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| 11 Theory slides |
| 11 Exercises - Grade E - A |
| Each lesson is meant to take 1-2 classroom sessions |
Here are a few recommended readings before getting started with this lesson.
Find the ratio of the length of a diagonal and a side of a regular pentagon.
Two polygons are similar if corresponding angles are congruent and corresponding sides are proportional. For triangles, the congruence of two angles already implies similarity.
If two angles of a triangle are congruent to two angles of another triangle, then the triangles are similar.
If ∠ A ≅ ∠ D and ∠ B ≅ ∠ E, then △ ABC ~ △ DEF.
Consider two triangles △ ABC and △ DEF, whose two corresponding angles are congruent.
These triangles can be proven to be similar by identifying a similarity transformation that maps one triangle onto the other. First, △ DEF can be dilated with the scale factor k= ABDE about D, forming the new triangle △ DE'F'.
Since a dilation is a similarity transformation, it can be concluded that △ DE'F' and △ DEF are similar triangles. Next, it has to be proven that a rigid motion that maps △ DE'F' onto △ ABC exists. The corresponding angles of similar figures are congruent, so ∠ E' and ∠ E are congruent angles. ∠ E'≅ ∠ E Additionally, since ∠ E is congruent to ∠ B, by the Transitive Property of Congruence, ∠ E' is congruent to ∠ B. ∠ E'≅ ∠ B The ratios of the corresponding side lengths of similar polygons are the same and equal to the scale factor. DE'/DE=k In this case, the scale factor k is ABDE. Applying the Transitive Property of Equality, an equation can be formed and simplified. DE'/DE=AB/DE ⇕ DE'=AB It has been obtained that the two angles and the included side of △ DE'F' are congruent to the corresponding two angles and the included side of △ ABC.
Therefore, by the Angle-Side-Angle (ASA) Congruence Theorem, the two triangles are congruent. △ ABC≅△ DE'F' Since congruent figures can be transformed into each other using rigid motions, and △ ABC and △ DE'F' are congruent triangles, there is a rigid motion placing △ DE'F' onto △ ABC.
Therefore, it can be concluded that △ ABC and △ DEF are similar triangles.
△ ABC~ △ DEF
The proof is now complete.
The Grim Reaper, who is 5 feet tall, stands 16 feet away from a street lamp at night. The Grim Reaper's shadow cast by the streetlamp light is 8 feet long. How tall is the street lamp?
Both the lamp post and the Grim Reaper stand vertically on horizontal ground.
A sketch of the situation is helpful for finding the solution. Under the assumption that the lamp post and the Grim Reaper make right angles in relation to the ground, two right triangles can be drawn. The unknown height of the lamp post is labeled as x.
As these triangles both have a right angle and share the angle on the right-hand side, they are similar by the Angle-Angle (AA) Similarity Theorem. Notice that the base of the larger triangle measures to be 24 feet.
Since the triangles are similar, the ratios between corresponding side lengths are the same. \begin{gathered} \dfrac x 5 = \dfrac{24}8 \end{gathered} The solution of this equation defines the value of x — the height of the street lamp.
The street lamp at 15 feet high towers over The Grimp Reaper.
For the given diagram, find the missing length.
A second theorem allows for determining triangle similarity when only the lengths of corresponding sides are known.
If corresponding sides of two triangles are proportional, then the triangles are similar.
If AB/DE=BC/EF=CA/FD, then △ ABC ~ △ DEF.
Consider two triangles △ ABC and △ DEF, whose corresponding sides are proportional.
These triangles can be proven to be similar by identifying a similarity transformation that maps one triangle onto the other. First, △ DEF can be dilated with the scale factor k= ABDE about D, forming the new triangle △ DE'F'.
Because dilation is a similarity transformation, it can be concluded that △ DE'F' and △ DEF are similar triangles. Now, it has to be proven that a rigid motion that maps △ DE'F' onto △ ABC exists. The ratios of the corresponding side lengths of similar polygons are the same and equal to the scale factor. DE'/DE= DF'/DF= E'F'/EF=k In this case, the scale factor k is ABDE. Since all of the sides of △ ABC and △ DEF are proportional, the scale factor can be expressed by any of the following ratios. k= AB/DE= BC/EF= CA/DF Applying the Transitive Property of Equality, three equations can be formed and simplified. DE'/DE&=AB/DE [0.3cm] DF'/DF&=CA/DF [0.3cm] E'F'/EF&=BC/EF ⇒ DE' &= AB DF' &= CA E'F' &= BC These relations imply that the three sides of △ DE'F' are congruent to the three sides of △ ABC. Therefore, by the Side-Side-Side (SSS) Congruence Theorem, the two triangles are congruent. △ ABC≅△ DE'F' Since congruent figures can be transformed into each other using rigid motions, and △ ABC and △ DE'F' are congruent triangles, there is a rigid motion placing △ DE'F' onto △ ABC.
The combination of this rigid motion and the dilation performed earlier forms a similarity transformation that maps △ DEF onto △ ABC.
Therefore, it can be concluded that △ ABC and △ DEF are similar triangles.
△ ABC~ △ DEF
The proof is now complete.
There are four congruent angles in the figure. Try to identify them.
∠ DBA ≅ ∠ BCE ≅ ∠ BEC ≅ ∠ DBE
Look for similar triangles and an isosceles triangle.
First, notice that segments BE and BC are equal in length.
These are two sides of △ BCE, so by the Isosceles Triangle Theorem, the opposite angles are congruent. ∠ BCE≅∠ BEC
Two of the triangles, △ ABD and △ ACE look similar.
Because the lengths of the sides are given, the ratio of corresponding sides can be calculated.
Ratio | Expression | Simplification |
---|---|---|
AC/AB | 459+1275/459=1734/459 | 34/9 |
AE/AD | 405+1125/405=1530/405 | 34/9 |
EC/DB | 1360/360 | 34/9 |
The last column of the table shows that the corresponding sides of △ ACE and △ ABD are proportional. AC/AB=AE/AD=EC/DB According to the Side-Side-Side (SSS) Similarity Theorem, the two triangles are similar. △ ACE~△ ABD Corresponding angles of similar triangles are congruent.
One pair of these angles is marked on the figure. ∠ BCE≅∠ DBA
In addition to the proportions in Step 2 showing that △ ACE and △ ABD are similar, they also show the two triangles are dilations of each other from the common vertex A. Since dilations map a segment to a parallel segment, segments DB and EC are parallel.
Furthermore, since EB is a transversal to two parallel lines, the Alternate Interior Angles Theorem guarantees that the angles at E and B are congruent. ∠ BEC≅∠ DBE These angles are marked on the figure.
The previous three steps showed three pairs of congruent angles. The transitive property of congruence shows that all four angles mentioned in these pairs are congruent to each other. ∠ DBA ≅ ∠ BCE ≅ ∠ BEC ≅ ∠ DBE The congruent angles are marked on the figure.
Two theorems have been covered, now a third theorem that can be used to prove triangle similarity will be investigated. This third theorem allows for determining triangle similarity when the lengths of two corresponding sides and the measure of the included angles are known.
If two sides of a triangle are proportional to two sides of another triangle and the included angles are congruent, then the triangles are similar.
If AB/DE=AC/DF and ∠ A ≅ ∠ D, then △ ABC ~ △ DEF.
Consider two triangles △ ABC and △ DEF, whose two pairs of corresponding sides are proportional and the included angles are congruent.
These triangles can be proven to be similar by identifying a similarity transformation that maps one triangle onto the other. First, △ DEF can be dilated with the scale factor k= ABDE about D, forming the new triangle △ DE'F'.
Because dilation is a similarity transformation, it can be concluded that △ DE'F' and △ DEF are similar triangles. Now, it has to be proven that a rigid motion that maps △ DE'F' onto △ ABC exists. The ratios of the corresponding side lengths of similar polygons are the same and equal to the scale factor. DE'/DE= DF'/DF=k In this case, the scale factor k is ABDE. Since AB and AC are proportional to DE and DF respectively, the scale factor can be expressed by any of the following ratios. k= AB/DE= AC/DF Applying the Transitive Property of Equality, three equations can be formed and simplified. l c r DE'/DE=AB/DE & ⇒ & DE'=AB [0.3cm] DF'/DF=AC/DF & ⇒ & DF'=AC These relations imply that the two sides of △ DE'F' are congruent to the corresponding two sides of △ ABC. Moreover, the included angles ∠ A and ∠ D are also congruent.
Therefore, by the Side-Angle-Side (SAS) Congruence Theorem, the two triangles are congruent. △ ABC≅△ DE'F' Since congruent figures can be transformed into each other using rigid motions, and △ ABC and △ DE'F' are congruent triangles, there is a rigid motion placing △ DE'F' onto △ ABC.
The combination of this rigid motion and the dilation performed earlier forms a similarity transformation that maps △ DEF onto △ ABC.
Therefore, it can be concluded that △ ABC and △ DEF are similar triangles.
△ ABC~ △ DEF
The proof is now complete.
The diagram shows the distances between points on a figure.
Show that △ ABC and △ AED are similar triangles. Then find DE.
Triangles △ ABC and △ AED have a common angle at A.
The table below contains the ratios of two pairs of corresponding sides of the two triangles.
Ratio | Expression | Simplified Form |
---|---|---|
AB/AE | 7/5+9=7/14 | 1/2 |
AC/AD | 5/7+3=5/10 | 1/2 |
Through applying the theorems of similar triangles, the ratio of the lengths of a diagonal and the sides of a regular pentagon can be found.
Begin by determining the angle measures of the figure.
The Polygon Angle Sum Theorem identifies the sum of the interior angle measures of a pentagon. (5-2)180=540 In a regular pentagon, all five angles are equal measures. Therefore, one of the measures of the angles is a fifth of the sum of the five angle measures. m∠ A = 540/5=108 Furthermore, since the sides of a regular pentagon are congruent, △ ABE is an isosceles triangle.
m∠ AEB= m∠ ABE, m∠ A= 108
LHS-108=RHS-108
a+a=2a
.LHS /2.=.RHS /2.
Substitute values
LHS-(36+36)=RHS-(36+36)
Next, focus on △ ACE. In this triangle, AC and EC are diagonals of the pentagon, and AE is a side.
In the diagram, a smaller triangle labeled △ AFE is also present. These two triangles share a common angle at A and congruent angles at C and E. ∠ CAE&≅∠ EAF ∠ ACE&≅∠ AEF According to the Angle-Angle (AA) Similarity Theorem, that means the two triangles are similar. Therefore, the corresponding sides are proportional. CA/AE=AE/AF Continuing forward, notice that triangles △ AFE and △ CEF are isosceles. Therefore, their legs have equal lengths. AE=EF=FC Using s for the length of the sides, AE=EF=FC=s, as indicated on the figure. Also, using d for the length of the diagonal, CE=d and FA=CA-FC=d-s.
LHS * (x-1)=RHS* (x-1)
Multiply parentheses
LHS-1=RHS-1
Use the Quadratic Formula: a = 1, b= - 1, c= - 1
- (- a)=a
Calculate power and product
a-(- b)=a+b
Add terms
Length of diagonal/Length of side=1+sqrt(5)/2≈ 1.618
The ratio of the diagonal to the side of a regular pentagon can be used to prove that the following construction creates a regular pentagon. This is a construction created by Yosifusa Hirano in the 19th century.
Are the following pair of triangles similar? If yes, state by which triangle similarity theorem?
Examining the diagram, we see that we have two right triangles. Additionally, we can identify one pair of acute angles that are also congruent since the triangles share an angle.
Therefore, by the Angle-Angle Similarity Theorem, these triangles are similar.
For similar figures, the ratio of two corresponding sides is equal to the ratio of two other corresponding sides. Notice that in similar figures, sides with the same relative lengths are corresponding. With this information, we can identify what would the corresponding sides be if the triangles are similar.
Let's write the ratios of corresponding sides. 6/15? =8/20? =11/33 If the triangles are similar, all of the ratios should be equal. Let's check if the ratios give the same quotient. 0.4= 0.4≠ 0.33... * Since the ratio of the longest sides is not equal to the ratio of the other two sides, these cannot be similar triangles.
Two triangles are similar if at least two pairs of angles are congruent. Since none of the given angles are congruent, we know that the triangles cannot be similar.
Which of the triangles B, C, and D are similar to A?
Examining the three triangles, we notice that for A and B we have two pairs of congruent angles. This means we can claim similarity by the Angle-Angle Similarity Theorem.
In △ C, we only have information about the side lengths. Since we also know the length of every side in △ A, we can investigate if the triangles are similar by the Side-Side-Side Similarity Theorem. In similar figures, sides with the same relative lengths are corresponding. With this information, we can write and evaluate the ratios of corresponding sides. 2/2.6? =3/4? =3.9/5.4 [0.9em] ⇓ [0.2em] 0.76923...≠0.75≠0.7222 ... * Since the ratios are not equal, we know that the triangles are not similar.
As for D, we have one pair of congruent angles. Since we have also been given the lengths of the sides that have an included angle, we can investigate if D and A are similar by the Side-Angle-Side Similarity Theorem. Again, since sides with the same relative length are corresponding, we can set up ratios to evaluate them. 4.68/3.9? =2.4/2 [0.9em] ⇓ [0.2em] 1.2 =1.2... ✓ The sides have matching ratios. Therefore, D is similar to A.
Consider the following triangles ABC and ADE.
Examining the diagram, we see that △ ABC and △ ADE have a pair of congruent angles. We also see that they share an angle at A.
Since the triangles have two pairs of congruent angles, we can claim similarity by the Angle-Angle Similarity Theorem.
To find the length of DE, we can use the similarity between the triangles. Let's separate the triangles to help us identify corresponding sides.
The ratio of corresponding sides in similar shapes is always the same. Let's use this to write an equation. x/10=24/16 Let's solve this equation for x.
As we can see, DE is 15 units long.
Consider △ ABC.
Is the following triangle similar to △ ABC?
Any equilateral triangle is also equiangular, meaning all of its angles are congruent. Since △ DEF has three angles of 60^(∘), we know that it is equiangular and, therefore, equilateral. Since △ ABC has three congruent sides, it is equilateral and, therefore, equiangular.
Now we can claim similarity by either Side-Side-Side Similarity Theorem or by the Angle-Angle Similarity Theorem.
Notice that all sides have a length of x. Like the previous section, this also fits the description of an equilateral triangle. Therefore, △ GHI must also be similar to △ ABC.
All equilateral triangles are similar because all of these triangles have the same three pairs of angles. Therefore, as long as we know that two triangles are equilateral or equiangular, we know they are similar.