What Is the Altitude of a Triangle?
The altitude of a triangle is a line segment drawn from a vertex perpendicular to the line containing the opposite side. The point where it meets that side is the foot of the altitude, and the length of the segment from vertex to foot is the triangle's height.
Every triangle has three altitudes, one from each vertex, because any of the three sides can be treated as the base. The altitude always meets its base at a right angle (90°). Depending on the triangle, an altitude can fall inside the triangle, lie along a side, or fall outside it, which is a point we come back to for obtuse triangles.
Properties of the Altitude of a Triangle
The perpendicular-from-a-vertex definition forces a clear set of properties:
Three altitudes per triangle, one to each side.
Each altitude is perpendicular to its base, meeting it at exactly 90°.
The three altitudes meet at one point, the orthocentre (covered below).
An altitude need not bisect the base — this is the key difference from a median, which always does.
Position depends on the triangle's type. In an acute triangle all three altitudes are inside; in a right triangle two of them are the legs themselves; in an obtuse triangle two altitudes fall outside the triangle.
The General Altitude Formula
Because an altitude is the height in the area formula, the two are tied together directly. The area of a triangle is half the base times the height:
$$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}.$$
Rearranging for the height (the altitude to that base) gives the general formula that works for any triangle:
$$h = \frac{2 \times \text{Area}}{\text{base}}.$$
So if you know a triangle's area and the side you want the altitude to, one division gives the altitude. This is also why each side has its own altitude: a different base means a different height for the same fixed area.
Altitude Formulas by Triangle Type
Different triangle types give the altitude a cleaner closed form. Each formula below is just the general $h = 2A/b$ with the area worked out for that shape, so they are not separate rules to memorise but the same idea specialised.
Triangle type | Altitude formula | What the variables mean |
|---|---|---|
Scalene (any) | $h = \dfrac{2\sqrt{s(s-a)(s-b)(s-c)}}{b}$ | $a, b, c$ are the sides; $s = \tfrac{a+b+c}{2}$ is the semi-perimeter; altitude is to side $b$ (Heron's area) |
Isosceles | $h = \sqrt{a^2 - \dfrac{b^2}{4}}$ | $a$ is each equal side; $b$ is the base; altitude is from the apex |
Equilateral | $h = \dfrac{a\sqrt{3}}{2}$ | $a$ is the side length |
Right (to a leg) | the other leg | the two legs are altitudes of each other |
Right (to hypotenuse) | $h = \sqrt{xy}$ | $x, y$ are the two segments the foot makes on the hypotenuse |
The scalene case uses Heron's formula to get the area first, then divides by the base, which is why it looks the most involved. The equilateral case is the tidiest: a fixed multiple, $\tfrac{\sqrt{3}}{2}$, of the side.
What Is the Orthocentre?
A natural question once you have three altitudes: do they meet anywhere special? They do. The three altitudes of any triangle always pass through a single common point called the orthocentre (often written $H$). Where that point sits tells you about the triangle:
Acute triangle — the orthocentre lies inside the triangle.
Right triangle — the orthocentre sits exactly at the right-angle vertex, because two of the altitudes are the legs themselves.
Obtuse triangle — the orthocentre falls outside the triangle, since two altitudes have to be extended beyond the figure to meet.
Altitude vs Median: What Is the Difference?
These two are the most-confused pair of triangle segments, so it is worth pinning the distinction. Both run from a vertex to the opposite side, but they are built on different rules. An altitude is defined by an angle (it must be perpendicular); a median is defined by a point (it must hit the midpoint).
Feature | Altitude | Median |
|---|---|---|
Goes from a vertex to | the opposite side, at 90° | the midpoint of the opposite side |
Always perpendicular? | Yes | No, not usually |
Always bisects the base? | No | Yes, by definition |
Stays inside the triangle? | No (outside for obtuse) | Yes, always |
Three of them meet at | the orthocentre | the centroid |
They coincide only in special cases, for example the altitude from the apex of an isosceles triangle to its base is also the median to that base. (See our sibling article on the median of a triangle for the full treatment of medians.)
Examples of Altitude of a Triangle
With the formulas and the orthocentre in hand, here is the altitude doing real work. The problems build from the general formula up to Heron-based scalene and right-triangle cases.
Example 1: A triangle has an area of 24 cm² and a base of 6 cm. Find the altitude to that base
Use the general formula:
$$h = \frac{2A}{b} = \frac{2 \times 24}{6} = \frac{48}{6} = 8 \text{ cm}.$$
Final answer: 8 cm.
Example 2: Find the altitude to the base of an isosceles triangle with equal sides 10 cm and base 12 cm
A tempting first move is to use a side directly as the height, taking the altitude as 10 cm. Check that against the picture: the altitude is the vertical drop from the apex to the base, and it is one leg of a right triangle whose hypotenuse is the 10 cm equal side. A leg is always shorter than its hypotenuse, so the altitude must be less than 10. Calling it 10 treats the slant side as if it were vertical.
Done correctly with the isosceles formula:
$$h = \sqrt{a^2 - \frac{b^2}{4}} = \sqrt{10^2 - \frac{12^2}{4}} = \sqrt{100 - 36} = \sqrt{64} = 8 \text{ cm}.$$
Final answer: 8 cm.
Example 3: Find the altitude of an equilateral triangle with side 6 cm
The equilateral altitude is a fixed multiple of the side:
$$h = \frac{a\sqrt{3}}{2} = \frac{6\sqrt{3}}{2} = 3\sqrt{3} \approx 5.20 \text{ cm}.$$
Final answer: $3\sqrt{3} \approx 5.20$ cm.
Example 4: A scalene triangle has sides 7 cm, 8 cm, and 9 cm. Find the altitude to the 8 cm side
First the semi-perimeter and area (Heron's formula):
$$s = \frac{7 + 8 + 9}{2} = 12, \qquad A = \sqrt{12(12-7)(12-8)(12-9)} = \sqrt{12 \cdot 5 \cdot 4 \cdot 3} = \sqrt{720} \approx 26.83 \text{ cm}^2.$$
Then the altitude to the 8 cm base:
$$h = \frac{2A}{b} = \frac{2 \times 26.83}{8} \approx 6.71 \text{ cm}.$$
Final answer: about 6.71 cm.
Example 5: In a right triangle, the altitude from the right angle meets the hypotenuse and splits it into segments of 4 cm and 9 cm. Find the altitude
For the altitude to the hypotenuse, the height is the geometric mean of the two segments:
$$h = \sqrt{xy} = \sqrt{4 \times 9} = \sqrt{36} = 6 \text{ cm}.$$
Final answer: 6 cm.
Example 6: A triangle has sides 5 cm, 12 cm, and 13 cm. Find the altitude to the 13 cm side
These sides satisfy $5^2 + 12^2 = 13^2$, so it is a right triangle with legs 5 and 12 and hypotenuse 13. Its area is $\tfrac{1}{2}(5)(12) = 30$ cm². The altitude to the hypotenuse:
$$h = \frac{2A}{b} = \frac{2 \times 30}{13} = \frac{60}{13} \approx 4.62 \text{ cm}.$$
Final answer: about 4.62 cm.
Why the Altitude of a Triangle Matters
The altitude is more than a line in a diagram, it is the measurement that makes area possible and the structure that engineers lean on.
Area, for any triangle. Without a height, the area formula has nothing to multiply. The altitude is what turns "three sides" into "this much surface", whether for a plot of land, a sail, or a triangular gable.
Load paths in structures. In a truss or a triangular frame, the altitude marks the line along which a vertical load travels straight to the base. Roof trusses and bridge supports are sized using this vertical drop.
The orthocentre and triangle centres. The orthocentre joins the centroid, circumcentre, and incentre as one of a triangle's defining points, and three of them (orthocentre, centroid, circumcentre) always lie on one straight line, the Euler line, a result that surprised mathematicians when Euler proved it.
Coordinate geometry and surveying. Finding the shortest distance from a point to a line is the same operation as finding an altitude, the perpendicular drop, which is how surveyors compute offsets and how navigation systems find the nearest point on a route.
For a Grade 9 student, the altitude is the hinge between area, the Pythagorean theorem, and the family of triangle centres, learn it well and three later topics stop feeling separate.
Where Students Trip Up on Altitudes
Mistake 1: Using a slant side as the height
Where it slips in: In an isosceles or scalene triangle, the student takes one of the slanted sides as the altitude.
Don't do this: Plug a side length straight into the area formula as the height.
The correct way: The altitude is the perpendicular drop, not a side. It is always shorter than the slant sides that lean over it. Use $h = \sqrt{a^2 - b^2/4}$ for the isosceles case, or $h = 2A/b$ in general.
Mistake 2: Confusing the altitude with the median
Where it slips in: The student assumes the altitude hits the midpoint of the base, or that the median is perpendicular.
Don't do this: Treat altitude and median as the same segment.
The correct way: An altitude is perpendicular but need not bisect the base; a median bisects the base but need not be perpendicular. They coincide only in symmetric cases (the altitude from the apex of an isosceles or equilateral triangle). The second-guesser who knows both definitions still has to ask which rule the problem is using.
Mistake 3: Expecting every altitude to stay inside the triangle
Where it slips in: For an obtuse triangle, the student looks for the altitude inside the figure and cannot find it.
Don't do this: Assume the foot of every altitude lands on the drawn side.
The correct way: In an obtuse triangle, two altitudes fall outside: the base must be extended as a dashed line, and the perpendicular drops to that extension. The altitude still exists, the foot just sits beyond the original side.
Key Takeaways
The altitude of a triangle is the perpendicular segment from a vertex to the opposite side, and its length is the height in the area formula.
The general formula is $h = 2A/b$; special types have cleaner forms (equilateral $\tfrac{a\sqrt{3}}{2}$, isosceles $\sqrt{a^2 - b^2/4}$).
Every triangle has three altitudes that meet at the orthocentre — inside an acute triangle, on the vertex of a right triangle, outside an obtuse one.
An altitude is perpendicular but need not bisect the base, which is what separates it from a median.
The most common error is using a slant side as the height; the altitude is always shorter than the sides that lean over it.
Practice These Problems to Solidify Your Understanding
A triangle has area 45 cm² and base 9 cm. Find the altitude to that base.
Find the altitude of an equilateral triangle with side 10 cm.
An isosceles triangle has equal sides 13 cm and base 10 cm. Find the altitude to the base.
Answer to Question 1: altitude = 10 cm. Answer to Question 2: altitude $= 5\sqrt{3} \approx 8.66$ cm. Answer to Question 3: altitude = 12 cm. If Question 3 gave 13 cm, check that you did not use the equal side as the height (see Mistake 1).
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