GTU Technical Representation Drawing I (2X15002) Semester 1 Summer 2026 Solved Paper
GTU Bachelor of Architecture Semester 1 Technical Representation Drawing - I (TRD-I) Summer 2026 Exam Paper - Detailed Step-by-Step Solved Solutions
Introduction
This document contains a student-friendly, step-by-step worked solution for the GTU Semester 1 B.Arch Examination in Technical Representation Drawing - I (T.R.D.-I) from Summer 2026. This resource is AI-generated. Students are highly encouraged to cross-reference these construction steps and definitions with their class notes and standard textbooks. Accuracy in technical drawing is best achieved through hands-on drafting with physical instruments.
Paper Information
- University
- Gujarat Technological University
- Department
- Architecture
- Semester
- 1
- Subject
- T.R.D.-I (Technical Representation Drawing - I)
- Subject Code
- 2X15002
- Exam
- Bachelor of Architecture - Semester 1 - Examination - Summer 2026
- Year
- 2026
Questions and Solutions
Q.1(a)
Question
A square prism, base 4 cm side and height 6.5 cm has its axis inclined at 45 to the H.P. and has an edge of its base, on the H.P. and perpendicular to V.P. Draw the orthographic projections.
Solution
Analysis of the Problem
Solid : Square Prism Base Dimensions : 4 cm × 4 cm Prism Height (Axis) : 6.5 cm Condition 1 : One edge of the base is on the Horizontal Plane (H.P.) and is perpendicular to the Vertical Plane (V.P.). Condition 2 : The axis of the prism is inclined at 45° to the H.P.
Geometric Principles
Since the axis is inclined at 45° to the H.P., the base of the prism is inclined at 90° - 45° = 45° to the H.P. One edge of this base rests on the H.P. and is perpendicular to the V.P.
Step-by-Step Construction Procedure
Step 1: Initial Position (Simple Position)
Draw a reference line XY . Assume the prism is resting flat on the H.P. with its axis vertical (perpendicular to H.P.) and one base edge perpendicular to the V.P. Draw the Top View (Plan) : Draw a square abcd of side 4 cm below XY . Position it such that sides ad and bc are perpendicular to the XY line. Draw the Front View (Elevation) : Project the points upwards to the XY line. Since the base is on the H.P., the front view of the base lies on XY . Draw a rectangle of width 4 cm and height 6.5 cm . Label the base vertices as a1'b1'c1'd1' (on XY ) and the top face vertices as a'b'c'd' (at a height of 6.5 cm ). Let the axis of this initial front view be vertical.
Step 2: Tilted Position (Final Projections)
We need to tilt the prism such that the base edge a1'd1' remains on the H.P. (on the XY line) while the axis is inclined at 45° to the H.P. This means the base itself is inclined at 45° to the H.P. Locate a point a1'(d1') on the XY line to represent the resting edge. Draw a line representing the tilted base through a1'(d1') at an angle of 45° to XY . Mark a length of 4 cm along this line to locate the opposite base corner b1'(c1') . From points a1'(d1') and b1'(c1') , draw perpendicular lines (at 90° to the tilted base line) of length 6.5 cm to represent the longer lateral edges. These lines will make an angle of 45° with XY . Connect the top endpoints to complete the tilted rectangle a'b'c'd'-a1'b1'c1'd1' . This is the final Front View . Project the final Top View :
Draw vertical projectors downwards from all eight vertices of the tilted front view. Draw horizontal projectors from the initial top view square abcd . Locate the intersecting points (e.g., the intersection of the vertical projector from a' and the horizontal projector from a yields the final plan point a ). Connect the points to complete the final plan. Use solid lines for visible boundaries and dashed lines for hidden features (such as the base edge resting on the ground which is covered by the body of the prism).
Q.1(b)
Question
A pentagonal prism, base of 4 cm side and axis of 7 cm is resting on one of the base edges on the H.P. The longer surface containing that edge is inclined at 45 to the H.P. and parallel to the V.P. Draw the projections.
Solution
Clarification of Ambiguity
The problem statement contains a minor geometric contradiction: "The longer surface containing that edge is inclined at 45 to the H.P. and parallel to the V.P." In orthographic projection, any plane/surface that is parallel to the V.P. must be perpendicular to the H.P. (i.e., making a 90° angle with the H.P.). It cannot be inclined at 45° to the H.P.
Assumption to proceed : We assume the question implies that the resting base edge is on the H.P. and is parallel to the V.P. , and the rectangular (longer) face containing this edge is inclined at 45° to the H.P. This means the axis of the prism is parallel to the V.P. and inclined to the H.P. at 45° .
Step-by-Step Construction Procedure
Step 1: Auxiliary End View (Side View) of Base
Draw reference line XY . Draw an auxiliary projection plane X1Y1 perpendicular to XY to draw the side view first. In the side view, the pentagon will show its true shape because the axis is parallel to the V.P. Draw a regular pentagon of side 4 cm resting on one of its edges on the ground line ( XY ). Since the face containing this edge is inclined at 45° , rotate this pentagon in the side view such that the bottom edge remains on the XY line and the adjacent longer face is inclined at 45° to the H.P.
Step 2: Front View (Elevation)
Project the vertices horizontally from the side view to the main front view region. The length of the prism along the axis is 7 cm . Since the axis is parallel to the V.P., draw parallel lines of length 7 cm at an angle of 45° (or corresponding projected height) to form the lateral edges of the prism. Complete the front view by drawing the two parallel pentagonal end-faces.
Step 3: Top View (Plan)
Project the vertices from the front view downwards. Project the widths from the side view using a 45° miter line from X1Y1 . Intersect these projection lines to find the final vertices of the pentagonal prism in the top view. Connect the vertices, using solid lines for visible outlines and dashed lines for hidden edges.
Q.2(a)
Question
Draw orthographic projection (plan and elevation and side elevation) of a rectangular plane(parallel to VP) of a size 4 cm x 6 cm, kept in such a way that the 4 cm side is making an angle of 30 with the H.P(horizontal plane).
Solution
Analysis of the Plane
Type : Rectangular plane of 4 cm × 6 cm Orientation : Parallel to the V.P. This means its Front View (Elevation) will show the true shape and size of the rectangle. Condition : The 4 cm side is inclined at 30° to the H.P. (which translates to 30° to the reference XY line in the front view). Distance Assumption : Assume the plane is located 2 cm in front of the V.P. and its lowest point is 1.5 cm above the H.P. for clarity.
Step-by-Step Drawing Steps
- Front View (Elevation)
Draw reference line XY . Draw a line a'b' of length 4 cm making an angle of 30° with the XY line. Point a' is the lower vertex. From a' and b' , draw perpendicular lines of length 6 cm upwards to locate points d' and c' respectively (such that angle angle d'a'b' = 90° and angle c'b'a' = 90° ). Join c'd' to complete the rectangle a'b'c'd' of size 4 cm × 6 cm . This represents the true shape of the rectangle.
- Top View (Plan)
Since the plane is parallel to the V.P., its projection on the H.P. (Top View) will be a straight line parallel to the XY line. Draw vertical projectors downwards from all four vertices ( a', b', c', d' ) of the front view. Draw a horizontal line parallel to XY at the chosen distance of 2 cm below XY . The intersections of the vertical projectors with this line give the points a, d (which coincide) and b, c (which coincide). The plan is represented by the line segment ad - bc , parallel to XY .
- Side View (Side Elevation)
Draw a profile plane line X1Y1 perpendicular to XY . Project horizontal lines from the front view vertices ( a', b', c', d' ) toward the profile plane. Project the top view line segment to the profile plane and transfer it vertically using a 45° miter line. Since the plane is perpendicular to the Profile Plane, the side elevation will be a straight vertical line of height equal to the vertical span of the tilted rectangle in the front view.
Q.2(b)
Question
Draw projection (plan and elevation and side elevation) of a hexagonal plane of each side 4 cm, kept in such a way that the plane is parallel to H.P. and perpendicular to V.P. The plane is kept 3 cm above H.P. and one side of the plane is 5 cm away from the V.P.
Solution
Analysis of the Plane
Shape : Regular hexagon, side = 4 cm Orientation : Parallel to the H.P. (hence perpendicular to the V.P.). Its true shape will be visible in the Top View (Plan) . Position 1 : 3 cm above the H.P. Position 2 : One edge of the hexagon is 5 cm away from the V.P. (We assume this edge is parallel to the V.P.).
Step-by-Step Drawing Steps
- Top View (Plan)
Draw the reference line XY . Draw a horizontal line parallel to XY at a distance of 5 cm below it. This represents the edge closest to the V.P. On this line, mark a segment ab = 4 cm . Construct a regular hexagon abcdef of side 4 cm starting from the edge ab , such that the rest of the hexagon lies further below the line (away from XY ). Each interior angle of a regular hexagon is 120° .
- Front View (Elevation)
Since the plane is parallel to the H.P., its front view will be a straight line parallel to XY . Draw a horizontal line at a height of 3 cm above the XY line. Project vertical lines upwards from all six vertices ( a, b, c, d, e, f ) of the top view. The intersections with the horizontal line give the points a'(b') , f'(c') , and e'(d') . The resulting line segment represents the elevation.
- Side View (Side Elevation)
Draw a vertical line X1Y1 as the profile reference line. Project the heights from the front view (which is a horizontal line at 3 cm ) to the side view. Project the widths from the top view to the side view using a 45° miter line. The side view will be a horizontal line segment of length equal to the total depth of the hexagon ( 4 × cos(30°) × 2 ≈ 6.93 cm ), located at a height of 3 cm above the XY line.
Q.2(b) (OR)
Question
A circle of dia 5 cm is perpendicular to VP, but making an angle of 45* with HP. Draw the orthographic projections.
Solution
Analysis of the Geometry
Object : Circle of diameter 5 cm Orientation : Perpendicular to V.P., making an angle of 45° with H.P. Visual outcome : Since the circle is perpendicular to V.P., its Front View is a straight line of length 5 cm (the diameter) inclined at 45° to the XY line. Its Top View will be an ellipse because the circular plane is inclined to the H.P.
Step-by-Step Drawing Steps
- Front View
Draw the reference line XY . Draw a line segment a'b' of length 5 cm inclined at 45° to the XY line. Point a' can lie on the XY line or a short distance above it. Divide the line segment a'b' into 8 equal parts (or 12 parts for greater precision). Let's label the intermediate division points as 1', 2', 3', 4', 5', 6', 7' .
- Top View (Plan)
Draw an auxiliary circle of diameter 5 cm below XY to act as a generator tool. Divide this circle into 8 equal parts. Project vertical lines downwards from the division points on the tilted line a'b' in the front view. From the center line of the circle generator, measure the lateral offsets (widths) of each of the 8 division points. Transfer these offsets onto the corresponding vertical projector lines in the top view. Connect the plotted points with a smooth curve to draw the resulting ellipse . The major axis of this ellipse remains 5 cm (parallel to XY ), and the minor axis is compressed to 5 × cos(45°) ≈ 3.54 cm .
Q.3(a)
Question
What is Orthographic Projection?
Solution
Definition
Orthographic projection is a technical drawing method used to represent a three-dimensional object on a two-dimensional plane.
Key Principles
Parallel Projectors : The projection lines (projectors) originating from the object are parallel to each other. Orthogonal to Plane : The projectors intersect the projection plane at a right angle ( 90° ). Multi-View System : Since a single view cannot fully define a 3D object, multiple views are projected on mutually perpendicular planes (Horizontal Plane for Plan, Vertical Plane for Elevation, and Profile Plane for Side Elevation).
Q.3(b)
Question
Name at-least 6 drafting instruments.
Solution
The essential instruments used in manual technical drafting include:
Drawing Board : A flat, rigid wooden board providing a smooth surface to support the drawing sheet. T-Square : An instrument used to draw accurate horizontal lines and align other tools like set-squares. Set-Squares ( 30°-60° and 45° ) : Used to draw vertical, perpendicular, and precise angular lines. Compass : Used for drawing circles, arcs, and transferring curved dimensions. Dividers : Used to step off equal distances and transfer dimensions from a scale to the drawing. Protractor : A semi-circular tool used to measure and lay out angles on the sheet.
(Alternative: Mini-Drafter, French Curves, Drawing Pencils of varying hardness like 2H, HB, H)
Q.3(a) (OR)
Question
Define First Angle Method in Orthographic Projections
Solution
Definition of First Angle Projection
In the First Angle Projection method, the object is assumed to be placed in the First Quadrant (above the Horizontal Plane and in front of the Vertical Plane).
Key Characteristics
Position of Object : The object is placed between the observer and the plane of projection. View Arrangement : The Front View (Elevation) is drawn above the XY reference line. The Top View (Plan) is drawn below the XY line. The Left-Hand Side View is drawn on the right side of the front view. Standardization : This is the standard projection method widely adopted in India (BIS), Europe, and ISO systems.
Q.3(b) (OR)
Question
Define Third Angle Method in Orthographic Projections
Solution
Definition of Third Angle Projection
In the Third Angle Projection method, the object is assumed to be placed in the Third Quadrant (below the Horizontal Plane and behind the Vertical Plane).
Key Characteristics
Position of Plane : The projection plane is assumed to be transparent and is placed between the observer and the object. View Arrangement : The Front View (Elevation) is drawn below the XY reference line. The Top View (Plan) is drawn above the XY line. The Left-Hand Side View is drawn on the left side of the front view. Standardization : This method is primarily used in North America (ANSI/US standards) and Japan.
Q.4(a)
Question
Bisect a 60* angle.
Solution
Step-by-Step Geometrical Construction
To bisect a given angle of 60° to create two equal 30° angles:
Draw the Angle : Draw two lines OA and OB meeting at a common vertex O such that the angle angle AOB = 60° using a protractor. Draw First Arc : With O as the center and any convenient radius on your compass, draw an arc that cuts the line OA at point P and the line OB at point Q . Draw Intersecting Arcs :
Set the compass radius to a value greater than half of the distance PQ . With P as the center, draw an arc in the interior region of the angle. Maintaining the exact same radius, place the compass needle on Q and draw another arc intersecting the first arc at point R .
Complete the Bisector : Draw a straight line from the vertex O through the intersection point R . Result : The line OR is the bisector. Angle angle AOR = angle BOR = 30° .
Q.4(b)
Question
Divide a 75 mm line in 9 equal parts
Solution
Step-by-Step Geometrical Construction (Acute Angle Method)
To divide a line of length 75 mm into exactly 9 equal segments without using decimal calculations:
Draw the Main Line : Draw a horizontal line segment AB = 75 mm using a scale. Draw Auxiliary Line : From end A , draw an inclined line AC making an acute angle (around 20° to 30° ) with AB extending downwards. Mark Equal Divisions :
Adjust your compass to a small, convenient fixed radius. Starting at point A , mark 9 consecutive points along the line AC . Label these points 1, 2, 3, 4, 5, 6, 7, 8, 9 such that A1 = 1-2 = 2-3 = dots = 8-9 .
Join the Endpoints : Draw a straight line connecting point 9 to point B . Draw Parallel Lines :
Using set-squares or a parallel rolling ruler, draw lines parallel to the line 9-B starting from each of the intermediate points 8, 7, 6, 5, 4, 3, 2, 1 . These parallel lines will intersect the original line segment AB at points 8', 7', 6', 5', 4', 3', 2', 1' .
Result : The segment AB is now divided into 9 mathematically equal parts, each measuring exactly 8.33 mm .
Q.4(a) (OR)
Question
Trisect a 65 mm horizontal line.
Solution
Step-by-Step Geometrical Construction (Division into 3 Parts)
To divide a horizontal line AB = 65 mm into 3 equal segments (trisection):
Draw the Line : Draw a horizontal line segment AB = 65 mm using a scale. Draw Auxiliary Line : From point A , draw an inclined line AC making an acute angle (approx. 30° ) downwards. Mark 3 Divisions : Using a compass with a fixed convenient radius, mark 3 consecutive points along AC starting from A . Label these points 1, 2, 3 . Join End Points : Draw a straight line from point 3 to point B . Project Parallel Lines : Draw lines parallel to 3-B passing through points 2 and 1 to intersect the original line AB at 2' and 1' respectively. Result : The points 1' and 2' divide AB into three equal parts: A-1' , 1'-2' , and 2'-B . Each part measures precisely 21.67 mm .
Q.4(b) (OR)
Question
Construct a pentagon having side of 40 mm.
Solution
Method 1: Using Interior Angles (Protractor Method)
Draw Base : Draw a horizontal line segment AB = 40 mm . Calculate Angles : The interior angle of a regular pentagon is calculated as: θ = ((5-2) × 180° / 5) = 108° \n3. Plot Angles :
At point A , measure an angle of 108° and draw a line segment AE = 40 mm . At point B , measure an angle of 108° and draw a line segment BC = 40 mm .
Locate Top Vertex D :
Set your compass to a radius of 40 mm . With C as the center, draw an arc. With E as the center, draw another arc intersecting the first one at point D .
Complete Pentagon : Join CD and ED with straight lines to get the regular pentagon ABCDE .
Method 2: Geometrical Construction (Three-Circle Compass Method)
Draw a line segment AB = 40 mm . With A as the center and radius AB , draw a circle. With B as the center and radius AB , draw a second circle. These circles intersect at points P (top) and Q (bottom). Join PQ with a light vertical line. With Q as the center and radius AB , draw a third circle. This circle cuts the vertical axis at R and the first two circles at points S and T . Draw a line starting from S passing through R to intersect the second circle at point C . Draw a line starting from T passing through R to intersect the first circle at point E . With C and E as centers and radius 40 mm , draw intersecting arcs above to locate D . Join A-B-C-D-E-A to complete the regular pentagon.
Frequently Asked Questions
What is the key difference between First Angle and Third Angle projections?
In First Angle Projection, the object is placed between the observer and the projection plane (Front View is above Plan). In Third Angle Projection, the plane is placed between the observer and the object (Front View is below Plan).
How do you handle contradictions or ambiguities in exam questions?
If a question has a geometrical impossibility (e.g., a surface parallel to V.P. but inclined to H.P. at 45° ), state your assumption clearly on the drawing sheet. Usually, the instructor intends for the resting edge to be parallel to the V.P. while the face is inclined.
Why is it recommended to divide a circle into 8 or 12 parts for projections?
Dividing a circle into 8 or 12 equal parts provides sufficient control points to accurately plot the resulting ellipse in tilted projections using a french curve.