MAT8 Q2W7D1: Exploring Triangle Inequality Theorems

Learning Goals

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

1. State and illustrate the Triangle Inequality Theorem in your own words and with sketches, including all three inequalities $a+b>c,b+c>a,a+c>b$, and correctly identify the included assumptions (positive lengths, non-degenerate triangles) in at least 3 examples.
2. Decide whether three given lengths can form a triangle using the theorem, justifying each decision with a brief inequality check, for at least 8 sets of side lengths with ≥ 90% accuracy.
3. Determine the valid range of the third side given two sides (for example, for sides $7$ and $12$, find all $x$ satisfying $|12-7|<x<12+7$), and solve at least 6 such problems accurately.

Key Ideas & Terms

• Triangle - A polygon with three sides and three interior angles. Side lengths are positive real numbers and must satisfy the triangle inequalities to form a valid triangle.
• Side length - The measure of a side of a triangle, expressed as a positive real number with units.
• Inequality / strict inequality (<, >) - A comparison statement between two numbers. A strict inequality uses < or > to indicate one quantity is smaller or larger than another without equality.
• Triangle Inequality Theorem - In any triangle with side lengths $a$, $b$, and $c$, the sum of the lengths of any two sides is greater than the length of the remaining side: $$a+b>c,b+c>a,a+c>b$$
• Feasible triangle - A set of three positive lengths that satisfies all three inequalities above. These lengths can form a triangle.
• Non-feasible triangle - A set of lengths that fails at least one of the triangle inequalities. These lengths cannot form a triangle.
• Degenerate triangle - A boundary case where the sum of two sides equals the third, for example $a+b=c$. This collapses into a straight segment and is not considered a triangle in this lesson.
• Third-side range - Given two sides of lengths $b$ and $c$, the possible third side $a$ must satisfy $$|b-c|<a<b+c$$. The same rule applies cyclically for the other sides.
• Absolute value - The distance of a number from zero on the number line. For any real $x$, $\left|x\right|$ is nonnegative. It is used to express differences like $|b-c|$ in the third-side range.
• Perimeter - The total length around the triangle: $P=a+b+c$. Useful for reasonableness checks when choosing side lengths.
• Counterexample - A specific example that shows a general statement is false. For triangle claims, it can be a set of three lengths that violates at least one triangle inequality.

Quick Recall / Prior Knowledge

A. Symbol Sense - Compare using <, >, or =

1. $7+5$ ___ $13$
2. $9+4$ ___ $12$
3. If $x>3$ and $y>x$, then $y$ ___ $3$
4. $18-5$ ___ $12$

Show Answer

1) < because $7+5=12<13$.
2) > because $9+4=13>12$.
3) > by transitivity: if $y>x$ and $x>3$, then $y>3$.
4) = because $18-5=13$ and $13=13$.

B. Absolute Value Refresher

1. $|9-14|$
2. $|14-9|$

Show Answer

1) $5$, because distance on the number line is nonnegative: $|9-14|=5$.
2) $5$ as well: $|14-9|=5$.

C. Angle Sum in a Triangle

In $\Delta ABC$, if $m\angle A=50°$ and $m\angle B=61°$, find $m\angle C$.

Show Answer

$$m\angle C=180°-(50°+61°)=69°$$ Triangles have angle sum $180°$.

D. Perimeter Check

A triangle has sides $a=4$, $b=6$, $c=8$. Compute the perimeter.

Show Answer

$P=a+b+c=4+6+8=18$.

E. Included Angle Identification

1. Which angle is included between sides $\mathrm{AB}$ and $\mathrm{AC}$?
2. Which angle is included between sides $\mathrm{BC}$ and $\mathrm{BA}$?

Show Answer

1) $\angle A$ - the common endpoint is at $A$.
2) $\angle B$ - the common endpoint is at $B$.

F. Complete Me Table

You have three strings of lengths $4$, $7$, and $10$. Complete the sums and compare with the third number.

Sum of two	Value	Compare to third
$4+7$	?	vs. $10$
$7+10$	?	vs. $4$
$4+10$	?	vs. $7$

Show Answer

1) $4+7=11$, and $11>10$.
2) $7+10=17$, and $17>4$.
3) $4+10=14$, and $14>7$.

Explore the Lesson

0. Welcome and road map

Today you will explore a powerful idea that quietly controls whether three lengths can meet to form a triangle. You will start with hands-on thinking, discover the rule through guided questions, and then learn to use it to test side lengths and to find the full range of a third side that makes sense. You will also see how this rule shows up in real tasks such as bracing a shelf, passing a ball on the court, planning a short-cut in a park, or choosing struts for a triangular frame.

By the end of this Explore section you should be able to: explain the triangle inequality in your own words, decide quickly whether three lengths can form a triangle, compute the complete interval of possible values for a third side when two sides are fixed, and apply these ideas to real-life designs and decisions. Throughout, you will see guiding questions with answers hidden inside collapsibles. Try first, then open to check. You will also encounter checkpoints with short mini-summaries, so you always know what you have learned.

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1. From experience to idea

Thought experiment A - two sticks and a band

Imagine you have two plastic strips and a rubber band. The strips meet at point A, like a hinge, and the rubber band connects their other ends B and C. When you open the hinge wider, what seems to happen to the distance BC? Hold AB and AC fixed. You can picture that as two side lengths that do not change. The distance BC is the third side of triangle ABC.

Guiding questions

1. If the angle at A gets larger, what happens to the stretch needed to connect B and C with the rubber band?

Show Answer

As the angle at A increases, points B and C move farther apart, so the rubber band needs to stretch more. That means the length BC increases. This is a natural hint that the side opposite a larger angle is longer in a triangle, and it motivates comparisons among sides.

2. If you close the hinge so much that the strips almost lie on the same straight line, what happens to BC compared with AB and AC?

Show Answer

When AB and AC nearly form a straight line, BC nearly equals the sum of AB and AC. If they were exactly collinear in opposite directions, you would not have a triangle at all. That boundary case is called degenerate and uses equality of two sides with the third.

Thought experiment B - walking around a corner

Suppose you want to go from point X to point Z in a city grid. One way is to walk first from X to Y, then from Y to Z. Another way is to walk directly from X to Z through the park. Which path is shorter? The idea here is that going directly is never longer than taking a detour. In a triangle, traveling along two sides is never shorter than jumping across the third side.

1. Can the two-leg path ever be exactly the same length as the direct path if the route truly turns at a corner?

Show Answer

No. If there is a genuine turn, the detour has positive bend, and the two-leg path is strictly longer than the straight path. Equality happens only when the corner disappears and the route is a straight line. That is the degenerate boundary.

Mini-summary: Two sides win over one side because going around a corner is a detour. Equality appears only when the shape collapses into a straight segment.

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2. Stating the Triangle Inequality Theorem

The core rule

For any triangle with side lengths $a$, $b$, and $c$ (all positive), the sum of the lengths of any two sides is greater than the length of the remaining side:

$$a+b>c,b+c>a,a+c>b$$

These are strict inequalities. Each one must hold if the three lengths form a nondegenerate triangle. When the sum of two sides equals the third, for example $a+b=c$, the points lie on a straight line. That flat case is not considered a triangle in this lesson. All three inequalities together are both necessary and sufficient for three positive lengths to form a triangle in the plane.

Quick reason based on travel

The direct segment between two points is the shortest connection. In triangle ABC, the straight side BC is shorter than the two-step route BA then AC. That gives $\mathrm{BA}+\mathrm{AC}>\mathrm{BC}$. Apply the same idea by relabeling vertices to obtain the other two inequalities.

Checkpoint 1 - decide triangle or not

Determine whether each set of lengths can form a triangle.

1. $4,5,9$
2. $7,11,3$
3. $8,5,2$
4. $6,10,15$

Show Answer

A) Test sums: $$4+5=9;5+9=14>4;4+9=13>5$$ The first sum is equal to the third side, not greater. That is degenerate. Not a triangle for this lesson.
B) $$7+11=18>3;11+3=14>7;7+3=10>11?$$ Last comparison is false since $10>11$ is not true. Fails. Not a triangle.
C) $$8+5=13>2;5+2=7>8?;8+2=10>5$$ The middle comparison fails since $7>8$ is false. Not a triangle.
D) $$6+10=16>15;10+15=25>6;6+15=21>10$$ All three are true. Triangle.

Mini-summary: To test a triple, check all three sums. Every sum must be strictly greater than the remaining side. Equality means degenerate. Less means impossible.

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3. Deepening the rule - the third side range

Very often you know two sides and need to find what the third side can be. Let two known sides have lengths $b$ and $c$, and the unknown third side be $a$. From the triangle inequality we obtain the compact interval:

$$|b-c|<a<b+c$$

Checkpoint 2 - build the range

1. $b=7,c=12$
2. $b=3.5,c=3.9$
3. $b=15,c=8$

Show Answer

A) $$|7-12|<a<7+12$$ $$5<a<19$$ B) $$|3.5-3.9|<a<3.5+3.9$$ $$0.4<a<7.4$$ C) $$|15-8|<a<23$$ $$7<a<23$$

Mini-summary: Third side must be strictly bigger than the difference and strictly smaller than the sum of the other two. The interval is always open at both ends to avoid the degenerate boundary.

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4. Worked examples with full solutions

Example 1 - test a triple

Do the lengths $6,8,13$ form a triangle?

Solution: Check all three sums.

$$6+8=14>13$$ $$8+13=21>6$$ $$6+13=19>8$$

All true. Triangle.

Example 2 - detect a boundary failure

Test $5,10,15$.

$$5+10=15$$

This is equality. Degenerate for this lesson.

Example 3 - find the third side range

Given two sides 11 cm and 7 cm, find all possible lengths for the third side.

$$|11-7|=4$$ $$11+7=18$$ $$4<a<18$$

Integer choices: $5,6,7,8,9,10,11,12,13,14,15,16,17$.

Example 4 - solve an inequality with a variable

Three planks form a triangle. Two planks measure $9$ and $14$. The third plank is $x$ centimeters. Find the range of $x$.

$$|14-9|<x<23$$ $$5<x<23$$

Example 5 - count integer solutions

You have two sides 13 m and 20 m. How many whole-number lengths for the third side are allowed?

$$|20-13|<a<33$$ $$7<a<33$$

Allowed integers: $8$ through $32$, inclusive. Count: $32-8+1=25$.

Example 6 - real-life bracing

You are building a triangular brace with two sides fixed at 1.2 m and 1.8 m. The third side must be a steel bar sold only at increments of 0.1 m. Which bar lengths fit?

$$|1.8-1.2|<a<3.0$$ $$0.6<a<3.0$$

Allowed bars by 0.1 m steps: $0.7,0.8,0.9,1.0,1.1,1.2,1.3,1.4,1.5,1.6,1.7,1.8,1.9,2.0,2.1,2.2,2.3,2.4,2.5,2.6,2.7,2.8,2.9$.

Mini-summary: Use difference and sum to get a clean interval. For counts, convert the interval to the appropriate step size, then count.

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5. Real-life applications

A. Construction and carpentry

When building a triangular brace, shelf support, or roof truss, pieces must meet without slack or overlap. If two beams are fixed, the third beam has only a certain range that will fit. Too short and it leaves a gap. Too long and it bulges the frame.

1. If two sides of a triangular frame are 2.4 m and 3.1 m, what is the minimum and maximum possible length for the third side to keep a proper triangle?

Show Answer

Minimum just above the difference: $|3.1-2.4|=0.7$. Maximum just below the sum: $3.1+2.4=5.5$. So $0.7<a<5.5$.

2. Why is equality not allowed if you want a rigid triangular brace?

Show Answer

Equality means the pieces lie straight, so the figure collapses. It will not lock angles in place. A useful brace needs a true triangle with strictly greater sums.

B. Sports passing lanes

Three players form a triangle of passing options. If one pass distance is fixed by position, the other two distances must make sense. If two players are 18 m apart and one of those players is 12 m from the third, then the remaining distance must satisfy $|18-12|<d<18+12$, so $6<d<30$. That helps a team reason about spacing that keeps all lanes open.

C. Navigation

Two legs of a walking or hiking route can be viewed as two sides of a triangle. The third side is the shortcut distance. The triangle inequality explains why the shortcut is shorter than the two-leg path and gives bounds for how short it could be.

Checkpoint 3 - apply the bounds

1. Two hiking legs are 4.2 km and 3.7 km. What are all possible straight-line distances between the starting point and the endpoint?
2. A soccer team keeps two players 20 m apart, and each is 15 m from a third player. What are the possible distances between the two 15 m players?

Show Answer

1) $$|4.2-3.7|<d<4.2+3.7$$ $$0.5<d<7.9$$ 2) $$|20-15|<d<35$$ $$5<d<35$$

Mini-summary: Sports and hiking situations are direct models of triangle side bounds. The difference and sum rule guides spacing and distance planning.

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6. Precision about strictness, units, and mistakes to avoid

Strictness of the inequality

Always use greater than, not greater than or equal to, when you compare the sum of two sides to the third side. The strict symbol prevents degenerate cases.

Units and consistency

If one side is in centimeters and another is in meters, convert to the same unit before testing or creating ranges. Otherwise, your comparisons are not meaningful.

Common pitfalls

• Checking only one inequality
• Forgetting the lower bound that uses absolute value
• Including the endpoints of the range
• Ignoring measurement error when building something physical

Checkpoint 4 - spot the mistake

1. With sides 9 and 4, the third side satisfies $9-4<a<9+4$, so $5\le a\le 13$.
2. With sides 2.5 m and 70 cm, compute the range for the third side as $1.8<a<3.2$.
3. Since $8+3>10$ is true, the triple 8, 3, 10 forms a triangle.

Show Answer

A) The mistake is using non-strict inequalities. The correct range is open: $5<a<13$.
B) Mixed units. Convert 70 cm to 0.70 m first. Then: $|2.5-0.7|<a<3.2$, which is $1.8<a<3.2$. The arithmetic was fine but the reasoning must state that the units were converted. Precision matters.
C) You must test all three. Check: $$8+10=18>3;3+10=13>8;8+3=11>10$$ Here all three are true, so this one does form a triangle. The mistake was the habit of checking only one inequality. This time it happened to work, but that is risky.

Mini-summary: Strict signs only. Unify units before calculating. Always check all three inequalities if you are testing a triple.

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7. Discovery path to the range formula

You can re-discover the range formula by logical steps without memorizing.

1. Start with the three core inequalities: $$a+b>c,b+c>a,c+a>b$$
2. From $b+c>a$, get $a<b+c$ for the upper bound.
3. From $a+c>b$ and $a+b>c$, get $a>b-c$ and $a>c-b$, hence $a>|b-c|$.
4. Combine to write the full interval: $$|b-c|<a<b+c$$.

Checkpoint 5 - recover the range yourself

Given $b=9$ and $c=16$, show the steps to reach the final interval for $a$.

Show Answer

1) Start with three inequalities: $$a+9>16,9+16>a,a+16>9$$ 2) From the second: $a<25$. 3) From the first: $a>7$. 4) From the third: $a>-7$ which is weaker. 5) Combine lower bounds to get $a>7$. 6) Final interval: $$7<a<25$$ which matches $|9-16|<a<25$.

Mini-summary: You can rebuild the interval from the three originals by simple algebra. Absolute value is a compact way to combine two lower bounds.

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8. Extended practice inside the Explore

Task 1 - quick tests

Which of the following triples can be triangle side lengths?

1. $10,12,3$
2. $9,9,17$
3. $5,7,9$

Show Answer

i) All three comparisons pass, so it does form a triangle. ii) $9+9=18>17$, others pass, triangle. iii) All sums pass: $$5+7=12>9;7+9=16>5;5+9=14>7$$ Triangle.

Task 2 - create an interval

Two sides are 4.6 and 6.2. Write the interval for the third side.

Show Answer

$|6.2-4.6|<a<10.8$, so $1.6<a<10.8$.

Task 3 - integer count

Two sides are 8 cm and 21 cm. How many integer lengths for the third side are allowed?

Show Answer

Range: $|21-8|<a<29$, so $13<a<29$. Allowed integers are 14 through 28 inclusive. Count: $28-14+1=15$.

Mini-summary: Practice turns understanding into speed. Intervals and integer counts follow consistent steps.

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9. Connecting to angles and tomorrow’s learning

Although today is about side lengths, you already noticed a pattern in the opening thought experiment. When you opened the hinge wider, the opposite side got longer. That relationship between sides and angles prepares you for future theorems that compare sides by comparing angles and vice versa. For now, keep the side inequality itself precise and automatic.

Checkpoint 6 - reasoning about changes

If two sides stay fixed and the included angle increases, what happens to the length of the side opposite that angle, and why does that still respect today’s inequalities?

Show Answer

The opposite side increases. The triangle inequality remains valid because the direct connection is still shorter than the two-step path. Changing the angle shifts how long the opposite side needs to be, but the sum of the two fixed sides still exceeds the third side, and the third side still exceeds the difference of the two fixed sides.

Mini-summary: Side and angle comparisons fit together consistently. Today’s theorem sets the ground for comparisons among sides and angles later.

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10. Multi-step real-world problems

Problem A - garden path shortcut

A gardener walks 120 m along one edge of a rectangular plot, then 80 m along the adjacent edge to reach a shed. Estimate the straight-line distance from the starting corner to the shed and show that the detour is longer.

Solution: The triangle inequality predicts that the two-leg path is longer than the straight path. The straight path here would be the diagonal. The inequality states $120+80>d$, so the diagonal $d$ is strictly less than 200 m.

Problem B - designing a triangular support

You have two beams of 2.0 m and 2.8 m fixed to the wall studs. The third connecting strut must make a triangle. The supplier sells struts between 0.5 m and 3.5 m. Which of those are usable?

Solution:

$$|2.8-2.0|<a<4.8$$ $$0.8<a<4.8$$

Supplier range is 0.5 to 3.5. The overlap with 0.8 to 4.8 is 0.8 to 3.5. Any strut longer than 0.8 m and up to but not equal to 3.5 m works.

Problem C - robot arm reach

A two-segment robotic arm has segments 35 cm and 20 cm. The reach from base to gripper depends on the angle at the elbow. What is the possible range of reach?

Solution: $15<r<55$ cm.

Checkpoint 7 - interpret ranges

1. If a design requires the third side to be at least 2.4 m, and two sides are 1.6 m and 1.1 m, is the design feasible?
2. A drone must keep three beacon distances forming a triangle to avoid collinear alignment. Distances are 40 m, 25 m, and variable $d$. What are safe values of $d$?

Show Answer

1) Range with sides 1.6 and 1.1: $0.5<a<2.7$. Values strictly between 2.4 m and 2.7 m are feasible. Equality at 2.7 m is not allowed. 2) Range with sides 40 and 25: $15<d<65$. Any distance in that open interval prevents collinearity in this simplified model.

Mini-summary: Interpreting an interval in context helps you choose feasible options. Always remember the open ends when strict triangles are required.

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11. Strategy toolbox

• For testing a triple: quickly add the two smallest numbers and compare with the largest.
• For the third side interval: compute difference first, sum second, then write the open interval and interpret it.
• For counting integers: translate the open interval to the smallest and largest integers inside it and count with the inclusive counting formula.

Checkpoint 8 - use the toolbox quickly

1. Can 12, 5, 6 be sides of a triangle?
2. Two sides are 12 and 5. How many integer lengths for the third side exist?

Show Answer

1) Largest is 12. The sum of the two smaller is 5 + 6 = 11, which is less than 12. Not a triangle. 2) Range: $7<a<17$. Integers 8 through 16. Count: $16-8+1=9$.

Mini-summary: A quick largest-versus-sum-of-others check is efficient. Difference and sum produce instant ranges.

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12. Reflection prompts inside Explore

Choose at least one prompt to think through silently before moving on.

• In your own words, describe why the third side cannot be as long as the sum of the other two in a real triangular frame.
• Someone claims that if two sides are fixed, any positive number works for the third side. Provide a counterexample using the difference bound.

Show Answer

The third side equal to the sum would lay flat with the other two, so there is no enclosed area and no angle to lock a frame. A counterexample: sides 5 and 8. The third side 1 is not allowed because the difference bound requires $3<a$. Here $a=1$ fails.

Mini-summary: Explaining ideas in your own words strengthens mastery. Counterexamples help test and refine claims.

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13. Synthesis

You have practiced seeing triangles as systems with limits. The triangle inequality is the key. It says that a path that turns is longer than the straight path. That principle yields three inequalities that must all be true in any triangle. From them you built the complete interval for an unknown third side. You applied the interval to construction, sports spacing, navigation, and robotics. You learned to avoid pitfalls such as equality, unit mismatch, and partial checks. You also connected today’s side-based thinking to upcoming angle comparisons. With these ideas you can now judge feasibility and ranges with confidence.

14. References

• BYJU’S. Triangle Inequality Theorem.
• Encyclopedia Britannica. Triangle inequality.
• Math is Fun. Triangle Inequality Theorem.
• MathBitsNotebook. Triangle Inequality and related theorems.
• Big Ideas Math. Inequalities in Two Triangles, reference section.

Example in Action

Worked Examples (5)

Example 1 – Test if a triple forms a triangle

Do the lengths $7,9,17$ form a triangle?

Show Answer

Step 1: Largest side is $17$.
Step 2: Sum of the two smaller: $$7+9=16$$ Compare: $16>17$? False since $16<17$.
Conclusion: Does not form a triangle.

Example 2 – Detect a degenerate boundary

Do the lengths $10,24,34$ form a triangle?

Show Answer

Largest side $34$. Sum of others: $$10+24=34$$ Equality gives a degenerate case. Not a triangle for this lesson.

Example 3 – Find the full third-side range

Two sides are $8$ cm and $13$ cm. Find all possible values of the third side $x$. Also list the integer values that work.

Show Answer

Difference $|13-8|=5$, sum $13+8=21$ → interval $$5<x<21$$ Integers: $6,7,8,9,10,11,12,13,14,15,16,17,18,19,20$.

Example 4 – Variable inequality from the triangle condition

A triangle has two sides $5$ and $12$. The third side is $x$. Write and solve the inequality for $x$.

Show Answer

$$|12-5|<x<17$$ Therefore $7<x<17$.

Example 5 – Real-world feasibility with a set of choices

Two beams are fixed at $25$ cm and $32$ cm. Available third-piece lengths are $\{10,15,57,60\}$. Which choices create a proper triangle?

Show Answer

Interval: difference $|32-25|=7$, sum $57$ → $7<a<57$.
Valid: $10$, $15$. Invalid: $57$ (endpoint), $60$ (exceeds).

Try It Out

1. Triangle or not? $9,14,25$
   Show Answer

   $$9+14=23<25$$ Not a triangle.
2. Triangle or not? $7,7,13$
   Show Answer

   All sums are greater than the remaining side. Triangle.
3. Third-side interval with $b=11$, $c=4$
   Show Answer

   $7<a<15$
4. Count integer third sides when two sides are $15$ and $28$
   Show Answer

   Interval $13<a<43$ → integers 14 to 42 → count $29$.
5. Choose feasible values for $a$ with sides $10$ and $17$ from $\{5,7,8,26,27\}$
   Show Answer

   Interval $7<a<27$. Valid: $8$, $26$.
6. Smallest possible integer third side for sides $19$ and $3$
   Show Answer

   Interval $16<a<22$ → smallest integer $17$.
7. Largest possible integer third side for the same pair
   Show Answer

   Largest integer less than 22 is $21$.
8. Range for $x$ with sides $12$ and $15$
   Show Answer

   $3<x<27$.
9. Extreme perimeters with integer third side for sides $9$ and $10$
   Show Answer

   Third side integers 2 to 18. Min perimeter $21$, max $37$.
10. Spot the errors for sides $6$ and $7$ with proposed third sides 1 and 13
    Show Answer

    Interval $1<a<13$. Neither endpoint allowed. Both invalid.

Check Yourself

1. Which statement best expresses the Triangle Inequality Theorem?
   1. $a+b\ge c$ only
   2. $a+b>c$, $b+c>a$, and $a+c>b$
   3. $a>b+c$ for some side
   4. $a+b=c$ whenever the triangle is flat
   Show Answer

   B. All three strict inequalities must hold for a nondegenerate triangle.
2. Triangle or not? $8,15,23$
   Show Answer

   Equality at 8 + 15 = 23 → degenerate → not a triangle.
3. Triangle or not? $4.5,7.2,12.0$
   Show Answer

   4.5 + 7.2 = 11.7 < 12.0 → not a triangle.
4. Third-side interval for $b=9$, $c=14$
   Show Answer

   $5<a<23$.
5. Variable range for $x,11,18$
   Show Answer

   $7<x<29$.
6. Integer count with sides $10$ and $19$
   Show Answer

   Integers 10 to 28 inside $9<a<29$ → 19 values.
7. Set selection for sides $6$ and $9$ from {2, 4, 5, 6, 8, 14, 15}
   Show Answer

   Valid: {4, 5, 6, 8, 14}. Interval $3<a<15$.
8. City walking shortcut with legs 1.2 km and 0.9 km
   Show Answer

   $0.3<d<2.1$ km.
9. Error analysis for $13,5,7$
   Show Answer

   5 + 7 = 12 < 13 → not a triangle. The claim is correct.
10. Minimal adjustment for $20,3,16$
    Show Answer

    Need k > 1 → smallest integer 2. Examples: (20, 5, 16) or (20, 3, 18).
11. Extreme perimeters with integer third side for sides $12$ and $17$
    Show Answer

    Third side 6 to 28. Min perimeter 35. Max 57.
12. Identify the incorrect inequality among the four given
    Show Answer

    The non-strict version is incorrect for nondegenerate triangles.
13. Edge case flat arrangement: 1.8, 0.6, 1.2
    Show Answer

    0.6 + 1.2 = 1.8 → degenerate → not a triangle.
14. Variable with decimals for sides 5.5, x, 10.4
    Show Answer

    $4.9<x<15.9$.
15. Feasibility with external constraints: sides 0.9 m and 1.1 m, third side in [1.0, 1.8]
    Show Answer

    Triangle interval (0.2, 2.0). Intersection [1.0, 1.8) → feasible.

Go Further

1. Design Optimization – Shelf Bracket With Stock Lengths

   Two sides: $b=21.3\mathrm{cm}$, $c=14.7\mathrm{cm}$. Third side $a$ from 0.5 cm steps.

   Show Answer

   Continuous range: $6.6<a<36.0$ cm. Stock: 7.0 to 35.5 by 0.5. Nearest to 18.2 is 18.0 cm.
2. Integer Triples Under a Perimeter Cap

   Two sides 11 cm and 17 cm. Perimeter ≤ 40 cm.

   Show Answer

   Triangle integers 7 to 27. Perimeter cap ≤ 12. Survive: 7 to 12. Count 6.
3. Reverse Engineering – What Must a Known Side Be?

   Require 4 < a < 10, one known side 6, choose $b$ so an $a$ exists.

   Show Answer

   Need overlap of (|b − 6|, b + 6) with (4, 10). Gives 0 < b < 16. Integer range 1 to 15 works.
4. Measurement Uncertainty – Safe Third-Side Interval

   Nominal b = 7.0 ± 0.2 cm, c = 11.0 ± 0.2 cm. Choose $a$ to always form a triangle.

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   Safe interval: $4.4<a<17.6$ cm.
5. Maximize the Largest Side Under a Fixed Perimeter

   Integers with perimeter 30. Find maximum largest side.

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   Largest L satisfies 2L < 30 → L < 15. Max is 14. Example (14, 7, 9).

My Reflection

Task type: 3-2-1 reflection. Where to answer: Write in your notebook.

1. Write 3 insights you gained today about triangle inequality.
2. Describe 2 real-life applications where the triangle inequality helps you make a decision or check feasibility.
3. Ask 1 question you still have.

Formatting tip: Title your page “Day 1 Reflection”, include the date, and number your 3-2-1 entries clearly.