BE Civil Engineering (IOE, TU) Engineering Drawing II (IOE, CE 451) Question Paper 2079 Nepal

Given: base diameter $D = 60\ \text{mm}$, so base radius $r = 30\ \text{mm}$; axis height $h = 80\ \text{mm}$.

Step 1 — Slant height (true length of generator).

$$l = \sqrt{r^2 + h^2} = \sqrt{30^2 + 80^2} = \sqrt{900 + 6400} = \sqrt{7300} = 85.44\ \text{mm}$$

Step 2 — Development of a cone is a circular sector of radius equal to the slant height $l$ and arc length equal to the base circumference. The included angle of the sector:

$$\theta = \frac{r}{l}\times 360^\circ = \frac{30}{85.44}\times 360^\circ = 0.3511\times360^\circ = 126.4^\circ$$

Check via arc length: base circumference $= 2\pi r = 2\pi(30) = 188.50\ \text{mm}$; sector arc $= l\theta_{rad} = 85.44 \times (126.4\times\pi/180) = 85.44\times2.206 = 188.5\ \text{mm}$. ✓

Development sketch (sector):

            O (apex)
           / \
          /   \        radius O-1 = O-1' = l = 85.44 mm
         /     \       included angle = 126.4 deg
        /  126  \      arc 1..1' divided into 12 equal parts
       /   deg   \
      1 --------- 1'   (arc = base circumference = 188.5 mm)

Divide the base circle into 12 equal parts; on the development the sector arc is also divided into 12 equal parts, each subtending $126.4/12 = 10.53^\circ$.

Step 3 — (b) Locate point $P$. $P$ is on the nearest generator (say generator O-1) at slant distance $50\ \text{mm}$ from apex $O$. On the development, mark $P$ on radial line $O\!-\!1$ at $50\ \text{mm}$ from $O$. Since $50 < 85.44$, $P$ lies between apex and base. ✓

Step 4 — (c) Shortest string wound once around and back to start. On the development, the string from the starting point (a base point, say $1$) around the full surface and back to the SAME point becomes a STRAIGHT chord joining $1$ to $1'$ (the two coincident edges of the sector, both at radius $l$ from $O$, separated by the sector angle $\theta$).

Using the cosine rule in triangle $O\,1\,1'$ with $O1 = O1' = l = 85.44\ \text{mm}$ and angle $1\,O\,1' = \theta = 126.4^\circ$:

$$L = \sqrt{l^2 + l^2 - 2l^2\cos\theta} = l\sqrt{2(1-\cos\theta)}$$ $$\cos126.4^\circ = -0.5976$$ $$L = 85.44\sqrt{2(1-(-0.5976))} = 85.44\sqrt{2(1.5976)} = 85.44\sqrt{3.1952} = 85.44\times1.7875 = 152.7\ \text{mm}$$

Final answers: slant height $l = \mathbf{85.44\ mm}$; sector angle $\theta = \mathbf{126.4^\circ}$; point $P$ marked at $50\ \text{mm}$ from apex on generator O-1; true length of the shortest wound string $= \mathbf{152.7\ mm}$.

Given: vertical cylinder radius $R = 35\ \text{mm}$; horizontal (penetrating) cylinder radius $r = 25\ \text{mm}$; axes intersect, so the smaller cylinder is fully engulfed laterally because $r < R$.

Step 1 — Method. The curve of intersection is found by the generator (cutting-plane) method: take cutting planes parallel to both axes... but the standard approach for two cylinders with intersecting axes at $90^\circ$:

• Draw the top view: the small cylinder appears as a $50\ \text{mm}$ wide rectangle; the big cylinder as a $70\ \text{mm}$ circle.
• Divide the circular end of the SMALL cylinder (seen as a circle of radius $25$ in the side/end view) into 12 equal parts: points $1,2,\dots,12$.
• Each division corresponds to a generator of the small cylinder at horizontal offset $y$ from the common axis.
• Project each generator onto the big cylinder; the point where it pierces the big cylinder gives the intersection point. In the front view its height follows the big cylinder's curvature.

Step 2 — Geometry of a point. Let the common axes intersection define origin. A generator of the small cylinder lies at horizontal offset

$$y_k = r\sin\phi_k \quad\text{and vertical offset}\quad z_k = r\cos\phi_k$$

where $\phi_k = (k-1)\times30^\circ$. The generator (a horizontal line) pierces the big cylinder where the big cylinder surface is, i.e. at the same $y$ on the circle of radius $R$. But the curve of intersection height in the FRONT view equals the small cylinder generator height $z_k$ measured on its circular end, because the generators are horizontal and the front view shows their true height. The front-view curve is therefore obtained by transferring $z_k$ at horizontal positions determined by the big cylinder.

The penetration point horizontal coordinate (distance from axis, in the direction of the small cylinder's motion) where generator at offset $y_k$ meets big cylinder:

$$x_k = \sqrt{R^2 - y_k^2}$$

Step 3 — Table (taking $\phi_k$, $y_k = 25\sin\phi_k$, $z_k = 25\cos\phi_k$, and $x_k = \sqrt{35^2 - y_k^2}$):

(Generators 8–12 mirror 6–2 by symmetry.)

Sample check, $k=3$: $y_3 = 25\sin60^\circ = 25(0.8660)=21.65$; $x_3=\sqrt{1225-468.75}=\sqrt{756.25}=27.50$. ✓

Step 4 — Front view. The intersection points are plotted at height $(55 + z_k)$ above the base of the vertical cylinder, horizontally at $\pm x_k$ from the vertical axis. Joining them with a smooth curve gives two symmetrical curves of intersection (the small cylinder enters on the left, exits on the right).

Highest point: at $z_{max} = +25\ \text{mm}$ (generators 1), so the curve peaks at $55 + 25 = \mathbf{80\ mm}$ above the base; lowest at $55 - 25 = 30\ \text{mm}$.

 front view (schematic)
  __________            curve dips toward centre where small
 |  /\  /\  |  <- top  cylinder axis crosses (x small), rises
 | /  \/  \ |          at the contour generators 1 and 7.
 |/        \|

Final answer: intersection curves are symmetric; highest point of the curve = 80 mm above base, lowest = 30 mm above base.

Given: internal $4.5\times3.6\ \text{m}$; wall thickness $t = 0.230\ \text{m}$; superstructure wall height $H = 3.0\ \text{m}$.

(a) Plan (1:50) — key dimensions:

• Internal: $4500 \times 3600\ \text{mm}$.
• External overall: $(4500 + 2\times230)\times(3600 + 2\times230) = 4960 \times 4060\ \text{mm}$.
• Door $1000$ wide on long wall; windows $1200$ wide centred on each short wall.

  <----------- 4960 ----------->
 +=============================+   ^
 |  +-----------------------+  |   |
 |W |                       | W|  4060
 |  |     4500 x 3600       |  |   |
 |  |       (internal)      |  |   |
 |  +---------[door]--------+  |   v
 +=============================+
   wall t = 230 mm all round

(b) Cross-section through window (description):

                 ___________________  <- 125 mm RCC roof slab
                |###################|
   lintel ----> |===================|  (over window, 230x150 RCC, 150 bearing)
                |   | window 1200 | |
   sill 0.9m -> |___|  opening   |_|
                |###################|  brick wall 230 thick
   GL ~~~~~~~~~~|###################|~~~~~~~~~
  plinth 450 -> |###################|
                |######[ DPC ]######|  damp-proof course at plinth
                |####### plinth #####|
  foundation -> |  ___________________ |
                | / stepped footing  \ | (e.g. 2 courses, 600 wide base)
                |/____________________\|
                |  P.C.C. bed (1:4:8)  |

Key: foundation depth ~$0.9\ \text{m}$ below GL, footing base spread, DPC at plinth, lintel with min $150\ \text{mm}$ bearing each side, $125\ \text{mm}$ RCC roof slab on top.

(c) Centre-line method. Centre-line perimeter of external walls = perimeter measured along the wall centre line. Centre-line plan dimensions $= \text{internal} + t$ (one $t$ added since centre line is half-thickness in from each face... actually centre-line length per side $=$ internal $+ t$):

• Long centre-line length $= 4500 + 230 = 4730\ \text{mm} = 4.730\ \text{m}$.
• Short centre-line length $= 3600 + 230 = 3830\ \text{mm} = 3.830\ \text{m}$.

Total centre-line length $L_{cl} = 2(4.730 + 3.830) = 2(8.560) = \mathbf{17.12\ m}$.

Gross brickwork volume (superstructure):

$$V_{gross} = L_{cl}\times t \times H = 17.12 \times 0.230 \times 3.0 = 11.813\ \text{m}^3$$

Deductions:

• Door: $1.0\times2.1\times0.230 = 0.483\ \text{m}^3$.
• Two windows: $2\times(1.2\times1.2\times0.230) = 2\times0.3312 = 0.662\ \text{m}^3$.
• Total deduction $= 0.483 + 0.662 = 1.145\ \text{m}^3$.

Net brickwork volume:

$$V_{net} = 11.813 - 1.145 = \mathbf{10.67\ m^3}$$

Final answers: centre-line length $= \mathbf{17.12\ m}$; net superstructure brickwork $= \mathbf{10.67\ m^3}$.

Given: block $5\times3\times4\ \text{m}$; faces at $30^\circ$ and $60^\circ$ to PP (they are mutually perpendicular: $30^\circ + 60^\circ = 90^\circ$ ✓); distance of station point from PP $d = 9\ \text{m}$; horizon height $1.6\ \text{m}$.

(a) Principle of two-point perspective. When a rectangular object is placed with its vertical edges parallel to the PP but its two sets of horizontal edges INCLINED to the PP, the two horizontal sets converge to TWO separate vanishing points on the horizon line. Vertical edges remain vertical (no vanishing point). The vanishing point of any set of parallel horizontal lines is found by drawing from the station point (SP) a line PARALLEL to that set; where it meets the PP, drop to the horizon line.

(b) Locating the vanishing points. Draw from SP two lines parallel to the two edge directions:

• Line parallel to the set making $30^\circ$ with PP meets PP at the trace for $VP_1$.
• Line parallel to the set making $60^\circ$ with PP meets PP at the trace for $VP_2$.

The horizontal distance of each VP from the centre of vision (CV, the foot of perpendicular from SP to PP) along the horizon equals:

$$x_{VP} = d\,\tan(\text{angle between that edge-set and the line SP–CV})$$

The SP–CV line is perpendicular to PP. A set making angle $\alpha$ with the PP makes $(90^\circ-\alpha)$ with the SP–CV line. Hence the parallel ray from SP makes $(90^\circ-\alpha)$ with SP–CV, giving:

For the $30^\circ$ set ($\alpha=30^\circ$): ray makes $60^\circ$ with SP–CV:

$$x_{VP1} = 9\tan60^\circ = 9\times1.7321 = 15.59\ \text{m (one side of CV)}$$

Wait — re-express using the standard rule $x_{VP}=d\tan\alpha$ where $\alpha$ is the angle the edges make with PP (the parallel ray from SP makes angle $\alpha$ with PP, i.e. $(90-\alpha)$ with the normal). Distance along PP from CV $= d\tan(90-\alpha)=d/\tan\alpha$... To avoid ambiguity use geometry directly:

The parallel ray from SP to an edge-set inclined at $\alpha$ to PP intersects the PP at a point whose distance from CV (measured along PP) is

$$x_{VP} = d\,\cot\alpha = \frac{d}{\tan\alpha}.$$

• $VP_1$ (for $30^\circ$ set): $x = 9/\tan30^\circ = 9/0.5774 = \mathbf{15.59\ m}$ on one side of CV.
• $VP_2$ (for $60^\circ$ set): $x = 9/\tan60^\circ = 9/1.7321 = \mathbf{5.196\ m}$ on the other side of CV.

Check: the two VPs lie on opposite sides; separation $= 15.59 + 5.196 = 20.78\ \text{m}$. The right angle at SP subtended by the two VPs is confirmed since $\tan^{-1}(d/x_{1}) + \tan^{-1}(d/x_2) = \tan^{-1}(9/15.59)+\tan^{-1}(9/5.196)=30^\circ+60^\circ=90^\circ$. ✓

(c) Perspective height of near vertical edge. Because this vertical edge actually TOUCHES the picture plane, it lies IN the PP. Any line lying in the PP is drawn to its TRUE length (no diminution). Therefore the near edge is drawn at its true height $= 4.0\ \text{m}$ (to the chosen scale). Its top end is $4.0 - 1.6 = 2.4\ \text{m}$ above the horizon line and its base is $1.6\ \text{m}$ below the horizon line.

Final answers: $VP_1 = \mathbf{15.59\ m}$ and $VP_2 = \mathbf{5.196\ m}$ from CV on opposite sides (sum $20.78\ \text{m}$); near edge drawn to true height $= \mathbf{4.0\ m}$ because it lies in the picture plane.

Given: square side $a = 200\ \text{mm}$; circle diameter $d = 300\ \text{mm}$ (radius $R = 150\ \text{mm}$); vertical height $H = 250\ \text{mm}$; coincident axes.

(a) Composition of the surface. A square-to-round transition piece is not a developable single surface; it is approximated by breaking it into:

• 4 plane triangles, one against each side of the square. The base of each triangle is a side of the square (length $200\ \text{mm}$) and its apex is a point on the circle.
• 4 conical (lobe) portions at the corners, each approximated by several small triangles whose common apex is a corner of the square and whose bases are short chords of the circle.

The whole surface is thus developed by triangulation: find the TRUE LENGTH of every triangle edge (corner-to-circle lines), then lay the triangles side by side.

(b) True-length calculations. Set origin at the centre of the bottom circle. Square corners are at height $H=250$ directly above points $(\pm100, \pm100)$ in plan (since half-side $=100\ \text{mm}$).

Divide each quarter of the circle into (say) 3 parts → 12 points around the circle at $30^\circ$ spacing. Circle points: $(R\cos\phi, R\sin\phi, 0)$ with $R=150$.

Line A — corner of square to the circle point on the diagonal (the corner lobe's central line). Take square corner $C=(100,100,250)$. The nearest diagonal circle point is at $\phi=45^\circ$: $P_{45}=(150\cos45^\circ,150\sin45^\circ,0)=(106.07,106.07,0)$. Plan (horizontal) distance:

$$\Delta = \sqrt{(106.07-100)^2+(106.07-100)^2}=\sqrt{6.07^2+6.07^2}=\sqrt{36.84+36.84}=\sqrt{73.68}=8.58\ \text{mm}$$

True length:

$$TL_A=\sqrt{\Delta^2+H^2}=\sqrt{8.58^2+250^2}=\sqrt{73.6+62500}=\sqrt{62573.6}=250.15\ \text{mm}$$

Line B — corner of square to the mid-point of the adjacent side's plane triangle apex. Take the same corner $C=(100,100,250)$ to circle point at $\phi=30^\circ$: $P_{30}=(150\cos30^\circ,150\sin30^\circ,0)=(129.90,75.00,0)$. Plan distance:

$$\Delta=\sqrt{(129.90-100)^2+(75.00-100)^2}=\sqrt{29.90^2+(-25.00)^2}=\sqrt{894.0+625.0}=\sqrt{1519.0}=38.97\ \text{mm}$$

True length:

$$TL_B=\sqrt{38.97^2+250^2}=\sqrt{1518.7+62500}=\sqrt{64018.7}=253.02\ \text{mm}$$

Line C — the plane-triangle apex line: mid-point of a square side to the circle point directly below it on the axis of that side. Mid-point of side $M=(0,100,250)$ to circle point at $\phi=90^\circ$: $P_{90}=(0,150,0)$. Plan distance:

$$\Delta=\sqrt{0^2+(150-100)^2}=50.00\ \text{mm}$$

True length:

$$TL_C=\sqrt{50^2+250^2}=\sqrt{2500+62500}=\sqrt{65000}=254.95\ \text{mm}$$

Summary table of characteristic true lengths:

Development procedure: using these true lengths, start with one plane triangle (base $200\ \text{mm}$, two sides $=TL_C$), then attach successive corner-lobe triangles by swinging arcs of the computed true lengths and the small chord lengths of the circle, building outward symmetrically.

Final answers: representative true lengths — diagonal corner line $\mathbf{250.15\ mm}$, lobe line $\mathbf{253.02\ mm}$, side-apex line $\mathbf{254.95\ mm}$.

Given: regular pentagon side $s = 25\ \text{mm}$; full height $60\ \text{mm}$; cutting plane at $45^\circ$ starting at $20\ \text{mm}$ on one edge.

Step 1 — Development of a prism is a series of rectangles laid flat. The base perimeter $= 5\times25 = 125\ \text{mm}$ is the total width; each vertical edge is spaced $25\ \text{mm}$ apart. Label vertical edges $1,2,3,4,5,1'$.

Step 2 — Heights of cut points along each edge. The plane is at $45^\circ$, so for every unit of horizontal travel the cut rises $1$ unit (slope $\tan45^\circ = 1$). It starts at edge 1 at height $20\ \text{mm}$.

We need the horizontal distance (in the direction of the plane's steepest rise) of each edge from edge 1. For simplicity assume the plane rises across the prism width so that consecutive edges in the development are separated by their plan offsets. Using a clean representative set (heights measured in the front view at each edge):

(Edges 3 and 4 are the farthest face, highest; edge 1 is the entry, lowest, by symmetry of a pentagon about the cutting direction.)

Step 3 — Construct the development.

 h(mm)
 43.5 |        ___3___4___
      |      /            \
 31.0 |    2/              \5
      |   /                 \
 20.0 | 1/                   \1'
      |__|____|____|____|____|__
        1    2    3    4    5   1'
        |<-- 25 each, total 125 -->|
  base line (bottom of truncated solid)

Draw a base line $125\ \text{mm}$ long, mark off five $25\ \text{mm}$ segments for the six edge lines $1,2,3,4,5,1'$. Erect verticals at each. Mark the cut height from the table on each vertical. Join the cut points with straight lines (the section edges are straight since each prism face is planar). The lower closed figure is the required development of the truncated prism's lateral surface.

Final answer: development is the rectangle strip $125\ \text{mm}$ wide with the top edge stepping through heights 20, 31, 43.5, 43.5, 31, 20 mm across edges 1→1', joined by straight lines.

(a) Coordinate input systems in AutoCAD.

• Absolute: measured from the drawing origin (0,0).
• Relative (incremental): the @ symbol means measured FROM the last picked point.
• Relative polar: @ distance < angle (degrees, CCW from positive X-axis), also from the last point.

(b) Computation. Start $P_1=(40,30)$. Displacement $\Delta X = +60$, $\Delta Y = +25$.

Relative rectangular entry: @60,25

Resulting absolute coordinate:

$$P_2 = (40+60,\ 30+25) = (100,\ 55)$$

True length of the line:

$$L = \sqrt{60^2 + 25^2} = \sqrt{3600 + 625} = \sqrt{4225} = 65.00\ \text{mm}$$

Inclination above horizontal:

$$\theta = \tan^{-1}\!\left(\frac{25}{60}\right) = \tan^{-1}(0.4167) = 22.62^\circ$$

Relative polar entry: @65<22.62

Final answers: relative rectangular @60,25; relative polar @65<22.62; new absolute point $\mathbf{(100, 55)}$; length $\mathbf{65.00\ mm}$; inclination $\mathbf{22.62^\circ}$.

Given: floor-to-floor height $h = 3.30\ \text{m} = 3300\ \text{mm}$; riser $R = 165\ \text{mm}$; tread $T = 250\ \text{mm}$.

(a) Number of risers:

$$N_R = \frac{h}{R} = \frac{3300}{165} = 20\ \text{risers}$$

For a dog-legged stair with two equal flights, each flight has

$$\frac{20}{2} = 10\ \text{risers per flight}.$$

Number of treads: the number of treads in a flight is one less than the number of risers in that flight (the last riser lands on the floor/landing):

$$N_T = 10 - 1 = 9\ \text{treads per flight}.$$

Total treads in the staircase $= 2\times9 = 18$.

Going of one flight (horizontal run):

$$G = N_T \times T = 9 \times 250 = 2250\ \text{mm} = \mathbf{2.25\ m}.$$

(b) Comfort (rule of) $2R + T$:

$$2R + T = 2(165) + 250 = 330 + 250 = 580\ \text{mm}.$$

The recommended comfortable range is $560\text{–}630\ \text{mm}$ (commonly target $\approx 600\ \text{mm}$). Since $580\ \text{mm}$ lies within this range, the proportions are comfortable and acceptable.

Also check riser+tread sum: $R + T = 165 + 250 = 415\ \text{mm}$ (good, near the $400\text{–}450\ \text{mm}$ guideline), and $R\times T = 165\times250 = 41250 \approx 41000$–$45000\ \text{mm}^2$ guideline. ✓

Final answers: total 20 risers (10 per flight), 18 treads (9 per flight), going per flight $= \mathbf{2.25\ m}$; $2R+T = \mathbf{580\ mm}$ → within comfort range, acceptable.

(a) Two methods of finding lines of intersection.

Line (generator/edge) method: The piercing points are found where the EDGES (or generators) of one solid meet the surface (faces) of the other. For each edge of the penetrating solid, its top and front views are used to find exactly where it enters and leaves the other solid; these piercing points are joined. Preferred when at least one solid has flat faces / straight edges (prisms, pyramids) so that edges are clearly defined.

Cutting-plane (auxiliary section) method: A series of auxiliary cutting planes are passed through both solids (usually parallel to a principal plane or containing the common axis). Each plane cuts a SECTION on each solid; the intersection of these two sections gives points on the required curve. Preferred when one or both solids are curved (cylinders, cones) so that there are no edges to use directly.

(b) Square prism penetrated by triangular prism.

• The triangular prism has 3 long edges. Each edge that passes through the square prism enters through one face and exits through the opposite/adjacent face → 2 piercing points per edge (one entry, one exit).
• With 3 edges fully penetrating: $3\times2 = 6$ piercing points total, joined into a closed line of intersection (since the triangular prism is bounded by flat faces, the line of intersection is made of straight segments, not a curve).

Projection procedure (outline):

1. Draw the top view and front view of both prisms in the given position.
2. In the top view, the horizontal triangular prism's edges appear as horizontal lines; note where each crosses the square prism's faces — these give the horizontal positions of piercing points.
3. Project these crossing points down/up to the corresponding faces in the front view, where the square prism's vertical faces appear as vertical lines, to fix the heights of the piercing points.
4. Identify entry and exit point for each triangular-prism edge (6 points).
5. Join the points in correct sequence with straight lines, observing visibility (solid lines for visible, dashed for hidden), to complete the line of intersection.

Final answer: 2 points of intersection per penetrating edge, 6 points total, joined by straight segments via the line (edge-piercing) method.

Given: persons $P = 8$; flow $q = 90\ \text{L/person/day}$; detention $t = 2\ \text{days}$; sludge $= 30\ \text{L/person}$; liquid depth $D = 1.2\ \text{m}$; $L{:}B = 3{:}1$; free-board $= 0.3\ \text{m}$.

Step 1 — Sewage (detention) volume:

$$V_{sewage} = P\times q\times t = 8\times90\times2 = 1440\ \text{litres} = 1.440\ \text{m}^3$$

Step 2 — Sludge storage volume:

$$V_{sludge} = P\times30 = 8\times30 = 240\ \text{litres} = 0.240\ \text{m}^3$$

Step 3 — Total required liquid capacity:

$$V = V_{sewage}+V_{sludge} = 1.440+0.240 = 1.680\ \text{m}^3$$

Step 4 — Plan area (liquid depth $1.2\ \text{m}$):

$$A = \frac{V}{D} = \frac{1.680}{1.2} = 1.40\ \text{m}^2$$

Step 5 — Length and breadth with $L = 3B$:

$$A = L\times B = 3B\times B = 3B^2 = 1.40 \;\Rightarrow\; B^2 = 0.4667 \;\Rightarrow\; B = 0.683\ \text{m}$$ $$L = 3B = 3\times0.683 = 2.049\ \text{m}$$

Adopt practical sizes: $L \approx 2.05\ \text{m}$, $B \approx 0.70\ \text{m}$.

Step 6 — Overall depth = liquid depth + free-board:

$$D_{total} = 1.2 + 0.3 = 1.5\ \text{m}$$

Final answer (provide internal dimensions): length $\mathbf{\approx 2.05\ m}$, breadth $\mathbf{\approx 0.70\ m}$, liquid depth $1.2\ \text{m}$, overall depth $\mathbf{1.5\ m}$ (incl. $0.3\ \text{m}$ free-board); total liquid capacity $\mathbf{1.68\ m^3}$.

(a) Scale conversion. Scale $1{:}200$ means $1\ \text{mm}$ on paper $= 200\ \text{mm}$ on ground.

$$\text{Actual length} = 46\ \text{mm}\times200 = 9200\ \text{mm} = \mathbf{9.20\ m}$$

(b) CAD vs manual drafting — two advantages (any two):

1. Speed and easy editing: objects can be copied, mirrored, arrayed and modified without redrawing; revisions are fast.
2. Accuracy and precision: exact coordinate/dimension input eliminates manual measurement error. (Other valid: easy storage/retrieval and reuse of standard blocks; layer management; automatic dimensioning and area/length queries.)

Standard line types (IS / ISO convention):

• (i) Centre line: a long chain / dash-dot thin line (long dash – short dash repeating).
• (ii) Hidden edge: a short-dash (dashed) thin line.

Final answers: actual length $= \mathbf{9.20\ m}$; centre line = chain (dash-dot) line; hidden edge = dashed line.