How to Find the Missing Side of a Triangle From Its Perimeter

The short answer

c = P − a − b

Subtract the two known sides from the perimeter. A perimeter of 30 with sides of 12 and 9 leaves 30 − 12 − 9 = 9 for the third side. Then, before using the answer, confirm the three lengths satisfy the triangle inequality: here 9 + 9 = 18 is greater than 12, so the triangle closes.

The perimeter is just the three sides added up, so recovering a missing side is one subtraction. What makes this worth a full page is the step people skip afterward. A subtraction always produces a number, but a number is not automatically a triangle. The three lengths must also be able to close, and checking that takes five seconds against the cost of designing around a shape that cannot exist.

The subtraction and its silent assumption

P = a + b + c is the whole theory. Rearranged, c = P − a − b, and there is genuinely nothing more to the arithmetic. The silent assumption is that a triangle with these three sides exists at all. The subtraction does not know geometry; it will cheerfully report a third side of 2 for a perimeter of 30 with sides 20 and 8, and no triangle with sides 20, 8, 2 can close, because 8 + 2 falls short of 20.

The gate is the triangle inequality: every pair of sides must sum to more than the third, and in practice you only need to test the longest side against the other two. Pass the gate and the shape exists. Fail it and the perimeter is inconsistent with the two given sides, full stop.

1

Subtract

c = P − a − b. For P = 30, a = 12, b = 9, the third side is 9.

2

Gate

Test the longest side: 9 + 9 = 18 exceeds 12, so the lengths close.

3

Reuse

Only now treat 12, 9, 9 as a real triangle: Heron gives its area, cosines give its angles.

The perimeter as a bar

The diagram lays the perimeter of 30 out as a bar and carves off the two known sides, leaving the third as whatever remains. Underneath, the recovered 12, 9, 9 triangle is drawn closed, with the inequality check that entitles it to exist.

A bar of length 30 split into segments of 12, 9 and an unknown that works out to 9, above the finished 12, 9, 9 triangle and the triangle inequality check that 9 plus 9 is greater than 12.
The bar makes the subtraction obvious. The inequality line underneath is the part that separates arithmetic from geometry.

A number is not a triangle

The subtraction succeeds even when the geometry fails. The inequality gate is what tells the two cases apart.

30 − 20 − 8givesno gate check2

Sides 20, 8, 2 cannot close: 8 + 2 = 10 falls short of 20.

30 − 12 − 9givesgate passed9

Sides 12, 9, 9 close comfortably: 18 exceeds 12.

A worked check you can repeat on paper

A fenced triangular yard uses 30 m of fencing. Two runs measure 12 m and 9 m; the third was never taped. The missing run is 30 − 12 − 9 = 9 m. Gate check: the longest side is 12, and 9 + 9 = 18 is greater, so the yard is a legitimate triangle, an isosceles one as it happens.

Push on to make the answer earn its keep. Heron on 12, 9, 9: s = 15, A = √(15 × 3 × 6 × 6) = √1620 = 40.25 m². The subtraction found a length; the gate certified it; Heron turned it into an area for seeding or paving. That chain is the normal life of this calculation.

Now a failing case, deliberately. Perimeter 30, sides 20 and 8. The subtraction says 2. The gate says 8 + 2 = 10 against 20: fail. The correct conclusion is not a triangle with a side of 2, it is that at least one of the three given numbers is wrong.

Special perimeters worth recognizing

Two patterns save time. An equilateral triangle has all sides equal, so each is the perimeter over three, no subtraction needed: perimeter 30 means sides of 10. An isosceles triangle with a known base and unknown equal legs splits the remainder in half: perimeter 30 with a base of 12 leaves 18 for two legs, 9 each, which is exactly the worked example above approached from a different given.

The isosceles version carries its own gate: each leg must exceed half the base, or the two legs cannot reach each other over it. Legs of 9 against half a base of 6 pass easily.

Mistakes that catch people out

  • Skipping the inequality gate. The subtraction always answers; the geometry does not always exist. Test the longest side against the sum of the other two before using the result.
  • Subtracting only one side. P minus one side is the sum of the other two, not the missing one. Both known sides come off.
  • Mixing units. A perimeter in meters with one side quoted in feet produces a confidently wrong third side. Convert first, subtract second.
  • Treating a degenerate result as valid. If the third side comes out exactly equal to the sum or difference of the others, the “triangle” is a flat line segment. Zero area, no interior, not usable.
  • Forgetting the perimeter includes the unknown. Occasionally a problem quotes the sum of only the two known sides. Read what the number actually covers before rearranging.

Try these

  1. P = 42, sides 15 and 13. Find the third side and run the gate.
  2. P = 24, equilateral. Find each side.
  3. P = 40, isosceles with base 16. Find each leg.
  4. P = 30, sides 20 and 8. What is the correct conclusion?
  5. P = 36, sides 10 and 8. Find the third side, then the area by Heron.
Show the answers
  1. 42 − 28 = 14; 13 + 14 = 27 exceeds 15, closes
  2. 24 / 3 = 8
  3. (40 − 16)/2 = 12 each; 12 exceeds 8, closes
  4. third side 2 fails the gate, the data is inconsistent
  5. third side 18; s = 18 makes s − c = 0, degenerate, flat line, no area

Common questions

How do you find the missing side of a triangle from the perimeter?

Subtract the two known sides from the perimeter. The perimeter is simply the three sides added together, so what remains after removing two of them is the third.

Do I need to check anything after subtracting?

Yes, the triangle inequality. The two shorter sides must sum to more than the longest one, or no triangle with those three lengths exists and the given data is inconsistent.

What if the subtraction gives zero or a negative number?

Then the two known sides already meet or exceed the whole perimeter, which is impossible for a real triangle. One of the three given numbers is wrong.

How does it work for an equilateral triangle?

Every side is the perimeter divided by three. No subtraction is needed because all three sides are the same length.

How does it work for an isosceles triangle?

Subtract the base from the perimeter and halve what remains to get each equal leg. Each leg must also exceed half the base for the shape to close.

Can I get the area from the perimeter alone?

No. Triangles with the same perimeter can have very different areas. You need all three sides, and then Heron turns them into the area.

Related lessons and calculators