How to Find All Three Heights of a Triangle

Geometry Lessons

The short answer

h = 2A / base

Compute the area once from the three sides with Heron, then divide twice that area by each side in turn. The 13, 14, 15 triangle has area 84, so its heights are 168/13 = 12.923, 168/14 = 12.0 and 168/15 = 11.2. The longest side always carries the shortest height.

A triangle has three altitudes, one from each vertex perpendicular to the opposite side, and problems that need one of them often end up needing all three. The wasteful approach solves three separate right triangle problems. The efficient one notices that all three altitudes are tied together by a single number, the area, and computes that number once.

One area, three heights

The area rule A = ½ × base × height holds no matter which side plays the base. Rearranged, h = 2A / base. So the moment the area is known, every height is one division away. The area is the exchange rate between bases and heights, and it never changes for a given triangle.

This also explains a pattern worth internalizing: heights are inversely proportional to their bases. The longest side gets the shortest altitude and the shortest side gets the tallest one, because each pairing must multiply out to the same doubled area. If your computed heights do not sort opposite to the sides, something is already wrong.

1

Heron

From sides 13, 14, 15: s = 21, A = √(21 × 8 × 7 × 6) = √7056 = 84.

2

Double

Twice the area is 168, and this number feeds all three divisions.

3

Divide

168/13 = 12.923, 168/14 = 12.0, 168/15 = 11.2. Three heights, one area.

The same triangle, three ways up

The diagram draws the 13, 14, 15 triangle three times, resting on a different side each time with that altitude dashed in. The triangle never changes, only its orientation does, and each base and height pair rebuilds the same 84.

The 13, 14, 15 triangle drawn three times, each time resting on a different base with its dashed altitude labeled 12.0, 11.2 and 12.92, under the note that the Heron area of 84 is used for all three.
Rotating the triangle changes which side is down, nothing else. Every base and height pair multiplies back to the same doubled area of 168.

Double before dividing

The area rule carries a half, so recovering a height needs the area doubled first. Dividing the raw area by the base gives exactly half the true height.

84 / 14giveshalf missing6.0

That is half the altitude. The half in the area formula was never undone.

2 × 84 / 14givesarea doubled12.0

Undo the half first, then divide by the base you care about.

A worked check you can repeat on paper

Heron first. The semi perimeter of the 13, 14, 15 triangle is (13 + 14 + 15)/2 = 21. The area is √(21 × (21 − 13) × (21 − 14) × (21 − 15)) = √(21 × 8 × 7 × 6) = √7056 = 84 exactly.

Now the three divisions: 168/13 = 12.923, 168/14 = 12.000, 168/15 = 11.200. Cross check one of them independently by coordinates. Put the side of length 14 on the ground. The foot of the apex sits at (14² + 13² − 15²)/(2 × 14) = 5 from the left end, and the apex height is √(13² − 5²) = √144 = 12. It matches the middle height on the nose, which vouches for the other two, since all three came from the same area.

Final habit check: rebuild the area from each pairing. ½ × 13 × 12.923 = 84.0, ½ × 14 × 12 = 84, ½ × 15 × 11.2 = 84. Three closings, no surprises.

Where the three altitudes meet

Drawn as full lines, the three altitudes pass through one common point, the orthocenter. In an acute triangle it sits inside; in an obtuse one it falls outside, because two of the altitude feet land on extensions of the sides rather than the sides themselves. The heights computed here are unaffected: h = 2A/base measures the perpendicular distance whether or not the foot lands inside the segment.

That external foot case matters in practice. For an obtuse triangle, physically measuring an altitude with a tape can require projecting a side beyond its endpoint. The arithmetic never needs the projection, which is a genuine advantage of computing heights from the area instead of measuring them.

Mistakes that catch people out

  • Forgetting the doubling. h = 2A/base, not A/base. The half in the area formula must be undone, or every height comes out exactly half size.
  • Pairing a height with the wrong side. Each altitude belongs to the side it lands on, not the vertex it leaves. Sorting the answers is the check: longest side, shortest height, always.
  • Recomputing the area three times. Not wrong, just wasteful, and three chances to mistype Heron instead of one. Compute 2A once and reuse it.
  • Expecting the altitude foot inside the side. In an obtuse triangle two feet fall on extensions. The height value is still correct as the perpendicular distance.
  • Using a rounded area for all three heights. Round at the end, not in the middle. A truncated 84.0 costs nothing here, but a Heron area rounded from 83.97 to 84 in a scalene case pushes a consistent bias into all three heights.

Try these

  1. A triangle has area 30 and a side of 12. Find the height onto that side.
  2. Sides 3, 4, 5. Find all three heights.
  3. Sides 13, 14, 15. Which side carries the shortest altitude?
  4. A triangle has heights 12 and 8 onto sides of 10 and x. Find x.
  5. Area 84, height 11.2. Which base does that height belong to?
Show the answers
  1. 2 × 30 / 12 = 5
  2. area 6, so 4.0, 3.0 and 2.4
  3. the 15 side, at 11.2
  4. same area both ways, so x = 120/8 = 15
  5. 168 / 11.2 = 15

Common questions

What is the formula for the height of a triangle onto a given side?

Twice the area divided by that side. Every side has its own matching height, and all three share the same area.

How do I get the area if I only know the three sides?

Heron. Halve the perimeter to get s, then the area is the square root of s times s minus a times s minus b times s minus c. For sides 13, 14 and 15 that gives exactly 84.

Why is the tallest height on the shortest side?

Because every base and height pair must multiply to the same doubled area. A smaller base needs a larger height to keep the product fixed, and the sort orders always run opposite.

Do the three altitudes really meet at one point?

Yes, at the orthocenter. It sits inside an acute triangle, on the right angle vertex of a right triangle, and outside an obtuse triangle.

Can an altitude land outside the triangle?

Its foot can. In an obtuse triangle the altitudes to the two shorter sides meet the extensions of those sides. The computed height is still the true perpendicular distance.

Is there a height version for the hypotenuse of a right triangle?

Yes, and it has an extra shortcut: the altitude to the hypotenuse equals the product of the legs divided by the hypotenuse, which agrees with twice the area over the hypotenuse.

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