An altitude is the perpendicular segment from a vertex of a polygon to the opposite side. In a triangle, the altitude is also called the height when the side is taken as the base.
The Formal Definition of Altitude
In geometry, an altitude of a triangle is a line segment from a vertex perpendicular to the line containing the opposite side. The foot of the altitude is the point where this segment meets the opposite side (or its extension, in the case of obtuse triangles).
Every triangle has three altitudes — one from each vertex. The three altitudes (or their extensions) meet at a single point called the orthocenter.
For obtuse triangles, the foot of one or two altitudes falls outside the triangle — the altitude is drawn to the extension of the opposite side, and the orthocenter lies outside the triangle too. For right triangles, the two legs are altitudes of each other, and the orthocenter is at the right-angle vertex.
Quick reference.
Definition: perpendicular segment from a vertex to the opposite side.
Symbol: $h$ (for height) or $h_{a}, h_{b}, h_{c}$ for altitudes to sides $a, b, c$.
Every triangle has: $3$ altitudes.
Concurrence point: the orthocenter.
Area formula: $A = \tfrac{1}{2} \times b \times h$.
Grade introduced: CCSS-M 6.G.A.1 (triangle area); NCERT Class 7 — Triangle Geometry.
The Three Altitudes and the Orthocenter
For a triangle $ABC$:
The altitude from $A$ is perpendicular to side $BC$.
The altitude from $B$ is perpendicular to side $AC$.
The altitude from $C$ is perpendicular to side $AB$.
These three altitudes meet at a single point — the orthocenter of the triangle. The orthocenter's position tells you a lot about the triangle:
Triangle type | Orthocenter location |
|---|---|
Acute (all angles $< 90°$) | Inside the triangle |
Right (one angle $= 90°$) | At the right-angle vertex |
Obtuse (one angle $> 90°$) | Outside the triangle |
Altitude Formulas for Common Triangles
These three come up most often in CBSE and Common Core problems.
Equilateral triangle (side $a$). All three altitudes have the same length:
$$h = \frac{a\sqrt{3}}{2}.$$
Isosceles triangle (equal sides $a$, base $b$). The altitude from the apex (the vertex between the equal sides) to the base:
$$h = \sqrt{a^{2} - \frac{b^{2}}{4}}.$$
Right triangle (legs $a, b$ and hypotenuse $c$). The two legs are altitudes of each other ($h_{a} = b$ and $h_{b} = a$). The altitude from the right-angle vertex to the hypotenuse is:
$$h_{c} = \frac{ab}{c}.$$
General triangle (sides $a, b, c$). Use the area in two ways. If the area is $A$ (computed by Heron's formula or otherwise), the altitude to side $a$ is:
$$h_{a} = \frac{2A}{a}.$$
Altitude in the Area Formula
The most common use of altitude:
$$\text{Area of a triangle} = \frac{1}{2} \times \text{base} \times \text{altitude}.$$
The "base" can be any of the three sides; the "altitude" is the perpendicular from the opposite vertex to that chosen side. The area is the same regardless of which side you pick as the base — but the altitude changes accordingly.
Three Worked Examples of Altitude — Quick, Standard, Stretch
Quick. Find the altitude of an equilateral triangle with side $6$ cm.
Use $h = \dfrac{a\sqrt{3}}{2}$ with $a = 6$:
$$h = \frac{6\sqrt{3}}{2} = 3\sqrt{3} \approx 5.20 \text{ cm}.$$
Final answer: $h = 3\sqrt{3} \approx 5.20$ cm.
Standard (Wrong Path First — The Mistake Worth Making Once). A triangle has a base of $10$ cm and an altitude of $8$ cm. Find its area.
The wrong path. A student multiplies $10 \times 8 = 80$ cm² and stops.
The flaw: the area formula is $A = \tfrac{1}{2} \times b \times h$, not $b \times h$. Multiplying without the $\tfrac{1}{2}$ gives the area of the rectangle with the same base and height — which is exactly twice the triangle's area.
The rescue. Apply the correct formula:
$$A = \tfrac{1}{2} \times 10 \times 8 = 40 \text{ cm}^{2}.$$
Final answer: $40$ cm².
The lesson — the $\tfrac{1}{2}$ in the triangle area formula is non-negotiable. Two right triangles glued along the hypotenuse make a rectangle — so a single triangle is half the rectangle's area.
Stretch. A right triangle has legs of $6$ cm and $8$ cm. Find the altitude from the right-angle vertex to the hypotenuse.
Hypotenuse $c = \sqrt{6^{2} + 8^{2}} = \sqrt{100} = 10$ cm.
Use $h_{c} = \dfrac{ab}{c}$:
$$h_{c} = \frac{6 \times 8}{10} = \frac{48}{10} = 4.8 \text{ cm}.$$
Sanity check: area via legs $= \tfrac{1}{2} \times 6 \times 8 = 24$ cm². Area via hypotenuse and altitude $= \tfrac{1}{2} \times 10 \times 4.8 = 24$ cm². ✓
Final answer: $h_{c} = 4.8$ cm.
This is the version of altitude problem that shows up in CBSE Class 9–10 and CCSS-M HSG-SRT geometry.
Where Altitude Appears — Beyond the Triangle
A few places this idea is the foundation under a larger result:
Trigonometry. In any triangle, $h = a \sin B = b \sin A$ — the relationship between altitude and angle is what builds the sine rule.
Surveying. A surveyor with a theodolite measures heights of buildings and mountains by triangulation — the altitude is what's being computed.
Astronomy. The "altitude" of a star above the horizon is also a triangle altitude — measured from the observer to the star.
Mt Everest's height. First measured in $1856$ by George Everest's Great Trigonometrical Survey using a chain of triangle altitudes from baseline measurements in the Indian plains.
Computer graphics. Z-buffering in 3D rendering uses altitudes from each pixel to triangles in the scene.
The earliest systematic treatment of triangle altitudes appears in Euclid's Elements Book I (c. 300 BCE). The orthocenter — the concurrent point of the three altitudes — was studied by Archimedes (c. 287–212 BCE) and is one of the four classical "triangle centres" along with the centroid, incenter, and circumcenter.
Tripping Points to Avoid with Altitude
Mistake 1: Forgetting the $\tfrac{1}{2}$ in the area formula
Where it slips in: Computing area as base $\times$ height.
Don't do this: Treat the triangle area like a rectangle area.
The correct way: $A = \tfrac{1}{2} \times b \times h$. A triangle is half its surrounding rectangle.
Mistake 2: Confusing altitude with side length
Where it slips in: In an isosceles triangle with equal sides $5$ cm and base $6$ cm, student says "altitude $= 5$."
Don't do this: Use a side as the altitude when it isn't perpendicular to the base.
The correct way: Apply $h = \sqrt{a^{2} - (b/2)^{2}}$: $h = \sqrt{25 - 9} = 4$ cm. The altitude is shorter than the equal side because it's the leg of a right triangle whose hypotenuse is the equal side.
Mistake 3: Drawing the altitude inside an obtuse triangle when it lies outside
Where it slips in: Obtuse triangle, altitude from one of the acute-angle vertices.
Don't do this: Force the altitude inside the triangle.
The correct way: For obtuse triangles, two of the three altitudes fall outside — the perpendicular drops to the extension of the opposite side, not the side itself. The orthocenter ends up outside the triangle.
Conclusion
An altitude is the perpendicular line segment from a vertex to the opposite side of a polygon.
A triangle has three altitudes; they meet at the orthocenter.
The triangle area formula $A = \tfrac{1}{2} \times b \times h$ uses altitude as the height.
For an equilateral triangle of side $a$, the altitude is $\tfrac{a\sqrt{3}}{2}$.
For obtuse triangles, two altitudes lie outside the triangle and the orthocenter is outside too.
The most common slip is dropping the $\tfrac{1}{2}$ in the area formula or confusing altitude with the equal side of an isosceles triangle.
Quick Self-Check — Three Problems
Find the altitude of an equilateral triangle with side $10$ cm.
A triangle has base $14$ cm and altitude $9$ cm. Find its area.
An isosceles triangle has equal sides $13$ cm and base $10$ cm. Find the altitude to the base.
If problem 2 gave $126$ cm², return to Mistake 1 above.
Want a live Bhanzu trainer to walk your child through triangle geometry, altitudes, and area? Book a free demo class — online globally.
Was this article helpful?
Your feedback helps us write better content