Similar Triangles

Definition

Two similar triangles

Generally, two triangles are said to be similar if they have the same shape, even if they are scaled, rotated or even flipped over.

The mathematical presentation of two similar triangles A1B1C1 and A2B2C2 as shown by the figure beside is:

ΔA1B1C1 ~ ΔA2B2C2

Two triangles are similar if:

1. Each angle in one triangle is congruent with (equal to) its corresponding angle in the other triangle i.e.:
∠A1 = ∠A2, ∠B1 = ∠B2 and ∠C1 = ∠C2

2. The ratio of the length of one side of one triangle to the corresponding side in the other triangle is the same i.e.:
$\frac{A_1B_1}{A_2B_2}=\frac{A_1C_1}{A_2C_2}=\frac{B_1C_1}{B_2C_2}$

3. The ratio of the length of two sides of one triangle to the corresponding sides in the other triangle is the same and
the angles between these sides are equal i.e.:
$\frac{B_1A_1}{B_2A_2}=\frac{A_1C_1}{A_2C_2}$ and $\angle A_1 = \angle A_2$
or
$\frac{A_1B_1}{A_2B_2}=\frac{B_1C_1}{B_2C_2}$ and $\angle B_1 = \angle B_2$
or
$\frac{B_1C_1}{B_2C_2}=\frac{C_1A_1}{C_2A_2}$ and $\angle C_1 = \angle C_2$

Be careful not to mix similar triangles with identical triangle. Identical triangles are those having the same corresponding sides’ lengths. Therefore, for identical triangles:

$\frac{A_1B_1}{A_2B_2}=\frac{A_1C_1}{A_2C_2}=\frac{B_1C_1}{B_2C_2}=1$

Therefore, all identical triangles are similar. However, not all similar triangles are identical.

Although the above shows that we need to know the measures of the three angles or the lengths of the three sides of each triangle in order to decide whether the two triangles are similar or not, it would be sufficient, for solving problems involving similar triangles, to know only three of the above measures for each triangle. These measures can be any of the following combinations:

1) the three angles of each triangle (without the need to know the lengths of their sides).

Or at least 2 angles of the first triangle are equal to 2 angles of the second triangle.
Because if 2 angles are equal the third angles are equal too.(The third angles are 180 - angle1 - angle2)

2) the lengths of the sides of each triangle (without the need to know the measures of their angles);

3) the lengths of two sides and the measure of one angle of each triangle. This angle should be the one formed by the two known sides.

In what follows we shall explain the solution of some problems involving similar triangles. We will start with those problems that can be solved with the direct application of the above rules, and then we will upgrade our discussion to explain some practical problems which use the similar triangles principle to be solved.

Direct Application of Problems Involving Similar Triangles

Example 1: Show that the two triangles given in the figure below are similar.
2 similar triangles with given sides

Solution:
Since the lengths of the sides of both triangles are known, the second condition can be checked:

$\frac{PQ}{AB}=\frac{6}{2}=3$ $\frac{QR}{CB}=\frac{12}{4}=3$ $\frac{PR}{AC}=\frac{15}{5}=3$

Example 2: Show that the two triangles given beside are similar and calculate the lengths of sides PQ and PR.

Solution:
∠A = ∠P and ∠B = ∠Q, ∠C = ∠R(because ∠C = 180 - ∠A - ∠B and ∠R = 180 - ∠P - ∠Q)

Therefore, the two triangles ΔABC and ΔPQR are similar. Consequently:
$\frac{AB}{PQ}=\frac{BC}{QR}=\frac{AC}{PR}$

$\frac{BC}{QR}=\frac{6}{12}=\frac{AB}{PQ}=\frac{4}{PQ} \Rightarrow PQ=\frac{4\times12}{6} = 8$ and
$\frac{BC}{QR}=\frac{6}{12}=\frac{AC}{PR}=\frac{7}{PR} \Rightarrow PR=\frac{7\times12}{6} = 14$

Example 3: Find the length AB in the triangle shown beside.

Solution:

∠ABC = ∠ADE, ∠ACB = ∠AED and ∠A is common => the two triangles ΔABC and ΔADE are similar.

$\frac{BC}{DE} = \frac{3}{6} = \frac{AB}{AD} = \frac{AB}{AB + BD} = \frac{AB}{AB + 4} = \frac{1}{2} \Rightarrow 2\times AB = AB + 4 \Rightarrow AB = 4$

Example 4: Given the shape shown by the figure beside. Find the length AD (x).

The two triangles ΔABC and ΔCDE appear to be similar since AB || DE and they have the same apex angle C.
It appears that one triangle is a scaled version of the other. However, we need to prove this mathematically.

AB || DE, CD || AC and BC || EC
∠BAC = ∠EDC and ∠ABC = ∠DEC

Considering the above and the common angle C, we may conclude that the two triangles ΔABC and ΔCDE are similar.

Therefore:
$\frac{DE}{AB} = \frac{7}{11} = \frac{CD}{CA} = \frac{15}{CA} \Rightarrow CA = \frac{15 \times 11}{7} = 23.57$
x = AC - DC = 23.57 - 15 = 8.57

Practical Examples

Example 5: A factory is using an inclined conveyor belt to transport its products from Level 1 to Level 2 which is 3m above level 1 as shown by the figure below. The inclined conveyor is supported from one end to Level 1 and from the other end to a post located 8m away from Level 1 support point.

The factory wants to extend its conveyor to reach a new Level 2 which is 9m above Level 1 while maintaining the inclination angle of the conveyor.

Find the distance at which a new post is to be installed to support the conveyor at its new end at Level 2. Also, calculate the additional distance that the product has to travel to reach the new level.

Solution:

First, let us denote each intersection point by a letter as shown in Red on the figure above.

Following the same explanation provided in the examples above, we can conclude that the two triangles ΔABC and ΔADE are similar. Therefore,

$\frac{DE}{BC} = \frac{3}{9} = \frac{AD}{AB} = \frac{8}{AB} \Rightarrow AB = \frac{8 \times 9}{3} = 24m$
x = AB - 8 = 24 - 8 = 16m

Hence, the new post should be placed at a distance of 16m from the existing post.

Since the construction is forming right-angle triangles, we can calculate the travel distance of the product as follows:

$AE = \sqrt{AD^2 + DE^2} = \sqrt{8^2 + 3^2} = 8.54m$

Similarly, $AC = \sqrt{AB^2 + BC^2} = \sqrt{24^2 + 9^2} = 25.63m$
which is the distance the product is currently travelling to reach the existing level.

y = AC - AE = 25.63 - 8.54 = 17.09m
which is the additional distance that the product has to travel to reach the new level.

Example 6: Steve wants to visit his friend who recently moved to a new house. The road map between Steve’s home and his friend’s as well as the distances known to Steve are as shown in the figure below. Guide Steve to reach his friend’s house using the shortest path.

Solution:

The road map can be geometrically expressed as shown by the figure below.

You may notice that the two triangles ΔABC and ΔCDE are similar and therefore:
$\frac{AB}{DE} = \frac{BC}{CD} = \frac{AC}{CE}$

From the problem description, we have:

AB = 15km, AC = 13.13km, CD = 4.41km and DE = 5km

From the above, we can calculate the following lengths:

$BC = \frac{AB \times CD}{DE} = \frac{15 \times 4.41}{5} = 13.23km$
$CE = \frac{AC \times CD}{BC} = \frac{13.13 \times 4.41}{13.23} = 4.38km$

In order for Steve to reach his friend’s house, he may follow any of the following routes:

A -> B -> C -> E -> G which has a total length of 7.5+13.23+4.38+2.5=27.61km

F -> B -> C -> D -> G which has a total length of 7.5+13.23+4.41+2.5=27.64km

F -> A -> C -> E -> G which has a total length of 7.5+13.13+4.38+2.5=27.51km

F -> A -> C -> D -> G which has a total length of 7.5+13.13+4.41+2.5=27.54km

Therefore, Route number 3 is to shortest one to be recommended to Steve.

Example 7:
Trisha wants to measure the height of a building but she does not have the tools to do so. She noticed that there is a tree located in front of the building so she decided to use her smartness and the geometry knowledge that she got at school to measure the building height. She measured the distance between the tree and the building and found that it is 30m. She stood in front of the tree and started backing until she could see the top edge of the building from above the tree top. She marked her place and measured it from the tree. It was 5m.

Knowing that the tree height is 2.8m and Trisha’s eyes height is 1.6m, help Trisha to do the math and calculate the building height.

Solution:

This problem can be geometrically represented as in the figure below.

First, let us make use of the similarity between the triangles ΔABC and ΔADE.

$\frac{BC}{DE} = \frac{1.6}{2.8} = \frac{AC}{AE} = \frac{AC}{5 + AC} \Rightarrow 2.8 \times AC = 1.6 \times (5 + AC) = 8 + 1.6 \times AC$

$(2.8 - 1.6) \times AC = 8 \Rightarrow AC = \frac{8}{1.2} = 6.67$

We can then use the similarity between triangles ΔACB and ΔAFG or between the triangles ΔADE and ΔAFG. Let us take the first option.

$\frac{BC}{FG} = \frac{1.6}{H} = \frac{AC}{AG} = \frac{6.67}{6.67 + 5 + 30} = 0.16 \Rightarrow H = \frac{1.6}{0.16} = 10m$


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