Geometrical Constructions and Conic Sections

 

BISECT A LINE AND AN ARC

1.     Bisect A 90 mm long line and A circular arc AB.

·        Draw either a straight-line AB measuring 90 mm in length and an arc labeled AB.

·        Adjust the compass to a radius longer than half the distance of AB. Using point, A as the center, draw arcs above and below AB.

·        Without changing the compass radius, place the compass at point B and draw arcs above and below AB that intersect the arcs at points C and D.

·        Connect points C and D with a straight line. This line is the perpendicular bisector that divides AB into two equal parts.

 

PERPENDICULAR TO A LINE

2.     Draw a perpendicular line to an 80 mm long straight-line AB, at a point P lying on the line at a distance of 30 mm from the end A.

 

Method-I

1.     Draw a straight-line AB measuring 80 mm and locate a point P on it, positioned 30 mm from point A.

2.     Using a compass set to any suitable radius, place the compass at point P and draw an arc that intersects line AB at two points, C and D.

3.     Adjust the compass to a radius greater than half the distance between C and D. With the compass centered at points C and D, draw two arcs that cross each other at point E.

4.     Connect point P to point E using a straight line. The resulting line PE will be perpendicular to line AB.

Method-II

1.     Draw a straight-line AB, 80 mm in length, and mark a point P on it such that P is 30 mm from point A.

2.     Select a point O at any location outside the line AB. Using O as the center, draw a circle with a radius equal to the distance OP, ensuring it intersects the line AB at point C.

3.     Connect point C to O and extend the line to meet the circle again at point D.

4.     Join point P to D. The line PD will be perpendicular to the line AB.

 

3.     Draw a perpendicular to an 80 mm long line AB, from a point P lying at a distance 50 mm from end A and 60 mm from end B.

 

Construction

 

a.      Draw an 80 mm long line AB.

b.     With centre A and radius 50 mm draw an arc.

c.      With centre B and radius 60 mm draw another arc to intersect previous drawn arc at point P.

 

Method 1

 

1.     Set the compass to any convenient radius and with centre P, draw an arc CD to meet the line AB at points C and D.

2.     Reset the compass to a radius greater than half of CD and centres C and D respectively, draw arcs to intersect at point E.

3.     Join P to E. Line PE is perpendicular to AB.

 

PARALLEL LINES

4.    Draw a line parallel to a given straight-line AB through a point 50 mm away from it.

1.     Mark any point D on the line AB. With centre D and radius DC, draw arc CE to meet AB at point E.

2.     With centre E and the same radius DC, draw an arc DF.

3.     With centre D and radius of chord length CE, draw an arc to intersect arc DF at point F.

4.     Join C to F. Line CF is parallel to the line AB.

 

DIVIDE A LINE

5.     Divide an 80 mm long straight line into seven equal parts.

 

a.      Draw an 80 mm long straight-line AB.

b.     Draw another line AC at any convenient acute angle with AB.

c.      Set the divider to a convenient length and mark off seven equal spaces on AC. Let these points be 1’, 2’, 3’, 4’, 5’, 6’, and 7’.

d.     Join 7’ to B.

e.      Set the drafter along 7’ B and draw parallel lines through points 1’, 2’, 3’, 4’, 5’, 6’ to meet AB at points 1, 2, 3, 4, 5 and 6 respectively.

f.       These points divide AB in seven equal parts.

 

ANGLE BISECTOR

6.     Draw an angle of 75° and bisect it with the help of a compass.

 

1.     Draw an angle AOB which is equal to 75°.

2.     Set compass to any convenient radius and with centre O, draw an arc CD to meet lines OA and OB at points C and D respectively.

3.     Reset the compass to any radius greater than half of chord length CD. With centres C and D respectively, draw arcs to intersect each other at point P.

4.     Join O to P. Line OP is the bisector of the angle AOB.

 

CENTRE OF AN ARC OR CIRCLE

 

7.     Locate the centre of the arc shown in Fig.

                 

1.      Draw two chords CD and EF of any convenient lengths in the arc.

2.     Draw PQ as the bisector of chord CD.

3.     Similarly, draw RS as the bisector of chord EF.

4.     Let the lines PQ and RS, produce if required, to intersect each other at point O. The point O is the required centre of the arc.

5.     This method can also be used to determine the centre of a circle.

CIRCLE THROUGH THREE POINTS

8.     Draw a circle passing through three points A, B and C not lying in a straight line.

 

 

1.     Mark points A, B and C. Join AB and BC.

2.     Draw PQ as the perpendicular bisector of line AB. Similarly, draw RS as the perpendicular bisector of line BC.

3.     Let the lines PQ and RS, produced if required, to intersect each other at point O. This point O is the required centre of the circle.

4.     Draw a circle with centre O and radius OA (= OB =OC). This circle passes through points A, B and C.

 

DIVIDE A CIRCLE

9.     Divide a 50 mm diameter circle into 12 equal segments.

 

a.      Draw a circle with centre O and 50 mm diameter.

b.     Using drafter, draw diameters AG and DJ, perpendicular of each other.

c.      Draw arcs of radius equal to the radius of the circle (= 25 mm) and centre A to meet the circumference of the circles at points C and K.

d.     Similarly, draw arcs of the same radius (= 25 mm) and centres D, G and J respectively, to meet the circumference of the circle at points B, F, E, I, H and L. The points divide the circumference of the circle into 12 equal segments.

 

TANGENT TO TWO CIRCLES

 

10.Draw exterior and interior tangents connecting two circles of radii 25 mm and 40 mm having their centres 100 mm apart.

 

Tangent BC common to two circles (a) Exterior (b) Interior

 

a.      Draw a 100 mm long line O1O2.

b.     Draw a circle with centre O1 and radius R1 (= 25 mm).

c.      Draw another circle with centre O2 and radius R2 (= 40 mm).

 

Exterior Tangent

a.      Draw an arc with centre O2 and radius R2 – R1 (= 15 mm).

b.     Through point O1, draw a tangent to this arc to touch at point A.

c.      Join O2A and produce it to meet the circle at point B.

d.     Through B, draw a line BC parallel to O1A, to touch the smaller circle at point C. Line BC is the required tangent exterior to the circles.

 

           Interior Tangent

a.      Draw an arc with centre O2 and radius R2 + R1 (= 65 mm).

b.     Through point O1, draw a tangent to this arc to touch at point A.

c.      Join O2 to A which intersect the bigger circle at point B.

d.     Through B, draw a line BC parallel to O1A, to touch the smaller circle at point C. Line BC is the required tangent interior to the circles.

 

CIRCLE TO CONNECT ANOTHER CIRCLE AND A POINT

 

11.Draw a circle of 60 mm diameter and mark a point P 70 mm away from its centre. Draw another circle which passes through the point P and is tangential to that circle at point Q.

 

a.      Draw a circle of diameter 60 mm with centre O. Mark a point Q on the circle.

b.     Mark point P at a distance 70 mm from O.

c.      Join PQ. Draw line AB as the bisector of line PQ.

d.     Let the line AB, produced if required, intersect the line OQ produced at point C. Draw a circle with centre C and radius CQ (= CP). This is the required circle.

 

Polygons

• A polygon is a closed curve made of a set of line segments connected end to end, with no two segments crossing.
• The straight-line segments are called sides (or edges).
• The points where sides meet are called vertices.
• Types of polygons by sides:
• 3 sides → Triangle
• 4 sides → Quadrilateral
• 5 sides → Pentagon
• Convex polygon: Any two points inside can be joined by a line segment that does not intersect any side.
• Concave polygon: If a line joining two interior points intersects a side.
• Equilateral polygon: All sides are equal in length.
• Equiangular polygon: All interior angles are equal.
• Regular polygon: Both equilateral and equiangular (e.g., equilateral triangle, square).
• Applications: Construction of regular polygons is common in engineering drawings (e.g., nut-bolt assemblies, spanners, gears, etc.).

 

 

CONSTRUCTION OF REGULAR POLYGONS

 

12.Draw a regular pentagon and a regular heptagon of 40 mm sides, using general method.

 

General method 1 for (a) pentagon (b) heptagon

 

 

Method 1 (a) for pentagon and (b) for heptagon.

 

1.     Draw a 40 mm long line OA.

2.     Draw a semicircle AP with centre O and radius OA.

3.     Divide the semi-circular AP into n equal parts, where n is the number of sides of the polygon (n = 5 for pentagon and n = 7 for heptagon). Starting from point P, mark these divisions as 1, 2, 3, 4, etc.

4.     Join O2, O3, O4, etc.

5.     Draw an arc with centre 2 and radius OA to cut the line O3 produced at B.

6.     Draw another arc with centre B and the same radius OA to cut line O4 produced at C. For heptagon, proceed to draw arcs with centres C and D, radius OA to meet lines O5 and O6 produced at points D and E, respectively.

7.     Join the points O, 2, B, C, …, A and obtain the required polygon.

General method 2 for (a) pentagon (b) heptagon

 

Method 2: Refer to Fig. (a) for pentagon and Fig.  (b) for heptagon.

1.     Draw a 40 mm long line AB.

2.     Draw PQ as perpendicular bisector to AB (Problem 3.1).

3.     Describe a semicircle with diameter AB, to intersect PQ at 4.

4.     With either A or B as the centre, draw an arc of radius AB to meet PQ at 6.

5.     Bisect the line 4-6 and obtain point 5 on PQ.

6.     Mark points 7 and 8 on PQ such that the lengths 4–5, 5–6, 6–7, 7–8 are equal.

7.     Draw a circle with centre 5 for pentagon and 7 for heptagon, to pass through points A and B.

8.     Taking AB as radius cut the circle at C, D, E, etc., and join them to get the required polygon.

 

CONSTRUCTION OF A REGULAR HEXAGON

 

13.Draw a regular hexagon of 40 mm sides, keeping a side (a) vertical (b) horizontal

Hexagon with a side (a) Vertical (b) Horizontal

 

a.      Draw a circle with centre O and radius 40 mm.

b.     Mark a diameter AD in vertical position for case (a) and in horizontal position for case (b).

c.      With radius OA and centres A and D, draw arcs to cut the circle at points B, F, C and E.

d.     Join ABCDEF to get the required hexagon.

 

 

CONE

A cone is a three-dimensional geometric shape with a circular base that tapers smoothly to a single point called the apex or vertex.

A cone is formed if a right-angled triangle with an apex angle α is rotated about its altitude as the axis. The apex angle of the cone is 2α.

 

ELLIPSE

When a cone is cut by a plane inclined to its axis in such a way that it intersects all the generators, the resulting section is an ellipse. For the section to form an ellipse, the cutting plane's inclination must exceed half of the cone's apex angle. θ > α.

 

Applications:

·        Concrete arches

·        Dams

·        Monuments structure, etc.

 

PARABOLA

When a cone is cut by a plane inclined to its axis and parallel to one of its generators, the resulting section is a parabola. In this case, the cutting plane's inclination is equal to half of the cone's apex angle. θ = α.

Applications:

·        Bridges

·        Antenna etc.

 

HYPERBOLA

When a double cone is sliced by a plane that goes through both halves, the section formed is a hyperbola. This happens when the cutting plane does not intersect the apex and is inclined at an angle less than half of the cone's apex angle. θ < α

Applications:

·        Water channel

·        Cooling Tower

·        Radio Astronomy

·        Telescope reflectors etc.

 

CONSTRUCTION OF ELLIPSE

An ellipse can be construction by the following methods:

 

1.     Eccentricity method (general method)

2.     Intersecting arcs method or arcs of circles method

3.     Concentric circles method

4.     Oblong method

a.      Rectangle method

b.     Parallelogram method

 

ECCENTRICITY METHOD

An ellipse is defined as the locus of a point P moving in a plane in such a way that the ratio of its distance from a fixed point F1 to the fixed straight-line DD’ is a constant and is always less than unity.

 

14.Draw an ellipse when the distance of its focus from its directrix is 50 mm and eccentricity is 2/3. Also, draw a tangent and a normal to the ellipse at a point 70 mm away from the directrix.

 

Construction

1.     Draw a vertical line DD’ to represent directrix.

2.     Draw the principal axis AB perpendicular to DD’.

3.     Mark focus F on the axis AB such that AF = 50 mm.

4.     Divide AF into five equal divisions. Mark the vertex V1 on the third division point from A. Therefore, V1F/V1A = 2/3 and thereby the vertex V1 represents a locus point of the ellipse.

5.     Draw a vertical line V1E equal to V1F. Join A to E and produce it to some distance. Therefore, in the triangle AV1E, V1E/V1A = V1F/V1A = 2/3

6.     Draw a line from the focus F, inclined at 45° to the axis AB to intersect the line AE produced at point C. Draw a perpendicular line from point C to meet the axis AB at point V2. The line V1V2 represents the major axis of the ellipse.

7.     7. Mark a point 1 anywhere on the major axis V1V2. Draw a line through point 1, perpendicular to the axis AB to meet line AE produced at point 1’. Therefore, 11’/1A = V1E/V1A = 2/3.

8.     With centre F and radius equal to 1-1’, draw arcs to intersect the perpendicular line 1-1’ at points P1 and P1’. These are the loci points of the ellipse because FP1/1A = FP1’/1A = 11’/1A = 2/3.

9.     Similarly, mark some more points, say 2, 3, 4, etc., on the major axis V1V2 which need not be equidistant and repeat steps 7 and 8. This will give some more loci points of the ellipse like; P2 and P2’, P3 and P3’, P4 and P4’, etc.

10.  Join all the loci points of the ellipse and obtain the required ellipse.

 

Tangent and normal to an ellipse

1.     Mark a point P on the ellipse at a distance of 70 mm from the directrix and Join PF.

2.     Draw a line FT perpendicular to the line PF to meet the directrix DD’ at point T.

3.     Join TP and produce to some point T’. The line TT’ is the required tangent.

4.     Through point P, draw a line NN’ perpendicular to TT’. The line NN’ is the required normal.

 

 

CONSTRUCTION OF PARABOLA

A parabola can be constructed by the following methods:

 

1.     Eccentricity method

2.     Offset method

3.     Tangent method

4.     Oblong method

a.      Rectangle method

b.     Parallelogram method

 

ECCENTRICITY METHOD

A parabola is defined as the locus of a point P moving in a plane in such a way that the ratio of its distance from a fixed point F to the fixed straight line DD’ is a constant and is always equal to unity.

 

15.Draw a parabola when the distance between its focus and directrix is 50 mm. Also, draw a tangent and a normal at a point 70 mm from the directrix

 

Construction

 

1.     Draw a vertical line DD’ to represent the directrix.

2.     Draw the principal axis AB perpendicular to the directrix DD’.

3.     Mark focus F on the axis AB such that AF = 50 mm.

4.     Mark vertex V at the mid of AF. This vertex V is the locus point of the parabola because e = VF/AV = 1

5.     Mark a point 1 anywhere on the axis AB and draw a perpendicular line through it.

6.     With centre F and radius equal to A1, draw arcs to intersect the perpendicular line through point 1 at points P1 and P1’. These are the loci points of the parabola because FP1/A1 = FP1’/A1 = 1

7.     Similarly, mark some more points, say 2, 3, etc., on the axis AB which need not be at equidistant and repeat Step 6. This will give some more loci points of the parabola like; P2 and P2’, P3 and P3’ etc.

8.     Join all the loci points P2’, P1’, V, P1, P2, etc., by a smooth curve and obtain the required parabola.

 

Tangent and normal to a parabola

1.     Mark a point P on the parabola at a distance of 70 mm from the directrix and join PF.

2.     Draw a line FT perpendicular to the line PF to meet the directrix DD’ at point T.

3.     Join T to P and produce to some point T’. The line TT’ is the required tangent.

4.     Through point P, draw a line NN’ perpendicular to TT’. The line NN’ is the required normal.

 

 

CONSTRUCTION OF HYPERBOLA

 

A hyperbola can be constructed by the following methods:

 

1.     Eccentricity method

2.     Intersecting arcs method

3.     Oblong method

4.     Intercept method

5.     Asymptotes method

(a)  Orthogonal asymptotes method

(b)  Oblique asymptotes method

 

ECCENTRICITY METHOD

A hyperbola is defined as the locus of a point P moving in a plane in such a way that the ratio of its distance from a fixed point F to the fixed straight line DD’ is a constant and is always greater than unity.

 

 

 

16.Draw a hyperbola when the distance of its focus from its directrix is 50 mm and eccentricity is 3/2. Also, draw a tangent and a normal to the hyperbola at a point 25 mm from the directrix.

 

 

Construction

1.     Draw a directrix DD’.

2.     Draw the principal axis AB perpendicular to the directrix DD’.

3.     Mark focus F on the axis AB such that AF = 50 mm.

4.     Divide AF into five equal divisions. Mark the vertex V on the second division-point from A. Therefore, VF/VA = 3/2 and thereby the vertex V represents a locus point of the hyperbola.

5.     Draw a vertical line VE equal to VF. Join A to E and produce it to some distance. Therefore, in the triangle AVE, VE/VA = VF/VA = 3/2.

6.     Mark a point 1 anywhere on the axis AB. Draw a line through point 1, perpendicular to the axis AB to meet line AE produced at point 1’. Therefore, 11’/1A = VE/VA = 3/2.

7.     With centre F and radius equal to 1-1’, draw arcs to intersect the perpendicular line 1-1’ at points P1 and P1’. These are the loci points of the hyperbola because FP1/1A = FP1’/1A = 11’/1A = 3/2.

8.     Similarly, mark some more points, say 2, 3, 4, etc., on the axis AB which need not be equal and repeat Steps 6 and 7. This will give some more loci points of the ellipse like; P2 and P2’, P3 and P3’, P4 and P4’, etc.

9.     Join points P4’, P3’, P2’, P1’, V, P1, P2, P3, P4 by a smooth curve. This is the required hyperbola.

 

Tangent and normal to a hyperbola

1.     Mark a point P on the hyperbola at a given distance, 25 mm from the directrix. Join PF.

2.     Draw a line FT perpendicular to PF to meet the directrix at point T.

3.     Join TP and extend it to some point T’. The line TT’ is the required tangent.

4.     Through point P, draw a line NN’ perpendicular to TT’. The line NN’ is the required normal.