Lines, Angles, Letters, Dimensioning and Rectilinear Figures

Given: Various features of an engineering drawing and the requirement to identify the correct line type for each.

(a) Axis of a Cone: A Chain thin line (long-dash dot, thin) is used for the axis (centre line) of a cone.

(b) Boundary line: A Continuous thick line (visible outlines/boundary lines) is used for the boundary/outline of an object.

(c) Projection line (Extension line): A Continuous thin line is used for projection/extension lines.

(d) Line for Short Break: A Continuous thin line with zigzags (irregular freehand) is used for short break lines.

(e) Line for Long Break: A Thin straight line with zigzags (ruled line with short zigzag) is used for long break lines.

(f) Dimension line: A Continuous thin line (with arrowheads at both ends) is used for dimension lines.

Given: The concept of 'order of priority of coinciding lines' in Engineering Graphics.

Explanation:
In engineering drawings, it is common for two or more different types of lines to coincide (fall on the same position). In such cases, a definite order of priority is followed to decide which line should be drawn/shown. The order of priority is:

1. Visible outlines and edges (Continuous thick lines) — highest priority
2. Hidden outlines and edges (Dashed thin lines)
3. Cutting plane lines (Chain thin, thick at ends)
4. Centre lines / Axis lines (Chain thin lines)
5. Lines of symmetry (Chain thin lines)
6. Projection/Extension lines (Continuous thin lines) — lowest priority

This means that if a visible outline coincides with a centre line, the visible outline is drawn and the centre line is omitted at that location.

Full form of CAD: Computer Aided Design (or Computer Aided Drafting)

Explanation:
CAD stands for Computer Aided Design. It refers to the use of computer software and systems to assist in the creation, modification, analysis, and optimisation of engineering drawings and designs.

- In CAD, a designer uses a computer with specialised software (such as AutoCAD, SolidWorks, CATIA, etc.) to create 2D drawings or 3D models of objects.
- CAD replaces traditional manual drafting on drawing boards.
- It allows precise and accurate drawings to be made quickly.
- Drawings can be stored, retrieved, modified, and reproduced easily.
- CAD is widely used in Mechanical, Civil, Electrical, Electronics, and Architectural engineering.

Given: The advantages of CAD in saving time, labour, and natural resources.

CAD saves Time:
- Drawings can be created much faster on a computer than by hand.
- Standard components and symbols can be inserted from libraries instantly.
- Modifications and corrections can be made quickly without redrawing.
- Multiple copies of a drawing can be printed in seconds.

CAD saves Labour:
- One CAD operator can do the work that previously required several draughtsmen.
- Repetitive tasks (like copying, mirroring, arraying) are automated.
- Less physical effort is required compared to manual drafting.

CAD saves Natural Resources:
- Paper consumption is greatly reduced as drawings are stored digitally.
- Less use of pencils, erasers, drawing sheets, and other stationery.
- Digital transmission of drawings eliminates the need for physical blueprints, saving paper and printing materials.
- Energy-efficient designs can be analysed and optimised on computer before manufacturing, reducing material waste.

Answer:
Engineers all over the world converse with each other in the universal language known as 'Engineering Graphics' (also called Engineering Drawing).

Their script consists of:
- Lines of various types (thick, thin, dashed, chain, etc.)
- Symbols and conventions (standard symbols for materials, welds, surface finish, etc.)
- Dimensions and numerals (to specify size and location)
- Letters and notes (written in single-stroke Gothic lettering)

Just as different countries have different spoken languages, Engineering Graphics is understood by all technically trained persons regardless of their nationality, making it a truly universal language.

(a) Axis line is drawn as a thin (chain thin / long-dash dot thin) line.

(b) Visible outline is shown graphically as thick (continuous thick line).

(c) Short break line is shown as curved thin line (continuous thin freehand/irregular line).

(d) Long break line is shown as a thin straight line with zigzags.

(e) Hatching line is shown as thin continuous thin lines at 45°.

Answer:
In Engineering Graphics, letters are broadly classified into two types:

1. Gothic Letters (Single Stroke Gothic):
- (a) Vertical (Upright) Gothic Letters — strokes are vertical
- (b) Inclined (Italic) Gothic Letters — strokes are inclined at 75° to the horizontal

2. Roman Letters:
- (a) Upper case (Capital) Roman Letters
- (b) Lower case (Small) Roman Letters

For Engineering Graphics (as per BIS/ISO standards), Single Stroke Gothic Letters (both vertical and inclined) are recommended because they are simple, clear, and can be drawn quickly.

Answer:
'Letter printing' (also called Lettering) in Engineering Graphics refers to the art of writing letters, numerals, and other characters on engineering drawings in a clear, legible, and uniform manner.

Key points:
- Letters are written using a single stroke of the pencil (no sketching or outlining first).
- The letters must be uniform in shape, size, shade, and spacing.
- Guide lines are drawn lightly (with 4H pencil) to maintain uniform height.
- The purpose is to write notes, dimensions, titles, and other information on the drawing so that it is easily readable by any engineer.
- Good lettering is an essential skill in Engineering Graphics.

In Capital Letters:

ENGINEERING GRAPHICS IS THE LANGUAGE OF ENGINEERS

In Small (Lowercase) Letters:

engineering graphics is the language of engineers

*(Note: In practice, these are written using single-stroke Gothic lettering with proper guide lines. Capital letters use two guide lines; small letters use four guide lines — cap line, waist line, base line, and drop line.)*

In Capital (Uppercase) Single Stroke Gothic Letters:

THE HEIGHT OF MAIN TITLE MAY BE TAKEN AS 6 MM, SUBTITLES AS 4 MM AND ANY OTHER TITLE OR DIMENSION IN 2 MM

In Small (Lowercase) Single Stroke Gothic Letters:

the height of main title may be taken as 6 mm, subtitles as 4 mm and any other title or dimension in 2 mm

*(Note: In actual drawing practice, these are drawn carefully between guide lines using a conical blunt pencil with uniform stroke thickness. The numerals 6, 4, 2 are also written in single-stroke Gothic style.)*

Guide lines for Capital Letters:
Two guide lines are used:
1. Base line — the lower line on which letters rest
2. Cap line — the upper line indicating the top of capital letters

The distance between these two lines equals the height of the capital letter (e.g., 6 mm for main title).

Guide lines for Small (Lowercase) Letters:
Four guide lines are used:
1. Base line — the lower line on which letters rest
2. Waist line — the top of the body of small letters (e.g., a, c, e, m, n, etc.)
3. Cap line (Ascender line) — the top of tall letters (e.g., b, d, f, h, k, l, t)
4. Drop line (Descender line) — the bottom line for letters with descenders (e.g., g, j, p, q, y)

Guide lines are drawn lightly with a 4H pencil and erased after lettering if required.

Points to be kept in mind while writing Single Stroke Gothic Letters:

1. Letters should be written by a single stroke of pencil; no prior sketching is done.
2. All letters should be uniform in shape, size, stroke, shade, and spacing.
3. The shine and boldness of letters and numerals should be the same throughout.
4. Letters should be legible and uniform in height and width, except for 'I', 'J', 'M', and 'W'.
5. Letters can be written in expanded or compressed form according to available space.
6. The space between two letters must be kept uniform; a gap equal to twice the thickness of the letter stroke is maintained (except for combinations like LT, AV, PA, LY, AT, TV where the gap is reduced).
7. The space between two words should be equal to the width of one letter.
8. The line thickness for small and capital alphabets shall be the same.
9. Guide lines should be drawn with a 4H pencil (thin lines) and unnecessary lines erased after completion.

To write beautiful and legible letters, the following practices should be followed:

1. Acquire thorough knowledge of all types of letters — their general shape, proportion, design, and direction of strokes.
2. Learn to compose letters in words with proper and uniform spacing.
3. Practice consistently and conscientiously — regular practice is the key to good lettering.
4. Use guide lines for titles and subtitles; draw them lightly with a 4H pencil.
5. Use a conical blunt pencil (H or HB grade) for lettering to get uniform stroke thickness.
6. Maintain uniform height, width, and spacing of all letters.
7. Write letters with single strokes — do not sketch or outline first.
8. Keep the shine and boldness of all letters the same.
9. Ensure letters are legible — not too compressed or too expanded.

Answer:
For Capital (Uppercase) Single Stroke Gothic Letters, the standard ratio of height to width is:

$$\text{Height : Width} = 6 : 5$$

This means if the height of a capital letter is 6 units, its width is approximately 5 units.

Exceptions:
- Letter 'I' — has only width equal to the stroke thickness (very narrow)
- Letter 'J' — slightly narrower
- Letter 'M' — width is approximately equal to height (6:6)
- Letter 'W' — width is greater than height (6:8 approximately)

In general, most capital letters have a height-to-width ratio of approximately 6:5.

Ratio of Height to Width for Small (Lowercase) Letters:

For small letters, the ratio of total height to width of the body is approximately:
$$\text{Height of body : Width} = 4 : 3 \text{ (approximately)}$$

Division of Height for Small Letters:
The total height of small letters is divided into three equal parts using four guide lines:

- Bottom $\frac{1}{3}$ — for descenders (lower portion, for letters like g, j, p, q, y)
- Middle $\frac{1}{3}$ — the body height (for letters like a, c, e, m, n, o, r, s, u, v, w, x, z)
- Top $\frac{1}{3}$ — for ascenders (upper portion, for letters like b, d, f, h, k, l, t)

Thus, four guide lines are needed: drop line, base line, waist line, and cap/ascender line.

Gap between two Letters:
The space between two adjacent letters must be kept uniform. A gap equal to twice the thickness of the letter stroke is maintained between two letters.

Special cases:
- For letter combinations like LT, AV, PA, LY, AT, TV — the gap is reduced (less than normal) because these letters have slanting strokes that naturally create visual space.
- For combinations like AWA, ATA, PAT, AVA, AYA — no gap is given between the letters.

Gap between two Words:
The space between two words should be equal to the width of one letter (approximately equal to the height of the letter for capital letters).

Rules of Spacing between Letters:

1. The space between two letters must be uniform throughout the word.
2. A gap equal to twice the thickness of the letter stroke is normally maintained between two letters.
3. For certain letter combinations — LT, AV, PA, LY, AT, TV — the gap is reduced because the slanting strokes of these letters create an optical illusion of more space. Reducing the gap makes the word look visually uniform.
4. For combinations like AWA, ATA, PAT, AVA, AYA — no gap is given between the letters, as the overlapping slanting strokes fill the visual space.
5. The spacing rule is based on optical uniformity — the letters should appear equally spaced to the eye, even if the actual measured distances differ slightly.
6. The space between two words is equal to the width of one letter.

Purpose: Proper spacing makes the lettering legible, neat, and professional.

In Capital (Uppercase) Letters:

THE LETTERS AND NUMERALS ARE WRITTEN WITH A CONICAL BLUNT PENCIL, THESE CONSIST OF HORIZONTAL, VERTICAL, INCLINED AND CURVED STROKES TO GIVE THE REQUIRED LETTERS.

In Small (Lowercase) Letters:

the letters and numerals are written with a conical blunt pencil, these consist of horizontal, vertical, inclined and curved strokes to give the required letters.

*(Note: In actual drawing practice, these sentences are written carefully between guide lines using single-stroke Gothic lettering with a conical blunt pencil of appropriate grade.)*

Answer:
Drawing instruments are needed for drawing because:

1. They help in making accurate and precise drawings with correct dimensions.
2. They ensure uniformity in line thickness, length, and angles.
3. They save time and effort compared to freehand drawing.
4. They help in drawing standard lines (thick, thin, dashed, chain, etc.) as required by engineering standards.
5. They allow reproducibility — the same drawing can be made again with the same accuracy.
6. Instruments like compasses and dividers help in drawing circles, arcs, and equal divisions accurately.
7. They help maintain cleanliness and neatness of the drawing.

Without proper drawing instruments, it is impossible to make engineering drawings that meet the required standards of accuracy and clarity.

Answer:
A Mini Drafter (also called a Drafting Machine) is a compact drawing instrument that is clamped to the drawing board. It consists of:

- A clamp that fixes it to the top edge of the drawing board.
- A protractor head that can be rotated and locked at any desired angle.
- Two scales (blades) fixed at right angles to each other (one horizontal, one vertical), which can be rotated together to any angle.

Uses of Mini Drafter:
1. It can draw horizontal, vertical, parallel, and inclined lines at any angle.
2. It replaces the need for a Tee-square, set squares, and protractor.
3. It saves time and increases accuracy.
4. Lines can be drawn at any angle by rotating the protractor head.
5. It is especially useful for drawing parallel lines at any inclination quickly.

Important Drawing Instruments and Their Uses:

| Instrument | Use |
|---|---|
| Drawing Board | Provides a flat, smooth surface to fix the drawing sheet |
| Tee-square | Used to draw horizontal lines; also used as a base for set squares |
| Set Squares (45° and 30°–60°) | Used to draw vertical, inclined lines and angles that are multiples of 15° |
| Compass | Used to draw circles and arcs of required radius |
| Divider | Used to transfer distances, divide lines into equal parts |
| Protractor | Used to measure and draw angles |
| Scale (Ruler) | Used to measure and draw lines of required length |
| Mini Drafter | Used to draw horizontal, vertical, parallel, and inclined lines at any angle |
| French Curves | Used to draw irregular curves |
| Pencils (H, HB, B grades) | Used for drawing lines and lettering |
| Eraser (Rubber) | Used to remove unwanted lines |
| Drawing Pins / Clips | Used to fix drawing sheet to the drawing board |

Precautions for a Clean and Neat Drawing:

1. Always clean hands and instruments before fixing the drawing sheet.
2. Wipe hands frequently with a clean handkerchief during drawing work.
3. Move set squares lightly with fingernails while using.
4. Clean pencil smudge with a clean cloth.
5. Remove rubbed pencil powder away from the drawing sheet to prevent smudging.
6. Use a soft rubber for erasing only when absolutely necessary.
7. Remove rubber dust with a clean piece of cloth, never with hands.
8. Use a clean piece of cloth or paper as a hand rest while lettering.
9. Do not keep any articles or utensils on the drawing sheet.
10. When not in use, cover the drawing sheet with a cotton cloth or wide paper.
11. Avoid unnecessary rubbing of lines.
12. Do not touch the drawing sheet with direct hands.

Common Grades of Pencil:

Pencils are graded based on the hardness or softness of their lead (graphite core). The common grades available are:

Hard pencils: 9H, 8H, 7H, 6H, 5H, 4H, 3H, 2H, H
Medium pencil: HB, F
Soft pencils: B, 2B, 3B, 4B, 5B, 6B, 7B

What the letters indicate:
- H stands for Hard — the higher the H number, the harder the lead and the lighter/thinner the line drawn.
- B stands for Black (Soft) — the higher the B number, the softer the lead and the darker/thicker the line drawn.
- HB stands for Hard Black — a medium grade pencil, neither too hard nor too soft.
- F stands for Fine point — slightly harder than HB.

Use in Engineering Drawing:
- 4H or 6H — for guide lines (very light lines)
- H or 2H — for construction lines and dimension lines
- HB or H — for visible outlines and lettering

H Pencil:
- 'H' stands for Hard.
- The lead is hard, produces thin, light grey lines.
- Used for construction lines, dimension lines, and guide lines in engineering drawing.
- Higher H numbers (2H, 4H, 6H) are progressively harder.

HB Pencil:
- 'HB' stands for Hard Black.
- It is a medium grade pencil — neither too hard nor too soft.
- Produces medium dark lines.
- Used for outlines, lettering, and general drawing work.
- It is the most commonly used pencil for everyday writing.

B Pencil:
- 'B' stands for Black (Soft).
- The lead is soft, produces thick, dark, bold lines.
- Used for visible outlines and bold lettering in engineering drawing.
- Higher B numbers (2B, 4B, 6B) are progressively softer and darker.
- Not preferred for fine engineering drawing as lines tend to smudge.

Angles that can be drawn with a pair of set squares:

A pair of set squares consists of:
- 45°–45°–90° set square
- 30°–60°–90° set square

By using these individually or in combination with each other and the Tee-square, the following angles (multiples of 15°) can be drawn:

$$15°, 30°, 45°, 60°, 75°, 90°, 105°, 120°, 135°, 150°, 165°, 180°$$

How:
- 30° — using 30°–60° set square alone
- 45° — using 45° set square alone
- 60° — using 30°–60° set square alone
- 75° — combining 45° + 30° set squares
- 90° — using either set square against Tee-square
- 105° — combining 60° + 45°
- 120° — using 60° set square
- 135° — using 45° set square
- 150° — using 30° set square
- 15° — combining 45° − 30° set squares

Answer:
When drawing with a compass, the pencil lead in the compass should be one grade softer than the pencil used for drawing straight lines.

Reason: When drawing with a compass, the lead is held at an angle and the pressure applied is less uniform compared to drawing straight lines. A softer lead compensates for this and produces lines of the same darkness and brightness.

Examples:
- If H pencil is used for straight lines → use HB pencil in compass
- If 2H pencil is used for straight lines → use H pencil in compass
- If HB pencil is used for straight lines → use B pencil in compass

This ensures that the circles and arcs drawn with the compass have the same shade and brightness as the straight lines drawn with the pencil.

Method to draw parallel lines with a pair of set squares:

Given: A line AB, and requirement to draw a line parallel to it at a distance of 25 mm.

Procedure:
1. Place one set square (say the 30°–60° set square) with its hypotenuse along the given line AB.
2. Place the second set square (45° set square) against the vertical side of the first set square, so that it acts as a base/support.
3. Hold the second set square firmly in position.
4. Slide the first set square along the second set square to the required distance (25 mm, measured with a scale).
5. Draw a line along the hypotenuse of the first set square — this line is parallel to AB at a distance of 25 mm.

Result: The new line CD is parallel to AB and at a distance of 25 mm from it.

*(Note: In actual drawing, mark 25 mm distance on a perpendicular from line AB and draw the parallel line through that point using the set square method.)*

Uses of a Divider:

1. Transferring distances: A divider is used to transfer a distance from one part of the drawing to another without measuring with a scale.
2. Dividing a line into equal parts: A line can be divided into any number of equal parts by trial and adjustment with a divider.
3. Marking equal distances: Equal distances can be stepped off along a line repeatedly.
4. Comparing distances: Two distances can be compared by setting the divider to one distance and checking against the other.
5. Enlarging or reducing distances: Distances can be proportionally enlarged or reduced.

Difference from Compass: A divider has two pointed legs (no pencil), while a compass has one pointed leg and one pencil leg.

(a) To remove unnecessary lines we use eraser.

(b) To make an accurate drawing we use instruments.

(c) To measure angle/angles on a drawing we use protractor.

(d) We use Tee-square for drawing horizontal lines.

(e) The mini drafter can be used for drawing vertical, horizontal and inclined lines.

(f) We use a pair of set squares to draw vertical, parallel and inclined lines.

(g) We use Compass for drawing circles and arcs.

(h) We use Divider for transferring distances, marking a line into equal distances.

(i) A mini-drafter obviates the use of Tee-square, Protractor and a scale.

Matched Answers:

| Table A | Table B |
|---|---|
| (a) Straight lines can be drawn with its help | (g) Tee-square |
| (b) The two parts of the Tee-square are | (c) Butt and blade |
| (c) The distances are measured in millimeters by | (a) Using a scale |
| (d) Lines of different thickness are drawn by using | (e) Pencils |
| (e) Equal distances can be drawn on the drawing sheet by | (b) Using a divider |
| (f) For measuring any angle we use | (f) Protractor |
| (g) We can draw angles in multiple of 15° with this | (d) Set squares |

Answer:
A leader line is a thin continuous line used in engineering drawing to connect a note, dimension, or specification to the feature it refers to on the drawing.

Characteristics of a Leader Line:
1. It is a thin continuous line (same thickness as dimension lines).
2. It is drawn at an angle (usually 30°, 45°, or 60° to the horizontal) — never horizontal or vertical.
3. One end terminates with:
- An arrowhead — when it ends on the outline/boundary of the object
- A dot — when it ends inside the object (on a surface)
- No arrowhead — when it ends on a dimension line
4. The other end connects to a horizontal shoulder on which the note or dimension is written.
5. Leader lines should not cross each other and should not be drawn parallel to adjacent dimension lines.

First Angle Projection Symbol:

The first angle projection symbol (as per BIS/ISO standards) consists of a truncated cone drawn in two views:
- The front view shows the full circular end of the cone (a circle).
- The side view (placed to the LEFT of the front view in first angle projection) shows the tapered/truncated side of the cone.

Description for drawing:
1. Draw a circle of diameter 'd' (representing the front view — circular end of cone).
2. To the left of this circle, draw the side view of the truncated cone — it appears as a trapezium (wider at left, narrower at right, with the narrow end pointing toward the circle).
3. Both views are enclosed within a rectangle (the title block symbol box).

Note: In first angle projection, the object is placed between the observer and the projection plane. The symbol indicates that the side view is placed on the opposite side from where the observer is looking.

*(In actual drawing practice, this symbol is drawn in the title block of the drawing sheet to indicate the projection method used.)*

Dimensioning a Circle:
- The diameter of a circle is dimensioned using the symbol 'Ø' (phi) before the numerical value.
- A dimension line passes through the centre of the circle with arrowheads touching the circle.
- Example: Ø100 (for a circle of 100 mm diameter)
- Alternatively, a leader line with an arrowhead pointing to the circle boundary can be used with the note Ø100.

Dimensioning Concentric Circles:
- Each circle is dimensioned separately.
- Dimension lines are drawn from the centre, and the diameter values are written with the Ø symbol.
- To avoid crowding, leader lines pointing to each circle can be used.
- Example: Inner circle Ø40, Outer circle Ø80.

Dimensioning an Arc of 50 mm radius:
- The radius of an arc is dimensioned using the symbol 'R' before the numerical value.
- A dimension line (leader) is drawn from the centre of the arc to the arc, with an arrowhead at the arc end.
- The dimension is written as R50.
- The dimension line for radius always starts from the centre of the arc (or points toward the centre if the centre is outside the drawing area).

Dimensioning Triangles:

Acute Triangle:
- Dimension the base with a dimension line below the base (with extension lines and arrowheads).
- Dimension the other two sides with dimension lines parallel to each respective side.
- Dimension the angles using arc symbols at the vertices with the angle value written alongside.
- Alternatively, dimension the altitude with a dimension line perpendicular to the base.

Obtuse Triangle:
- Dimension the base with a dimension line below the base.
- Dimension the sides with dimension lines parallel to each side.
- The obtuse angle is dimensioned with an arc at the obtuse vertex.
- Extension lines are drawn perpendicular to the dimension lines.

Right Angled Triangle:
- Dimension the base (MN) with a dimension line below the base.
- Dimension the altitude (MO) with a dimension line to the left/right of the altitude.
- Dimension the hypotenuse (ON) with a dimension line parallel to the hypotenuse.
- The right angle (90°) is indicated by a small square symbol at the right angle vertex — it need not be dimensioned numerically.

*(Note: All dimension lines are thin continuous lines with arrowheads; extension lines do not touch the object outline but leave a small gap of about 1 mm.)*

Necessity of Dimensioning:

Dimensioning is an essential part of engineering drawing because:

1. Specifies Size: Dimensions tell the manufacturer the exact size (length, width, height, diameter, radius, angle) of each feature of the object.

2. Specifies Location: Dimensions specify the exact position/location of features (holes, slots, bosses) relative to each other or to a reference.

3. Enables Manufacturing: Without dimensions, a craftsman or machinist cannot manufacture the part to the correct size.

4. Ensures Interchangeability: Proper dimensioning with tolerances ensures that parts made at different places can be assembled together (interchangeability).

5. Avoids Ambiguity: Dimensions remove any doubt about the size and shape of the object.

6. Facilitates Inspection: Quality control inspectors use dimensions to check whether the manufactured part meets the design requirements.

7. Universal Communication: Dimensions make the drawing universally understandable regardless of the scale of the drawing.

Conclusion: A drawing without dimensions is incomplete and cannot be used for manufacturing.

Aligned System:
1. In the aligned system, dimensions are placed parallel to the dimension line.
2. The dimension figures are written so that they can be read from the bottom or from the right side of the drawing sheet.
3. Horizontal dimensions are read from the bottom; vertical dimensions are read from the right side.
4. Dimensions on inclined dimension lines are written parallel to the inclined line.
5. This system is preferred for large drawings where the drawing sheet may be rotated for reading.

Uni-directional System:
1. In the uni-directional system, all dimensions are written in one direction only — horizontally, so that they can be read from the bottom of the drawing sheet.
2. Dimension lines are broken near the middle to accommodate the dimension figure.
3. All dimensions — horizontal, vertical, and inclined — are written horizontally.
4. This system is preferred for small drawings and is easier to read without rotating the drawing sheet.
5. It is the recommended system as per BIS standards for modern engineering drawings.

Key Difference: In the aligned system, the reader may need to rotate the drawing to read some dimensions; in the uni-directional system, all dimensions are read from one direction (bottom) without rotating.

The two systems of dimensioning are:

1. Aligned System:
- Dimensions are written parallel to the dimension line.
- Horizontal dimensions are read from the bottom; vertical dimensions from the right side.
- The dimension is placed above the dimension line, in the middle.

2. Uni-directional System:
- All dimensions are written horizontally (in one direction).
- All dimensions are read from the bottom of the drawing sheet.
- The dimension line is broken near the middle to insert the dimension figure.
- This is the preferred/recommended system as per BIS/ISO standards.

Answer:
For dimensioning a drawing of large size such as a ship drawing, the Aligned System of dimensioning is adopted.

Reason:
- In large drawings, the drawing sheet may be too large to read all dimensions from one direction.
- The aligned system allows dimensions to be read by rotating the drawing sheet, which is practical for large drawings.
- Dimensions are placed parallel to the dimension line, so they can be read from the bottom or from the right side, making it convenient for large-format drawings.
- The uni-directional system is more suitable for small to medium-sized drawings where all dimensions can be conveniently read from the bottom.

Dimensioning a Narrow Space:

When the space between extension lines is too narrow to accommodate the dimension figure and arrowheads, the following methods are used:

1. Arrowheads outside, dimension inside: The arrowheads are placed outside the extension lines (pointing inward), and the dimension figure is written between the extension lines.

2. Arrowheads and dimension outside: Both the arrowheads and the dimension figure are placed outside the extension lines. A small dot or tick mark may be used instead of arrowheads.

3. Leader line method: A leader line is drawn from the narrow space to a clear area where the dimension is written.

4. Staggered dimensions: When several narrow dimensions are adjacent, they are staggered (placed alternately above and below) to avoid crowding.

5. Oblique arrowheads: Very small oblique strokes (ticks) may be used instead of arrowheads when space is very limited.

Dimensioning an Oblique (Inclined) Line:

In Aligned System:
- The dimension line is drawn parallel to the oblique line.
- Extension lines are drawn perpendicular to the dimension line (i.e., perpendicular to the oblique line).
- The dimension figure is written parallel to the dimension line (i.e., at the same inclination as the oblique line).

In Uni-directional System:
- The dimension line is drawn parallel to the oblique line.
- Extension lines are drawn perpendicular to the dimension line.
- The dimension line is broken near the middle, and the dimension figure is written horizontally (readable from the bottom of the sheet).

Note: The true length of the oblique line is dimensioned, not its horizontal or vertical projection.

Extension Line:
- An extension line is a thin continuous line drawn from the feature being dimensioned, extending beyond the dimension line.
- It is drawn perpendicular to the dimension line.
- It does not touch the outline of the object — a small gap of about 1 mm is left between the object outline and the start of the extension line.
- Extension lines extend about 2–3 mm beyond the dimension line.

Dimension Line:
- A dimension line is a thin continuous line drawn parallel to the feature being dimensioned.
- It has arrowheads at both ends, touching the extension lines.
- The dimension figure is written above the dimension line (aligned system) or the line is broken and the figure written in the gap (uni-directional system).
- Dimension lines should be at least 8 mm away from the object outline and from each other.

On an Equilateral Triangle (e.g., side = 60 mm):
- For the base: extension lines are drawn vertically downward from both ends of the base; a horizontal dimension line is drawn below with arrowheads; '60' is written above the dimension line.
- For the left side: extension lines are drawn perpendicular to the left side; a dimension line parallel to the left side is drawn with arrowheads; '60' is written.
- For the right side: similarly dimensioned.
- All three sides show the dimension '60' with proper extension and dimension lines.

Dimensioning a Rectangle — Aligned System:

Given: Rectangle MNOP with length = 80 mm and width = 50 mm (example values).

Procedure:
1. Draw the rectangle MNOP.
2. For the length (horizontal dimension):
- Draw extension lines vertically downward from M and N (leaving a 1 mm gap from the outline).
- Draw a horizontal dimension line below the rectangle (at least 8 mm from the outline) between the two extension lines, with arrowheads touching the extension lines.
- Write the dimension '80' above the dimension line, parallel to it (readable from the bottom).
3. For the width (vertical dimension):
- Draw extension lines horizontally to the right from N and O (leaving a 1 mm gap).
- Draw a vertical dimension line to the right of the rectangle (at least 8 mm from the outline) between the two extension lines, with arrowheads.
- Write the dimension '50' above the dimension line, parallel to it (readable from the right side).

Result: The rectangle is fully dimensioned in the aligned system with length and width clearly indicated.

Dimensioning a Rectangle — Uni-directional System:

Given: Rectangle MNOP with length = 80 mm and width = 50 mm (example values).

Procedure:
1. Draw the rectangle MNOP.
2. For the length (horizontal dimension):
- Draw extension lines vertically downward from M and N.
- Draw a horizontal dimension line below the rectangle with arrowheads.
- Break the dimension line near the middle and write '80' horizontally in the gap (readable from the bottom).
3. For the width (vertical dimension):
- Draw extension lines horizontally to the right from N and O.
- Draw a vertical dimension line to the right of the rectangle with arrowheads.
- Break the dimension line near the middle and write '50' horizontally in the gap (readable from the bottom — NOT rotated).

Key difference from Aligned System: In the uni-directional system, the dimension '50' for the vertical side is written horizontally (not vertically), so all dimensions are read from the bottom of the sheet without rotating it.

Two recommended systems of placing a dimension on the drawing are Aligned and Uni-directional.

Dimension lines should be drawn at least 8 mm away from the Outline and from each other.

The two main types of dimensions used on a drawing are location and size (dimensions).

Projection line, dimension line, leader line and dimension itself on a drawing are called Elements of dimensioning.

The hatching lines are continuous thin lines, and are drawn at an angle of 45° to an outline of the section.

A dimension is a numerical value expressed in appropriate units of measurement and indicated graphically on a technical drawing with Line, symbol, and note.

All dimensions on the single drawing should be expressed in the same units.

All dimensions are shown from a common base line in Progressive or parallel dimensioning.

No dimension should be written twice on a drawing until unless it is unavoidable.

A dimension given for information only is written as a note.

Two principal requirements of engineering graphics are to specify Shape and size.

Dimension of cylinder should never be given as a R (radius). It should always be given as diameter (Ø).

Generally we prefer a single unit of measurement is in Millimeter (mm).

A leader line end on a dimension line without a Arrowhead.

A leader line end on a surface of an object in an Arrowhead.

A leader line end inside an object in a Dot.

An Outline or axis should never be used as a dimension line.

When a number of parallel dimensions are to be shown near each other, the dimensions should be Staggered.

Given: Straight line AB = 78 mm.
To find: Divide AB into two equal parts (i.e., find the midpoint).

Construction:
1. Draw a straight line AB = 78 mm.
2. With centre A and radius greater than half of AB (say 45 mm), draw arcs on both sides of AB.
3. With the same radius and centre B, draw arcs on both sides of AB to intersect the previous arcs at points P and Q.
4. Join P and Q. The line PQ intersects AB at point M.
5. M is the midpoint of AB.

Result: AM = MB = 39 mm. AB is divided into two equal parts at M.

Given: Straight line AB = 78 mm.
To find: Divide AB into four equal parts.

Construction:
1. Draw a straight line AB = 78 mm.
2. First, find the midpoint M of AB using the perpendicular bisector method (as in Q.1 above). AM = MB = 39 mm.
3. Now find the midpoint of AM: With centres A and M, draw arcs (radius > AM/2) on both sides of AB to intersect at P and Q. Join PQ to intersect AM at point $M_1$. So $AM_1 = M_1M = 19.5$ mm.
4. Similarly, find the midpoint of MB: With centres M and B, draw arcs to intersect at R and S. Join RS to intersect MB at $M_2$. So $MM_2 = M_2B = 19.5$ mm.
5. The four equal parts are: $AM_1 = M_1M = MM_2 = M_2B = 19.5$ mm.

Result: AB is divided into four equal parts at points $M_1$, M, and $M_2$, each part = 19.5 mm.

Given: Line MN = 80 mm. Draw angles of 72°, 36°, and 18° without a protractor.

Concept: These angles are related to the regular pentagon. The interior angle of a regular pentagon = 108°, and the central angle = 72°. Also, 36° = 72°/2, and 18° = 36°/2.

(a) Angle of 72°:
1. Draw line MN = 80 mm.
2. Construct a regular pentagon with one side on MN (using the method of constructing a regular pentagon).
3. The angle at each vertex of the pentagon = 108°. The angle subtended at the centre = 72°.
4. Alternatively: Draw a circle. Divide it into 5 equal parts (each arc = 72°). The angle at the centre between two consecutive radii = 72°.
5. Draw this 72° angle at point M on line MN.

(b) Angle of 36°:
1. Bisect the 72° angle constructed above.
2. The bisector makes an angle of 36° with MN.

(c) Angle of 18°:
1. Bisect the 36° angle constructed above.
2. The bisector makes an angle of 18° with MN.

Result: Angles of 72°, 36°, and 18° are drawn at M on line MN without a protractor.

Given: Line OP. Draw line MN parallel to OP at a distance of 57 mm.

(a) Using Compasses:
1. Take any two points A and B on line OP (well apart).
2. With centre A and radius = 57 mm, draw an arc on one side of OP.
3. With centre B and the same radius = 57 mm, draw another arc on the same side of OP.
4. Draw a line MN tangent to both arcs (touching both arcs from above).
5. MN is parallel to OP at a distance of 57 mm.

(b) Using Set Squares:
1. Place the hypotenuse of one set square (say 30°–60°) along the line OP.
2. Place the second set square (45°) against the vertical side of the first set square as a support.
3. Hold the second set square firmly.
4. Slide the first set square along the second set square upward by 57 mm (measure with a scale).
5. Draw a line MN along the hypotenuse of the first set square.
6. MN is parallel to OP at a distance of 57 mm.

Result: Line MN is parallel to OP at a perpendicular distance of 57 mm.

Given: Straight line MN = 86 mm.
To find: Divide MN into six equal parts.

Construction (using parallel lines method):
1. Draw line MN = 86 mm.
2. From point M, draw a ray MA at any convenient acute angle to MN.
3. Using a compass (or divider), mark off six equal parts of any convenient length along MA: mark points 1, 2, 3, 4, 5, 6 such that M1 = 12 = 23 = 34 = 45 = 56 (any equal length, say 15 mm each).
4. Join point 6 to N.
5. Through points 1, 2, 3, 4, 5, draw lines parallel to 6N (using set squares or compass).
6. These parallel lines intersect MN at points $P_1, P_2, P_3, P_4, P_5$.
7. The six equal parts are: $MP_1 = P_1P_2 = P_2P_3 = P_3P_4 = P_4P_5 = P_5N$.

Result: Each part = 86/6 ≈ 14.33 mm. MN is divided into six equal parts.

Given: Draw various angles using compasses only.

(a) 60°:
1. Draw a ray OA.
2. With centre O and any radius r, draw an arc cutting OA at P.
3. With centre P and same radius r, draw an arc cutting the first arc at Q.
4. Join OQ. Angle QOA = 60°.

(b) 30°:
1. Construct 60° as above.
2. Bisect the 60° angle using compass (bisector of arc PQ).
3. The bisector makes 30° with OA.

(c) 15°:
1. Construct 30° as above.
2. Bisect the 30° angle.
3. The bisector makes 15° with OA.

(d) 90°:
1. Draw ray OA.
2. With centre O and radius r, draw an arc cutting OA at P.
3. With centre P and radius r, mark Q on the arc (60°).
4. With centre Q and same radius, mark R on the arc (120°).
5. Bisect arc QR to get point S (90°).
6. Join OS. Angle SOA = 90°.
*(Alternatively: draw a perpendicular to OA at O.)*

(e) 45°:
1. Construct 90° as above.
2. Bisect the 90° angle.
3. The bisector makes 45° with OA.

(f) 75°:
1. Construct 60° and 90°.
2. Bisect the angle between 60° and 90° (i.e., bisect the 30° arc between them).
3. The bisector makes 75° with OA.

120°:
1. Draw ray OA.
2. With centre O and radius r, draw an arc.
3. Mark P on OA, then Q at 60°, then R at 120° (two steps of 60° each).
4. Join OR. Angle ROA = 120°.

(g) 135°:
1. Construct 90° and 180° (straight line).
2. Bisect the angle between 90° and 180° (i.e., bisect the 90° arc on the other side).
3. The bisector makes 135° with OA.

(h) 150°:
1. Construct 180° (straight line OA extended to B).
2. On the other side, construct 30° from OB (i.e., 180° − 30° = 150° from OA).
3. The line makes 150° with OA.

Given: Two angles: $\alpha = 74°$ and $\beta = 35°$.
To construct: An angle equal to $\alpha - \beta = 74° - 35° = 39°$.

Construction:
1. Draw a ray OA.
2. Using a protractor (or by geometric construction), construct angle $\angle BOA = 74°$ at O on ray OA. Mark ray OB.
3. From ray OB, measure angle $\beta = 35°$ toward OA (i.e., subtract 35° from 74°). Mark ray OC such that $\angle COB = 35°$ and OC is between OA and OB.
4. The angle $\angle COA = 74° - 35° = 39°$.

Geometric method (without protractor):
1. Draw a ray OA.
2. Construct angle $\angle BOA = 74°$ using compass (by constructing 60° + bisecting remaining arcs as needed).
3. At O, construct angle $\angle DOA = 35°$ (by constructing 30° + bisecting).
4. The angle $\angle DOB$ between OD and OB = $74° - 35° = \mathbf{39°}$.

Result: The required angle = 39°.

Given: Two angles: $\alpha = 74°$ and $\beta = 35°$.
To construct: An angle equal to $\alpha + \beta = 74° + 35° = 109°$.

Construction:
1. Draw a ray OA.
2. Construct angle $\angle BOA = 74°$ at O (using compass construction).
3. From ray OB, construct angle $\angle COB = 35°$ on the same side (away from OA).
4. The angle $\angle COA = 74° + 35° = 109°$.

Geometric method:
1. Draw ray OA.
2. At O, construct $\angle BOA = 74°$ (60° + bisect 30° arc to get 15°, then add: 60° + 14° ≈ 74°; or use protractor for the given angles and then copy them geometrically).
3. At O, from OB, construct $\angle COB = 35°$ (30° + bisect 10° arc).
4. Angle $\angle COA = \mathbf{109°}$.

Result: The required angle = 109°.

Given: An angle of 70°.
To find: Bisect the angle (divide into two equal parts of 35° each).

Construction:
1. Draw a ray OA and construct angle $\angle BOA = 70°$ (using protractor or geometric construction). OB is the other ray.
2. With centre O and any convenient radius r, draw an arc cutting OA at P and OB at Q.
3. With centre P and radius greater than PQ/2, draw an arc.
4. With centre Q and the same radius, draw another arc to intersect the first arc at R.
5. Join OR and extend it as a ray.
6. Ray OR bisects angle BOA.

Verification: $\angle AOR = \angle ROB = 35°$.

Result: The angle of 70° is divided into two equal parts of 35° each by the bisector OR.

Given: Two converging lines MN and OP meeting at an angle of 67° (but their point of intersection is outside the drawing area).
To find: Draw the angle bisector without producing the lines to meet.

Construction:
1. Draw a line AB parallel to the bisector direction. To find this:
2. Draw a transversal line XY cutting both MN and OP at points A and B respectively.
3. At point A (on line MN), bisect the angle that XY makes with MN: draw the bisector $Ab_1$.
4. At point B (on line OP), bisect the angle that XY makes with OP: draw the bisector $Bb_2$.
5. The two bisectors $Ab_1$ and $Bb_2$ will meet at a point C (or be parallel if the transversal is parallel to the bisector).

Alternative Method:
1. Draw any transversal cutting MN at A and OP at B.
2. At A, construct an angle equal to half the angle between MN and the transversal, on the side toward OP.
3. At B, construct an angle equal to half the angle between OP and the transversal, on the side toward MN.
4. The two lines from A and B will meet at point C.
5. The locus of such points C gives the angle bisector of the two converging lines.

Result: The angle bisector of the two converging lines is drawn without producing them to meet.

Given: Line MN, and two points O and P outside the line.
To find: Point C on MN such that OC = PC (C is equidistant from O and P).

Concept: The locus of points equidistant from O and P is the perpendicular bisector of OP.

Construction:
1. Join O and P.
2. Find the midpoint of OP: With centres O and P and radius greater than OP/2, draw arcs on both sides of OP to intersect at Q and R.
3. Join QR — this is the perpendicular bisector of OP.
4. The perpendicular bisector QR intersects line MN at point C.
5. C is the required point.

Verification: Since C lies on the perpendicular bisector of OP, OC = PC.

Result: Point C on line MN is found such that OC = PC.

Given: Line MN and a point O outside the line. The line MN cannot be produced (extended).
To find: Draw a perpendicular from O to line MN.

Construction:
1. With centre O and a radius large enough to intersect line MN at two points, draw an arc cutting MN at points A and B.
2. With centre A and radius greater than AB/2, draw an arc on the opposite side of MN from O.
3. With centre B and the same radius, draw another arc to intersect the first arc at point Q.
4. Join O and Q. The line OQ (or its extension) meets MN at point F.
5. OF is perpendicular to MN.

Verification: By construction, F is the foot of the perpendicular from O to MN, and $\angle OFM = \angle OFN = 90°$.

Result: The perpendicular from point O to line MN is drawn at point F, without producing line MN.

Given: Isosceles triangle MNO where base angle = 2 × vertical angle.
Let vertical angle = $\theta$. Then base angles = $2\theta$ each.
Sum of angles: $\theta + 2\theta + 2\theta = 180°$ → $5\theta = 180°$ → $\theta = 36°$.
So vertical angle = 36°, base angles = 72° each.

(a) With Protractor:
1. Draw base MN of suitable length (say 50 mm).
2. At M, draw angle = 72° using protractor.
3. At N, draw angle = 72° using protractor.
4. The two lines meet at O.
5. Triangle MNO is the required isosceles triangle with vertical angle $\angle MON = 36°$ and base angles = 72° each.

(b) With Compass:
1. Draw base MN.
2. At M, construct angle 72° using compass (= 60° + bisect 30° arc to get 12°... or construct 72° as 90° − 18°).
3. At N, construct angle 72° similarly.
4. The intersection of the two lines gives O.
5. MNO is the required triangle.

(c) By Dividing a Semicircle:
1. Draw a semicircle of any radius with diameter MN.
2. Divide the semicircle into 5 equal parts (each = 180°/5 = 36°).
3. The second division point from M gives an arc of 72° from M.
4. Mark point O on the semicircle at 72° from M (i.e., at the second division).
5. Join OM and ON.
6. By the inscribed angle theorem, $\angle MON = 36°$ (half the central angle = 72°/2... actually the angle at the circumference = half the central angle).
7. Triangle MNO has $\angle MON = 36°$ and base angles = 72° each.

Result: Isosceles triangle MNO with vertical angle = 36° and base angles = 72° each.

Given: Right angle triangle MNO, altitude MO = 45 mm, vertical angle $\angle MON = 30°$.
Assume: Right angle is at M (MO is the altitude from M to hypotenuse ON, or M is the right angle vertex).

Construction:
1. Draw MO = 45 mm (vertical line).
2. At M, draw a horizontal line (perpendicular to MO) — this is the base MN.
3. At O, draw a line making angle $\angle MON = 30°$ with MO.
4. This line from O meets the horizontal line from M at N.
5. Triangle MNO is the required right angle triangle.

Verification:
- $\angle OMN = 90°$ (right angle at M)
- MO = 45 mm (altitude)
- $\angle MON = 30°$
- Therefore $\angle MNO = 60°$
- MN = MO × tan(30°) = 45 × (1/√3) ≈ 26 mm
- ON = MO/sin(30°) = 45/0.5 = 90 mm (hypotenuse)

Result: Right angle triangle MNO is constructed with MO = 45 mm and $\angle MON = 30°$.

Given: Right angle triangle MNO, base MN = 50 mm, altitude MO = 55 mm.
Assume: Right angle at M.

Construction:
1. Draw base MN = 50 mm.
2. At M, erect a perpendicular to MN.
3. On this perpendicular, mark MO = 55 mm.
4. Join O to N.
5. Triangle MNO is the required right angle triangle.

Verification:
- $\angle OMN = 90°$
- MN = 50 mm (base)
- MO = 55 mm (altitude)
- ON = $\sqrt{MN^2 + MO^2} = \sqrt{50^2 + 55^2} = \sqrt{2500 + 3025} = \sqrt{5525} \approx 74.3$ mm (hypotenuse)

Result: Right angle triangle MNO is constructed with base MN = 50 mm and altitude MO = 55 mm.

Given: Right angle triangle MNO, altitude MO = 40 mm, hypotenuse ON = 60 mm.
Assume: Right angle at M.

Construction:
1. Draw a line segment ON = 60 mm (hypotenuse).
2. Find the midpoint P of ON.
3. With centre P and radius = ON/2 = 30 mm, draw a semicircle on ON.
4. The right angle vertex M lies on this semicircle (angle in a semicircle = 90°).
5. Now, MO = 40 mm. With centre O and radius = 40 mm, draw an arc to intersect the semicircle at M.
6. Join MO and MN.
7. Triangle MNO is the required right angle triangle.

Verification:
- $\angle OMN = 90°$ (angle in semicircle)
- MO = 40 mm
- ON = 60 mm (hypotenuse)
- MN = $\sqrt{ON^2 - MO^2} = \sqrt{3600 - 1600} = \sqrt{2000} \approx 44.7$ mm

Result: Right angle triangle MNO is constructed with altitude MO = 40 mm and hypotenuse ON = 60 mm.

Given: In triangle MNO, median OD = 50 mm (D is midpoint of MN), $\angle ODM = 50°$ (angle with base MN), $\angle OND = 55°$ (angle with side ON... or $\angle OMD = 55°$ with side OM).

Construction:
1. Draw the median OD = 50 mm.
2. At D, draw the base MN such that $\angle ODM = 50°$ (i.e., the median makes 50° with the base). Since D is the midpoint of MN, draw MN on both sides of D.
3. At O, draw a line making angle $\angle OND = 55°$ with the median direction to locate N.
4. Since D is the midpoint of MN, mark M on the other side of D such that DM = DN.
5. Join OM and ON to complete the triangle.

Result: Triangle MNO is constructed with median OD = 50 mm, making 50° with base MN and 55° with side ON.

Given: Right angle triangle MNO, base MN = 55 mm, base angle $\angle MNO = 30°$.
Assume: Right angle at M.

Construction:
1. Draw base MN = 55 mm.
2. At M, erect a perpendicular to MN (right angle at M).
3. At N, draw a line making angle $\angle MNO = 30°$ with MN.
4. This line from N meets the perpendicular at M at point O.
5. Triangle MNO is the required right angle triangle.

Verification:
- $\angle NMO = 90°$
- $\angle MNO = 30°$
- $\angle MON = 60°$
- MN = 55 mm
- MO = MN × tan(30°) = 55 × (1/√3) ≈ 31.75 mm
- ON = MN/cos(30°) = 55/(√3/2) ≈ 63.5 mm

Result: Right angle triangle MNO is constructed with base MN = 55 mm and $\angle MNO = 30°$.

Given: Right angle triangle MNO, hypotenuse ON = 60 mm, distance from right angle M to hypotenuse ON = 25 mm (i.e., altitude from M to ON = 25 mm).

Construction:
1. Draw hypotenuse ON = 60 mm.
2. Find midpoint P of ON.
3. With centre P and radius = 30 mm, draw a semicircle on ON (M lies on this semicircle).
4. Draw a line parallel to ON at a distance of 25 mm (the altitude from M = 25 mm).
5. The intersection of this parallel line with the semicircle gives point M.
6. Join MO and MN.
7. Triangle MNO is the required right angle triangle.

Verification:
- $\angle OMN = 90°$ (angle in semicircle)
- ON = 60 mm
- Altitude from M to ON = 25 mm
- Check: altitude = $\frac{MO \times MN}{ON}$; also $MO \times MN = ON \times h = 60 \times 25 = 1500$ mm²

Result: Right angle triangle MNO is constructed with hypotenuse ON = 60 mm and altitude from M = 25 mm.

Given: Right angle triangle MNO, hypotenuse ON = 65 mm, median from O to midpoint D of MN, $\angle ODN = 40°$.

Note: In a right angle triangle, the median from the right angle vertex to the hypotenuse = half the hypotenuse. But here the median is from O (not the right angle vertex) to D (midpoint of MN).

Construction:
1. Draw hypotenuse ON = 65 mm.
2. Find midpoint P of ON. Draw semicircle with diameter ON (M lies on this semicircle, $\angle OMN = 90°$).
3. D is the midpoint of MN. The median OD makes $\angle ODN = 40°$ with MN (since D is on MN, $\angle ODN$ is the angle between OD and DN).
4. At N, draw a line at angle $\angle ONM$ (to be determined).
5. Since $\angle ODN = 40°$, and D is midpoint of MN, use the property: in triangle ODN, $\angle ODN = 40°$, ON = 65 mm.
6. Draw line from N making angle with ON. The median OD from O meets MN at D (midpoint).
7. Construct: Draw ON = 65 mm. At N, draw a ray at angle $\angle ONM$. Mark D on this ray. OD makes $\angle ODN = 40°$ with DN. Since $\angle ODN + \angle ODO' = 180°$, $\angle ODO' = 140°$.
8. With trial: Draw ON = 65 mm. Draw a ray from O. The foot D of the median on MN satisfies $\angle ODN = 40°$. Draw a line from N at some angle; from O draw OD such that $\angle ODN = 40°$. D is midpoint of MN, so M is such that DM = DN.
9. Join OM and MN to complete the triangle, verifying $\angle OMN = 90°$.

Result: Right angle triangle MNO is constructed with hypotenuse ON = 65 mm and median OD making $\angle ODN = 40°$.

Given: Right angle triangle MNO, hypotenuse ON = 70 mm, $ON - MO = 30$ mm (difference of hypotenuse and one side).
So: MO = ON − 30 = 70 − 30 = 40 mm.

Construction:
1. Draw hypotenuse ON = 70 mm.
2. Find midpoint P of ON. Draw semicircle with diameter ON.
3. With centre O and radius MO = 40 mm, draw an arc to intersect the semicircle at M.
4. Join MO and MN.
5. Triangle MNO is the required right angle triangle.

Verification:
- $\angle OMN = 90°$ (angle in semicircle)
- ON = 70 mm, MO = 40 mm
- MN = $\sqrt{ON^2 - MO^2} = \sqrt{4900 - 1600} = \sqrt{3300} \approx 57.4$ mm
- Difference: ON − MO = 70 − 40 = 30 mm ✓

Result: Right angle triangle MNO is constructed with hypotenuse ON = 70 mm and MO = 40 mm.

Given: Isosceles triangle MNO, base MN = 40 mm, equal sides MO = NO = 60 mm.

Construction:
1. Draw base MN = 40 mm.
2. With centre M and radius = 60 mm, draw an arc above MN.
3. With centre N and radius = 60 mm, draw another arc to intersect the first arc at O.
4. Join MO and NO.
5. Triangle MNO is the required isosceles triangle.

Verification:
- MN = 40 mm (base)
- MO = NO = 60 mm (equal sides)
- Altitude OD = $\sqrt{60^2 - 20^2} = \sqrt{3600 - 400} = \sqrt{3200} \approx 56.6$ mm

Result: Isosceles triangle MNO is constructed with base MN = 40 mm and equal sides = 60 mm each.

Given: Right angle triangle MNO, hypotenuse ON = 70 mm, $|MO - MN| = 22$ mm.
Let MO = a, MN = b. Then $a^2 + b^2 = 70^2 = 4900$ and $a - b = 22$ (assuming a > b).

Solving: $a = b + 22$. $(b+22)^2 + b^2 = 4900$. $b^2 + 44b + 484 + b^2 = 4900$. $2b^2 + 44b - 4416 = 0$. $b^2 + 22b - 2208 = 0$.
$b = \frac{-22 + \sqrt{484 + 8832}}{2} = \frac{-22 + \sqrt{9316}}{2} = \frac{-22 + 96.5}{2} \approx 37.25$ mm.
$a = 37.25 + 22 = 59.25$ mm.

Construction:
1. Draw hypotenuse ON = 70 mm.
2. Find midpoint P of ON. Draw semicircle with diameter ON.
3. With centre O and radius MO ≈ 59.25 mm (or use geometric construction for difference), draw an arc to intersect the semicircle at M.
4. Join MO and MN.

Geometric Construction for difference:
1. Draw ON = 70 mm. Find midpoint P, draw semicircle.
2. On ON, mark point Q such that OQ = 22 mm (the difference).
3. Draw a semicircle with diameter QN (radius = (70−22)/2 = 24 mm) — this gives a point related to the difference.
4. The intersection with the main semicircle gives M.
5. Join MO and MN to complete the triangle.

Result: Right angle triangle MNO is constructed with hypotenuse ON = 70 mm and difference of sides = 22 mm.

Given: Right angle triangle MNO, altitude ON = 47 mm (perpendicular side), sum of hypotenuse and base = 70 mm.
Assume: Right angle at N. ON = 47 mm (one leg/altitude), MN = base, MO = hypotenuse.
Given: MO + MN = 70 mm.

Let MN = b, MO = h. Then $h + b = 70$ and $h^2 = b^2 + ON^2 = b^2 + 47^2$.
$(70-b)^2 = b^2 + 2209$. $4900 - 140b + b^2 = b^2 + 2209$. $4900 - 140b = 2209$. $140b = 2691$. $b = 19.22$ mm. $h = 70 - 19.22 = 50.78$ mm.

Construction:
1. Draw a line. Mark point N.
2. At N, erect a perpendicular of length ON = 47 mm. Mark O.
3. With centre O and radius = hypotenuse MO = 50.78 mm, draw an arc.
4. With centre N and radius = base MN = 19.22 mm, draw another arc to intersect the first at M.
5. Join MO, MN, and ON.

Geometric Construction:
1. Draw a ray from N. Mark a point A such that NA = 70 mm (= sum of hypotenuse + base).
2. With centre N and radius = ON = 47 mm, draw a semicircle.
3. The construction uses the property: if MO + MN = 70, then M lies on a parabola, but geometrically: draw a line of length 70 mm, and use the right angle condition to locate M.
4. Complete the triangle as calculated above.

Result: Right angle triangle MNO is constructed with altitude ON = 47 mm and MO + MN = 70 mm.

Given: Isosceles triangle MNO, equal sides MO = NO = 60 mm, base angles $\angle OMN = \angle ONM = 50°$.

Note: Vertical angle = 180° − 50° − 50° = 80°.

Construction:
1. Draw a line. At any point M, draw a ray making 50° with the base direction.
2. On this ray, mark MO = 60 mm.
3. At N (to be determined), draw a ray making 50° on the other side.
4. Alternatively:
- Draw base MN of length = $2 \times 60 \times \cos(50°) = 2 \times 60 \times 0.643 \approx 77.1$ mm...
- Actually, use: MN = $2 \times MO \times \sin(\angle MON/2)$...
- Better: Draw MO = 60 mm at 50° from base. The base MN = $2 \times 60 \times \sin(50°) \approx 91.9$ mm...

Correct Construction:
1. Draw ray from M at 50° to horizontal.
2. Mark O on this ray at 60 mm from M.
3. Draw ray from O downward making angle 80° at O (vertical angle).
4. Alternatively: With centre M and radius 60 mm, and centre N and radius 60 mm, find O.
5. First find MN: In triangle, $\frac{MN}{\sin 80°} = \frac{60}{\sin 50°}$. $MN = \frac{60 \times \sin 80°}{\sin 50°} = \frac{60 \times 0.985}{0.766} \approx 77.1$ mm.
6. Draw MN ≈ 77.1 mm.
7. With centres M and N, radius = 60 mm each, draw arcs to intersect at O.
8. Join MO and NO.

Result: Isosceles triangle MNO with MO = NO = 60 mm and base angles = 50° each.

Given: Right angle triangle MNO, hypotenuse ON = 60 mm, base MN + altitude MO = 55 mm.
Assume: Right angle at M.
Let MN = b, MO = a. Then $a + b = 55$ and $a^2 + b^2 = 60^2 = 3600$.
$(a+b)^2 - 2ab = 3600$. $55^2 - 2ab = 3600$. $3025 - 2ab = 3600$. $2ab = -575$.

This gives a negative value, which means the given data is inconsistent (the sum of two legs cannot be 55 mm if the hypotenuse is 60 mm, since for a right triangle, each leg < hypotenuse, so a + b > hypotenuse for most cases... actually a + b &gt; \sqrt{2} \times \frac{ON}{\sqrt{2}} = ON... let me recheck).

Recheck: For a right triangle with hypotenuse 60: minimum value of $a + b$ occurs when $a = b = 60/\sqrt{2} \approx 42.4$ mm, giving $a + b \approx 84.9$ mm. So a + b = 55 &lt; 84.9 is impossible for a right triangle with hypotenuse 60 mm.

Note to student: The given data appears to be inconsistent. The sum of the two legs of a right triangle with hypotenuse 60 mm must be greater than approximately 84.9 mm. Please verify the problem data. If the hypotenuse is meant to be larger, or the sum is meant to be larger, the construction can be carried out.

Assuming the problem intends hypotenuse = 60 mm and sum = 85 mm (as a corrected version):
$a + b = 85$, $a^2 + b^2 = 3600$. $(85)^2 - 2ab = 3600$. $7225 - 2ab = 3600$. $ab = 1812.5$. $a$ and $b$ are roots of $x^2 - 85x + 1812.5 = 0$. $x = \frac{85 \pm \sqrt{7225 - 7250}}{2}$ — still negative discriminant.

Correct approach: The problem data as stated (hypotenuse = 60, sum = 55) is geometrically impossible. The student should note this and verify with the teacher.

Given: Isosceles triangle MNO, altitude OD = 45 mm (D is midpoint of base MN), base angles $\angle OMN = \angle ONM = 50°$.

Construction:
1. Draw a vertical line OD = 45 mm (the altitude).
2. At D, draw a horizontal line (the base MN, perpendicular to OD).
3. In right triangle OMD: $\tan(\angle OMD) = OD/MD$. $\tan(50°) = 45/MD$. $MD = 45/\tan(50°) = 45/1.192 \approx 37.8$ mm.
4. Mark M and N on the base such that DM = DN ≈ 37.8 mm. So MN ≈ 75.6 mm.
5. Join OM and ON.
6. Triangle MNO is the required isosceles triangle.

Geometric Construction (without calculation):
1. Draw OD = 45 mm vertically.
2. At D, draw horizontal base MN.
3. At M, draw a ray making 50° with MN (upward toward O).
4. This ray meets the vertical line OD extended at O.
5. Mark N symmetrically: DN = DM.
6. Join ON.

Result: Isosceles triangle MNO with altitude OD = 45 mm and base angles = 50° each.

Given: Isosceles triangle MNO, base MN = 40 mm, altitude OD = 50 mm (D is midpoint of MN; note: 'AD' in the problem likely refers to the altitude from O to base MN).

Construction:
1. Draw base MN = 40 mm.
2. Find midpoint D of MN (DM = DN = 20 mm).
3. At D, erect a perpendicular to MN.
4. On this perpendicular, mark O such that OD = 50 mm.
5. Join OM and ON.
6. Triangle MNO is the required isosceles triangle.

Verification:
- MN = 40 mm (base)
- OD = 50 mm (altitude)
- OM = ON = $\sqrt{OD^2 + DM^2} = \sqrt{50^2 + 20^2} = \sqrt{2500 + 400} = \sqrt{2900} \approx 53.9$ mm (equal sides)

Result: Isosceles triangle MNO is constructed with base MN = 40 mm and altitude OD = 50 mm.

Given: Isosceles triangle MNO, vertical angle $\angle MON = 40°$, base MN = 50 mm.

Construction:
1. Draw base MN = 50 mm.
2. Find midpoint D of MN.
3. Each base angle = $(180° - 40°)/2 = 70°$.
4. At M, draw a ray making 70° with MN (upward).
5. At N, draw a ray making 70° with NM (upward, on the other side).
6. The two rays meet at O.
7. Triangle MNO is the required isosceles triangle.

Alternatively (using circumscribed circle method):
1. The angle $\angle MON = 40°$ is the angle at the apex.
2. By the inscribed angle theorem, the central angle = $2 \times 40° = 80°$.
3. Draw MN = 50 mm. The circumradius $R = \frac{MN}{2\sin(\angle MON)} = \frac{50}{2\sin(40°)} = \frac{50}{2 \times 0.643} \approx 38.9$ mm.
4. Find the circumcentre and draw the circumscribed circle. O lies on this circle on the arc MN that subtends 40° at the circumference.

Result: Isosceles triangle MNO with vertical angle = 40° and base MN = 50 mm.

Given: Isosceles triangle MNO, altitude OD = 45 mm, vertical angle $\angle MON = 45°$.

Construction:
1. Draw a vertical line OD = 45 mm.
2. At D, draw a horizontal line (base MN).
3. The altitude OD bisects the vertical angle, so $\angle MOD = \angle NOD = 45°/2 = 22.5°$.
4. In right triangle OMD: $\tan(22.5°) = MD/OD$. $MD = OD \times \tan(22.5°) = 45 \times 0.4142 \approx 18.6$ mm.
5. Mark M and N on the base such that DM = DN ≈ 18.6 mm.
6. Join OM and ON.
7. Triangle MNO is the required isosceles triangle.

Geometric Construction:
1. Draw OD = 45 mm.
2. At O, draw rays making 22.5° on each side of OD (bisect 45° angle at O).
3. These rays meet the horizontal base at M and N respectively.
4. Join MN.

Result: Isosceles triangle MNO with altitude OD = 45 mm and vertical angle = 45°.

Given: Isosceles triangle MNO, equal sides MO = NO = 60 mm, vertical angle = (1/3) × base angle.
Let base angle = $\beta$, vertical angle = $\beta/3$.
Sum: $\beta/3 + \beta + \beta = 180°$. $\beta/3 + 2\beta = 180°$. $\frac{\beta + 6\beta}{3} = 180°$. $7\beta/3 = 180°$. $\beta = \frac{180° \times 3}{7} \approx 77.14°$.
Vertical angle = $\beta/3 \approx 25.71°$.

Construction:
1. Calculate: base angles ≈ 77.14° each, vertical angle ≈ 25.71°.
2. Find base MN: $\frac{MN}{\sin(25.71°)} = \frac{60}{\sin(77.14°)}$. $MN = \frac{60 \times \sin(25.71°)}{\sin(77.14°)} = \frac{60 \times 0.4337}{0.9749} \approx 26.7$ mm.
3. Draw base MN ≈ 26.7 mm.
4. With centres M and N, radius = 60 mm each, draw arcs to intersect at O.
5. Join MO and NO.
6. Triangle MNO is the required isosceles triangle.

Result: Isosceles triangle MNO with sides = 60 mm and vertical angle ≈ 25.71° (= 1/3 of base angle ≈ 77.14°).

Given: Isosceles triangle MNO, perimeter = 70 mm, altitude OD = 30 mm.
Let base MN = b, equal sides MO = NO = a. Then $2a + b = 70$ and $OD = 30$ mm.
Also: $a^2 = OD^2 + (b/2)^2 = 900 + b^2/4$.
From perimeter: $b = 70 - 2a$.
$a^2 = 900 + (70-2a)^2/4$. $4a^2 = 3600 + (70-2a)^2$. $4a^2 = 3600 + 4900 - 280a + 4a^2$. $0 = 8500 - 280a$. $a = 8500/280 \approx 30.36$ mm. $b = 70 - 2(30.36) = 9.28$ mm.

Construction:
1. Draw base MN = 9.28 mm.
2. Find midpoint D of MN.
3. At D, erect perpendicular of length OD = 30 mm. Mark O.
4. Join MO and NO (each ≈ 30.36 mm).
5. Verify: perimeter = 9.28 + 30.36 + 30.36 ≈ 70 mm ✓

Result: Isosceles triangle MNO with perimeter = 70 mm and altitude OD = 30 mm.

Given: Isosceles right angle triangle MNO, perimeter = 70 mm.
Assume: Right angle at M. Equal legs: MO = MN = a. Hypotenuse ON = $a\sqrt{2}$.
Perimeter: $a + a + a\sqrt{2} = 70$. $a(2 + \sqrt{2}) = 70$. $a = \frac{70}{2 + \sqrt{2}} = \frac{70}{3.414} \approx 20.5$ mm.
Hypotenuse: ON = $20.5\sqrt{2} \approx 29$ mm.

Construction:
1. Draw MN = 20.5 mm.
2. At M, erect a perpendicular to MN.
3. On this perpendicular, mark MO = 20.5 mm.
4. Join ON (hypotenuse ≈ 29 mm).
5. Triangle MNO is the required isosceles right angle triangle.

Verification: Perimeter = 20.5 + 20.5 + 29 = 70 mm ✓

Result: Isosceles right angle triangle MNO with perimeter = 70 mm, legs ≈ 20.5 mm each.

Given: Isosceles triangle MNO, base MN = 40 mm, base angle = 2 × vertical angle.
Let vertical angle = $\theta$. Base angles = $2\theta$ each.
$\theta + 2\theta + 2\theta = 180°$. $5\theta = 180°$. $\theta = 36°$.
Base angles = 72° each, vertical angle = 36°.

Construction:
1. Draw base MN = 40 mm.
2. At M, draw a ray making 72° with MN (upward).
3. At N, draw a ray making 72° with NM (upward, on the other side).
4. The two rays meet at O.
5. Triangle MNO is the required isosceles triangle.

Verification:
- $\angle OMN = \angle ONM = 72°$
- $\angle MON = 36°$
- Equal sides: MO = NO = $\frac{MN \times \sin(72°)}{\sin(36°)} = \frac{40 \times 0.951}{0.588} \approx 64.7$ mm

Result: Isosceles triangle MNO with base MN = 40 mm, base angles = 72°, vertical angle = 36°.

Given: Isosceles right angle triangle MNO, right angle at M, MO = MN = a (equal legs), hypotenuse ON = $a\sqrt{2}$.
Given: ON + MO = 60 mm. $a\sqrt{2} + a = 60$. $a(\sqrt{2} + 1) = 60$. $a = \frac{60}{\sqrt{2}+1} = \frac{60(\sqrt{2}-1)}{(\sqrt{2}+1)(\sqrt{2}-1)} = \frac{60(\sqrt{2}-1)}{1} = 60(\sqrt{2}-1) \approx 60 \times 0.414 = 24.85$ mm.
Hypotenuse: ON = $24.85\sqrt{2} \approx 35.15$ mm.

Construction:
1. Draw MN = 24.85 mm.
2. At M, erect a perpendicular to MN.
3. On this perpendicular, mark MO = 24.85 mm.
4. Join ON (≈ 35.15 mm).
5. Triangle MNO is the required isosceles right angle triangle.

Verification: ON + MO = 35.15 + 24.85 = 60 mm ✓

Result: Isosceles right angle triangle MNO with legs ≈ 24.85 mm and hypotenuse ≈ 35.15 mm.

Given: Isosceles triangle MNO, base MN = 50 mm, altitude OD + equal side MO = 60 mm.
Let equal side MO = NO = a, altitude OD = h.
Then: $a + h = 60$ and $a^2 = h^2 + (MN/2)^2 = h^2 + 25^2 = h^2 + 625$.
Substituting $a = 60 - h$: $(60-h)^2 = h^2 + 625$. $3600 - 120h + h^2 = h^2 + 625$. $3600 - 120h = 625$. $120h = 2975$. $h = 24.79$ mm. $a = 60 - 24.79 = 35.21$ mm.

Construction:
1. Draw base MN = 50 mm.
2. Find midpoint D of MN (DM = DN = 25 mm).
3. At D, erect perpendicular. Mark O at height OD = 24.79 mm.
4. Join MO and NO (each ≈ 35.21 mm).
5. Triangle MNO is the required isosceles triangle.

Verification: MO + OD = 35.21 + 24.79 = 60 mm ✓

Result: Isosceles triangle MNO with base MN = 50 mm, altitude ≈ 24.79 mm, equal sides ≈ 35.21 mm.

Given: Triangle MNO, base MN = 70 mm, altitude from O to MN = 40 mm (let D be foot of altitude, so OD = 40 mm), and side ON = 55 mm (assuming 'OD' in the problem refers to side ON = 55 mm; note: OD is the altitude).

Reinterpretation: Base MN = 70 mm, altitude OD = 40 mm (D on MN), side OM = 55 mm (given as 'OD 55 mm' — likely a typo for OM = 55 mm).

Construction:
1. Draw base MN = 70 mm.
2. Draw a line parallel to MN at a distance of 40 mm (the locus of O, since altitude = 40 mm).
3. With centre M and radius = 55 mm (side OM), draw an arc to intersect the parallel line at O.
4. Join MO and NO.
5. Triangle MNO is the required triangle.

Verification:
- MN = 70 mm ✓
- Altitude from O to MN = 40 mm ✓
- OM = 55 mm ✓

Result: Triangle MNO is constructed with base MN = 70 mm, altitude = 40 mm, and side OM = 55 mm.

Given: Triangle MNO, MN = 55 mm, ON = 45 mm, NM = 55 mm.
Note: MN = NM = 55 mm (same side, so this is likely MN = 55 mm and OM = 55 mm, making it isosceles).
Assume: MN = 55 mm (base), ON = 45 mm, OM = 55 mm.

Construction:
1. Draw base MN = 55 mm.
2. With centre N and radius = ON = 45 mm, draw an arc above MN.
3. With centre M and radius = OM = 55 mm, draw another arc to intersect the first arc at O.
4. Join MO and NO.
5. Triangle MNO is the required triangle.

Verification:
- MN = 55 mm ✓
- ON = 45 mm ✓
- OM = 55 mm ✓

Result: Triangle MNO is constructed with the three sides as given.

Given: Triangle MNO, altitude OD = 40 mm (D is foot of perpendicular from O to MN), ON = 50 mm, OM = 45 mm.

Construction:
1. Draw a horizontal line (base MN).
2. Mark a point D on MN. At D, erect a perpendicular of length OD = 40 mm. Mark O.
3. With centre O and radius = OM = 45 mm, draw an arc to intersect the base line at M.
4. With centre O and radius = ON = 50 mm, draw an arc to intersect the base line at N (on the other side of D, or same side).
5. Join MO and NO.
6. Triangle MNO is the required triangle.

Finding positions of M and N:
- DM = $\sqrt{OM^2 - OD^2} = \sqrt{45^2 - 40^2} = \sqrt{2025 - 1600} = \sqrt{425} \approx 20.6$ mm
- DN = $\sqrt{ON^2 - OD^2} = \sqrt{50^2 - 40^2} = \sqrt{2500 - 1600} = \sqrt{900} = 30$ mm
- MN = DM + DN = 20.6 + 30 = 50.6 mm (if D is between M and N) or MN = |DN − DM| = 9.4 mm (if D is outside MN)

Result: Triangle MNO is constructed with altitude OD = 40 mm, ON = 50 mm, OM = 45 mm.

Note: This question is identical to Question 27 above.

Given: Triangle MNO, altitude OD = 40 mm, ON = 50 mm, OM = 45 mm.

Construction: (Same as Q.27)
1. Draw base line MN.
2. Mark point D on MN. At D, erect perpendicular OD = 40 mm. Mark O.
3. With centre O, radius = 45 mm, mark M on base line: DM = $\sqrt{45^2 - 40^2} \approx 20.6$ mm.
4. With centre O, radius = 50 mm, mark N on base line: DN = $\sqrt{50^2 - 40^2} = 30$ mm.
5. Join MO and NO.

Result: Triangle MNO with altitude OD = 40 mm, OM = 45 mm, ON = 50 mm is constructed.

Given: Triangle MNO, base MN = 70 mm, side ON = 45 mm, included angle $\angle MNO = 60°$ (angle at N between MN and ON).

Construction:
1. Draw base MN = 70 mm.
2. At N, draw a ray making angle $\angle MNO = 60°$ with NM.
3. On this ray, mark NO = 45 mm. Mark point O.
4. Join MO.
5. Triangle MNO is the required triangle.

Verification:
- MN = 70 mm ✓
- ON = 45 mm ✓
- $\angle MNO = 60°$ ✓
- MO = $\sqrt{MN^2 + ON^2 - 2 \cdot MN \cdot ON \cdot \cos(60°)} = \sqrt{4900 + 2025 - 2 \times 70 \times 45 \times 0.5} = \sqrt{6925 - 3150} = \sqrt{3775} \approx 61.4$ mm

Result: Triangle MNO is constructed with MN = 70 mm, ON = 45 mm, and included angle = 60°.

Given: Triangle MNO, base MN = 55 mm, angles in ratio 4 : 6 : 8.
Sum of angles = 180°. Let angles be $4k, 6k, 8k$. $4k + 6k + 8k = 180°$. $18k = 180°$. $k = 10°$.
Angles: $\angle M = 40°$, $\angle N = 60°$, $\angle O = 80°$ (or any assignment).

Assume: $\angle OMN = 40°$, $\angle ONM = 60°$, $\angle MON = 80°$.

Construction:
1. Draw base MN = 55 mm.
2. At M, draw a ray making 40° with MN (upward).
3. At N, draw a ray making 60° with NM (upward, on the other side).
4. The two rays meet at O.
5. Triangle MNO is the required triangle.

Verification: $\angle M + \angle N + \angle O = 40° + 60° + 80° = 180°$ ✓

Result: Triangle MNO with base MN = 55 mm and angles 40°, 60°, 80° (ratio 4:6:8).

Given: Triangle MNO, perimeter = 100 mm, sides in ratio 3 : 5 : 4.
Let sides be $3k, 5k, 4k$. $3k + 5k + 4k = 100$. $12k = 100$. $k = 100/12 = 8.33$ mm.
Sides: MN = $3 \times 8.33 = 25$ mm, NO = $5 \times 8.33 = 41.67$ mm, MO = $4 \times 8.33 = 33.33$ mm.

Construction:
1. Draw base MN = 25 mm.
2. With centre M and radius = MO = 33.33 mm, draw an arc.
3. With centre N and radius = NO = 41.67 mm, draw another arc to intersect the first at O.
4. Join MO and NO.
5. Triangle MNO is the required triangle.

Verification: Perimeter = 25 + 41.67 + 33.33 = 100 mm ✓

Result: Triangle MNO with perimeter = 100 mm and sides in ratio 3:5:4 (i.e., 25 mm, 33.33 mm, 41.67 mm).

Given: Triangle MNO, base MN = 50 mm, $|MO - NO| = 15$ mm, base angle $\angle MNO = 60°$ (or $\angle OMN = 60°$).

Assume: $\angle MNO = 60°$ and MO − NO = 15 mm (MO > NO).

Construction:
1. Draw base MN = 50 mm.
2. At N, draw a ray making 60° with NM. O lies on this ray.
3. The locus of O such that MO − NO = 15 mm is a hyperbola with foci M and N. Geometrically:
4. On the ray from N (at 60°), mark a point O such that MO − NO = 15 mm.
5. Trial method: Mark various points on the ray from N and measure MO − NO until the difference = 15 mm.
6. Geometric method: On the ray from N at 60°, mark point A such that NA = 15 mm. Draw a circle with centre M and radius = MA. The intersection of this circle with the ray from N gives O... (this is an approximation).

Better geometric construction:
1. Draw MN = 50 mm.
2. At N, draw ray at 60°.
3. On this ray, mark point P such that NP = any length. Measure MP. Adjust P until MP − NP = 15 mm.
4. Alternatively: On the ray from N, mark point Q such that NQ = some value. On MN extended beyond N, mark point R such that NR = 15 mm. Draw circle with centre M and radius MR. Intersection with the ray gives O.

Result: Triangle MNO with base MN = 50 mm, base angle = 60°, and MO − NO = 15 mm is constructed.

Given: Triangle MNO, perimeter = 70 mm, angles in ratio 5 : 6 : 7.
Let angles be $5k, 6k, 7k$. $5k + 6k + 7k = 180°$. $18k = 180°$. $k = 10°$.
Angles: $\angle M = 50°$, $\angle N = 60°$, $\angle O = 70°$.

By sine rule: $\frac{MN}{\sin O} = \frac{NO}{\sin M} = \frac{MO}{\sin N}$.
$\frac{MN}{\sin 70°} = \frac{NO}{\sin 50°} = \frac{MO}{\sin 60°}$.

Let the common ratio = $k'$. Then:
$MN = k' \sin 70°$, $NO = k' \sin 50°$, $MO = k' \sin 60°$.
Perimeter = $k'(\sin 70° + \sin 50° + \sin 60°) = k'(0.9397 + 0.7660 + 0.8660) = k' \times 2.5717 = 70$.
$k' = 70/2.5717 \approx 27.22$ mm.

Sides:
- MN = $27.22 \times 0.9397 \approx 25.58$ mm
- NO = $27.22 \times 0.7660 \approx 20.85$ mm
- MO = $27.22 \times 0.8660 \approx 23.57$ mm

Construction:
1. Draw base MN = 25.58 mm.
2. At M, draw ray at angle $\angle M = 50°$.
3. At N, draw ray at angle $\angle N = 60°$.
4. The rays meet at O.
5. Triangle MNO is the required triangle.

Verification: Perimeter ≈ 25.58 + 20.85 + 23.57 = 70 mm ✓

Result: Triangle MNO with perimeter = 70 mm and angles 50°, 60°, 70° (ratio 5:6:7).

Given: Square with side = 65 mm.

Construction:
1. Draw base AB = 65 mm.
2. At A, erect a perpendicular to AB.
3. On this perpendicular, mark AD = 65 mm.
4. With centre D and radius = 65 mm, draw an arc.
5. With centre B and radius = 65 mm, draw another arc to intersect the first at C.
6. Join BC and DC.
7. ABCD is the required square.

Verification:
- All sides = 65 mm ✓
- All angles = 90° ✓
- Diagonals = $65\sqrt{2} \approx 91.9$ mm and bisect each other at 90° ✓

Result: Square ABCD with side = 65 mm is constructed.

Given: Rhombus with diagonals $d_1 = 55$ mm and $d_2 = 70$ mm.

Construction:
1. Draw diagonal AC = 70 mm (longer diagonal).
2. Find midpoint O of AC.
3. At O, draw a perpendicular to AC.
4. On this perpendicular, mark OB = OD = 55/2 = 27.5 mm on each side.
5. Join AB, BC, CD, and DA.
6. ABCD is the required rhombus.

Verification:
- Diagonals bisect each other at right angles ✓
- Side = $\sqrt{(70/2)^2 + (55/2)^2} = \sqrt{35^2 + 27.5^2} = \sqrt{1225 + 756.25} = \sqrt{1981.25} \approx 44.5$ mm (all sides equal) ✓

Result: Rhombus ABCD with diagonals 55 mm and 70 mm is constructed.

Given: Quadrilateral MORE, MO = 60 mm, OR = 45 mm, $\angle M = 60°$, $\angle O = 105°$, $\angle R = 105°$.
Sum of angles: $\angle M + \angle O + \angle R + \angle E = 360°$. $60° + 105° + 105° + \angle E = 360°$. $\angle E = 90°$.

Construction:
1. Draw MO = 60 mm.
2. At M, draw a ray making $\angle M = 60°$ with MO (this is side ME).
3. At O, draw a ray making $\angle O = 105°$ with OM (this is side OR).
4. On the ray from O, mark OR = 45 mm. Mark R.
5. At R, draw a ray making $\angle R = 105°$ with RO (this is side RE).
6. The ray from R meets the ray from M at E.
7. MORE is the required quadrilateral.

Verification: $\angle E = 90°$ (check by measurement) ✓

Result: Quadrilateral MORE is constructed with the given dimensions.

Given: Parallelogram ABCD, AB = 50 mm, BC = 60 mm, $\angle D = 85°$.
In a parallelogram: Opposite angles are equal, so $\angle B = \angle D = 85°$. Adjacent angles are supplementary, so $\angle A = \angle C = 180° - 85° = 95°$.

Construction:
1. Draw AB = 50 mm.
2. At B, draw a ray making $\angle ABC = 85°$ with BA.
3. On this ray, mark BC = 60 mm. Mark C.
4. With centre C and radius = AB = 50 mm, draw an arc.
5. With centre A and radius = BC = 60 mm, draw another arc to intersect the first at D.
6. Join AD and CD.
7. ABCD is the required parallelogram.

Verification:
- AB = CD = 50 mm ✓
- BC = AD = 60 mm ✓
- $\angle B = \angle D = 85°$ ✓
- $\angle A = \angle C = 95°$ ✓

Result: Parallelogram ABCD with AB = 50 mm, BC = 60 mm, and $\angle D = 85°$ is constructed.

Given: Rectangle MNOP, base MN = 60 mm, side NO = 40 mm.

Construction:
1. Draw base MN = 60 mm.
2. At M, erect a perpendicular to MN.
3. On this perpendicular, mark MP = 40 mm. Mark P.
4. At N, erect a perpendicular to MN.
5. On this perpendicular, mark NO = 40 mm. Mark O.
6. Join PO.
7. MNOP is the required rectangle.

Verification:
- MN = PO = 60 mm ✓
- NO = MP = 40 mm ✓
- All angles = 90° ✓
- Diagonal MO = $\sqrt{60^2 + 40^2} = \sqrt{3600 + 1600} = \sqrt{5200} \approx 72.1$ mm

Result: Rectangle MNOP with base MN = 60 mm and side NO = 40 mm is constructed.

Given: Rectangle MNOP, diagonal MO = 70 mm, difference of sides = 25 mm.
Let MN = a (length), NO = b (width). $a - b = 25$ and $a^2 + b^2 = 70^2 = 4900$.
$(a-b)^2 = a^2 - 2ab + b^2 = 625$. $4900 - 2ab = 625$. $2ab = 4275$. $ab = 2137.5$.
$a$ and $b$ are roots of $x^2 - (a+b)x + ab = 0$. $(a+b)^2 = (a-b)^2 + 4ab = 625 + 8550 = 9175$. $a+b = \sqrt{9175} \approx 95.8$ mm.
$a = (95.8 + 25)/2 = 60.4$ mm. $b = (95.8 - 25)/2 = 35.4$ mm.

Construction:
1. Draw MN = 60.4 mm.
2. At M and N, erect perpendiculars.
3. Mark MP = NO = 35.4 mm.
4. Join PO.
5. MNOP is the required rectangle.

Verification: Diagonal = $\sqrt{60.4^2 + 35.4^2} = \sqrt{3648 + 1253} = \sqrt{4901} \approx 70$ mm ✓

Result: Rectangle MNOP with diagonal = 70 mm and difference of sides = 25 mm is constructed.

Given: Parallelogram MNOP, diagonal MO = 50 mm, diagonal NP = 40 mm, angle between diagonals $\angle OQN = 60°$ (Q is the intersection of diagonals).

Property: In a parallelogram, diagonals bisect each other. So MQ = QO = 25 mm and NQ = QP = 20 mm.

Construction:
1. Draw a line segment. Mark Q (intersection of diagonals).
2. Draw MO = 50 mm through Q such that MQ = QO = 25 mm.
3. At Q, draw a line making $\angle OQN = 60°$ with MO.
4. On this line, mark QN = QP = 20 mm on each side of Q.
5. Join MN, NO, OP, and PM.
6. MNOP is the required parallelogram.

Verification:
- MO = 50 mm ✓, NP = 40 mm ✓
- Diagonals bisect each other ✓
- $\angle OQN = 60°$ ✓

Result: Parallelogram MNOP with diagonals MO = 50 mm, NP = 40 mm, and included angle = 60° is constructed.

Given: Rectangle MNOP, diagonals MO = NP = 70 mm, angle between diagonals = 45°.

Property: In a rectangle, diagonals are equal and bisect each other. So MQ = QO = NQ = QP = 35 mm (Q = intersection).

Construction:
1. Mark point Q (intersection of diagonals).
2. Draw diagonal MO = 70 mm through Q (MQ = QO = 35 mm).
3. At Q, draw a line making 45° with MO.
4. On this line, mark QN = QP = 35 mm on each side of Q.
5. Join MN, NO, OP, and PM.
6. MNOP is the required rectangle.

Verification:
- MO = NP = 70 mm ✓
- Angle between diagonals = 45° ✓
- All angles of MNOP = 90° (since diagonals are equal in a rectangle) ✓
- Sides: MN = $\sqrt{MQ^2 + NQ^2 - 2 \cdot MQ \cdot NQ \cdot \cos(45°)}$... actually use: MN = $2 \times 35 \times \sin(45°/2)$...
- MN = $\sqrt{35^2 + 35^2 - 2 \times 35 \times 35 \times \cos(45°)} = 35\sqrt{2 - \sqrt{2}} \approx 26.8$ mm
- NO = $35\sqrt{2 + \sqrt{2}} \approx 64.9$ mm

Result: Rectangle MNOP with diagonals = 70 mm and included angle = 45° is constructed.

Given: Rhombus MNOP, side = 50 mm, $\angle PMN = 60°$.

Construction:
1. Draw MN = 50 mm.
2. At M, draw a ray making $\angle PMN = 60°$ with MN.
3. On this ray, mark MP = 50 mm. Mark P.
4. With centre P and radius = 50 mm, draw an arc.
5. With centre N and radius = 50 mm, draw another arc to intersect the first at O.
6. Join NO and PO.
7. MNOP is the required rhombus.

Verification:
- All sides = 50 mm ✓
- $\angle PMN = 60°$ ✓
- Opposite angle $\angle PON = 60°$ ✓
- Adjacent angles $\angle MNO = \angle MPO = 120°$ ✓
- Diagonals: MO = $2 \times 50 \times \sin(60°) = 86.6$ mm; NP = $2 \times 50 \times \sin(30°) = 50$ mm

Result: Rhombus MNOP with side = 50 mm and $\angle PMN = 60°$ is constructed.

Given: Trapezium MNOP, difference of diagonals MO − NP = 30 mm (or |MO − NP| = 30 mm).

Note: A trapezium has one pair of parallel sides. Additional data is needed to fully define the trapezium (such as the lengths of the parallel sides, the non-parallel sides, or the angles). With only the difference of diagonals given, the problem is under-constrained.

Assumption: Let MN ∥ OP (parallel sides). Assume some standard values to illustrate the construction. Let MN = 70 mm, OP = 40 mm, and the legs MP = NO = 35 mm (isosceles trapezium). Then the diagonals can be calculated and the difference checked.

General Construction Method:
1. Draw the longer parallel side MN.
2. Draw the shorter parallel side OP parallel to MN at a suitable height.
3. Connect M to P and N to O (the legs).
4. Measure diagonals MO and NP.
5. Adjust the position of OP until the difference of diagonals = 30 mm.

Result: Trapezium MNOP is constructed with the difference of its diagonals = 30 mm. (Additional data required for a unique solution.)

Given: Trapezium MNOP, MP = 40 mm, PO = 30 mm, ON = 40 mm, difference of parallel sides MN − PO = 25 mm (but PO = 30 mm is given as a side, so the parallel sides are MN and PO... wait, PO = 30 mm is given as a side).

Reinterpretation: In trapezium MNOP, MN ∥ OP (parallel sides). The non-parallel sides (legs) are MP = 40 mm and ON = 40 mm. The side PO = 30 mm. The difference of parallel sides MN − PO = 25 mm. So MN = PO + 25 = 30 + 25 = 55 mm.

Construction:
1. Draw MN = 55 mm (longer parallel side).
2. Draw a line parallel to MN. The position of this line is determined by the legs.
3. At M, draw a ray. At N, draw a ray. The parallel side PO = 30 mm connects P and O.
4. Since MP = ON = 40 mm (isosceles trapezium), the trapezium is symmetric.
5. The height $h = \sqrt{MP^2 - ((MN-PO)/2)^2} = \sqrt{40^2 - 12.5^2} = \sqrt{1600 - 156.25} = \sqrt{1443.75} \approx 38$ mm.
6. Draw MN = 55 mm. Find midpoint of MN. Draw perpendicular of height 38 mm.
7. Mark P and O symmetrically: each at horizontal distance 12.5 mm from the ends of MN, at height 38 mm.
8. PO = 55 − 2 × 12.5 = 30 mm ✓
9. Join MP and NO.

Result: Trapezium MNOP with MN = 55 mm, PO = 30 mm, MP = ON = 40 mm is constructed.

Given: Regular pentagon with side = 25 mm.
Interior angle of pentagon = $\frac{(5-2) \times 180°}{5} = \frac{540°}{5} = 108°$.

Method 1: Using Protractor
1. Draw side AB = 25 mm.
2. At B, draw a ray making interior angle = 108° with BA. Mark BC = 25 mm on this ray.
3. At C, draw a ray making 108° with CB. Mark CD = 25 mm.
4. At D, draw a ray making 108° with DC. Mark DE = 25 mm.
5. Join EA. Verify EA = 25 mm.
6. ABCDE is the required regular pentagon.

Method 2: Using Compass (Geometric Construction)
1. Draw side AB = 25 mm.
2. Extend AB beyond A. At A, construct an angle of 108° with AB (= 180° − 72°; construct 72° = 60° + 12°, or use the pentagon construction).
3. Mark AE = 25 mm on this ray.
4. Repeat at each vertex.
5. The fifth side should close automatically.

Method 3: Using Circumscribed Circle
1. The circumradius $R = \frac{a}{2\sin(\pi/5)} = \frac{25}{2\sin 36°} = \frac{25}{2 \times 0.5878} \approx 21.3$ mm.
2. Draw a circle of radius 21.3 mm.
3. Divide the circle into 5 equal parts (each central angle = 72°).
4. Join consecutive division points.
5. Each side = 25 mm.

Result: Regular pentagon with side = 25 mm is constructed.

Given: Regular hexagon with side = 30 mm.
Interior angle of hexagon = $\frac{(6-2) \times 180°}{6} = \frac{720°}{6} = 120°$.
Property: In a regular hexagon, the circumradius = side length.

Method 1: Using Compass (Most Common)
1. Draw a circle of radius = 30 mm (circumradius = side for hexagon).
2. Mark any point A on the circle.
3. With the same radius (30 mm) and centre A, mark point B on the circle.
4. With centre B and same radius, mark C. Continue: D, E, F.
5. Join A-B-C-D-E-F-A.
6. ABCDEF is the required regular hexagon.

Method 2: Using Set Squares
1. Draw a horizontal line. Mark centre O.
2. Draw a circle of radius 30 mm.
3. Using 30°–60° set square, draw lines at 60° intervals through the centre to intersect the circle at 6 points.
4. Join consecutive points.

Method 3: Using Protractor
1. Draw side AB = 30 mm.
2. At each vertex, draw the next side at 120° interior angle.
3. Each side = 30 mm.

Result: Regular hexagon with side = 30 mm is constructed.

Given: Regular octagon with side = 25 mm.
Interior angle of octagon = $\frac{(8-2) \times 180°}{8} = \frac{1080°}{8} = 135°$.

Method 1: Using Protractor
1. Draw side AB = 25 mm.
2. At B, draw a ray making interior angle = 135° with BA. Mark BC = 25 mm.
3. At C, draw a ray making 135° with CB. Mark CD = 25 mm.
4. Continue at D, E, F, G similarly.
5. The eighth side HA should close automatically.
6. ABCDEFGH is the required regular octagon.

Method 2: Using Set Squares (45° set square)
1. Draw a square with side equal to the distance across flats of the octagon.
2. Cut the corners at 45° to get the octagon.
3. For side = 25 mm: The square side = $25(1 + \sqrt{2}) \approx 25 \times 2.414 = 60.4$ mm.
4. Draw a square of side 60.4 mm.
5. Mark 25 mm from each corner along each side.
6. Cut the corners by joining these marks.
7. The resulting octagon has all sides = 25 mm.

Method 3: Using Circumscribed Circle
1. Circumradius $R = \frac{a}{2\sin(\pi/8)} = \frac{25}{2\sin 22.5°} = \frac{25}{2 \times 0.3827} \approx 32.7$ mm.
2. Draw a circle of radius 32.7 mm.
3. Divide into 8 equal parts (each central angle = 45°).
4. Join consecutive points.

Result: Regular octagon with side = 25 mm is constructed.