Front View = two concentric squares (90 mm and 60 mm). Step 2 Top View = inclined at 60° to XY. New Front View = foreshortened projection.

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5Draw the projections of a square prism having 30 mm edge of its base and the axis 55 mm long when it is resting on its base with its axis inclined at 30° to V.P.

Given:

Square prism: base edge = 30 mm, axis = 55 mm.
Resting on its base on H.P. (axis ⊥ H.P.).
Axis inclined at 30° to V.P.

Concept: Two-step method — axis parallel to H.P. (after tilting), inclined to V.P. But since the prism rests on its base, the axis is vertical initially, then tilted.

Actually: The prism rests on its base → axis is ⊥ H.P. initially. The axis is inclined at 30° to V.P. means the base is rotated so that the axis (when projected on H.P.) makes 30° with V.P.

Step 1 (Axis ⊥ H.P., base on H.P.):

Top View: Square, 30 mm side, with sides parallel to XY.
Front View: Rectangle, 30 mm × 55 mm.

Step 2 (Rotate base so axis makes 30° to V.P.):

Rotate the top view square so that the axis direction makes 30° with XY.
The top view square is rotated by 30°.
Project the new top view up to get the new front view.
New Front View = outline of the prism as seen from front after rotation.

Result: Top View = square (30 mm) rotated so that its diagonal or side makes appropriate angle. Front View = projected outline of prism, 55 mm tall.

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6Draw the projections of a hexagonal prism having 20 mm edge of its base and the axis 50 mm long when it is resting on its base, with its axis parallel to H.P. and inclined at 40° to the V.P.

Given:

Hexagonal prism: base edge = 20 mm, axis = 50 mm.
Resting on its base on H.P. — but axis is parallel to H.P. → the prism is lying on its side.
Axis parallel to H.P. and inclined at 40° to V.P.

Concept: Two-step method — axis parallel to H.P., inclined to V.P.

Step 1 (Axis ⊥ V.P.):

Assume axis ⊥ V.P. (base parallel to V.P.).
Front View: Regular hexagon, 20 mm side. One face on H.P. → one side of hexagon on XY.
Top View: Rectangle — length = 50 mm, width = across-flats of hexagon = $20\sqrt{3} \approx 34.6$ mm.

Step 2 (Tilt axis at 40° to V.P.):

Redraw the Top View by tilting the axis at 40° to XY.
Rotate the top view rectangle so that the axis makes 40° with XY.
Project the new top view points up to get the new front view.
New Front View = foreshortened projection of the hexagonal prism.

Result: Step 1 Front View = hexagon (20 mm side) with one side on XY. Step 2 Top View = rectangle inclined at 40° to XY. New Front View = foreshortened projection.

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7A triangular prism of 50 mm base edges of the base and axis 60 mm long, resting on its base and its axis inclined at 45° to the V.P. Draw the projections of the prism.

Given:

Triangular prism: base edge = 50 mm, axis = 60 mm.
Resting on its base on H.P. (axis ⊥ H.P. initially).
Axis inclined at 45° to V.P.

Concept: Two-step method — axis parallel to H.P. (after tilting), inclined to V.P.

Step 1 (Axis ⊥ H.P.):

Top View: Equilateral triangle, 50 mm side, with one side parallel to XY.
Front View: Rectangle — width = 50 mm, height = 60 mm (showing the rectangular face).

Step 2 (Rotate so axis makes 45° to V.P.):

Rotate the top view triangle so that the axis direction makes 45° with XY.
The top view triangle is rotated by 45°.
Project the new top view up to get the new front view.
New Front View = outline of the prism as seen from front.

Result: Top View = equilateral triangle (50 mm) rotated at 45° to XY. Front View = projected outline of prism, 60 mm tall.

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8A hexagonal pyramid of 25 mm edges of the base and 60 mm long axis, resting on its base, has its axis inclined to V.P. at 30°. Draw its projections.

Given:

Hexagonal pyramid: base edge = 25 mm, axis = 60 mm.
Resting on its base on H.P. (axis ⊥ H.P. initially).
Axis inclined at 30° to V.P.

Concept: Two-step method — axis ⊥ H.P. initially, then rotated so axis makes 30° to V.P.

Step 1 (Axis ⊥ H.P.):

Top View: Regular hexagon, 25 mm side, with one pair of sides parallel to XY.
Front View: Isosceles triangle — base = across-flats of hexagon = $25\sqrt{3} \approx 43.3$ mm (or across corners = 50 mm depending on orientation), height = 60 mm.

Step 2 (Rotate so axis makes 30° to V.P.):

Rotate the top view hexagon so that the axis direction (from centre to apex projection) makes 30° with XY.
The top view hexagon is rotated by 30°.
Project the new top view up to get the new front view.
New Front View = outline of the hexagonal pyramid as seen from front after rotation.

Result: Top View = regular hexagon (25 mm side) rotated so that the axis makes 30° to XY. Front View = projected outline of hexagonal pyramid, 60 mm tall.

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