Example 1

The diagram shows how a distance across a lake can be calculated using the concept of similar triangles. In the diagram, AB is parallel to ED while BE and AD intersect at the point C.
(a) Name the two triangles that are similar. Give reasons for your answer.
(b) Find the distance, DE, across the lake.
(a) Angle ACB = angle DCE (vert. opp. angles)
Since AB is parallel to ED, angle ABC = angle DEC (alternate angles)
Therefore, triangle ABC is similar to triangle DEC (AA).

(b) Since triangle ABC is similar to triangle DEC, AB : DE = BC : EC (corresponding sides of similar triangles)
Therefore, DE = AB * EC / BC = 652m

Example 2

In the diagram, AC is parallel to DF, DE is parallel to BC, AE = 4cm, AC = 6cm, DB = 3cm and BF = 6cm.
Find the lengths of AD and CF in the diagram.
Since AC is parallel to DF, angle ACB = angle DFB (corresponding angles)
Also, angle ABC = angle DBF (common angles)
Triangle ABC is similar to triangle DBF (AA)
Therefore, AB : DB = BC : BF = AC : DF (corresponding sides of similar triangles)
Since AC is parallel to DF, DE is parallel to BC, DECF is a parallelogram.
DF = EC = 6 - 4 = 2cm.
Therefore, AB = DB * AC / DF = 9cm, and AD = 9 - 3 = 6cm.
BC = BF * AC / DF = 18cm, and CF = 18 - 6 = 12cm.

Example 3

In the diagram (not drawn to scale), BDC is a straight line and angle BDA = angle BAC.
(a) Name an angle equal to BAD.
(b) Calculate AB if BD = 8 cm, AD = 6 cm and AC = 7 cm.
(a) Angle BCA is equal to angle BAD. (Can you explain why?)

(b) Angle BCA = angle BAD (from part (a))
Angle ABC = angle DBA (common angles)
Therefore, triangle BCA = triangle BAD (AA)
AB : DB = AC : DA = 7 : 6 (correponding sides of similar triangles)
AB = 8 * 7 / 6 = 9.33cm (3 s.f.)


Download the assignment here: Assignment 1_Similarity.doc
For Mr Colin Toh's classes, please download the assignment:
Complete and submit hardcopies of your work to your respective maths teachers. Deadlines will be given by your maths teachers too.