MAT8 Q2W6D2: Applying the Triangle Inequality Theorem and Exploring the Range of the Third Side

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

1. Find the range of the third side $x$ of a triangle when two sides $a$ and $b$ are given, using $|a-b|<x<a+b$, and express answers in both inequality and interval notation.
2. Test candidate side lengths against the derived range to decide whether a triangle is possible in real-number, decimal, and fractional contexts (with correct unit conversions).
3. Order sides and angles of a triangle from largest to smallest given the side lengths, and identify the largest/smallest angle by comparing opposite sides.

• Range form of the Triangle Inequality – For sides $a$, $b$, and unknown side $x$: $|a-b|<x<a+b$.
• Third side $x$ – The unknown side length whose feasible values must satisfy the range form.
• Strict inequality – Comparisons using $<$ or $>$; equality is not allowed for a valid (nondegenerate) triangle.
• Interval notation – Writing the solution set for $x$ as $(L,U)$, where $L=|a-b|$ and $U=a+b$.
• Feasible interval – The open interval of all values of $x$ that make a triangle possible.
• Angle–side relationship – In a triangle, the largest angle is opposite the longest side, and the smallest angle is opposite the shortest side.
• Ascending/descending order – Sorting sides to match sides with opposite angles quickly.

A. Triangle or not

Decide if each set of lengths can form a triangle.

1. 7, 11, 19
2. 6, 10, 15
3. 3.2, 4.1, 7.3
4. $\frac{5}{6},\frac{3}{4},\frac{7}{12}$

Show Answer

1) Largest 19; two smaller sum $7+11=18<19$ → not a triangle.
2) Largest 15; $6+10=16>15$ → triangle.
3) Largest 7.3; $3.2+4.1=7.3\le 7.3$ → not a triangle.
4) Order two smaller: $\frac{7}{12}$ and $\frac{5}{6}=\frac{10}{12}$; sum $\frac{17}{12}$. Largest $\frac{3}{4}=\frac{9}{12}$. Since $\frac{17}{12}>\frac{9}{12}$ → triangle.

B. Convert units then decide

1. 0.8 m, 30 cm, 60 cm
2. 18 in, 2 ft, 15 in

Show Answer

1) Convert 0.8 m = 80 cm → set 80, 60, 30. Largest 80; two smaller sum $60+30=90>80$ → triangle.
2) Convert 2 ft = 24 in → set 24, 18, 15. Largest 24; two smaller sum $18+15=33>24$ → triangle.

C. Angle–side relationship refresher

1. 7, 9, 12
2. 4.5, 5.1, 3.8

Show Answer

1) Longest side 12 → largest angle is opposite 12. Shortest side 7 → smallest angle is opposite 7.
2) Longest side 5.1 → largest angle is opposite 5.1. Shortest side 3.8 → smallest angle is opposite 3.8.

D. Bridge to today’s new idea

Fill in the range of possible values for the third side $x$. Use $|a-b|<x<a+b$.

1. Given sides $14$ and $11$
2. Given sides $6$ and $10$
3. Given sides $\frac{7}{5}$ and $\frac{9}{10}$

Show Answer

1) $|14-11|<x<14+11$ → $3<x<25$ → interval $(3,25)$.
2) $|6-10|<x<6+10$ → $4<x<16$ → interval $(4,16)$.
3) Lower bound $|\frac{7}{5}-\frac{9}{10}|$ = $|\frac{14}{10}-\frac{9}{10}|=\frac{5}{10}=\frac{1}{2}$; upper bound $\frac{7}{5}+\frac{9}{10}=\frac{23}{10}$. So $\frac{1}{2}<x<\frac{23}{10}$ → interval $(\frac{1}{2},\frac{23}{10})$.

Title: Range of the Third Side and Ordering Angles by Side Lengths

Big goal: Given two side lengths of a triangle, determine the complete set of valid values for the third side and use side comparisons to order angles. All comparisons and solutions use MathML. All answers are hidden.

1) Warm-up narrative: from single checks to full ranges

You are designing a wall shelf with two fixed boards already cut to length. Yesterday, you learned to test one candidate length by checking whether the sum of two sides is greater than the third. Today, you will stop guessing single numbers and learn how to write every possible value of the third side at once. Instead of “Does 17 work?”, you will answer “Which lengths work at all?” This saves time, avoids trial and error, and makes your decisions more reliable.

Guiding question: If two sides measure $9$ cm and $14$ cm, do you expect a single correct value for the third side or many correct values?

Show Answer

Many correct values. Any third side that satisfies the triangle inequalities will work.

Mini-summary: Today you will transform a yes/no test into a complete interval of allowed values.

2) From three inequalities to one open interval (derivation you can reuse)

Let the fixed sides be $a$ and $b$, and let the third side be the unknown $x$. A nondegenerate triangle must satisfy all three strict inequalities:

• $a+b>x$
• $a+x>b$
• $b+x>a$

Rearrange the second and third:

• From $a+x>b$, subtract $a$ to get $x>b-a$.
• From $b+x>a$, subtract $b$ to get $x>a-b$.

To satisfy both, $x$ must be greater than the larger of $b-a$ and $a-b$, which is the absolute value:

• Lower bound: $x>|a-b|$
• Upper bound: $x<a+b$

Define $L=|a-b|$ and $U=a+b$. Then the complete rule is the open interval: $L<x<U$ — that is, $(L,U)$.

Why strict? If $x=L$ or $x=U$, the three points lie on a straight line and the triangle has zero area, which is not allowed for this lesson.

Try it – name the interval ends and write the open interval.

1. $a=9$, $b=14$
2. $a=7.2$, $b=5.6$
3. $a=\frac{5}{4}$, $b=\frac{7}{6}$

Show Answer

1) $L=|9-14|=5$, $U=23$ → $(5,23)$.
2) $L=|7.2-5.6|=1.6$, $U=12.8$ → $(1.6,12.8)$.
3) $\frac{5}{4}=\frac{15}{12}$, $\frac{7}{6}=\frac{14}{12}$ → $L=\frac{1}{12}$, $U=\frac{29}{12}$ → $(\frac{1}{12},\frac{29}{12})$.

Mini-summary: The three inequalities compress into a single open interval for the third side: $(L,U)$.

3) Edge-case drill: mastering “open” ends

With $L=5$, $U=23$, classify each candidate as valid or invalid: $x=L$, $x=L+0.1$, $x=U-0.1$, $x=U$.

Show Answer

Use $L<x<U$. Invalid: $x=L$, $x=U$. Valid: $x=L+0.1$, $x=U-0.1$.

Mini-summary: Endpoints are never allowed; values strictly inside are allowed.

4) Candidate Test (one-line rule you can reuse all year)

Candidate Test: Accept a candidate $x$ iff $L<x<U$. Reject when $x=L$, $x=U$, $x<L$, or $x>U$.

Show Answer

Example 1 (9 and 14): accept 10 and 20; reject 5 and 23.
Example 2 (7.2 and 5.6): accept 8.0; reject 1.5, 1.6, and 12.8.

Mini-summary: One inequality with two numbers decides everything.

5) Reading the interval: number-line thinking

With $L=8$, $U=28$, decide quickly: 8, 8.01, 27.99, 28.

Show Answer

Reject 8 and 28; accept 8.01 and 27.99.

Mini-summary: Visualize open ends to avoid accepting equality by accident.

6) Fractions: exact arithmetic prevents borderline errors

Example (exact): $a=\frac{7}{5}$, $b=\frac{9}{10}$. Then $L=\frac{1}{2}$, $U=\frac{23}{10}$, so $(\frac{1}{2},\frac{23}{10})$.

Try it – fractions: (5/4, 2/3) and (7/8, 5/6) — compute $L$, $U$, and intervals.

Show Answer

(5/4, 2/3): $L=\frac{7}{12}$, $U=\frac{23}{12}$ → $(\frac{7}{12},\frac{23}{12})$.
(7/8, 5/6): $L=\frac{1}{24}$, $U=\frac{41}{24}$ → $(\frac{1}{24},\frac{41}{24})$.

Mini-summary: Exact fraction work is safest near the ends of the interval.

7) Mixed units: always “Convert → Compute → Decide”

Mini-card: 1) Convert to a single unit. 2) Compute $L=|a-b|$, $U=a+b$. 3) Decide with $(L,U)$.

Example: 0.6 m and 45 cm → in cm: $L=15$, $U=105$ → $(15,105)$ cm.

Show Answer

1.2 m and 75 cm → meters: $L=0.45$, $U=1.95$ → $(0.45,1.95)$.
18 in and 2 ft → inches: $L=6$, $U=42$ → $(6,42)$.

Mini-summary: Convert first, compute second, decide third.

8) Real-life applications: design, navigation, and sports

a) Furniture bracing: (48, 82) → $L=34$, $U=130$ → $(34,130)$. Test 34 and 35.

Show Answer

34 rejected (endpoint); 35 accepted.

b) Triangulating a signal: separation 2.4 km, one distance 1.9 km → $(0.5,4.3)$ km. Test 0.5 and 0.52.

Show Answer

0.5 rejected (endpoint); 0.52 accepted.

c) Sports layout: (18, 11) → $(7,29)$ m. Integer positions?

Show Answer

8 through 28 → $29-7-1=21$ positions.

Mini-summary: Different contexts, same interval rule.

9) Pitfalls and quick fixes

• Always use $L=|a-b|$.
• Keep strict $<$, not $\le$.
• Convert units before computing.
• Avoid rounding near endpoints.

Student Work A vs B: With (12, 19): A uses $x>-7$ and accepts $x=-3$. B uses $7<x<31$.

Show Answer

B is correct. Lengths are nonnegative; lower bound is 7.

Mini-summary: Absolute value eliminates sign mistakes.

10) Reverse reading: interval given, deduce side information

Interval (21, 35) → $|a-b|=21$, $a+b=35$. Possible pairs?

Show Answer

(28, 7) or (7, 28). Verify inequalities afterward.

Mini-summary: The interval reveals the sum and absolute difference.

11) Ordering angles by comparing sides (standalone micro-section)

Label triangle ABC with opposite sides $a$, $b$, $c$. If $c>b>a$, then $C>B>A$.

Show Answer

1) For sides 7, 11, 19 (assume valid), largest angle opposite 19; smallest opposite 7.
2) For 4.8, 4.8, 6.2, two base angles are equal and smaller than the angle opposite 6.2.

Mini-summary: Longest side ↔ largest angle; shortest side ↔ smallest angle.

12) Blended tasks: choose a valid third side, then rank angles

Given (12, 19): find (L, U); then for $x=20$ and $x=13$, identify largest/smallest angles.

Show Answer

(L, U) = (7, 31). With x=20, largest: opposite 20. With x=13, smallest: opposite 12.

Mini-summary: After choosing a valid $x$, ordering angles is immediate.

13) Counting integer solutions (with a one-line justification)

Valid integers run from $L+1$ to $U-1$. Count is $U-L-1$.

Show Answer

a) (8, 13) → L=5, U=21 → count 15.
b) (20, 5) → L=15, U=25 → count 9.

Mini-summary: Counting integers requires subtracting both endpoints.

14) Optional – Advanced design: tolerance and safety margins

With tolerances: a=60±0.2, b=85±0.2 (cm). Conservative bounds:

Lower bound (worst case): $L^{*}=|85.2-59.8|=25.4$ cm. Upper bound (worst case): $U^{*}=59.8+84.8=144.6$ cm. Interval: $(25.4,144.6)$ cm.

Show Answer

Yes, safety margins can exclude choices near original ends to guarantee validity.

Mini-summary: Safety margins shrink mathematical intervals to handle uncertainty.

15) Interval intersections with practical limits

Feasible = $(L,U)$ ∩ practical constraint. Example: (13, 31) ∩ [12, 25] = (13, 25].

Show Answer

(10, 20] when intersecting (5, 23) with [10, 20].

Mini-summary: Intersections keep the triangle rule intact while respecting external limits.

16) Capstone challenge: end-to-end application

Garden edges 7.4 m and 12.9 m; catalog 0.1 m steps, 2.0–21.0 m.

Show Answer

(L, U) = (5.5, 20.3) → (5.5, 20.3). Smallest valid 5.6 m; largest valid 20.2 m. With x=13.0, largest angle opposite 13.0. With x=8.0, smallest angle opposite 7.4.

Mini-summary: The open interval combines with discrete increments to guide exact purchases and angle reasoning.

17) Short geometric intuition: why endpoints fail

If $x=U$ or $x=L$, the three points are collinear and area is zero. A triangle requires area.

Show Answer

“Strictly greater” is required: two shorter sides together must be strictly greater than the longest.

18) Consolidated practice ladder (with unified notation every time)

A. Compute intervals for (15, 10) and (6.5, 2.4). B. Test 40, 30, 18 for (12, 29) and 0.7, 2.5, 1.0 for (0.9, 1.6). C. Fractions & angle link for (11/8, 5/6) and x=7/4.

Show Answer

A: (5, 25); (4.1, 8.9). B: (17, 41) → 40, 30, 18 valid; (0.7, 2.5) → reject endpoints, 1.0 valid. C: (13/24, 53/24); largest angle opposite 7/4.

19) Self-coaching prompts

Show Answer

Look for correct $|a-b|$, $a+b$, and strict inequalities.

20) Final takeaways

• Third side rule: $|a-b|<x<a+b$.
• Endpoints excluded.
• Convert units first.
• Integer count: $U-L-1$.
• Longest side ↔ largest angle.

References

• CK-12 Foundation. Triangle Inequality Theorem and applications.
• Encyclopedia Britannica. Triangle inequality.
• BYJU’S. Triangle Inequality Theorem and Converse – Range of the third side.
• AllMath. Inequality in Mathematics: Definition, Types, and Examples.

Worked Example 1 — Integer sides, interval + quick tests

Two sides: 14 cm and 9 cm.

1. Find the interval for $x$.
2. Decide if 5, 10, 23 are valid choices.
3. How many integer values of $x$ are possible?

Show Answer

a) $L=5$, $U=23$ → $(5,23)$.
b) 5, 23 invalid (endpoints); 10 valid.
c) Count = $23-5-1=17$.

Worked Example 2 — Decimals, interval + catalog

Two sides: 7.2 m and 3.9 m. Supplier step 0.1 m.

Show Answer

Interval (3.3, 11.1). Smallest supplier: 3.4 m. Largest supplier: 11.0 m.

Worked Example 3 — Fractions, exact arithmetic

Two sides: 7/5 m and 9/10 m.

Show Answer

(L, U) = (1/2, 23/10). x=1/2 invalid; x=7/4 valid.

Worked Example 4 — Mixed units

Two sides: 0.6 m and 45 cm. Interval in cm; test 15, 35, 105.

Show Answer

(15, 105) cm. 15, 105 invalid; 35 valid.

Worked Example 5 — Blend

Two sides: 12 cm and 19 cm. Find (L, U). With x=20 cm, which angle is largest? With x=13 cm, which is smallest?

Show Answer

(7, 31). Largest opposite 20; smallest opposite 12.

Now You Try — 5 Tasks

1. Integers: (8, 17) → interval and count.
2. Decimals: (2.6, 4.1), step 0.05 → smallest & largest supplier lengths.
3. Fractions: (11/8, 5/6) → interval.
4. Mixed units: (1.2 m, 65 cm) → interval in cm; test 105 cm.
5. Blend: (10, 18). With x=19, largest angle? With x=12, smallest angle?

Show Answer

1) (9, 25) → 15 integers. 2) (1.5, 6.7) → 1.55 m and 6.65 m. 3) (13/24, 53/24). 4) (55, 185) cm; 105 valid. 5) (8, 28); largest opposite 19; smallest opposite 10.

1. Integers — (11, 17): interval for $x$; test 6, 20, 28.
2. Integers — (13, 19): how many integer $x$?
3. Fractions — (3/4, 5/6): interval for $x$.
4. Decimals — (4.5, 2.7), step 0.1: interval; smallest & largest supplier lengths.
5. Mixed units — (90 cm, 1.4 m): interval in cm; are 50 cm and 229 cm valid?
6. Reverse — interval (8, 20): find one (a, b).
7. Angles — (9, 15), choose x=22: largest and smallest angles?
8. Intersection — (12, 25) with catalog [18, 30]: triangle interval and feasible set.
9. Candidate filter — (7.2, 10.6): which of 3.4, 3.5, 17.7, 17.8, 9.2 are valid?
10. Mixed challenge — (18, 10): interval; how many integer multiples of 3? With x=21, largest angle?

Show Answer

1) (6, 28); 6, 28 invalid; 20 valid. 2) 25 integers. 3) (1/12, 19/12). 4) (1.8, 7.2) → 1.9 m and 7.1 m. 5) (50, 230) → 50 invalid, 229 valid. 6) (14, 6) or (6, 14). 7) Largest opposite 22; smallest opposite 9. 8) (13, 37); feasible (18, 30]. 9) Valid: 3.5, 17.7, 9.2. 10) (8, 28); 7 multiples; largest opposite 21.

1. (13, 22): write interval.
2. (9, 14): is x=23 valid?
3. Fractions (5/3, 7/6): interval.
4. Mixed units (0.9 m, 65 cm): interval in cm.
5. Count integers for (11, 18).
6. Count odd integers for (8, 15).
7. Reverse from (4, 10): one (a, b).
8. Angles — (12, 19), x=20: largest and smallest.
9. Intersection — (5, 19) with [12, 20].
10. Supplier step — (7.2, 3.9), step 0.2: smallest & largest.
11. Tolerance — a=40±0.2, b=55±0.2: conservative interval.
12. Degenerate check — (24, 42), x=18.
13. Proposals — (9, 12): 2.8, 3.0, 21.0, 20.9?
14. Layout — (18, 26): interval; does 44 work?
15. Conceptual — why absolute value?

Show Answer

1) (9, 35). 2) No (endpoint U). 3) (1/2, 17/6). 4) (25, 155) cm. 5) 21. 6) 7. 7) (7, 3) or (3, 7). 8) Largest opposite 20; smallest opposite 12. 9) (14, 24) ∩ [12, 20] = (14, 20]. 10) 3.5 m and 11.0 m. 11) (15.4, 94.6). 12) Invalid (x=L). 13) Only 20.9 valid. 14) (8, 44); 44 invalid (endpoint). 15) Because $x>b-a$ and $x>a-b$ must both hold, so $x>|a-b|$.

1) Catalog Optimization — Maximize the Largest Angle

(14, 21), step 0.5 cm, catalog 7.0–35.0 cm.

Show Answer

Interval (7, 35). To maximize largest angle, choose the largest catalog value strictly less than 35 → 34.5 cm.

2) Safety-First Choice — Tolerance + Catalog

a=60±0.2, b=85±0.2 cm; integer x in [59, 145].

Show Answer

Safe interval (25.4, 144.6). Guaranteed integers: 26–144 inclusive.

3) Targeted Angle Control — Largest at A

Given b=18 cm, c=25 cm. Choose all integer a so angle A is largest.

Show Answer

Legal a: 7a<43 .="" a="" for="" largest="" require="">25. Integers: 26–42.

4) Minimal Adjustment — Fix a Near-Miss Triangle

Set (8, 15, 23). One 0.1 cm change allowed.

Show Answer

A) 8→8.1 → valid. B) 15→15.1 → valid. C) 23→22.9 → valid. Each uses minimal 0.1 cm change.

5) Reverse Puzzle — Interval Given + Angle Condition

Interval (13, 31). Largest angle opposite side 22 cm.

Show Answer

a) (22, 9) or (9, 22). b) 22 must be the longest side → x<22 .="" 22="" c="" details="" refined="" x:="">

3–2–1 Reflection

3 things I can confidently do now:

• Write the third-side range as $|a-b|<x<a+b$.
• Test any candidate $x$ using $L<x<U$.
• Order angles by side lengths.

2 spots that still feel tricky (explain briefly):

__________________________________________
__________________________________________

1 real-life scenario where I could use today’s skill (describe it and write the inequality in MathML):

Scenario: ________________________________
Inequality: $\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_$
Decision (valid or not) and why: ____________________

Exit check (one line): Rate your confidence (1–5) and one action for next class (e.g., endpoints with decimals, convert units first).