Construction I : Triangles & Angles SS1 Mathematics Lesson Note

CONSTRUCTION OF TRIANGLES

Example:

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/

Solutions:

1. i. First, sketch the triangle to be constructed.
2. Draw a line AB = 7cm

iii. Then construct an angle of 120° at B with a radius of 9.5cm and centre  A, draw an arc to cut the 120°

1. Draw line AC

From the diagram /BC/ = 3.6cm

2) i. First, sketch the triangle to be constructed.

5. Draw line PQ = 5.5cm

iii. Construct an angle of 75° at Q

8. With centre Q and radius 8.5cm, draw an arc to cut the angle 75° at R.
9. Draw line QR.
10. Bisect line PR

vii. From the diagram, /QM/ = 5.5cm

DRAWING AND BISECTION OF LINE SEGMENTS

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

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

8. Draw the  given line segment AB ( let AB = 8.6cm)
9. With centre A and a radius of about ¾ of the length of AB, draw an arc above and below the line AB

iii. With centre B and the same radius used in step 2 above, draw arcs to cut the previous arcs in step 2.

3. Draw a line through the 2 points of an intersection of the pair or arcs obtained from steps 2 and 3. The line drawn is the perpendicular bisector of line AB.
4. Thus AE  = EB = 4.3cm

CONSTRUCTION AND BISECTION OF ANGLES : 90°, 45°, 135°, 22 ½° , 67 ½°

To construct angle 90°, take the following steps:

1. Draw a line BC and mark a point A at which the angle of 90° is to be constructed.
2. With centre B and any suitable radius draw an arc above line BC.

iii. With centre C and the same radius used in step 2, draw an arc to cut the previous Arc at D.

1. Draw a line through points A and D. Thus < DAB = <DAC = 90°

NOTE:

1. Since 45° = ½ of 90°, angle 90° is bisected to obtain angle 45°.
2. Similarly 22 ½° = ½ of 45°, By bisecting angle 45°, we can obtain angle  22 ½°
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 can obtain an angle of 135°.
4. As explained above bisection of angle 135° will give an angle 67 ½°

CONSTRUCTION AND BISECTION OF ANGLES: 60°, 30°, 75°, 105°, 120°, 150°.

To construct an angle of 60°, the following steps must be taken:

1. Draw a line AB and mark the point A at which the angle 60o is  to be constructed
2. With centre A and any convenient radius, draw an arc to cut line AB at C.

iii. With centre C and the same radius used to draw the arc in step 2 above, draw another arc to cut the previous arc at D.

1. Draw line AD and extend it to E

Then, <EAB = 60°.

NOTE:

1. Since 30° = ½ of 60°, angle 60° is bisected to obtain angle 30°
2. To construct angle 75°, since 75° = 60°+ ½ of 30°, then first construct angle 90° at a point on a straight line. Next, construct an angle of 60° at the same point where an angle of 90° has been constructed.  Then the angle 30° difference between the angle 90° and 60° is bisected to give 15° on either side. Thus 60° + 15° = 75°.

To construct an angle of 120°

The following steps must be taken :

1. Draw a straight line AB and mark a point C on the line where the angle 120° is to be constructed.

ii With centre C and a suitable radius, draw a well-extended arc to cut line CB at point D.

iii. With centre D and the same radius used in step 2 above draw an arc to cut the extended arc in step 2

at point E.

1. With E as a centre and the same radius, draw an arc to cut the extended arc at point F.
2. Draw line CF. Thus <FCB = 120.

Construction of angle 150°.

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

EXERCISE

1. Construct ∆XYZ such that XY = 5cm, XYZ = 120o and YZ = 7cm. Measure the following
2. a) |XZ|
3. 10.4cm      B. 13cm        C. 8cm    D. 4cm
4. b) <YXZ
5. 25°      B. 30°     C. 35°       D. 40°
6. c) <XZY
7. 30°       B. 25°       C. 50°       D. 60°
8. Construct ∆ABC such that AB = 6cm, BC = 7.5cm and ABC = 75°. Bisect AB at P and AC at Q. Measure
9. a) |PQ|
10. 3.8cm      B. 10cm      C. 2cm      D. 8cm
11. b) |QC|
12. 5.1cm       B. 6.8cm       C. 4.1cm      D. 8.2cm

3(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.

4(a) Construct ABC such that /AB/ = 7cm, /BC/ = 6cm and ABC = 60o

(b) The bisector of C meets the perpendicular bisector of |AC| at X. Find the point X by construction

(c) Measure |BX|