Sign In
| 18 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.
Dylan and Zosia love to solve riddles and decided to have a competition to see who can solve them faster. In one riddle, they have to form a triangle with three matchsticks that are 4, 4.5, and 8.8 centimeters long.
Dylan says that such a triangle cannot be formed, while Zosia says it can. Who is right?
The second type of triangle groups all those triangles that have one right angle.
A right triangle is a triangle that has one right angle. The side opposite the right angle is always the longest and is known as the hypotenuse. The other sides are commonly called legs. Notice that in a right triangle, the legs are perpendicular to each other.
The next type of triangle includes those triangles that have one obtuse angle.
An obtuse triangle is a triangle that has one obtuse angle. In other words, one of the angles measures more than 90∘.
The last classification of triangles according to the measure of their angles consists of those triangles that have three angles of the same measure.
Classify the given triangle according to its angle measures as an acute, right, or obtuse triangle.
The same way two angles with the same measure are called congruent angles, there is a special term for two segments that have the same length.
In addition to their angle measures, triangles can also be classified by comparing the lengths of their three sides. There are three different types. The first includes those triangles in which all three sides have the same length.
The second way of classifying a triangle according to its side lengths includes all the triangles in which only two sides have the same length.
The last classification of triangles based on their side lengths is when the three sides have different lengths.
Classify the given triangle according to its side lengths as an equilateral, isosceles, or scalene triangle.
At lunch, Ali challenged Dylan and Zosia to build a triangle with the objects they had on hand. He offered a free dessert to whoever did it successfully.
Since Zosia brought Chinese food and a soft drink, she decided to use the two chopsticks and the straw to build her triangle. The straw is 18 centimeters long and the chopsticks are each 24 centimeters long.
For his part, Dylan decided to use a pen, a short pencil, and a small crayon from his backpack. The pen, pencil, and crayon are 15, 9, and 5 centimeters long, respectively.
Check if the lengths of the objects satisfy the Triangle Inequality Theorem.
In order for three segments to form a triangle, their lengths must satisfy the Triangle Inequality Theorem. In other words, the sum of the lengths of any two segments must be greater than the length of the third segment. Consider Zosia's and Dylan's attempts separately.
Start by writing the lengths of the objects that Zosia chose.
Object | Length (cm) |
---|---|
Straw | 18 |
Chopstick 1 | 24 |
Chopstick 2 | 24 |
Now, calculate the sums of the lengths of the possible pairs of objects and check if they are greater than the length of the third object.
Pair of Objects | Sum of Lengths | Object 3 | Comparison |
---|---|---|---|
Straw and Chopstick 1 | 18+24=42 | Chopstick 2 | 42>24✓ |
Straw and Chopstick 2 | 18+24=42 | Chopstick 1 | 42>24✓ |
Chopstick 1 and Chopstick 2 | 24+24=48 | Straw | 48>18✓ |
As shown, the sum of the lengths of any two objects is always greater than the length of the third object. Therefore, Zosia can form a triangle with the objects she chose, so she will get a free dessert!
Notice that the triangle has two sides of the same length, which means that Zozia built an isosceles triangle.
Dylan can build a triangle with the objects he chose only if their lengths meet the Triangle Inequality Theorem. Follow the same process to see if he can earn a free dessert. Begin by writing the lengths of the objects.
Object | Length (cm) |
---|---|
Pen | 15 |
Pencil | 9 |
Crayon | 5 |
Next, calculate the sums of the lengths of the possible pairs of objects and verify if they are greater than the length of the third object.
Pair of Objects | Sum of Lengths | Object 3 | Comparison |
---|---|---|---|
Pen and Pencil | 15+9=24 | Crayon | 24>5✓ |
Pen and Crayon | 15+5=20 | Pencil | 20>9✓ |
Pencil and Crayon | 9+5=14 | Pen | 14>15× |
The sum of the lengths of the pencil and the crayon in the last row of the table is not greater than the length of the pen, which means that the object's lengths do not meet the Triangle Inequality Theorem. Dylan cannot build a triangle with the objects he chose, so he will not get a free dessert. Bummer!
There is no need to verify all three inequalities. Just check whether the sum of the lengths of the two shorter segments is greater than the length of the longer segment.
At the beginning of the lesson, Dylan and Zosia were challenged to form a triangle using three matchsticks. However, they came to opposite conclusions. Dylan claimed that no triangle could be formed with the given matches, while Zosia said that a triangle could be formed.
The riddle can be solved with the information given in this lesson. The three matches can form a triangle if they satisfy the Triangle Inequality Theorem. In other words, check whether the sum of the lengths of any pair of matches is greater than the length of the third match.
Pair of Matches | Sum of Lengths | Third Match | Comparison |
---|---|---|---|
Match 1 & Match 2 | 8.8+4.5=13.3 | 4.0 | 13.3>4.0 ✓ |
Match 1 & Match 3 | 8.8+4.0=12 | 4.5 | 12>4.5 ✓ |
Match 2 & Match 3 | 4.5+4.0=8.5 | 8.8 | 8.5>8.8 × |
As shown, the sum of the lengths of the two shorter matches is not greater than the length of the longest match. Therefore, the given matchsticks cannot form a triangle. Dylan was right!
Select the most appropriate classification for the given triangle.
Let's begin by classifying the given triangle according to its side lengths.
We can see that two sides have the same length, 4 units long, while the third side is 5 units long. Therefore, the triangle is an isosceles triangle. Next, let's consider to the angle measures of △ ABC.
The three angle measures are less than 90^(∘), which means they are all acute. Because of this, △ ABC is also an acute triangle. The most appropriate classification combines the two classifications we made. △ ABC is an acute triangle. △ ABC is an isosceles triangle. ⇓ △ ABC is an acute isosceles triangle.
Let's start by classifying △ PQR according to the angle measures.
We can see that angles Q and R are acute. Although the measure of ∠ P is not written using numbers, the angle is denoted with a square angle marker. This means that ∠ P is a right angle. Therefore, △ PQR is a right triangle. Let's now focus on the side lengths.
The three sides have different lengths, which means that △ PQR is a scalene triangle. Finally, let's combine the two classifications. △ PQR is a right triangle. △ PQR is a scalene triangle. ⇓ △ PQR is a right scalene triangle.
Fill in the blank with the most appropriate word or phrase.
Let △ABC be a triangle in which one angle has a measure greater than 100∘. If AB≅AC and AC≆BC, then △ABC is a(n) triangle. |
In △JKL, one of the angles has a measure of 90∘ and the shorter sides have different lengths. Because of this, △JKL is a(n) triangle. |
Let's begin by classifying △ ABC according to its angle measures. We are given that one angle has a measure greater than 100^(∘). This means that the triangle has one obtuse angle. One angle is obtuse. ⇓ △ ABC is an obtuse triangle. Let's now classify △ ABC based on its side lengths. We know that AB≅AC, which means that these two segments have the same length. On the other hand, we know that AC≆BC, which means these segments have different lengths. AB = AC AC ≠ BC ⇒ AB ≠ BC From these statements, we can say that only two sides have the same length. Only two sides have the same length. ⇓ △ ABC is an isosceles triangle. The most appropriate phrase to fill in the blank is the one combining the two classifications of △ ABC.
Let △ ABC be a triangle in which one angle has a measure greater than 100^(∘). If AB≅ AC and AC≆BC, then △ ABC is an obtuse isosceles triangle.
Let's classify △ JKL according to the measure of its angles. We are given that one angle has a measure of 90^(∘), so △ JKL has one right angle. One angle is a right angle. ⇓ △ JKL is a right triangle. Next, let's analyze the side lengths of △ JKL. Since the triangle is a right triangle, we know that one side is longer than the other two. This longer side is the hypotenuse. Additionally, we know that the shorter sides have different lengths, which means that the three sides all have different lengths. The sides have different lengths. ⇓ △ JKL is a scalene triangle. Finally, let's combine the two classifications to get the most appropriate phrase that fills in the blank.
In △ JKL, one of the angles has a measure of 90^(∘) and the shorter sides have different lengths. Because of this, △ JKL is a right scalene triangle.
The Triangle Inequality Theorem says that in a triangle, the sum of the lengths of any two sides is greater than the length of the third side. The given three segments can form a triangle if their lengths satisfy such condition. AB = 7 BC = 5 AC = 3.5 Let's calculate the sums of the lengths of the possible pairs of segments and check if they are greater than the lengths of the third segment.
Pair of Segments | Sum of Lengths | Third Segment | Comparison |
---|---|---|---|
AB & BC | 7+5=12 | AC | 12 > 3.5 ✓ |
BC & AC | 5+3.5=8.5 | AB | 8.5 > 7 ✓ |
AB & AC | 7+3.5=10.5 | BC | 10.5 > 5 ✓ |
The lengths of the given segments meet the three inequalities. Therefore, the segments can form a triangle.
We are given the lengths of two sides of △ PQR. PQ = 5 QR = 8 PR = ? The points P, Q, and R already form a triangle. This means that the length of PR must be a number such that the three triangle inequalities are satisfied. Let x be the length of PR. Let's make a table like the one we made in Part A.
Pair of Segments | Sum of Lengths | Third Segment | Comparison |
---|---|---|---|
PQ & QR | 5+8=13 | PR | 13 > x |
QR & PR | 8+x | PQ | 8+x > 5 |
PQ & PR | 5+x | QR | 5+x > 8 |
The last column of the first row tells us that the length of PR is less than 13. Let's now solve the other two inequalities for x. 8+x &> 5 &⇒ x &> -3 [0.4em] 5+x &> 8 &⇒ x &> 3 Since the value of x represents the length of PR, it must be positive. This means that the inequality x> -3 does not give us any information. However, the last inequality tells us that the length of PR must be greater than 3. The length ofPRis greater than3. The length ofPRis less than13. ⇓ 3 < x < 13 Let's now recall the given four possible lengths for PR. 3 5 10 12 The value 3 is the only one that do not lie between 3 and 13. This means that the length of PR cannot be 3. The following diagram shows the triangles that can be formed with the other three given lengths.