Orthographic Projections of Simple Machine Blocks

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:

1. 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.

2. 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.

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.

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*.

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*.

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.

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.

*(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.

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).

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

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).

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

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

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

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

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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).

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.

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.

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.

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.

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.

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.

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).