EE-107 Engineering Drawing Mar-Apr 2025 paper solved: involute of a pentagon, hexagonal pyramid and true shape of a prism section

This post solves the AP SBTET C-23 first year engineering drawing question paper 2025 (EE-107, March/April 2025 board exam). It answers all 4 Part A questions and all 6 Part B questions. Each question starts with the idea in simple words, then the exam answer: numbered steps of construction and the finished drawing, drawn to scale from computed points. You can open the original question paper PDF next to this page and check every question and figure.

Engineering Drawing question paper 2025: pattern and syllabus chapters

The EE-107 Engineering Drawing paper of March 2025 is a common first year paper for many branches (its code line reads C23-AEI-BM-CH-CHST-CCN–EE-EEVT-TT–107). It is 3 hours long and carries 60 marks. Part A has 4 questions of 5 marks each (20 marks, all compulsory). Part B has 6 questions of 10 marks each, and you answer any four (40 marks).

The paper does not print marks beside the questions; they come from the instructions. The questions follow the chapter blueprint of the C-23 curriculum exactly: 2, 1, 1, 0 and 0 short questions and 0, 1, 2, 1 and 2 essay questions from Chapters 1 to 5.

Part A instructions as printed on the paper (PART—A, 5×4=20):

Instructions : (1) Answer all questions.

(2) Each question carries five marks.

(3) All the dimensions are in mm.

Part B instructions as printed on the paper (PART—B, 10×4=40):

Instructions : (1) Answer any four questions.

(2) Each question carries ten marks.

(3) All the dimensions are in mm.

Q no	Topic	Syllabus chapter	Marks
1	Single stroke vertical letters, 10 mm	Ch 1 Drawing instruments, lettering and dimensioning	5
2	Redraw a plate as per BIS	Ch 1	5
3	Hexagon of side 30 by the general method	Ch 2 Geometric constructions	5
4	Line parallel to both HP and VP	Ch 3 Projections of points, lines, planes and solids	5
5	Involute of a pentagon of side 25	Ch 2	10
6	Circular plane inclined to VP	Ch 3	10
7	Hexagonal pyramid resting on HP	Ch 3	10
8	Section of a square prism and true shape	Ch 4 Sectional views	10
9	Orthographic views of a stepped block	Ch 5 Orthographic projection	10
10	Orthographic views of a T-shaped block	Ch 5	10

The paper has a few printing slips. Q7 reads “side of base” with a word missing, and the Q2 figure uses old-style “10 φ” and “20 R”. The questions below copy them exactly. None of them changes what you draw.

How this post draws. Every figure is drawn to scale from computed points, and each caption gives the drawing scale. Scale 1:1 (full size) means 1 mm on paper shows 1 mm on the object; scale 1:2 would mean 1 cm on paper shows 2 cm on the object. Every answer drawing in this paper fits an A3 sheet at 1:1. In the drawings, thick lines are visible edges. Thin lines are construction lines and projectors. Dashed lines are hidden edges, and chain lines (long dash, dot) are centre lines and cutting planes.

My advice: in Part B, pick Q5, Q7, Q9 and Q10. The involute and the pyramid have short fixed steps, so every step earns marks. Q9 and Q10 are plain blocks with no hidden lines in the side views. Take Q6 if you can plot an ellipse by points. Leave Q8 unless you have practised auxiliary views, because it asks for three drawings in one answer.

You can see every Engineering Drawing session on the Engineering Drawing subject page.

Q1. Print the following in single stroke capital vertical letters of 10 mm size : (5 marks)

Question (as printed):

1. Print the following in single stroke capital vertical letters of 10 mm size : “GOVERNMENT OF ANDHRA PRADESH”

Understand it first

1. Lettering means writing titles, notes and numbers on a drawing by hand, in a fixed neat style.
2. Single stroke means each part of a letter is drawn with one stroke of the pencil. The line is as thick as the pencil point. You do not draw a hollow outline and fill it.
3. Capital means upper-case letters only. Vertical means the letters stand straight up, at 90° to the guide line. (Sloped letters lean to the right at 75°.)
4. 10 mm size means the height h of every capital letter is 10 mm.
5. Guide lines are two thin, light lines drawn 10 mm apart. Every letter sits on the lower line and touches the upper line. They keep all letters the same height.
6. The BIS code SP 46 gives the proportions of “type B” lettering as fractions of h. For h = 10 mm these are:
   • line width d = h/10 = 1 mm,
   • gap between letters a = 2h/10 = 2 mm,
   • gap between words e = 6h/10 = 6 mm,
   • pitch between two lines of capital text b = 13h/10 = 13 mm.
7. The width of a letter is not one fixed number. Most letters here are about 6 mm wide. M is wider (about 8 mm), and E and F are a little narrower (about 5 mm).

It works like a ruled notebook: the lines keep every letter the same size, even when you write fast. The same topic came in the EE-107 October 2024 paper with a different title.

Figure Q1a: lettering proportions for h = 10 mm (type B): height 10, line width 1, letter gap 2, word gap 6.

Exam answer (5 marks)

Steps:

1. Guide lines: draw two thin horizontal guide lines 10 mm apart with a sharp 2H pencil, long enough for the whole title.
2. Spacing: plan the title: about 6 mm per letter (8 mm for M), a 2 mm gap between letters and a 6 mm gap between words. The title then comes to about 210 mm, which fits on an A3 sheet in one line.
3. Letters: draw each capital letter upright (90° to the guide line), touching both guide lines, with single strokes. Draw the round letters G, O, D, R, P and S with smooth curves.
4. Line width: go over every letter with an HB pencil so that all strokes have the same thickness, about 1 mm (h/10).
5. Finish: check that the title reads “GOVERNMENT OF ANDHRA PRADESH” in one line, with uniform height, width and spacing, and leave the guide lines faint.

Figure Q1b: the title in single stroke capital vertical letters, h = 10 mm (drawn to scale; draw it full size (scale 1:1) on your sheet, all dimensions in mm).

Key points to write: guide lines 10 mm apart; capital, upright, single stroke; uniform 1 mm strokes; 2 mm between letters, 6 mm between words.

Q2. Redraw the following figure in full scale according to Bureau of Indian Standards : (5 marks)

Question (as printed): 2. Redraw the following figure in full scale according to Bureau of Indian Standards :

[Figure: a plate 90 long with a 20 R corner arc at the top left and a 10 R corner arc at the bottom left. Three holes marked 10 φ (one leader to all three) sit on a centre line 15 above the bottom edge. The right edge has three notches 10 deep, with the chain 5, 5, 10, 10, 10, 5, 5 printed beside it.]

Understand it first

1. Full scale means 1 : 1. A line that is 90 mm on the object is 90 mm on your sheet.
2. BIS is the Bureau of Indian Standards. Its code SP 46 sets the rules for lines, lettering and dimensions on school and college drawings.
3. Dimensioning means writing the sizes on a drawing with dimension lines, extension lines, arrowheads and numbers.
4. An extension line is a thin line that comes out from the object. A dimension line is a thin line with a filled arrowhead at each end. The number sits above it, and you read it from the bottom or from the right side of the sheet.
5. φ means diameter and R means radius. BIS writes the symbol before the number: φ10, R20. The paper’s figure writes them after the number (10 φ, 20 R), which is the old style.
6. A leader is a thin line from a note to the feature. It ends in an arrowhead on the edge of a hole or arc.
7. BIS also asks for a thick outline, thin chain lines for centre lines, and each size shown only once.

Think of a recipe card. It is easy to read only when every card writes the quantity and the unit in the same order.

Figure Q2a: old style (symbol after the number) and BIS style (symbol before the number).

Reading the figure. The chain on the right edge adds up to 5 + 5 + 10 + 10 + 10 + 5 + 5 = 50, so the plate is 90 × 50. The notches are 10 mm deep and sit at 5 to 10, 20 to 30 and 40 to 45 mm from the bottom. The left edge is straight between the two corner arcs. The spacing of the three holes is not printed. Scaled from the printed figure, their centres sit 20, 40 and 60 mm from the left edge, 20 mm apart.

Figure Q2b: given figure (redrawn) with the dimensions as printed; the hole spacing is not printed.

Exam answer (5 marks)

Steps:

1. Outline: draw the plate full size: bottom edge 90, right edge 50 high, top-left corner arc R20 (centre 20 from the left and 30 up) and bottom-left corner arc R10 (centre 10 from the left and 10 up). Make the outline thick.
2. Notches: mark the right edge 5, 5, 10, 10, 10, 5, 5 from the bottom and cut three notches 10 deep (at 5 to 10, 20 to 30 and 40 to 45).
3. Holes: draw a thin chain centre line 15 above the bottom edge and three φ10 holes on it, 20, 40 and 60 from the left edge (spacing scaled), each with its own cross centre lines.
4. Dimensions: with thin dimension and extension lines and filled arrowheads, write 90 (overall), 10 (notch depth), the chain 5, 5, 10, 10, 10, 5, 5, the height 15 and the hole spacing 20, 20, 20. Put every number above its line, reading from the bottom or the right.
5. Symbols and finish: write R20 and R10 on radial lines with the arrowhead on the arc, and “3 HOLES φ10” on one leader that touches a hole. Check that no size is given twice.

Figure Q2c: answer dimensioned as per BIS (drawn to scale; draw it full size (scale 1:1) on your sheet, all dimensions in mm).

Key points to write: full scale 1 : 1; φ and R before the number; thin dimension lines with filled arrowheads; numbers above the line; thick outline, chain centre lines; hole spacing read as 20 mm from the figure.

Q3. Construct a hexagon of side length 30 mm by general method. (5 marks)

Understand it first

1. A regular hexagon has 6 equal sides and 6 equal angles.
2. The exterior angle of any regular polygon is 360° ÷ n, where n is the number of sides. For a hexagon it is 360° ÷ 6 = 60°.
3. The interior angle is 180° minus the exterior angle: 180° − 60° = 120°.
4. The general method works for any number of sides. You draw a semicircle with radius equal to the side and divide it into n equal parts. Each part is 180° ÷ n.
5. Two parts make 2 × (180° ÷ n) = 360° ÷ n, which is the exterior angle. So the line to division 2 always makes the interior angle with the first side. That is why the rule is “always join to point 2”.
6. The other corners come from rays through divisions 3, 4 and 5 and arcs of the side length.

It is like walking round a six-sided park. At every corner you turn by the same 60°, and after 6 turns you are back where you started.

Figure Q3a: why the general method works: two of the six equal parts give the 60° exterior angle.

Exam answer (5 marks)

Steps:

1. Side AB: draw AB = 30 mm and extend BA beyond A to P.
2. Semicircle: with A as centre and radius 30, draw a semicircle from P to B. Divide it into 6 equal parts (30° each) and number them 1 to 5 from P.
3. Side AF: join A to point 2. A–2 is the side AF (30 long, at 120° to AB).
4. Rays: join A to points 3, 4 and 5 and extend these rays.
5. Corners: with B as centre and radius 30, cut ray A–5 at C. With C as centre, cut ray A–4 at D. With F as centre, cut ray A–3 at E. Join B, C, D, E and F to complete the hexagon ABCDEF.

Figure Q3b: hexagon ABCDEF of side 30 by the general method (drawn to scale; draw it full size (scale 1:1) on your sheet, all dimensions in mm).

Answer: a regular hexagon ABCDEF with all sides 30 mm and all angles 120°; across the flats it measures 51.96 mm (a check, not a step).

Q4. Draw the projections of a straight line 75 mm long when it is parallel to both HP and VP. (5 marks)

Question (as printed): 4. Draw the projections of a straight line 75 mm long when it is parallel to both HP and VP. It is 25 mm in front of VP and 50 mm above HP.

Understand it first

1. H.P. is the horizontal plane, like the floor. V.P. is the vertical plane, like a wall behind the object.
2. The two planes meet in a line called the XY line (or reference line).
3. In first angle projection, the object sits in front of the VP and above the HP. The object is between you and the plane.
4. The front view is drawn on the VP. Its points get a dash: a’, b’. The top view is drawn on the HP. Its points have no dash: a, b.
5. After drawing, the HP is turned down by 90° about XY. So on paper, the front view is above XY and the top view is below it.
6. A line parallel to a plane shows its true length (real length) on that plane.
7. This line is parallel to both planes. So both views show the true length, 75 mm, and both are parallel to XY.
8. “50 mm above HP” sets the height of the front view above XY. “25 mm in front of VP” sets the distance of the top view below XY.

Hold a pencil level, 25 cm in front of a wall and 50 cm above the floor, pointing left to right. From the front and from above, you see its full length.

Figure Q4a: the VP and HP, the XY line, and a line parallel to both planes in the first quadrant.

Exam answer (5 marks)

Steps:

1. XY line: draw the reference line XY.
2. Front view: draw a’b’ = 75 mm parallel to XY and 50 mm above it.
3. Top view: draw ab = 75 mm parallel to XY and 25 mm below it.
4. Projectors: join a’ to a and b’ to b with thin vertical projectors. They cross XY at 90°.
5. Finish: make a’b’ and ab thick, label a’, b’, a, b, the heights 50 and 25, and write “both views are true length (75 mm)”.

Figure Q4b: front view 50 above XY and top view 25 below XY, both true length (drawn to scale; draw it full size (scale 1:1) on your sheet, all dimensions in mm).

Answer: a’b’ = 75 mm, parallel to XY and 50 mm above it; ab = 75 mm, parallel to XY and 25 mm below it.

Q5. Draw an involute on a pentagon of side 25 mm. (10 marks)

Understand it first

1. An involute is the path traced by the end of a tight string as you unwind it from a shape.
2. Here the shape is a pentagon: 5 equal sides of 25 mm. Its interior angle is 108° and its exterior angle is 72°.
3. The string is as long as the perimeter of the pentagon: 5 × 25 = 125 mm.
4. While the string leaves one side, its end swings round one corner. So that part of the curve is a circular arc with the corner as centre.
5. After each corner, one more side of string is free. So the radius grows by 25 every time: 25, 50, 75, 100 and 125 mm.
6. Each arc turns through the exterior angle, 72°. Five arcs make 5 × 72° = 360°, one full turn.
7. The last point lies on the first side produced, 125 mm from the start. That is a built-in check.

Wind a thread round a pentagon-shaped pencil box. Keep it tight and unwind it. The end of the thread draws a spiral that opens wider at every corner.

Figure Q5a: the free part of the string swings about one corner at a time, so the involute is made of arcs.

Exam answer (10 marks)

Steps of construction:

1. Pentagon: draw a regular pentagon of side 25 mm with corners numbered 1 to 5 (general method, or 108° angles), with side 1–2 horizontal at the bottom.
2. Start point: name corner 1 as P, the starting point of the string.
3. Produce the sides: extend the sides 3–2, 4–3, 5–4, 1–5 and 2–1 beyond the second corner of each, as thin lines.
4. Arc 1: with 2 as centre and radius 2–P = 25 mm, draw an arc from P to cut side 3–2 produced at P₁.
5. Arc 2: with 3 as centre and radius 3–P₁ = 50 mm, draw an arc from P₁ to cut side 4–3 produced at P₂.
6. Arc 3: with 4 as centre and radius 4–P₂ = 75 mm, draw an arc to cut side 5–4 produced at P₃.
7. Arc 4: with 5 as centre and radius 5–P₃ = 100 mm, draw an arc to cut side 1–5 produced at P₄.
8. Arc 5: with 1 as centre and radius 1–P₄ = 125 mm, draw an arc to cut side 2–1 produced at P₅.
9. Check: P₅ must lie on side 2–1 produced, 125 mm from corner 1 (the perimeter, 5 × 25).
10. Finish: go over the arcs P–P₁–P₂–P₃–P₄–P₅ as one smooth thick curve, keep the produced sides thin, and label the corners and P₁ to P₅.

Figure Q5b: the involute of a pentagon of side 25: five arcs of radius 25 to 125 (drawn to scale; draw it full size (scale 1:1) on your sheet, all dimensions in mm).

Answer: the involute is five arcs of radius 25, 50, 75, 100 and 125 mm with centres at corners 2, 3, 4, 5 and 1, ending at P₅, 125 mm from P.

Exam tip: each arc must end exactly on the next produced side. If an arc ends early or late, the next radius will be wrong and the curve will kink.

Q6. A circular plane of diameter 60 mm is touching the VP with a point on its circumference. (10 marks)

Question (as printed): 6. A circular plane of diameter 60 mm is touching the VP with a point on its circumference. The plane is inclined at 45° to VP and perpendicular to HP. The centre of plane is 40 mm above HP, draw the projections.

Understand it first

1. A circular plane is a thin flat disc, like a coin, here 60 mm across.
2. When the disc is parallel to VP, its front view is a true circle and its top view is a straight line.
3. Perpendicular to HP means the disc stands upright. Its top view is always a straight line, 60 mm long.
4. Inclined at 45° to VP means that top view line makes 45° with XY.
5. Touching the VP with a point on its circumference means one point of the rim, p, is on the wall. With the disc upright, this is an end of the horizontal diameter. So p is on XY in the top view.
6. We draw in two stages. Stage 1: the disc lies in VP, so the front view is a true circle with centre 40 above XY. Stage 2: turn the top view about p until it makes 45° with XY.
7. Turning about a vertical axis does not change any height. So every point keeps its stage 1 height in the new front view.
8. The front view becomes an ellipse. The vertical diameter stays 60 (true length). The horizontal diameter shrinks to 60 × cos 45° = 42.43 mm.

Stand a coin on a table with its face towards you, then turn it 45° about its left edge. From the front the coin now looks like a tall oval.

Exam answer (10 marks)

Steps of construction:

1. XY line: draw XY.
2. Stage 1, front view: draw a circle of 60 mm diameter with centre 40 mm above XY. Divide it into 12 equal parts and number them 0’ to 11’ (6’ is the left end, p’).
3. Stage 1, top view: the disc lies in VP, so its top view is a line 60 mm long on XY. Project the 12 points down to it. Name the left end p.
4. Stage 2, top view: from p on XY draw a line 60 mm long at 45° to XY (in front of VP). Mark the 12 points on it at the same distances from p as in stage 1.
5. Projectors: from each stage 2 point draw a vertical projector upwards.
6. Heights: from each stage 1 front view point draw a horizontal line to the right. The heights do not change.
7. Front view points: where each projector meets the horizontal line of the same number, mark the point 0’ to 11’.
8. Ellipse: join the 12 points with a smooth curve. It is 60 mm high and 42.43 mm wide, with p’ at its left end.
9. Check: the top and bottom of the ellipse are 70 mm and 10 mm above XY (40 ± 30), and its centre is 40 mm above XY.
10. Finish: make the final top view line and the ellipse thick, keep stage 1 and the projectors thin, and label p, p’ and the 45° angle.

Figure Q6: stage 1 (disc in VP) and stage 2 (turned 45° about p); the front view becomes an ellipse (drawn to scale; draw it full size (scale 1:1) on your sheet, all dimensions in mm).

Answer: top view: a line 60 mm long at 45° to XY from p on XY; front view: an ellipse 60 mm high and 42.43 mm wide with its centre 40 mm above XY.

Q7. A hexagonal pyramid side of base 30 mm and height 60 mm rests with its base on HP such that one of the edges of the base is parallel to and 10 mm in front of VP. (10 marks)

Question (as printed): 7. A hexagonal pyramid side of base 30 mm and height 60 mm rests with its base on HP such that one of the edges of the base is parallel to and 10 mm in front of VP. Draw its projections.

“Side of base” has a word missing: it means a pyramid with a base side of 30 mm.

Understand it first

1. A hexagonal pyramid has a regular hexagon as its base and one point on top, the apex.
2. The axis joins the centre of the base to the apex. Its length is the height, here 60 mm.
3. The lines from the apex to the six base corners are the slant edges.
4. Base on HP means the pyramid stands on the floor. The axis is vertical. So the top view shows the true shape of the hexagon, and the front view is a triangle standing on XY.
5. “One edge of the base parallel to and 10 mm in front of VP” means the back edge (ef) of the hexagon is parallel to XY and 10 mm below it in the top view.
6. A hexagon of side 30 is 51.96 mm across the flats (30 × √3) and 60 mm across the corners. So the centre is 10 + 25.98 = 35.98 mm in front of VP.
7. In the front view the back corners hide behind the front corners. So b’ and f’ fall on one point, and so do c’ and e’.

Think of a six-sided tent seen from the front. You see a triangle. Seen from above, you see the six-sided floor with the poles meeting in the middle.

Figure Q7a: parts of a hexagonal pyramid: base, apex, axis and slant edges (hidden edges dashed).

Exam answer (10 marks)

Steps of construction:

1. XY line: draw XY.
2. Back edge: below XY draw the edge ef, 30 mm long, parallel to XY and 10 mm below it.
3. Hexagon: on ef complete a regular hexagon of side 30 (angles 120°), going towards the front. Name the corners a (left), b, c (front), d (right), e and f (back).
4. Centre: mark the centre o, 35.98 mm below XY, where the long diagonals ad, be and cf cross.
5. Slant edges in top view: join o to all six corners. These lines are the slant edges seen from above.
6. Projectors: draw vertical projectors up from a, b, c, d and o to XY.
7. Base in front view: the base is on HP, so its front view is a line on XY from a’ to d’, 60 mm long. b’(f’) and c’(e’) are on it.
8. Apex: on the projector from o, mark o’ 60 mm above XY.
9. Slant edges in front view: join o’ to a’, b’(f’), c’(e’) and d’. All edges are visible, because the back edges lie exactly behind the front ones.
10. Finish: make the outlines thick, keep projectors thin, label every corner, and write the 10 mm and 60 mm sizes.

Figure Q7b: top view (true hexagon) and front view (triangle) of the pyramid (drawn to scale; draw it full size (scale 1:1) on your sheet, all dimensions in mm).

Answer: top view: a regular hexagon of side 30 with ef 10 mm below XY and the slant edges meeting at o; front view: a triangle 60 mm wide and 60 mm high with apex o’.

Q8. A square prism of base side 30 mm and height 60 mm is resting on HP on one of its bases, with a base side inclined at 30° to VP. (10 marks)

Question (as printed): 8. A square prism of base side 30 mm and height 60 mm is resting on HP on one of its bases, with a base side inclined at 30° to VP. It is cut by a plane inclined at 45° to HP and perpendicular to VP and is bisecting the axis of the prism. Draw its front view, sectional top view and the true shape of section.

Understand it first

1. A square prism has two square ends and four rectangular side faces. Here the square side is 30 mm and the height is 60 mm.
2. Resting on HP on its base means it stands upright. Its top view is a true square. “A base side inclined at 30° to VP” means one side of that square makes 30° with XY.
3. A section plane (cutting plane) is an imaginary knife. Here it is at 90° to VP and at 45° to HP. So in the front view it shows as a straight line at 45°, called the VT (vertical trace).
4. Bisecting the axis means the plane passes through the midpoint of the axis, 30 mm above the base.
5. The VT cuts the four vertical edges. These four points, 1’, 2’, 3’ and 4’, are the corners of the cut face.
6. The sectional top view is the top view of the part left after removing the part above the plane. The cut face is shown with hatching: thin parallel lines at 45°, as the BIS code recommends. Here the cut points lie on the vertical edges, so the hatched area is the whole square.
7. The cut face is tilted, so the top view does not show its real size. That view is the apparent shape. The true shape is seen only from a direction square to the cutting plane, on an auxiliary plane X₁Y₁ drawn parallel to the VT.
8. On X₁Y₁, each point is as far from X₁Y₁ as the same point is from XY in the top view.

Slice a tall block of paneer with the knife held at 45°. From above, the cut looks like the square top. Tilt the slice to face you, and you see that it is longer.

The October 2024 paper also cut a square prism, but with a plane parallel to VP, where the front view already shows the true shape.

Exam answer (10 marks)

The paper does not give the distance from VP, so this answer keeps the axis 30 mm in front of VP and lets the plane rise to the left.

Steps of construction:

1. Top view: draw XY. Below it draw the square abcd of side 30 with one side at 30° to XY; a is the back corner (nearest XY).
2. Front view: project the corners up and draw the front view: a rectangle 40.98 mm wide and 60 mm high. The edge a’ is behind the prism, so draw it dashed.
3. Cutting plane: draw the VT at 45° to XY through the midpoint of the axis, 30 mm above XY.
4. Section points: mark 1’, 2’, 3’, 4’ where the VT cuts the vertical edges b’, a’, c’, d’. They are 50.49, 35.49, 24.51 and 9.51 mm above XY.
5. Cut prism: draw the part below the VT thick. Show the removed upper part in thin lines.
6. Sectional top view: the section points lie on the vertical edges, so their top views are the corners b, a, c, d. Hatch the whole square at 45° with thin lines about 3 mm apart.
7. Auxiliary line: draw X₁Y₁ parallel to the VT, a short distance away from it.
8. Projectors: from 1’, 2’, 3’, 4’ draw projectors at 90° to X₁Y₁.
9. True shape: on each projector mark the distance from X₁Y₁ equal to the distance of b, a, c, d from XY in the top view (35.49, 9.51, 50.49 and 24.51 mm). Join 1–3–4–2 and hatch it.
10. Finish: label FRONT VIEW, SECTIONAL TOP VIEW and TRUE SHAPE OF SECTION, VT, X₁Y₁ and all points.

Figure Q8: front view with the VT, sectional top view (hatched square) and true shape of the section on X₁Y₁ (drawn to scale; draw it full size (scale 1:1) on your sheet, all dimensions in mm).

Check your drawing: the true shape is a parallelogram with sides 39.69 mm and 33.54 mm, and diagonals 58.99 mm and 43.82 mm.

Answer: front view with VT at 45° through the axis midpoint; sectional top view: the square abcd fully hatched; true shape: a parallelogram 1–3–4–2 with sides 39.69 and 33.54 mm.

Exam tip: the distances on X₁Y₁ come from the top view (distance from XY), not from the front view. Mixing them up gives a wrong true shape.

Q9. Draw front view, top view and left side view for the figure shown below : (10 marks)

Question (as printed): 9. Draw front view, top view and left side view for the figure shown below :

[Figure: a pictorial (isometric) view of a stepped block with one arrow X. Printed dimensions: 130 (length), 90 (width, split 30, 30, 30), 45 (height, split 15, 15, 15), 30 (end block length) and 45 (end block height).]

Understand it first

1. Orthographic projection shows a solid object with flat views, each seen straight on from one side.
2. The front view is seen in the direction of the arrow X, here looking at the 130 mm long face. The top view is seen from above. The left side view is seen from the left end.
3. In first angle projection the top view goes directly below the front view, and the left side view goes to the right of the front view.
4. In the top view, the front of the object is at the bottom (farthest from XY). In the left side view, the front of the object is on the right (farthest from the front view).
5. Reading this figure: the block is 130 long, 90 wide and 45 high. On the left 100 mm there are three steps, each 30 deep: 15 high at the front, 30 in the middle and 45 at the back. The right end is a full-height block, 30 long and 45 high, across the whole width.
6. Only the arrow X is printed. The top and left views are taken in the usual directions.
7. Seen from the left, the steps are in front of the taller end block. Nothing hides behind material, so this object has no hidden lines.

Picture a staircase of three steps with a tall wall at its right end. From the front you see the three step faces one above another. From the left you see the step outline.

Figure Q9a: given figure (redrawn) from the paper’s dimensions, with the arrow X (isometric view, dimensions in mm).

Exam answer (10 marks)

Steps of construction:

1. Layout: draw XY. Leave room for the front view above it, the top view below it and the left side view to the right of the front view.
2. Front view outline (X): draw a rectangle 130 wide and 45 high.
3. Step lines in front view: draw horizontal lines at 15 and 30 mm height, from the left end to 100 mm.
4. End block in front view: draw a vertical line at 100 mm from the left, from 15 up to 45 mm.
5. Top view outline: below XY draw a rectangle 130 × 90. Its back edge is nearest XY and its front edge is at the bottom.
6. Step lines in top view: draw lines 30 and 60 mm from the front edge, from the left end to 100 mm.
7. End block in top view: draw a line at 100 mm from the left, from the front edge to the line 60 from the front.
8. Left side view outline: to the right of the front view draw a rectangle 90 wide and 45 high, back on the left and front on the right.
9. Steps in side view: draw the step outline: 45 high for the back 30 mm, 30 high for the middle 30 mm and 15 high for the front 30 mm.
10. Finish: draw thin projectors between the views, add the main dimensions (130, 90, 45, 30, 15), and label FRONT VIEW, TOP VIEW and LEFT SIDE VIEW.

Figure Q9b: front view, top view and left side view in first angle projection (drawn to scale; draw it full size (scale 1:1) on your sheet, all dimensions in mm).

Answer: three views as in Figure Q9b: front view 130 × 45 with two step lines; top view 130 × 90 with two step lines and the end block line; left side view 90 × 45 with three steps; no hidden lines.

Q10. Draw front view, top view and left side view for the figure shown below : (10 marks)

Question (as printed): 10. Draw front view, top view and left side view for the figure shown below :

[Figure: a pictorial (isometric) view of a T-shaped block on a base with one arrow X. Printed dimensions: base 60, 40 and 20 (height); 20, 20, 20 along the front edge of the base; 40 (stem height); top block 40 long, 20 wide and 20 high.]

Understand it first

1. The arrow X looks at the 60 mm long face of the base. That gives the front view.
2. Reading this figure: a base 60 × 40 and 20 high. On it stands a stem 20 × 20 and 40 high, in the middle of the 60 mm length (20, 20, 20). On the stem sits a top block 40 long, 20 wide and 20 high. So the object looks like a letter T on a slab.
3. The total height is 20 + 40 + 20 = 80 mm.
4. Not printed: where the T stands across the 40 mm width, and how far the top block overhangs the stem. Scaled from the figure, the T is centred (10 mm from each side of the base) and the top block overhangs 10 mm on each side of the stem. This answer uses those values.
5. Seen from above, the top block covers the stem completely. So the stem’s edges in the top view are hidden and drawn dashed.
6. Seen from the left, the top block and the stem have the same 20 mm width. They show as one tall rectangle, with a line at 60 mm where the top block’s underside starts.

Think of a T-shaped signboard fixed on a concrete block. From the front you see the T. From above you see only the top board and the block around it.

Figure Q10a: given figure (redrawn) from the paper’s dimensions (isometric view, dimensions in mm); the position across the width is scaled as centred.

Exam answer (10 marks)

Steps of construction:

1. Layout: draw XY, with room for the front view above, the top view below and the left side view to the right.
2. Base in front view (X): draw a rectangle 60 wide and 20 high, standing on XY.
3. Stem in front view: on the base, draw the stem 20 wide (from 20 to 40 mm) and 40 high.
4. Top block in front view: on the stem, draw the top block 40 wide (from 10 to 50 mm) and 20 high. The front view is a T on a base, 80 high.
5. Top view outline: below XY draw the base, a rectangle 60 × 40, back edge nearest XY.
6. Top block in top view: draw a rectangle 40 × 20 in the middle (from 10 to 50 along the length, 10 to 30 across the width).
7. Hidden stem in top view: draw the stem edges at 20 and 40 mm from the left as dashed lines across the top block.
8. Left side view: to the right, draw the base 40 wide and 20 high, and on it an upright 20 wide (10 mm from each side) rising to 80 mm.
9. Line in side view: draw a horizontal line across the upright at 60 mm height, where the top block’s underside starts.
10. Finish: add thin projectors, the main dimensions (60, 40, 20, 40 and 20), and label FRONT VIEW (X), TOP VIEW and LEFT SIDE VIEW.

Figure Q10b: front view (X), top view and left side view in first angle projection (drawn to scale; draw it full size (scale 1:1) on your sheet, all dimensions in mm).

Answer: three views as in Figure Q10b: front view a T on a base, 60 wide and 80 high; top view 60 × 40 with the top block and two hidden stem lines; left side view a base 40 × 20 with a 20 mm wide upright and a line at 60.

FAQ

Which four Part B questions should I choose in the EE-107 March 2025 paper?

Q5 (involute of a pentagon), Q7 (hexagonal pyramid), Q9 and Q10 (orthographic views) are the safest four. Q5 and Q7 have short fixed steps, and Q9 and Q10 are simple blocks with no hidden lines in the side views. Take Q6 (circular plane) if you can draw an ellipse by points. Leave Q8 unless you have practised auxiliary views, because it asks for three drawings.

Which chapters carry the most marks in this Engineering Drawing paper?

Chapter 3 (projections of points, lines, planes and solids) carries 25 marks: Q4, Q6 and Q7. Orthographic projection (Chapter 5) carries 20 marks (Q9, Q10), geometric constructions 15 (Q3, Q5), and Chapters 1 and 4 carry 10 each. With the Part B choice these add up to 80, the same as the C-23 weightage.

Some dimensions are missing in the Q2 and Q10 figures. What should I do?

Q2 does not print the spacing of the three holes, and Q10 does not print where the T stands across the 40 mm width. Scale the printed figure: the holes come out 20 mm apart, and the T comes out centred. Draw them that way and write one line saying you scaled them. Keep every printed dimension exactly as given.

Are there misprints in this paper?

Yes, small ones. Q7 reads “A hexagonal pyramid side of base 30 mm”, with a word missing; it means a pyramid with a base side of 30 mm. The Q2 figure writes “10 φ” and “20 R” in an old style; BIS writes φ10 and R20. The code line prints “CCN–” with a double hyphen. None of these changes what you draw.

Where do the top view and left side view go in first angle projection?

In first angle projection the top view goes directly below the front view, and the left side view goes to the right of the front view. The XY line separates the front view from the top view. All three views are joined by thin projectors, so every point lines up across the views.

Sources

• AP SBTET C-23 EE-107 Engineering Drawing question paper, Mar-Apr 2025
• C-23 Curriculum, Diploma in Electrical and Electronics Engineering, SBTET AP
• Reference books (C-23 syllabus):
  • P.I. Varghese: Engineering Graphics, McGraw-Hill
  • Basant Agarwal & C.M. Agarwal: Engineering Drawing, McGraw-Hill
  • N.D. Bhatt: Engineering Drawing
  • TTTI Madras: T.S.M. & S.S.M. on Technical Drawing
  • SP-46-1998, Bureau of Indian Standards
• SP 46:2003 Engineering Drawing Practice for Schools and Colleges, Bureau of Indian Standards (lettering, dimensioning, sections)