MAT8 Q2W6D1: Introduction to Triangle Inequality Theorem

🎯 Learning Goals

By the end of the lesson, you will be able to:

1. Recall and describe real-life situations showing inequalities.
2. Define and use key terms related to triangle inequality (theorem, corollary).
3. Determine whether three given side lengths can form a triangle using the Triangle Inequality Theorem.

🧩 Key Ideas & Terms

• Inequality - A mathematical statement that compares two values using symbols such as $<$, $>$, $\le$, or $\ge$.
• Triangle Inequality Theorem - In any triangle, the sum of the lengths of any two sides is always greater than the length of the remaining side.
• Theorem - A proposition or statement that has been proven true through logical reasoning, based on accepted assumptions and previously established theorems.
• Corollary - A statement that follows directly and easily from a theorem, requiring little or no additional proof.

🔄 Quick Recall / Prior Knowledge

Activity: Compare and Tell
Look at the following situations. Write the inequality that matches each description.

1. A basketball player (85 kg) is heavier than his younger brother (60 kg).
2. A pencil (15 cm) is shorter than a ruler (30 cm).
3. A tricycle fare (₱15) is cheaper than a jeepney fare (₱20).

Show Answer

1. $85>60$
2. $15<30$
3. $15<20$

📖 Explore the Lesson

Anchor scenario - a shelf brace that must be a triangle

You are improving a wall shelf. To stop wobble, you plan to add a triangular wooden brace that connects the shelf’s front edge to the wall. In your toolkit you find three slats with lengths 26 cm, 34 cm, and 60 cm. You want to know whether these three slats can be joined end to end to form a triangle for the brace.

Observe - Look at the three measurements.
List - Arrange from smallest to largest: 26, 34, 60.
Group - The two smaller slats together will have a combined length of $26+34=60$.
Name - Triangle inequality idea.
Explain - For a physical triangle, the two shorter sides together must be strictly longer than the longest side.
Apply - Compare the sum of the two smaller with the longest.

Comparison in MathML: $26+34=60\le 60$. The set fails. The shape collapses into a straight segment, not a real triangle.

Try it - quick check on your brace box
You also have a spare set: 28 cm, 37 cm, 52 cm. Will these form a triangle?

Show Answer

Order the sides: 28, 37, 52. Compare the two smaller with the largest: $28+37=65>52$. Yes, triangle is possible.

Mini-summary: A triangle closes only when the sum of the two shorter sides beats the longest side with a strict greater than comparison.

You already compare quantities every day. Mathematics expresses these comparisons with symbols:

• greater than - $>$
• less than - $<$
• greater than or equal to - $\ge$
• less than or equal to - $\le$

Try it - symbol match

1. 15 is greater than 9
2. 0.75 is less than or equal to 0.75
3. 3 is less than 8

Show Answer

1. $15>9$
2. $0.75\le 0.75$
3. $3<8$

Mini-summary: Inequality symbols compress long sentences into short math statements.

Imagine placing three sticks tip to tip on a table to close a triangular loop.

Observe - Test 4 cm, 5 cm, 20 cm.
List - Smallest to largest: 4, 5, 20.
Group - Two smaller together: $4+5=9$.
Name - Triangle inequality idea.
Explain - The joined pair must reach across the longest side.
Apply - Compare: $9<20$ so it fails.

Try it - thought experiment sets

1. 7, 9, 15
2. 6, 8, 15
3. 6, 10, 15

Show Answer

1) Order 7, 9, 15. $7+9=16>15$ - yes.
2) Order 6, 8, 15. $6+8=14<15$ - no.
3) Order 6, 10, 15. $6+10=16>15$ - yes.

Mini-summary: The two shorter sides together must be strictly greater than the longest.

Decision Test for Three Lengths - Triangle or Not

1. Identify the largest length.
2. Add the other two lengths.
3. Compare: if $\mathrm{\backslash \#} \text{sum of two smaller}$ $>$ $\mathrm{\backslash \#} \text{largest}$ - triangle is possible. If $\le$ - not a triangle.

Triangle Inequality Theorem

For a triangle with side lengths $a$, $b$, and $c$:

$\begin{array}{ccc}a+b& >& c\\ b+c& >& a\\ a+c& >& b\end{array}$

Try it - Decision Test ladder

• Start - integers: 9, 12, 22
• Stretch - decimals: 2.6, 4.1, 6.7
• Challenge - fractions: $\frac{5}{6},\frac{7}{8},\frac{13}{12}$

Show Answer

Start: order 9, 12, 22. $9+12=21<22$ - not a triangle.
Stretch: order 2.6, 4.1, 6.7. $2.6+4.1=6.7\le 6.7$ - not a triangle.
Challenge: to denominator 24, $\frac{5}{6}=\frac{20}{24}$, $\frac{7}{8}=\frac{21}{24}$, sum $\frac{41}{24}$ and largest $\frac{13}{12}=\frac{26}{24}$, so $\frac{41}{24}>\frac{26}{24}$ - triangle.

Mini-summary: One decisive comparison is enough if you first identify the largest side.

Place two pegs on a board to represent two vertices. Use a string of length $c$ to represent the long side. Now take two strings of lengths $a$ and $b$ and try to meet them tip to tip across the long side. If $a+b=c$, the two short strings reach exactly the endpoints and pull tight into a straight segment. There is no slack to lift into a shape with area. Only when $a+b>c$ can the strings bow inward and make a true triangle.

Try it - strictness check

• 8, 8, 16
• 8, 9, 16

Show Answer

$8+8=16\le 16$ - not a triangle.
$8+9=17>16$ - triangle.

Mini-summary: Equality collapses the shape. The inequality must be strict.

Walking path model. Walking from X to Y directly is shorter than walking from X to Z to Y, unless Z lies exactly on the straight line. In triangle language: $\mathrm{XZ}+\mathrm{ZY}>\mathrm{XY}$ for a true triangle.

Try it - route planning
A rectangular park has a straight diagonal path XY of 140 m. You consider walking around two edges XZ and ZY which total 160 m. Is a triangular route feasible?

Show Answer

Yes. Since $160>140$, a triangular route that bows inward is possible.

Mini-summary: Going around is longer than going straight in a triangle.

Construction and carpentry

• Bars: 120 cm, 180 cm, 290 cm - $120+180=300>290$ - triangle possible.
• Bars: 120 cm, 170 cm, 300 cm - $120+170=290<300$ - not a triangle.

Try it - unit consistency
A steel set lists 0.6 m, 0.9 m, and 1.45 m. Convert and test in centimeters.

Show Answer

Convert: 60 cm, 90 cm, 145 cm. Check: $60+90=150>145$ - triangle possible.

Sports and training routes

• Legs: 300 m, 400 m, 700 m - $300+400=700\le 700$ - not a triangle.
• Replace 700 m with 690 m - $300+400=700>690$ - triangle possible.

Try it - quick adjustment
If one leg is 250 m and another is 380 m, what is the largest value the third leg can be to still get a triangle?

Show Answer

It must be strictly less than the sum: $250+380=630$, so the third leg must be less than $630$ m.

Navigation and mapping

Set 1: $\mathrm{AB}=1.2$ km, $\mathrm{BC}=3.1$ km, $\mathrm{CA}=1.7$ km. Two smaller 1.2 and 1.7 sum 2.9, largest 3.1, so $2.9<3.1$ - inconsistent for a nondegenerate triangle.

Set 2: $\mathrm{AB}=1.2$ km, $\mathrm{BC}=3.1$ km, $\mathrm{CA}=4.5$ km - two smaller 1.2 and 3.1 sum 4.3 which is $<$ 4.5 - impossible.

Try it - data validation

1. 2.0, 3.5, 4.0
2. 1.1, 2.2, 3.3

Show Answer

2.0 and 3.5 sum 5.5 $>$ 4.0 - possible. 1.1 and 2.2 sum 3.3 $\le$ 3.3 - not a triangle.

Mini-summary: Triangle inequality is a quick data quality check.

Cycle A - integers

Observe - Set: 11, 12, 9
List - 9, 11, 12
Group - two smaller sum $9+11=20$
Name - Triangle Inequality Theorem
Explain - $20>12$ so triangle is possible
Apply - Yes

Try it: Will 6, 9, 16 form a triangle?

Show Answer

Order 6, 9, 16. Sum $6+9=15$ and $15<16$ - not a triangle.

Mini-summary: Always reduce to one decisive comparison.

Cycle B - decimals

Set: 2.5, 3.1, 5.4 - two smaller $2.5+3.1=5.6$ which is $>$ 5.4 - triangle.

Try it: 4.8, 1.9, 6.7

Show Answer

Order 1.9, 4.8, 6.7. $1.9+4.8=6.7$ and $6.7\le 6.7$ - not a triangle.

Mini-summary: Equality still fails at decimals.

Cycle C - fractions

Set: $\frac{3}{4},\frac{9}{7},\frac{2}{3}$. To denominator 84, two smaller are $\frac{56}{84}$ and $\frac{63}{84}$, sum $\frac{119}{84}$, largest is $\frac{108}{84}$, so triangle.

Try it: $\frac{5}{12},\frac{7}{12},\frac{1}{2}$

Show Answer

Two smaller 5/12 and 1/2=6/12. Sum 11/12 which is $>$ 7/12 - triangle.

Mini-summary: The same test works with fractions by common denominators.

Mistake 1 - checking the wrong pair

Fix: Label the largest, compare the sum of the other two against it.

Try it: 7, 12, 20 - which comparison is decisive?

Show Answer

Largest 20. Compare $7+12=19<20$ - not a triangle.

Mistake 2 - allowing equality

Fix: Use strict $>$.

Try it: 30, 40, 70

Show Answer

$30+40=70\le 70$ - not a triangle.

Mistake 3 - inconsistent units

Student Work A vs B

• Student A: 30 cm, 0.7 m, 40 cm - treats 0.7 m as 0.7 cm and accepts.
• Student B: converts 0.7 m = 70 cm - set 30, 70, 40 - compares $30+40=70\le 70$ and rejects.

Question: Whose work is correct and why?

Show Answer

Student B is correct because they converted units before applying the theorem.

Mini-summary: Correct unit conversion is part of correct reasoning.

Checkpoint 1 - identify the largest then decide

• A) 5, 12, 13
• B) 10, 25, 14

Show Answer

A) Largest 13. $5+12=17>13$ - triangle.
B) Largest 25. $10+14=24<25$ - not a triangle.

Mini-summary: One decisive inequality per set is enough.

Checkpoint 2 - decimals and fractions ladder

• Start: 1.2, 1.3, 2.6
• Stretch: 0.9, 1.8, 2.7
• Challenge: $\frac{4}{5},\frac{1}{2},\frac{9}{10}$

Show Answer

Start: $1.2+1.3=2.5<2.6$ - not a triangle.
Stretch: $0.9+1.8=2.7\le 2.7$ - not a triangle.
Challenge: two smaller 1/2=5/10 and 4/5=8/10, sum 13/10 which is $>$ 9/10 - triangle.

Mini-summary: Different number types use the same logic.

If the largest side is labeled $c$, then checking $a+b>c$ is decisive. The other two inequalities automatically hold if this is true because each left side includes the largest side $c$ plus another side.

Try it - write to explain: In two sentences, explain why checking $a+b>c$ with $c$ the largest is decisive.

Show Answer

If $c$ is the largest, then $c\ge a$ and $c\ge b$. When $a+b>c$ holds, then $b+c>a$ and $a+c>b$ are both true.

Mini-summary: Decisive checks reduce work and errors.

Anchor Revision 1 - Slats: 24 cm, 33 cm, 58 cm.

Show Answer

Order 24, 33, 58. $24+33=57<58$ - not a triangle.

Anchor Revision 2 - Replace 58 with 52 cm.

Show Answer

Order 24, 33, 52. $24+33=57>52$ - triangle possible.

Mini-summary: Designing with the inequality in mind saves time before cutting or drilling.

Set	Largest	Sum of two smaller	Decision
5, 8, 4	8	$5+4=9$	Triangle
3, 6, 2	6	$3+2=5$	Not a triangle
11, 12, 9	12	$11+9=20$	Triangle
2, 10, 8	10	$2+8=10$	Not a triangle

Try it - confirm the pattern - Explain in one short line why row 4 is not a triangle.

Show Answer

The sum of the two smaller equals the largest, so $\le$ holds and the shape collapses.

Mixed Set 1

1. 14, 18, 33
2. 21, 22, 44
3. 4.5, 5.8, 10.2
4. $\frac{7}{10},\frac{9}{10},\frac{3}{5}$

Show Answer

14+18=32 $<$ 33 - not a triangle. 21+22=43 $<$ 44 - not a triangle. 4.5+5.8=10.3 $>$ 10.2 - triangle. Two smaller 3/5=6/10 and 7/10 sum 13/10 $>$ 9/10 - triangle.

Mixed Set 2 - unit conversions

1. 25 cm, 0.4 m, 20 cm
2. 1.2 m, 65 cm, 40 cm

Show Answer

0.4 m=40 cm - set 25, 40, 20 - 25+20=45 $>$ 40 - triangle. 1.2 m=120 cm - set 120, 65, 40 - 65+40=105 $<$ 120 - not a triangle.

Consider sides as straight segments in a plane. If one side were at least as long as the sum of the other two, placing the two shorter end to end along the line of the longest would exactly reach or fail to reach the far endpoint without bending inward. A triangle requires bending inward to create area, which occurs only when the sum of the two shorter exceeds the longest.

Try it - verbalize the idea - Complete: A triangle has area only when the two shorter sides together are __________ than the longest side.

Show Answer

strictly greater

Tomorrow you will study how large or small the third side can be when two sides are fixed. If sides are $28$ and $7$, the third side $x$ must satisfy $28-7<x<28+7$, which is $21<x<35$. We will derive that range carefully on Day 2.

Try it - gentle teaser - Two sides are 10 and 18. Which could be the third side: 9, 28, 27?

Show Answer

It must be greater than $18-10=8$ and less than $18+10=28$. 9 works, 28 does not, 27 works.

You have three choices for your brace triangle. Choose all that can work.

• Set A: 24 cm, 33 cm, 52 cm
• Set B: 26 cm, 34 cm, 60 cm
• Set C: 28 cm, 37 cm, 52 cm

Show Answer

Set A: $24+33=57>52$ - works. Set B: $26+34=60\le 60$ - fails. Set C: $28+37=65>52$ - works.

Key takeaways: Decide quickly and correctly whether three lengths can close into a real triangle across contexts. Use the strict inequality and convert units first.

References:

• AllMath. Inequality in Mathematics: Definition, Types, and Examples.
• BYJU’S. Triangle Inequality Theorem.
• Encyclopedia Britannica. Triangle inequality.
• SplashLearn. Exterior Angle Theorem: Definition, Proof, Examples, Facts, FAQs.

💡 Example in Action

Worked Example 1

Problem: Do lengths 12, 14, and 21 form a triangle?

Steps: 1) Largest 21. 2) $12+14=26$. 3) $26>21$.

Show Answer

Yes.

Worked Example 2

Problem: Do lengths 5, 10, and 15 form a triangle?

Steps: 1) Largest 15. 2) $5+10=15$. 3) $15\le 15$.

Show Answer

No.

Worked Example 3

Problem: Do lengths 9, 11, and 21 form a triangle?

Steps: 1) Largest 21. 2) $9+11=20$. 3) $20<21$.

Show Answer

No.

Worked Example 4

Problem: Do lengths 2.7, 3.2, and 5.9 form a triangle?

Steps: 1) Largest 5.9. 2) $2.7+3.2=5.9$. 3) $5.9\le 5.9$.

Show Answer

No.

Worked Example 5

Problem: Do lengths $\frac{3}{5},\frac{1}{2},\frac{7}{10}$ form a triangle?

Steps: Largest $\frac{7}{10}$. Sum of other two $\frac{6}{10}+\frac{5}{10}=\frac{11}{10}$. Compare $\frac{11}{10}>\frac{7}{10}$.

Show Answer

Yes.

Now You Try - 5 items

1. 6, 10, 15
2. 7, 8, 9
3. 1.2, 1.3, 2.7
4. $\frac{2}{3},\frac{3}{4},\frac{5}{6}$
5. 25 cm, 0.4 m, 20 cm

Show Answer

1) $6+10=16>15$ - triangle.
2) $7+8=15>9$ - triangle.
3) $1.2+1.3=2.5<2.7$ - not a triangle.
4) $\frac{2}{3}+\frac{3}{4}=\frac{17}{12}>\frac{5}{6}$ - triangle.
5) 0.4 m=40 cm - $25+20=45>40$ - triangle.

📝 Try It Out

1. 4, 9, 14
2. 10, 15, 24
3. 2.4, 1.1, 3.6
4. 0.7, 0.8, 1.1
5. $\frac{5}{12},\frac{7}{12},\frac{1}{2}$
6. 18 cm, 0.5 m, 30 cm
7. 13, 18, 30
8. 5, 5, 9
9. 3.5, 6.4, 10.1
10. $\frac{9}{10},\frac{3}{5},\frac{7}{10}$

Show Answer

1) $4+9=13<14$ - not a triangle.
2) $10+15=25>24$ - triangle.
3) $2.4+1.1=3.5<3.6$ - not a triangle.
4) $0.7+0.8=1.5>1.1$ - triangle.
5) 1/2=6/12, so $\frac{5}{12}+\frac{6}{12}=\frac{11}{12}>\frac{7}{12}$ - triangle.
6) 0.5 m=50 cm, set 18, 30, 50 - $18+30=48<50$ - not a triangle.
7) $13+18=31>30$ - triangle.
8) $5+5=10>9$ - triangle.
9) $3.5+6.4=9.9<10.1$ - not a triangle.
10) 3/5=6/10, so $\frac{6}{10}+\frac{7}{10}=\frac{13}{10}>\frac{9}{10}$ - triangle.

✅ Check Yourself

1. Which single check is decisive if 23 is the largest of $a$, $b$, 23? A) $a+b>23$ B) $a+b\ge 23$ C) $a+b<23$
2. True or False: If two sides are 8 and 12, then a third side of 20 makes a triangle.
3. Decide: 1.5, 2.1, 3.6.
4. What is wrong: “Since 3 + 4 = 7 equals the largest 7, a triangle exists.”
5. Convert and decide: 0.9 m, 30 cm, 70 cm.
6. Pick the set that can form a triangle. A) 4, 10, 14 B) 5, 10, 12 C) 9, 10, 20
7. For $\frac{1}{4},\frac{1}{3},\frac{2}{3}$, identify the largest, then decide.
8. Choose the correct statement. A) Sum of two smaller is $\ge$ largest. B) Sum of two smaller is $>$ largest. C) Sum of two smaller is $<$ largest.
9. A path around two sides is 420 m and the diagonal is 420 m. Is a triangle formed by these distances?
10. Unit consistency: 18 in, 2 ft, 36 in. Convert and decide.
11. Diagnose data: 2.0 km, 3.5 km, 5.6 km.
12. Choose all that could be sides. A) 14, 19, 35 B) 14, 22, 35 C) 3.4, 6.6, 10.1
13. Fill the blank: Sum of the two shorter must be __________ than the longest side.
14. If the largest side is $c$, which comparison must hold?
15. Quick decide: 11, 20, 32.

Show Answer

1) A. 2) False - $8+12=20\le 20$. 3) $1.5+2.1=3.6\le 3.6$ - no. 4) Equality gives a straight line. 5) 0.9 m=90 cm - set 90, 30, 70 - $30+70=100>90$ - yes. 6) B only - $5+10=15>12$. 7) Largest 2/3, two smaller 1/4 and 1/3 sum 7/12 $<$ 8/12 - not a triangle. 8) B. 9) No - equality fails strictness. 10) 2 ft=24 in - set 18, 24, 36 - $18+24=42>36$ - yes. 11) $2.0+3.5=5.5<5.6$ - inconsistent. 12) Only B - $14+22=36>35$. 13) strictly greater. 14) $a+b>c$ where $c$ is largest. 15) $11+20=31<32$ - not a triangle.

🚀 Go Further

1. Design check: From {18, 22, 27, 31, 40}, list two triples that make a triangle and one that fails, with decisive comparisons.
2. Error hunt: A classmate claims 12, 19, 31 forms a triangle because 12+19=31 “touches.” Explain the flaw and correct the rule.
3. Mixed units: Choose two sets and convert to a single unit before deciding - A) 0.5 m, 45 cm, 60 cm B) 2 ft, 15 in, 30 in C) 1.1 m, 75 cm, 40 cm.
4. Reasoning write-up: In 3-4 sentences, explain why checking only the largest side is enough. Include a numeric example.
5. Design with constraint: Choose three lengths that form a triangle and keep the longest below 50 cm from {12, 17, 29, 34, 41, 50, 51}. Pick two valid triangles and justify.

Show Answer

1) Works: {22, 27, 40} - $22+27=49>40$; {18, 31, 40} - $18+31=49>40$. Fails: {18, 22, 40} - $18+22=40\le 40$.
2) The rule is strict. Equality collapses into a straight line. Correct rule: sum of the two shorter must be strictly greater than the longest.
3) A) 0.5 m=50 cm - 50, 45, 60 - 45+50=95 $>$ 60 - triangle. B) 2 ft=24 in - 24, 15, 30 - 24+15=39 $>$ 30 - triangle. C) 1.1 m=110 cm - 110, 75, 40 - 75+40=115 $>$ 110 - triangle.
4) Sample: labeling the largest makes one check decisive. Example: 7, 9, 15 - $7+9=16>15$ so other inequalities hold.
5) {12, 34, 41} - $12+34=46>41$; {17, 29, 41} - $17+29=46>41$.

🔗 My Reflection

Choose one and answer in your notebook.

Option A - Short writing: Explain how the single decisive comparison helps you quickly judge if three lengths make a triangle. Include one correct example and one near-miss example.

Option B - Checklist: Tick those you can do now.

• I can convert units before testing.
• I can identify the largest side quickly.
• I can use a strict comparison.
• I can explain why equality fails.
• I can apply the test to fractions and decimals.

Option C - 3-2-1: Write 3 insights you learned, 2 errors to avoid, and 1 question you still have.

Show Answer

Answers vary. Look for correct use of strict inequality, unit conversion, and a decisive comparison with a supporting example.