Alternative Format — Lesson 2: Similar Triangles

Let's Start Thinking

Similar Triangles in the Real-World

Many real-world problems, in everything from construction to astrophysics, can be solved by making use of similar triangles.

Wooden frame of a house under construction.

Source: Construction - DmitryPK/iStock/Getty Images

Observatory with shooting stars above.

Source: Observatory - Cylonphoto/iStock/Getty Images

Typically, these problems involve setting up a pair of similar triangles whose side lengths correspond to distances in the real world. In doing so, we can connect unknown distances to known ones using the scale factor of a similarity relation.

In this lesson, we will first review how to establish that two triangles are similar. Once we have done that, we will use similar triangle constructions to solve a variety of applications.


Lesson Goals

  • Demonstrate that two triangles are similar using similarity rules.
  • Use similarity between triangles to solve for an unknown side length.
  • Construct an appropriate pair of similar triangles to solve a real-world problem.

Try This

An architect has been asked to design a bridge that will span a large canyon.

How can the architect determine an appropriate length for the bridge by only making measurements on one side of the canyon? 


Similarity Rules


Similarity for Triangles

How do we show that a pair of triangles are similar to one another? Recall the definition of similarity:

Recall

Two polygons are similar when

  • all corresponding angles are equal and
  • all corresponding sides are proportional.

For a pair of triangles, does this mean that we then have to compare three pairs of angles and three pairs of sides to demonstrate similarity? Fortunately, the answer is no. In practice, one does not need to compare all corresponding angles and sides to demonstrate similarity.

We will see three similarity rules that can be used to establish similarity with a minimal number of comparisons.

Explore This 1

Question

In the diagram below, \(\angle A\) is fixed and shared between \(\triangle ABC\) and \(\triangle AB'C'\). The points \(B\) and \(B'\) lie on the same line extending from \(A\). In addition, the points \(C\) and \(C'\) lie on the same line extending from \(A\).

Triangles ABC and AB prime C prime share angle A.

Are the two triangles always similar if they share \(\angle A\)? What if they share two or three angles?

 

Description

We may move point \(B'\) along the line which extends from \(A\) and passes through \(B\) and \(B'\), while keeping all other points fixed. In this way, \(AB\) and \(AB'\) both lie on the same line, though they may differ in length. \(\angle A\) is thus unchanged. Equivalently, we may choose to move \(C'\) along the line which extends from \(A\) and passes through \(C\) and \(C'\), while keeping all other points fixed, leaving \(\angle A\) unchanged.

Consider moving \(B'\) such that \(AB'\) is extended in length.

Triangles ABC and AB prime C prime share angle A.

Alternatively, consider moving \(B'\) such that \(AB'\) is shortened in length, such that \(AB'\) is shorter than \(AB\), and \(\angle AB'C'\) is close to \(90^{\circ}\).

Triangles ABC and AB prime C prime share angle A.

Now consider setting \(\angle AB'C'\) equal to the value of \(\angle ABC\). 

Angle A B prime C prime is set equal to the value of angle ABC

Created with GeoGebra. Author: University of Waterloo. CC BY-NC-SA 4.0.

Note that angles \(\angle ACB\) and \(\angle AC'B'\) are also now equal.

Interactive Version

Similar Triangles AA Similarity

Explore This 1 Summary

In the Explore This, you should have found that two triangles with only one equal angle are not necessarily similar.

For example, \(\triangle ABC\) is not similar to \(\triangle AB'C'\), even though they share \(\angle A\):

Angle A is shared in triangles ABC and AB prime C prime. Triangle ABC has an acute angle at C. Triangle AB prime C prime has an obtuse angle at C prime.

On the other hand, once the triangles have two or three equal angles, you should find that they must be similar.

For example, \(\triangle ABC\) is similar to \(\triangle AB'C'\) since \(\angle A\) is shared and \(\angle C = \angle C'\).

You may have also noticed that setting two equal angles has the effect setting all three angles equal. This is because the third angle in any triangle is determined once the first two angles are set. We conclude from this activity that when a pair of triangles shares two angles, the triangles must be similar.

This demonstrates a result known as the angle-angle (AA) similarity rule.

Explore This 2

Question

In the diagram below, \(\angle A\) is fixed and shared between \(\triangle ABC\) and \(\triangle AB'C'\). The points \(B\) and \(B'\) lie on the same line extending from \(A\). In addition, the points \(C\) and \(C'\) lie on the same line extending from \(A\).

Triangles ABC and AB prime C prime are similar and share angle A.

How does fixing the ratio between a set of corresponding sides affect the triangles? 

Description

We may move point \(C'\) along the line which extends from \(A\) and passes through \(C\) and \(C'\), while keeping all other points fixed. In this way, \(AC\) and \(AC'\) both lie on the same line, though they may differ in length. \(\angle A\) is thus unchanged. Equivalently, we may choose to move \(B'\) along the line which extends from \(A\) and passes through \(B\) and \(B'\), while keeping all other points fixed, leaving \(\angle A\) unchanged.

Consider moving \(C'\) such that \(AC'\) is extended in length.

Triangles ABC and AB prime C prime are similar and share angle A.

Equaivalently, we could also have chosen to move \(C'\) such that \(AC'\) is shortened in length.

Now set \(AB'\) so that \(\dfrac{AB'}{AB}=\dfrac{AC'}{AC}\).

B prime is adjusted such that the ratio AB-prime over AB is equal to AC-prime over AC

Created with GeoGebra. Author: University of Waterloo. CC BY-NC-SA 4.0.

Equivalently, we could choose to move point \(B'\) so that \(AB\) is extended (or, shortened) in length, and then set \(AC'\) so that \(\dfrac{AB'}{AB}=\dfrac{AC'}{AC}\).

 

Interactive Version

Similar Triangles SAS

Explore This 2 Summary

As in the previous activity, while the two triangles only share a single angle, you should see that they are not necessarily going to be similar.

For example, \(\triangle ABC\) is not similar to \(\triangle AB'C'\) even though they share \(\angle A\).

Trinagles ABC and AB prime C prime share angle A. In triangle ABC, angle C is acute. In triangle AB prime C prime, angle C prime is obtuse.

However, once the corresponding sides adjacent to the shared angle are set to be in the same proportion, you should have found that the two triangles must be similar. Two triangles with two pairs of corresponding sides in the same proportion and with the angle adjacent to both of those sides equal are always similar.

For example, \(\triangle ABC\) is similar to \(\triangle AB'C'\), since \(\angle A\) is shared and  \(\dfrac{AB'}{AB} = \dfrac{AC'}{AC}\).

We conclude from this activity that it is sufficient to demonstrate similarity by showing that two pairs of corresponding sides are in the same proportion and the angles adjacent to both of those pairs of sides are equal.

This demonstrates a result known as the the side-angle-side (SAS) similarity rule.

The Problem With a Side-Side-Angle Comparison

When invoking the side-angle-side rule, it is crucial to keep in mind that you cannot compare any corresponding angle and any two corresponding sides. In particular, you must use the angle at the vertex formed by the sides that are used. For the side-angle-side (SAS) rule, we must use the angle at the vertex formed by the sides that are used. Because this angle is sandwiched between both sides, we call this angle the contained angle.

Why do we have to use the contained angle with the side-angle-side rule? Consider \( \triangle ABC\) and \(\triangle AB'C' \).

Observe:

  • \(\triangle ABC\) and \(\triangle AB'C' \) share \(\angle A\),
  • \(\dfrac{AB}{AB'}=\dfrac{BC}{B'C'}\)

Therefore, we could say that the two triangles satisfy a side-side-angle or angle-side-side comparison. And in fact, \(\triangle ABC \sim \triangle AB'C'\).

So what's the problem? The problem is that there exists another triangle, \(\triangle AB'D\), which satisfies the same conditions. 

Specifically, notice that, \(B'D = B'C'\) so \(\dfrac{AB}{AB'}=\dfrac{BC}{B'D}\), but \(\triangle ABC\) is clearly not similar to \(\triangle AB'D\).

Due to this ambiguity, we have the following important fact:

A side-side-angle comparison is not sufficient to demonstrate similarity.

Explore This 3

Question

How does forcing corresponding sides of \(\triangle ABC\) and \(\triangle AB'C'\) to be in ratio affect their relationship if the triangles are not forced to share equal angles? 

Description

To begin with, in the diagram below, \(\angle A\) is and shared between \(\triangle ABC\) and \(\triangle AB'C'\). 

To change the shape of \(\triangle AB'C'\), we may freely move \(B'\), while keeping other points fixed. 

Next, we may freely move \(C'\), while keeping other points fixed. 

Consider adjusting the length of \(AB'\) so that \(\dfrac{AB'}{AB}=\dfrac{AC'}{AC}\).

Alternatively, consider adjusting the length of \(AC'\) so that \(\dfrac{AB'}{AB}=\dfrac{B'C'}{BC}\).

Finally, consider adjusting the lengths of both \(AB'\) and \(AC'\), so that, \(\dfrac{AB'}{AB}=\dfrac{AC'}{AC}\) and \(\dfrac{AB'}{AB}=\dfrac{B'C'}{BC}\) simultaneously

Created with GeoGebra. Author: University of Waterloo. CC BY-NC-SA 4.0.

Interactive Version

Similar Triangles SSS Similarity

Explore This 3 Summary

In the Explore This, you should have found that \(\triangle AB'C'\) is initially free to take any shape.

If you set one pair of corresponding sides to be in the same proportion, you may have noticed that the relative placements of points \(B'\) and \(C'\) become constrained. But we still do not have similarity between the two triangles in general.

Triangle ABC and triangle AB prime C prime share the point A. C and C prime are both on the same line, which extends through A.

However, once all three corresponding sides are set to be in the same proportion, the two triangles will always be similar.

For example, \(\triangle ABC\) is similar to \(\triangle AB'C'\), since the ratios of the corresponding sides are all equal. That is, \(\dfrac{AB'}{AB}=\dfrac{AC'}{AC}=\dfrac{B'C'}{BC}\).

From this, we conclude that similarity can be demonstrated by showing that all of the corresponding sides in a pair of triangles are in the same proportion.

This demonstrates the result known as the side-side-side (SSS) similarity rule.


Check Your Understanding 1

Question — Version 1

Given \(\triangle ABC\) and the following information about \(\triangle PQR\), select the correct phrase to complete the statement below. Remember to check the vertex ordering.

Information about \(\triangle PQR\):

\(\angle PQR=75^{\circ}\)

\(\angle QRP=80^{\circ}\)

Diagram of \(\triangle ABC\):

Triangle ABC. Angle A = 25 degrees, Angle B = 75 degrees, Angle C = 80 degrees. Side lengths are a=18.3, b=41.9, and c=42.6

Statement: \(\triangle ABC\) and \(\triangle PQR\)

  1. are similar by the SSS rule.
  2. are similar by the SAS rule.
  3. are similar by the AA rule.
  4. are not similar.
  5. There is not enough information to answer this question.

Answer — Version 1

c. \(\triangle ABC\) and \(\triangle PQR\) are similar by the AA rule.

Feedback — Version 1

When identifying similar triangles, the order of the vertices in the triangle name is important.
Observe that when comparing \(\triangle ABC\) and \(\triangle PQR\):
\(\angle PQR\) corresponds to \(\angle ABC\) and \(\angle PQR=\angle ABC\)
\(\angle QRP\) corresponds to \(\angle BCA\) and \(\angle QRP=\angle BCA\).
Since all corresponding angles are equal, the AA similarity rule tells us that \(\triangle ABC\) is similar to \(\triangle PQR\).

Question — Version 2

Given \(\triangle ABC\) and the following information about \(\triangle PQR\), select the correct phrase to complete the statement below. Remember to check the vertex ordering.

Information about \(\triangle PQR\):

\(\angle RPQ=65^{\circ}\)

\(RP=41.6\) m

\( PQ=24.2 \) m

Diagram of \(\triangle ABC\):

Triangle ABC. Angle A = 65 degrees, Angle B = 80 degrees, Angle C = 35 degrees. Side lengths are a=19.1, b=20.8, and c=12.1

Statement: \(\triangle ABC\) and \(\triangle PQR\)

  1. are similar by the SSS rule.
  2. are similar by the SAS rule.
  3. are similar by the AA rule.
  4. are not similar.
  5. There is not enough information to answer this question.

Answer — Version 2

b. \(\triangle ABC\) and \(\triangle PQR\) are similar by the SAS rule.

Feedback — Version 2

When identifying similar triangles, the order of the vertices in the triangle name is important.
Observe that when comparing \(\triangle ABC\) and \(\triangle PQR\):
\(\angle RPQ\) corresponds to \(\angle CAB\) and \(\angle RPQ=\angle CAB\)
Side \(RP\) corresponds to side \(CA\) and \(RP=2\times CA\)
Side \(PQ\) corresponds to side \(AB\) and \(PQ=2\times AB\)
By the SAS similarity rule, two triangles are similar if the two sets of corresponding sides are in ratio, and the
contained angle is equal. since these conditions are satisfied, \(\triangle ABC\) is similar to \(\triangle PQR\) .

Question — Version 3

Given \(\triangle ABC\) and the following information about \(\triangle PQR\), select the correct phrase to complete the statement below. Remember to check the vertex ordering.

Information about \(\triangle PQR\):

\(PQ=53.7\) m

\(QR=58.2\) m

\(PR=63.2\) m

Diagram of \(\triangle ABC\):

Triangle ABC. Angle A = 60 degrees, Angle B = 70 degrees, Angle C = 50 degrees. Side lengths are a=17.9, b=19.4, and c=15.8

Statement: \(\triangle ABC\) and \(\triangle PQR\)

  1. are similar by the SSS rule.
  2. are similar by the SAS rule.
  3. are similar by the AA rule.
  4. are not similar.
  5. There is not enough information to answer this question.

Answer — Version 3

d. \(\triangle ABC\) and \(\triangle PQR\) are not similar.

Feedback — Version 3

When identifying similar triangles, the order of the vertices in the triangle name is important.
Observe that when comparing \(\triangle ABC\) and \(\triangle PQR\):
\(PQ\) corresponds to \(AB\) and \(PQ\approx3.4\times AB\)
\(QR\) corresponds to \(BC\) and \(QR\approx3.3\times BC\)
\(PR\) corresponds to \(AC\) and \(PR\approx3.3\times AC\)
The SSS rule for similarity states that two triangles are similar if all three sets of corresponding sides are in ratio. Since this condition is not satisfied, \(\triangle ABC\) is not similar to \(\triangle PQR\).


Let's Recap

To review, we have the following three similarity rules for triangles:

  • The angle-angle (AA) similarity rule says that when a pair of triangles have two corresponding angles equal, the triangles are similar.
  • The side-angle-side (SAS) similarity rule says that when two pairs of corresponding sides are in the same proportion and the contained angles are equal, the triangles are similar.
  • The side-side-side (SSS) similarity rule says that when all three pairs of corresponding sides are in the same proportion, the triangles are similar.
Angle-Angle (AA) Side-Angle-Side (SAS) Side-Side-Side (SSS)
Triangle with 3 angles labelled. Triangle with 1 angle labelled between two sides with lengths 10 and 8. Triangle with side lengths 10, 8, and 15.
Larger triangle with the same 3 angles. Triangle with same size angle labelled between two sides with lengths 15 and 12. Triangle with side lengths 15,12, and 22.5.

Example 1

Show that \(\triangle MNO \sim \triangle RST\).

Triangle MNO. Angle M=90, MN=3, MO=4.

Triangle RST. Angle R=90. RS=9 and RT=12.

Solution

We are given two lengths and a contained angle in each triangle. A quick calculation reveals that we can pair up the given lengths so that they are in equal proportions.

First, consider \(\dfrac{RS}{MN}\):

\[ \frac{RS}{MN} = \frac{9}{3}= 3 \]

Now, consider \(\dfrac{RT}{MO}\):

\[ \frac{RT}{MO} = \frac{12}{4} = 3\]

Both contained angles are also right angles and therefore equal.

Therefore, by the side-angle-side (SAS) similarity rule, triangle \( MNO\) is similar to triangle \(RST\).


Applications of Similar Triangles


Try This Revisited

An architect has been asked to design a bridge that will span a large canyon. How can the architect determine an appropriate length for the bridge by only making measurements on one side of the canyon?

Solution

The architect can solve this problem by using similar triangles as follows. First, they label the point where the bridge should begin by \(S\) and a visual reference point on the opposite side of the canyon, where the bridge should end, by \(T\).

Bridge spanning canyon from point S to point T. ST is an unknown length

Next, they walk along the edge of the canyon some distance, label another reference point by \(X\), and then proceed a little bit farther along the edge and label yet another reference point by \(A\).

Finally, they turn \(90^{\circ}\) and walk away from the canyon until, from their point of view, the reference points \(X\) and \(T\) are aligned. They label this point \(B\). The architect can now construct two triangles, \(\triangle STX\) and \(\triangle ABX\). And they are free to measure any side lengths on their side of the canyon.

Triangles STX and ABX. Angles BAX and TSX=90, Angles TXS and AXB are similar angles.

How does this help the architect determine length of side \(ST\)? For concreteness, let's suppose they measure sides \(SX\), \(AX\), and \(AB\) and find the following: 

  • \(SX=100\) metres, 
  • \(AX=25\) metres, and
  • \(AB=10\) metres.

SX=100 m, XA=25 m, AB=10 m.

Notice that \(\angle SXT\) and \(\angle AXB\) are opposite angles, and therefore equal:

\[ \angle SXT = \angle AXB \]

Also, \(\angle TSX\) and \(\angle BAX\) are both right angles:

\[\angle TSX = \angle BAX\]

Since the two triangles have two equal angles, they must be similar by the angle-angle (AA)

similarity rule. 

\[ \begin{align*} \triangle {S}{T}{X} &\sim \triangle ABX \end{align*}\]

Now, since the two triangles are similar, we know that the ratios of corresponding side lengths must be equal. By inspection, side \(ST\) corresponds to side \(AB\) and side \(SX\) corresponds to side \(AX\). This means that \(\dfrac{ST}{AB}=\dfrac{SX}{AX}\).

Equivalently, \(ST\) is equal to \(SX\) over \(AX\) all multiplied by \(AB\). The architect can use this equation along with their measured values to calculate the width of the canyon. Doing so, they find that \(ST\) is equal to \(40\) metres.

\(\quad \implies \quad ST = \dfrac{SX}{AX} \cdot AB =\dfrac{100 }{25 }\cdot(10 ) = 40\)

Therefore, the architect should design the bridge to be at least \(40\) metres long.

Example 2

Rachael, who is \(1.7\) metres tall, is standing \(9\) metres from the base of a tree. She notices that the top of her shadow matches up with the tip of the shadow from the tree \(1\) metre from her feet. Determine the height of the tree.

Solution

Let us form a triangle, \(\triangle ABC\), by connecting the point where the shadows meet, which we'll label \(A\); the top of Rachael's head, which we'll label \(B\); and Rachael's feet, which we'll label \(C\). Let's make a second triangle \(\triangle ADE\), by extending \(AB\) to the top of the tree, which we'll label \(D\); and extending \(AC\) to the base of the tree, which we'll label \(E\).

Rachael stands 1.7 metres tall 9 metres away from a tree. Her shadow meets up with tree shadow 1 metre from her feet.

Right-triangle ABC. BC=1.7 m AC=1 m. Right-triangle ADE. AE=10 m, ED=?

Sources: Chestnut Tree - ikryannikovgmailcom/iStock/Getty Images; Students - MicrovOne/iStock/Getty Images

Notice that these triangles share an angle at point \(A\). Observe that

\[\angle BAC=\angle DAE\]

Additionally, they each contain a right angle. These are also equal:

\[\angle BCA=\angle DEA\]

Since we have two equal angles, then by the angle-angle similarity rule,

\[\triangle ABC \sim \triangle ADE\]

Now, we can use the fact that corresponding sides of similar triangles are in the same proportions to calculate the length of the unknown side \(DE\), which corresponds to the height of the tree. By inspection, side \(AC\) corresponds to side \(AE\) and side \(BC\) corresponds to side \(DE\). Their ratios must be in the same proportion.

\[\frac{DE}{BC} = \frac{AE}{AC} \]

Recall that right-triangle ABC. BC=1.7 m AC=1 m. Right-triangle ADE. AE=10 m, ED=?

Sources: Chestnut Tree - ikryannikovgmailcom/iStock/Getty Images; Students - MicrovOne/iStock/Getty Images

Using this relation, we can solve for the length of \(DE\):

\[\begin{align*} DE &= \left( \dfrac{AE}{AC} \right) \cdot BC \\ &= \left( \dfrac{AC + CE}{AC} \right) \cdot BC\\ &= \left( \dfrac{1 + 9}{1} \right) \cdot (1.7)\\ &=17 \end{align*}\]

 

Therefore, the tree is approximately ‌\(17\) metres tall.

Check Your Understanding 2

Question

A circular crater has a diameter of \(150\) metres. You climb a watchtower situated \(40\) metres from the edge of the crater until you just see the centre of the bottom of the crater appear over the rim. At this point, you have climbed \(75\) metres up the watchtower. What is the depth below ground level at the centre of the crater? (Ground level is determined by the base of the watchtower.) Round your answer to the nearest metre.

A man stands at the top of a tower, looking down at a circular crater. A triangle is formed by the depth of the crater at its center, a radius of the crater on the line between the man and the center of the crater, and the man's line of sight from the crater's edge to the center of the bottom of the crater.

Answer

The depth of the crater is \(13\) metres.

Feedback

Let us label the point of observation as \(A\), the base of the tower as \(B\), the rim of the crater as \(C\), ground level above the centre of the crater as \(D\), and the centre of the bottom of the crater as \(E\). This allows us to construct a pair of triangles, \(\triangle A B C\) and \(\triangle CDE\).

A man stands at the top of a tower, looking down at a circular crater.

Note, in these triangles \(AB\) represents our height on the watchtower, \(BC\) is the distance from the base of the tower to the rim of the crater, and \(CD\) is equal to half of the diameter of the crater. Our goal is to determine the length of \(DE\) — the depth of the crater.

Observe that these two triangles are similar by the angle-angle similarity rule.

\(\begin{aligned} \angle A B C =\angle E D C &\quad \text { Right angles } \\ \angle A C B =\angle E C D &\quad \text { Opposite angles } \\ \Longrightarrow \triangle A B C \sim \triangle E D C &\quad \text { Angle-angle similarity } \end{aligned}\)

It follows that

\(D E=\left(\dfrac{C D}{B C}\right) \cdot A B=\left(\dfrac{75}{40}\right) \cdot(7)=13.125\)

Therefore, the crater is approximately \(13\) metres deep.

Example 3

Eve flies in a straight line from Waterloo to Toronto and back to Waterloo. She records the one-way distance from Waterloo to Toronto as \(93\) km. On her map, Waterloo and Toronto are \(7.8\) cm apart and Niagara Falls and Waterloo are \(10.3\) cm apart. She knows the map is drawn to scale, but the scale is missing. What is the real-world straight line distance from Niagara Falls to Waterloo?

(Note: You may ignore the curvature of the Earth.)

Solution

Let's label the city of Waterloo with the letter \(W\), Toronto with the letter \(T\), and Niagara Falls with the letter \(N\). These form a triangle, \(\triangle WTN\). On the map, these locations also form a triangle, which we'll label triangle \(W'\), \(T'\), \(N'\), where the point \(W'\) denotes Waterloo on the map, and similarly for the other two cities.

Triangle WTN. TW=93 km. WN is the desired unknown quantity.

Triangle W'T'N'. T'W'=7.8 cm N'W'=10.3 cm.

Since real-world distances are in the same proportion to corresponding distances on the map, \( \triangle WTN \sim \triangle W'T'N'\) by the side-side-side similarity rule.

Since the triangles are similar, the ratios of all three pairs of corresponding sides must be equal. In particular,

\[ \frac{WN}{W'N'} = \frac{WT}{W'T'} \]
 

Rearranging to solve for \(WN\), and plugging in known values, we find:

\[\begin{align*} WN &= \left( \dfrac{W'N'}{W'T'} \right) WT\\ &= \left(\dfrac{10.3 \mathrm{\;cm}}{7.8\mathrm{\;cm}} \right) \cdot (93\text{ km} )\\ &\approx 123\text{ km} \end{align*}\]

Therefore, the distance from Waterloo to Niagara Falls is approximately \(123\) kilometres.

Check Your Understanding 3

Question

On a scale model of the solar system, the model planets move around so as to accurately portray a downscaled model of the actual solar system in real time. You measure the distance between the models of the Sun and Earth to be separated by \(7.5\) cm and the distance between the models of Earth and Mars to be separated by \(13.5\) cm. Given that the actual distance from the Earth to the sun is \(150\) million km, by what distance are Earth and Mars currently separated?

Answer

\(270\) million km.

Feedback

Let the positions of Earth, Sun, and Mars in the real world be represented by the letters \(E\), \(S\), and \(M\) respectively.  Let the positions of Earth, Sun, and Mars in the model be represented by \(E'\), \(S'\), and \(M'\) respectively.

The triangles formed by the Sun, Earth, and Mars in the real world and on the scale model are similar by the side-side-side (SSS) similarity rule. Therefore, the ratio of the actual Earth-Sun separation distance and the corresponding distance on the model must be equal to the ratual Earth-Mars distance and the corresponding distance on the model.

\(\dfrac{E S}{E^{\prime} S^{\prime}}=\dfrac{E M}{E^{\prime} M^{\prime}}\)

We can now determine the actual Earth-Mars separation distance.

\(E M=\dfrac{E^{\prime} M^{\prime}}{E^{\prime} S^{\prime}} \cdot E S=\left(\dfrac{13.5 \mathrm{cm}}{7.5 \mathrm{cm}}\right) \cdot(150 \text { million } \mathrm{km})=270 \text{ million } \mathrm{km}\)

Therefore, Earth and Mars are separated by 270 million km.


Example 4

Alice, Bob, and Charlie are scuba diving in murky water and would like to determine the visibility (that is, the distance they could be separated before they can no longer see each other). To do this, Alice stays still while Bob and Charlie each swim directly away from her in different directions until they can both see Alice but cannot see each other. Initially, Bob and Charlie are both \(2\) metres from Alice and \(3\) metres from each other.

If Bob and Charlie signal to Alice that they can no longer see each other once they are each \(10\) metres from Alice, what is the visibility?

Solution

Alice, Bob, and Charlie initially form a triangle \(ABC\) with sides lengths \(AB = AC = 2\) m and \(BC = 3\) m.

After Bob and Charlie swim away, they form a new triangle \(AB'C'\) with side lengths \(AB' = AC' = 10\) m.

What is the length of \(B'C'\)?

We can make the following observations: 

  • the ratios of \(AB'\) to \(AB\) and \(AC'\) to \(AC\) are both equal to \(5\) and
  • these pairs of sides contain the shared angle \(A\).

Therefore, triangle \(AB'C'\) is similar to \(ABC\) with a scale factor of \(5\) by the side-angle-side (SAS) similarity rule.

Since triangles are similar, we also know \(B'C'\) should be equal to \(BC\) multiplied by the scale factor

\[ B'C' = 5 \cdot BC = 15 \]

Therefore, the visibility in the water is \(15\) meters.


Check Your Understanding 4

Question

While waiting at a bus stop, Vince notices that he can see the reflection of a street lamp in a puddle. From science class, Vince knows that the angle of reflection is equal to the angle of incidence when the light reflects off the puddle. The man is \(1.8\) m tall, and he is standing \(3.8\) m from the puddle. The lamp is \(12.8\) m on the opposite side of the puddle. Find the height of the street lamp. Round your answer to \(1\) decimal place.

Sources: Street Light - alex_profa/iStock/Getty Images; Person - panic_attack/iStock/Getty Images

Answer

The height of the lamp is \(6.1\) metres.

Feedback

Let's begin by labelling the puddle as point \(A\), Vince's feet as point \(B\), Vince's eyes as point \(C\), the base of the streetlamp as point \(D\), and the streetlamp light as point \(E\). This gives us the following picture.

Note that two triangles are formed. Triangle ABC has known side lengths AB=3.8 and BC=1.8. Triangle ADE has known side length AD=12.8. DE is the desired unknown quantity.

Sources: Street Light - alex_profa/iStock/Getty Images; Person - panic_attack/iStock/Getty Images

Observe that \(\angle A B C\) and \(\angle A D E\) are both right angles and therefore equal to one another. Also, we are given that the incident angle, \(\angle D A E,\) is equal to the angle of reflection, \(\angle B A C\). Therefore, \(\triangle A B C \sim \triangle A D E\) by the angle-angle similarity rule.

It follows now that the ratio of \(A B\) to \(A D\) must be equal to the ratio of \(B C\) to \(D E\). This allows us to solve for the length of \(D E\).

\(\begin{align*} D E &=\frac{A D}{A B} \cdot B C \\ &=\frac{12.8}{3.8} \cdot 1.8 \\ &=6.063 \ldots \end{align*}\)

Therefore, the height of the street lamp is approximately \(6.1\) metres.


Wrap-Up


Lesson Summary

In this lesson, we learned the following:

  • Two triangles are similar when they satisfy any of the following rules:
    • Angle-angle similarity (two equal angles)
    • Side-angle-side similarity (two sides in the same proportion and the contained angle is equal)
    • Side-side-side similarity (all three sides are in the same proportion)
    Angle-Angle (AA) Side-Angle-Side (SAS) Side-Side-Side (SSS)
    Triangle with 3 angles labelled. Triangle with 1 angle labelled between two sides with lengths 10 and 8. Triangle with side lengths 10, 8, and 15.
    Larger triangle with the same 3 angles. Triangle with same size angle labelled between two sides with lengths 15 and 12. Triangle with side lengths 15,12, and 22.5.
  • Similar triangles can be constructed to solve for unknown distances in real-world problems.

Take It With You

The first calculation of the circumference of the Earth was performed by Eratosthenes around \(240\) BCE based on the following observation.

If the Sun is directly overhead in the city of Syene, then at the same time in the city of Alexandria (\(850\) km away) a pole would have to be tilted roughly \(7\) degrees away from vertical (towards the Sun) to not cast a shadow.

Can you use this information to estimate the circumference of the Earth?