Construction II: Construction of Quadilaterals – SS1 Mathematics Lesson Note (Second Term)

Construction of a quadrilateral is the accurate drawing of a four-sided shape with ruler and compasses from given sides and angles.

Behavioural objectives

By the end of the lesson, students should be able to:

1. construct a quadrilateral or parallelogram from given sides and an angle, and measure a diagonal;
2. construct an equilateral triangle and bisect its angles;
3. describe the locus of a point at a given distance from a fixed point or from a straight line;
4. describe the locus of points equidistant from two given points or from two straight lines;
5. construct loci in a triangle and measure the distance between their points of intersection.

How do you construct a quadrilateral?

To construct a quadrilateral, sketch it first, draw one side, construct the given angle, then mark the other vertices with arcs of the given lengths.

General steps

1. Make a rough sketch and write the given lengths and angles on it.
2. Draw the longest known side as the base with a ruler.
3. Construct the given angle at one end of the base.
4. Mark the next vertex on the angle arm with a compass arc of the given length.
5. Find the last vertex where two arcs of the given lengths cross.
6. Join the vertices and measure what is asked.

How do you construct angles such as 105° and 135° with compasses?

• 60°: draw an arc from the end of the base, then an arc of the same radius from where it cuts the base.
• 90°: construct a perpendicular to the base at the given point.
• 105°: construct 90° and 120°, then bisect the 30° gap between them (90° + 15°).
• 135°: construct 90° and extend the base to 180°, then bisect the 90° gap (90° + 45°).

Examples

1. Construct a quadrilateral ABCD in which AB is parallel to DC, |AB| = 4 cm, |BC| = 5 cm, |DC| = 7 cm and ∠ADC = 105°. Measure the diagonal BD.
2. Use your ruler and compasses to construct the parallelogram PQRS in which |QR| = 5 cm, |RS| = 11 cm and ∠QRS = 135°. Measure the length of the shorter diagonal of PQRS.

Solution to example 1

First sketch the quadrilateral to be constructed, then follow the steps below.

1. Draw line DNC = 7 cm with DN = 3 cm and NC = 4 cm.
2. Construct ∠CDM = 105°.
3. With N as the centre and a radius of 5 cm, draw an arc to cut DM at A.
4. With A as the centre and a radius of 4 cm, draw an arc.
5. With C as the centre and a radius of 5 cm, draw a second arc to cut the first arc at B.
6. Join A to B and C to B to complete the quadrilateral ABCD.
7. Join B to D and measure it.

• Why N is used: |NC| = |AB| = 4 cm and AB is parallel to NC, so ABCN is a parallelogram and |AN| = |BC| = 5 cm.
• Result: by measurement, |BD| = 4.5 cm.

Solution to example 2

First sketch the parallelogram PQRS, then follow the steps of the construction below.

1. Draw line QR = 5 cm.
2. Construct ∠R = 135°.
3. With R as the centre and a radius of 11 cm, draw an arc to cut the 135° line at S.
4. With S as the centre and a radius of 5 cm, draw an arc.
5. With Q as the centre and a radius of 11 cm, draw a second arc to cut the arc in step 4. This is point P.
6. Draw lines to join S to P and P to Q.
7. Draw a dotted line through diagonal RP and measure it.

• Why PR is the shorter diagonal: the angle at Q is 45°, which is smaller than 135°, so the diagonal across it is shorter.
• Result: by measurement, the length of the shorter diagonal PR is 8.3 cm.

How do you construct an equilateral triangle?

An equilateral triangle is a triangle in which all the sides are of equal length and each of its angles is 60°.

Examples

1. Construct an equilateral triangle XYZ such that |XY| = 5 cm.
2. (a) Construct an equilateral triangle ABC such that |AB| = 7 cm.
   (b) Construct the bisectors of ∠A, ∠B and ∠C.
   (c) What did you observe?

Steps for example 1

1. Draw line XY = 5 cm.
2. With X as the centre and a radius of 5 cm, draw an arc above XY.
3. With Y as the centre and the same radius, draw an arc to cut the first arc at Z.
4. Join X to Z and Y to Z.
5. Check that each angle is 60° and each side is 5 cm.

Steps for example 2

1. Construct triangle ABC with |AB| = |BC| = |CA| = 7 cm, as in example 1.
2. At A, draw an arc to cut AB and AC. From those two points draw arcs of equal radius that cross inside the triangle.
3. Join A to the crossing point and extend it to meet BC.
4. Repeat steps 2 and 3 at B and at C.

What do you observe?

• One meeting point: the three bisectors pass through the same point inside the triangle.
• Equal parts: each 60° angle is divided into two angles of 30°.
• Opposite side: each bisector meets the opposite side at right angles and cuts it into two equal parts.

What is a locus?

A locus is the path traced by a point that moves according to a given rule.

The plural is loci. The four basic loci are below.

What is the locus of points at a given distance from a fixed point?

The locus of points at a constant distance x cm from a fixed point O is a circle of radius x cm, centre O.

• Figure: draw O as a fixed point, with P₁ and P₂ at a constant distance x cm from O.
• Construction: set the compasses to x cm, place the point at O and draw the circle.

What is the locus of points at a given distance from a straight line?

The locus of points at a distance x cm from a straight line AB is a pair of straight lines parallel to AB, one on each side.

• Figure: draw a straight line AB which continues indefinitely in both directions, with points P₁, P₂, P₃ and P₄ at a distance x cm from AB.
• Two lines: this locus consists of two separate lines, each x cm from AB.

What is the locus of points equidistant from two given points?

The locus of points equidistant from two fixed points X and Y is the perpendicular bisector of XY.

• Figure: P₁, P₂ and P₃ are such that |P₁X| = |P₁Y|, |P₂X| = |P₂Y| and |P₃X| = |P₃Y|.
• Position: P₁, P₂ and P₃ all lie on the perpendicular bisector of XY.

How to construct the perpendicular bisector of XY

1. With X as the centre and a radius greater than half of XY, draw arcs above and below XY.
2. With Y as the centre and the same radius, draw arcs to cut the first two.
3. Join the two crossing points with a straight line.

What is the locus of points equidistant from two straight lines?

The locus of points equidistant from two intersecting straight lines is the pair of bisectors of the angles between the lines.

• Figure: sketch straight lines AB and CD which intersect at O.
• Acute angle: P₁ and P₂ are each equidistant from AB and CD, and lie on the bisector of the acute angle between the two lines.
• Obtuse angle: P₃ is also equidistant from AB and CD, and lies on the bisector of the obtuse angle.
• Complete locus: both bisectors together make up the locus.
• Right angles: the two parts of the locus intersect at right angles.

Summary table of the four loci

Worked example: loci in a triangle

Using ruler and compasses only:

1. Construct triangle ABC such that |AB| = 6 cm, |AC| = 8.5 cm and ∠BAC = 120°.
2. Construct the locus l₁ of points equidistant from A and B.
3. Construct the locus l₂ of points equidistant from AB and AC.
4. Find the points of intersection, P₁ and P₂, of l₁ and l₂ and measure |P₁P₂|.

Solution

1. Construct ∠BAC = 120° at A, using two 60° arcs, with |AB| = 6 cm and |AC| = 8.5 cm. Join B to C.
2. l₁ is the perpendicular bisector of AB.
3. l₂ is in two parts. AP₁ is the bisector of ∠BAC.
4. AP₂ is perpendicular to AP₁. Points on AP₂ are equidistant from AB and CA produced.
5. Mark P₁ and P₂ where l₁ cuts the two parts of l₂.
6. Measure |P₁P₂|.

By measurement, |P₁P₂| is about 6.9 cm.

Practice questions

1. A circle, centre O, radius 5 cm, is drawn on a sheet of paper. A point P moves on the paper so that it is always 2 cm from the circle. The locus of P is:
   A. a circle, centre O, radius 3 cm
   B. two circles, centre O, radii 3 cm and 7 cm
   C. a circle, centre O, radius 6 cm
   D. two circles, centre O, radii 4 cm and 6 cm
   E. a circle, centre O, radius 3.5 cm
2. XYZ is a straight line such that |XY| = |YZ| = 3 cm. A point P moves in the plane of XYZ so that |PY| < |XY|. Which of the following describes the locus of P?
   A. a line through X perpendicular to XZ
   B. a line through Y perpendicular to XZ
   C. a line through Z perpendicular to XZ
   D. a circular disc, centre X, radius 3 cm
   E. a circular disc, centre Y, radius 3 cm
3. Describe the locus of a point which moves so that it is always 5 cm from a fixed point O in a plane.
   A. a rectangle which measures 10 cm by 5 cm
   B. a square of side length 5 cm
   C. a parallelogram whose diagonals are 10 cm and 5 cm
   D. a circle of radius 5 cm, centre O
   E. a circle of radius 10 cm, centre O
4. Describe the locus of a point which moves along a level floor so that it is 2 m from a wall of a room.
   A. One line, parallel to and 2 m from the wall
   B. Two lines, one on each side, parallel to and 2 m from the wall
   C. A circle of radius 2 m
   D. A semi-circle of radius ½ m
   E. Two perpendicular lines, each of length 2 m
5. Describe the locus of a point which moves so that it is 3 cm from a fixed line AB in a plane.
   A. 2 lines parallel to AB and 6 cm apart, joined by semi-circular ends
   B. 2 lines parallel to AB and 8 cm apart, joined by semi-circular ends
   C. 2 lines perpendicular to AB
   D. A circle of radius 6 cm
   E. A circle of radius 3 cm
6. (a) Using ruler and compasses only, construct triangle ABC such that |AC| = 8.5 cm and ∠ACB = 135°.
   (b) Using any geometrical instruments, find a point P within triangle ABC which is at a distance 2.8 cm from AC and 6 cm from B. Measure the length of AP.

Evaluation

1. State the locus of a point which is always x cm from a fixed point O.
2. Describe the locus of points at a distance x cm from a straight line AB that continues indefinitely in both directions.
3. What is the locus of points equidistant from two given points X and Y?
4. Describe the complete locus of points equidistant from two intersecting straight lines.
5. Construct an equilateral triangle XYZ such that |XY| = 5 cm, and construct the bisectors of its angles.

Assignment

1. Construct a trapezium ABCD in which AB is parallel to DC, |AB| = 4 cm, |BC| = 8 cm, |CD| = 11 cm and |DA| = 6 cm. (Hint: in a rough figure, divide the trapezium into parallelogram ABXD and triangle BCX. First construct triangle BCX.)