Any two sides must beat the third.
a + b > c, for every pairing.
Picture the two shorter sides swinging down like hinges. If together they are shorter than the base, they never touch. If they exactly equal it, they lie flat — a line, not a triangle.
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Triangle Inequality Theorems
Two short sides must still reach across the long one, or there is no triangle.
Find the range
Two sides of a triangle measure 7 cm and 10 cm. Between which values must the third side lie?
The third side must be less than the sum (10 + 7 = 17) and more than the difference (10 − 7 = 3), so 3 < x < 17. D uses ≤, but at exactly 3 or 17 the three sides lie flat in a line and form no triangle at all. B forgets the lower bound — a side of 1 cm could never reach between the ends of a 10 cm side and a 7 cm side.
Testing three given lengths.
Only one check matters: the two shortest against the longest.
4, 6, 9
4 + 6 = 10 > 9
a triangle ✓
3, 5, 9
3 + 5 = 8 < 9
impossible ✗
6, 6, 12
6 + 6 = 12
flat, still impossible ✗
If the two smallest beat the largest, the other two pairings pass automatically — so one comparison settles it.
The range for a missing side.
difference < third side < sum
sides 7 and 10
find the range for x
upper: 7 + 10 = 17
x must be less
lower: 10 − 7 = 3
x must be more
3 < x < 17
strictly between
Strictly between — never equal
x = 17 gives a flat line and x = 3 does too. The endpoints are excluded, which is why the signs are < and not ≤.
Bigger angle, longer side opposite.
∠A = 50°, ∠B = 58°, ∠C = 72°
which side is longest?
∠C is largest
so side c, opposite it, is longest
a < b < c
same order as the angles
The rule runs both ways: the longest side is opposite the largest angle, and the largest angle is opposite the longest side.
A side is named with the lower-case letter matching the angle it faces. Side a is opposite angle A — which is what makes "opposite" easy to track.
An exterior angle beats either remote interior angle.
exterior angle = sum of the two remote interiors
from the angle-sum rule
so it is greater than each of them
since both are positive
exterior 110° = 45° + 65°
and 110 > 65 > 45 ✓
This is not a separate fact to memorise. A sum of two positive numbers is bigger than either one, so the inequality follows immediately from the equality.
Counting whole-number sides, and naming the opposite angle.
The angle opposite a side is at the vertex whose letter is NOT in the side's name.
Two sides of a triangle are 5 and 13. How many whole-number lengths can the third side have?
13 − 5 = 8 < x < 18 = 13 + 5
the range
whole numbers: 9, 10, …, 17
ends excluded
17 − 9 + 1 = 9 values
count inclusively
18 − 8 is not the count
Subtracting the bounds gives 10, but neither 8 nor 18 is allowed, and 18 − 8 counts gaps, not numbers. List the first and last allowed values (9 and 17), then take last − first + 1.
In triangle PQR, PQ = 9, QR = 4 and RP = 7. Order the angles from smallest to largest.
PQ = 9 is longest → opposite ∠R
R is not in "PQ"
QR = 4 is shortest → opposite ∠P
P is not in "QR"
RP = 7 → opposite ∠Q
the middle one
∠P < ∠Q < ∠R
same order as the sides
Variations: decimals, a side with a variable, an exterior angle in x.
The same two rules with less friendly numbers.
Can 2.5 cm, 3.5 cm and 6.5 cm form a triangle?
2.5 + 3.5 = 6
the two shortest
6 < 6.5
short of the longest
No triangle. Decimals change nothing: still two shortest against the longest.
A triangle has sides x, x + 2 and 10, with 10 the longest. What must x be?
x + (x + 2) > 10
the two shorter sides must beat 10
2x + 2 > 10, x > 4
solve the inequality
Check x = 5: sides 5, 7, 10 and 5 + 7 = 12 > 10 ✓. At x = 4 the sides 4, 6, 10 lie flat.
An exterior angle measures (3x + 10)°. Its remote interior angles are 40° and (x + 20)°. Find the exterior angle.
3x + 10 = 40 + (x + 20)
exterior = sum of remote interiors
2x = 50, x = 25
exterior = 3(25) + 10 = 85°
Check: 40 + 45 = 85 ✓, and 85° is bigger than each remote angle, as the inequality says.
On the exam
Items ask whether three lengths can form a triangle, for the range or the number of whole-number values of a third side, or which side or angle is largest. The fastest route: two shortest against the longest; difference < x < sum; and the angle opposite a side is the vertex letter missing from its name. For a 10-second check, remember the endpoints are never allowed — equal to the sum means a flat line.
Now you try
Practice problem
Which set works?
Which of these could be the side lengths of a triangle?
Compare the two shortest with the longest each time. B: 5 + 7 = 12 > 11 ✓ — a triangle. A: 2 + 3 = 5 < 6, too short. C: 4 + 4 = 8, exactly equal, so the sides lie flat in a line. D: 1 + 9 = 10, also flat. Only B leaves any room to form a shape.
Hint: add the two smallest numbers in each set and compare with the biggest. Equal is not good enough.