Construction I: Triangles & Angles – SS1 Mathematics Lesson Note (Second Term)

Construction means drawing lines, angles and triangles accurately with a ruler and a pair of compasses. In this lesson, students construct triangles from given sides and angles, bisect line segments, and construct angles such as 90°, 45°, 60°, 30°, 75°, 120° and 150° by bisecting or combining basic angles.

Behavioural objectives

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

• construct a triangle given two sides and the angle between them, and measure the third side;
• bisect a line segment and construct its perpendicular bisector;
• construct angles of 90°, 60° and 120° using ruler and compasses;
• obtain angles of 45°, 22½°, 30°, 67½° and 15° by bisection;
• construct angles of 135°, 75° and 150° by combining or bisecting other angles.

How do we construct a triangle from two sides and the angle between them?

Examples

1. Construct ∆ABC in which |AB| = 7cm, |AC| = 9.5cm and ∠ABC = 120°. Measure |BC|.
2. Construct ∆PQR in which |PQ| = 5.5cm, |QR| = 8.5cm and ∠PQR = 75°. Construct M, the midpoint of PR. Measure |QM|.

Solution to example 1

1. First, sketch the triangle to be constructed.
2. Draw a line AB = 7cm.
3. Construct an angle of 120° at B.
4. With centre A and radius 9.5cm, draw an arc to cut the 120° line.
5. Draw line AC.

From the diagram, |BC| = 3.8cm.

Solution to example 2

1. First, sketch the triangle to be constructed.
2. Draw line PQ = 5.5cm.
3. Construct an angle of 75° at Q.
4. With centre Q and radius 8.5cm, draw an arc to cut the angle 75° at R.
5. Draw line QR.
6. Draw line PR and bisect it. Label the midpoint M and join QM.

From the diagram, |QM| = 5.6cm.

What does it mean to bisect a line segment?

To bisect a given line segment means to divide it into two parts of equal length.

The steps to bisect a given line segment are as follows:

1. Draw the given line segment AB (let AB = 8.6cm).
2. With centre A and a radius of about ¾ of the length of AB, draw an arc above and below the line AB.
3. With centre B and the same radius used in step 2, draw arcs to cut the previous arcs.
4. Draw a line through the two points of intersection of the pairs of arcs obtained in steps 2 and 3. The line drawn is the perpendicular bisector of line AB.
5. Label the point where the perpendicular bisector meets AB as E. Thus |AE| = |EB| = 4.3cm.

How do we construct and bisect angles of 90°, 45°, 135°, 22½° and 67½°?

Constructing 90°

To construct an angle of 90°, take the following steps:

1. Draw a line and mark a point A on it, where the angle of 90° is to be constructed.
2. With centre A and any suitable radius, draw arcs to cut the line at B and C, one on each side of A.
3. With centre B and a larger radius, draw an arc above the line.
4. With centre C and the same radius used in step 3, draw an arc to cut the previous arc at D.
5. Draw a line through points A and D. Thus ∠DAB = ∠DAC = 90°.

Obtaining 45°, 22½°, 135° and 67½°

1. Since 45° = ½ of 90°, angle 90° is bisected to obtain angle 45°.
2. Similarly, 22½° = ½ of 45°. By bisecting angle 45°, we obtain angle 22½°.
3. Also, 135° = 90° + 45°. Thus, by constructing an angle of 90° at a point on a line and bisecting the 90° on the other side, we obtain an angle of 135°.
4. As explained above, bisecting angle 135° gives angle 67½°.

How do we construct angles of 60°, 30°, 75°, 120° and 150°?

Constructing 60°

To construct an angle of 60°, take the following steps:

1. Draw a line AB and mark the point A at which the angle 60° is to be constructed.
2. With centre A and any convenient radius, draw an arc to cut line AB at C.
3. With centre C and the same radius used in step 2, draw another arc to cut the previous arc at D.
4. Draw line AD and extend it to E.

Then ∠EAB = 60°.

Constructing 30° and 75°

1. Since 30° = ½ of 60°, angle 60° is bisected to obtain angle 30°.
2. To construct angle 75°, note that 75° = 60° + ½ of 30°. First construct angle 90° at a point on a straight line. Next, construct an angle of 60° at the same point. Then bisect the 30° difference between the angles 90° and 60° to give 15° on either side. Thus 60° + 15° = 75°.

Constructing 120°

Take the following steps:

1. Draw a straight line AB and mark a point C on the line where the angle 120° is to be constructed.
2. With centre C and a suitable radius, draw a well-extended arc to cut line CB at point D.
3. With centre D and the same radius used in step 2, draw an arc to cut the extended arc at point E.
4. With E as centre and the same radius, draw an arc to cut the extended arc at point F.
5. Draw line CF. Thus ∠FCB = 120°.

Constructing 150°

Since 150° = 120° + ½ of 60°, first construct angle 120° on a straight line. Then bisect the adjacent 60° angle to get 30°. Thus 30° + 120° on the right-hand side gives the required angle, ∠DCB = 150°.

Evaluation

1. Construct ∆XYZ such that |XY| = 5cm, ∠XYZ = 120° and |YZ| = 7cm. Measure the following:
   • (a) |XZ|
     A. 10.4cm B. 13cm C. 8cm D. 4cm
   • (b) ∠YXZ
     A. 25° B. 30° C. 35° D. 40°
   • (c) ∠XZY
     A. 30° B. 25° C. 50° D. 60°
2. Construct ∆ABC such that |AB| = 6cm, |BC| = 7.5cm and ∠ABC = 75°. Bisect AB at P and AC at Q. Measure the following:
   • (a) |PQ|
     A. 3.8cm B. 10cm C. 2cm D. 8cm
   • (b) |QC|
     A. 5.1cm B. 6.8cm C. 4.1cm D. 8.2cm

Assignment

1. • (a) Use ruler and compasses to construct ∆PQR in which ∠Q = 90°, |QR| = 5cm and |PR| = 10cm.
   • (b) Measure |PQ|.
   • (c) Use Pythagoras’ theorem to check the result.
2. • (a) Construct ∆ABC such that |AB| = 7cm, |BC| = 6cm and ∠ABC = 60°.
   • (b) The bisector of ∠C meets the perpendicular bisector of |AC| at X. Find the point X by construction.
   • (c) Measure |BX|.