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| 16 Theory slides |
| 13 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.
To strengthen roof trusses, usually triangular shaped structures are used. In the diagram, AC=CF, AB=BD, and DE=EF. The beams BC, DF, CE, and AD are built in a way that BC∥DF and CE∥AD.
Considering the previous exploration, the sum of interior angles of a triangle can be derived.
m∠A+m∠B+m∠C=180∘
This theorem is also known as the Triangle Angle Sum Theorem.
Consider a triangle with vertices A, B, and C, and the parallel line to BC through A. Let ∠1 and ∠2 be the angles outside △ABC formed by this line and the sides AB and AC.
By the Alternate Interior Angles Theorem, ∠B is congruent to ∠1 and ∠C is congruent to ∠2.
Dylan is designing a wooden sofa made of oak wood for his local park. The sides of the sofa will have identical dimensions in the shape of a triangle. He already has decided on the angle measures of the top corner and bottom-right corner of each side.
To cut the sides of the sofa out of the board using a table saw, which can cut at angles, Dylan needs to find the measure of the third angle. Dylan's hands are full — help him find the measure of the third angle.
Use the Interior Angles Theorem.
Since the sides have a triangular shape and the measures of two angles are known, the Interior Angles Theorem can be used to find the missing angle measure. Let x be the measure of the missing angle.
Solving this equation for x, the measure of the missing angle can be found.
The previous exploration shows that there is a clear relation between an exterior angle of a triangle and its remote interior angles.
m∠PCA=m∠A+m∠B
Consider a triangle with vertices A, B, and C, and one of the exterior angles corresponding to ∠C.
m∠C=180∘−m∠PCA
Remove parentheses
LHS−180∘=RHS−180∘
LHS+m∠PCA=RHS+m∠PCA
Rearrange equation
Consider △ABC, where D and E are the midpoints of AB and AC, respectively. Let ∠PCA be one exterior angle of △ABC.
Dylan is almost ready to cut the sides of the sofa. Before doing so, he wants to be sure that people sitting on his sofa can lean back freely and feel comfortable. Therefore, he needs to find the measure of the angle exterior to the third angle.
Use the Triangle Exterior Angle Theorem.
Recall that by the Triangle Exterior Angle Theorem, the measure of a triangle's exterior angle is equal to the sum of the measures of its two remote interior angles. Therefore, the measure of the exterior angle x can be expressed.
Since the measure of the exterior angle is less than 90∘, the people sitting on this sofa can lean back and feel comfortable. Thanks for helping Dylan.
Reflecting a right triangle about either of its legs forms an isosceles triangle. Note that a reflection is a rigid motion, so the side lengths and the interior angles of the right triangle are preserved.
AB≅AC ⇒ ∠B≅∠C
Consider a triangle ABC with two congruent sides, or an isosceles triangle.
Statement | Reason |
---|---|
∠BAP ≅ ∠CAP | Definition of an angle bisector |
BA ≅ CA | Given |
AP ≅ AP | Reflexive Property of Congruence |
Consider an isosceles triangle △ABC.
A line passing through A and the midpoint of BC will be drawn. Let P be the midpoint.
Since BP and PC are congruent, the distance between B and P is equal to the distance between C and P. Therefore, B is the image of C after a reflection across AP. Also, because A lies on AP, a reflection across AP maps A onto itself. The same is true for P.
Reflection Across AP | |
---|---|
Preimage | Image |
C | B |
A | A |
P | P |
Dylan notices that he needs a support beam to support the seat. The bottoms of each side panel are 3 feet long. Therefore, if he places the support beam from the corner with the larger angle measure to the opposite side in a position where the endpoint of the support beam is 3 feet away from the bottom-right corner, then it will fit just right.
In this case, what should be the measure of the angle between the support beam and the bottom of the side panel?
Consider the Base Angles Theorem.
Placing the support beam as shown forms an isosceles triangle.
As it is seen in the previous exploration, using the rigid motions, the Triangle Midsegment Theorem can be proven.
DE∥BC and DE=21BC
This theorem can be proven by placing the triangle on a coordinate plane. For simplicity, vertex B will be placed at the origin and vertex C on the x-axis.
To prove this theorem, it must be proven that DE is parallel to BC and that DE is half of BC.
If the slopes of these two segments are equal, then they are parallel. The y-coordinate of both B(0,0) and C(a,0) is 0. Therefore, BC is a horizontal segment. Next, the coordinates of D and E will be found using the Midpoint Formula.
M(2x1+x2,2y1+y2) | |||
---|---|---|---|
Segment | Endpoints | Substitute | Simplify |
BA | B(0,0) and A(b,c) | D(20+b,20+c) | D(2b,2c) |
CA | C(a,0) and A(b,c) | E(2a+b,20+c) | E(2a+b,2c) |
Since both BC and DE are horizontal, their lengths are given by the difference of the x-coordinates of their endpoints.
Segment | Endpoints | Length | Simplify |
---|---|---|---|
BC | B(0,0) and C(a,0) | BC=a−0 | BC=a |
DE | D(2b,2c) and E(2a+b,2c) | DE=2a+b−2b | DE=21a |
Assume that E′, the image of E after the translation along DB, does not lie on BC. This means that E′ lies either above or below BC. The proof will be developed for only one case but is valid for both.
Based on the assumption, let F denote the point of intersection of BE′ and EC.
Next, it will be proven that △ADE is congruent to △EE′F.
It is given that AD=DB because D is the midpoint of AB. Additionally, since △ADE is translated along DB, it can be concluded that DB=EE′. Then, by the Transitive Property of Equality, AD=EE′.
Recall that AE=DE′. Additionally, DE is the common side of △ADE and △E′ED. All pieces of information can now be summarized.
Summarize the obtained information about the triangles △ADE and △EE′F.
Therefore, by the Angle-Side-Angle Congruence Theorem, △ADE is congruent to △EE′F.
Assumption of the Theorem | Indirect Assumption |
---|---|
AE=EF=EC | EF<EC |
EF=EC and EF<EC × |
Therefore, this contradiction verifies that the image of E must lie on BC.
By the Converse Corresponding Angles Theorem, BC is parallel to DE.
BC∥DE
DE=21BC
Finally, Dylan is ready to place the seat. He plans to place it just above the support beam such that it will be parallel to the bottom. Therefore, the corners of the seat will be at the midpoints of the sides.
How can he find the width of the seat knowing that the bottom of the side is 3 feet long.
The seat will be aligned with the midsegment of the triangular side.
In this lesson, the investigated theorems about triangles have been proven using a variety of methods. Furthermore, with the help of these theorems, the challenge provided at the beginning of the lesson can be solved. Recall the diagram.
Consider the given information about the beams of the roof.
Use the Triangle Midsegment Theorem.
By the definition of a midsegment, both BC and CE are midsegments of △ADF. By the Triangle Midsegment Theorem, CE is half of AD, and BC is half of DF.
Knowing that CE is 43 inches and DF is 68 inches, these values can be substitute into these equations to find AD and BC.
Therefore, the length of AD is 86 inches and the length of BC is 34 inches.
Find the value of x.
Examining the diagram, we can see that one of the angles is a right angle which makes this a right triangle. According to the Interior Angles Theorem, the sum of the three angles equal 180^(∘). 3x+2x+90^(∘)=180^(∘) If we solve this equation, we can find the value of x.
As in Part A, we will add the measures of the three angles and equate the sum with 180^(∘).
x+(3x+6^(∘))+90^(∘)=180^(∘)
If we solve this equation, we can find the value of x.
According to the Interior Angles Theorem, the sum of the measures of the interior angles of a triangle is 180^(∘). Therefore, if the angle measures of a set adds to 180^(∘), we know that the set could form a triangle.
Set | Sum of Angles | Evaluate |
---|---|---|
A | 102^(∘)+50^(∘)+40^(∘) | 192^(∘) * |
B | 96^(∘)+72^(∘)+12^(∘) | 180^(∘) ✓ |
C | 166^(∘)+113^(∘)+82^(∘) | 361^(∘) * |
D | 101^(∘)+40^(∘)+38^(∘) | 179^(∘) * |
E | 90^(∘)+45^(∘)+45^(∘) | 180^(∘) ✓ |
F | 84^(∘)+62^(∘)+34^(∘) | 180^(∘) ✓ |
As we can see, sets B, E, and F each add up to 180^(∘).
Consider the characteristics of △ABC.
Since DE is a midsegment of the triangle, we know by the Triangle Midsegment Theorem that it is half the length of AC. Since we know the coordinates of A and C, we can determine this distance by using the Distance Formula.
Now we can use the Triangle Midsegment Theorem to determine the length of DE.
Let's start by drawing EF.
Notice that because EF is a midsegment, we can find the coordinates of F by calculating the midpoint of AC by using the Midpoint Formula.
The coordinates of F are (3,-3).
Find the value of x if DE is a midsegment of △ABC.
According to the Triangle Midsegment Theorem, if a segment joins the midpoints of two sides of a triangle, then this segment is half the length of the third side. Since DE is the midsegment parallel to BC in △ ABC, we can, according to the theorem, write the following equation. DE=1/2BC Now, we can find the value of x by substituting the given values into this equation.
This time, we want to find the length of the third side when knowing the length of the midsegment parallel to the third side. Again, we will use the Triangle Midsegment Theorem.
Consider the angles and segments of the following figure.
The expression of AE≅ DE means that these sides are congruent. Let's mark that information in our diagram.
The Isosceles Triangle Theorem tells us that when two sides of a triangle are congruent, then the angles that are opposite of these sides are also congruent.
Now we can complete the statement. If AE≅ DE, then $∠ A$ ≅ $∠ D$
The statement ∠ EBC ≅ ∠ ECB states that these angles are congruent angles. Let's add that information to our diagram.
By the Converse Isosceles Triangle Theorem, we know that the sides that are opposite of ∠ EBC and ∠ ECB are congruent.
Now we can fill in the blanks in the statement. If ∠ EBC ≅ ∠ ECB, then $EB$ ≅ $EC$
A triangle is classified as obtuse if the measure of one of its angles is greater than 90^(∘). Let's draw an obtuse angle. We will arbitrarily choose an angle with a measure of 120^(∘).
If we extend the sides equally and connect the endpoints, we create an isosceles triangle.
Therefore, it is possible to draw an obtuse isosceles triangle
An equilateral triangle is a triangle with three congruent sides. This also makes the triangle equiangular. Since the sum of the measures of the interior angles equals 180^(∘), we can find the measure of each angle by dividing 180^(∘) by 3. 180^(∘)/3=60^(∘) The measure of each interior angle of an equilateral triangle is 60^(∘). Therefore, all the angles in an equilateral triangle are acute.
This means an obtuse equilateral triangle does not exist.
Find the measure of the exterior angle at A.
Examining the diagram, we see that the triangle has two interior angles of 75^(∘) and 64^(∘). According to the Triangle Exterior Angle Theorem, the measure of a triangle's exterior angle equals the sum of the two non-adjacent interior angles.
With this information, we can calculate the measure of the exterior angle, which we have labeled θ in the diagram above. m∠ θ=63^(∘)+75^(∘)=138^(∘)
As in Part A, we will use the Triangle Exterior Angle Theorem to relate the exterior angles with the triangle's non-adjacent interior angles. 2x-3^(∘)=45^(∘)+x Let's solve for m∠ x.
When we know the measure of x, we can find the measure of the exterior angle. 2( x)-3^(∘)=2( 48^(∘))-3^(∘)=93^(∘)
Again, as in previous parts, we will write an equation using the Triangle Exterior Angle Theorem. 3x+6^(∘)=25^(∘)+(2x+18^(∘)) Let's solve for x.
Let's calculate the measure of the exterior angle by substituting 37 for x in the expression for the angle measure. 3( x)+6^(∘)=3( 37^(∘))+6^(∘)=117^(∘)
The Triangle Exterior Angle Theorem tells us that the measure of an exterior angle of a triangle equals the sum of both nonadjacent interior angles.
Therefore, if the sum of the smaller angles equals the measure of the larger angle, we know that the three angles could represent the measures of one exterior angle and two interior angles of a triangle.
Set | Largest angle | Sum of Smaller Angles | Equals Largest Angle? |
---|---|---|---|
A | 99^(∘) | 61^(∘)+38^(∘) | 99^(∘) ✓ |
B | 80^(∘) | 57^(∘)+23^(∘) | 80^(∘) ✓ |
C | 90^(∘) | 51^(∘)+41^(∘) | 92^(∘) * |
D | 95^(∘) | 20^(∘)+75^(∘) | 95^(∘) ✓ |
As we can see, A, B, and D could represent the exterior angle and two interior angles of a triangle.