Altitude of a Triangle

An altitude of a triangle is the perpendicular segment from a vertex to the line that contains the opposite side. The point where it meets this line is the foot of the altitude.

In the figure, $CC_1$ is the altitude from the vertex $C.$ It is perpendicular to $AB,$ and $C_1$ is its foot. Its length is written $h_c,$ the altitude to the side $c.$ In the same way $h_a$ and $h_b$ are the altitudes to the sides $a$ and $b.$ Every triangle has three altitudes.

The altitude is the distance from the vertex to the line of the opposite side: it is the shortest of all segments that join the vertex to that line. That is why the altitude is also called the height of the triangle, and why it appears in the area formula

${S=\frac12\,a\,h_a=\frac12\,b\,h_b=\frac12\,c\,h_c}.$

In an acute triangle the three feet lie on the sides. In an obtuse triangle two of the feet lie on the extensions of the sides. That is why the definition speaks about the line that contains the side.

On this page: Orthocenter · Acute triangle · Right triangle · Obtuse triangle · Formulas · Solved problems · Common mistakes · Practice

Where Do the Altitudes Meet? The Orthocenter

The three altitudes of a triangle, or the lines that contain them, meet at one point. This point is the orthocenter of the triangle, usually marked $H.$

Where the orthocenter lies depends on the type of the triangle.

Altitudes of an acute triangle

All three altitudes lie inside the triangle, and the orthocenter $H$ is an interior point. Two useful facts about the angles:

• ${\angle BAA_1=90^\circ-\beta},$ because triangle $ABA_1$ has a right angle at $A_1.$ In the same way ${\angle CAA_1=90^\circ-\gamma},$ ${\angle ABB_1=90^\circ-\alpha}$ and so on.
• ${\angle AHB=180^\circ-\gamma},$ ${\angle BHC=180^\circ-\alpha}$ and ${\angle CHA=180^\circ-\beta}.$ Indeed, the quadrilateral $CB_1HA_1$ has right angles at $A_1$ and $B_1,$ so ${\angle A_1HB_1}={360^\circ-90^\circ-90^\circ-\gamma}={180^\circ-\gamma},$ and ${\angle AHB=\angle A_1HB_1}$ as vertical angles.

Altitudes of a right triangle

Let ${\gamma=90^\circ}.$ The legs are perpendicular, so the altitude from $A$ is the leg $AC,$ and the altitude from $B$ is the leg $BC.$ Both pass through $C,$ so the orthocenter is the vertex of the right angle: ${H=C}.$ Only the third altitude, $CC_1$ to the hypotenuse, has to be drawn. Its length is ${h_c=\frac{ab}{c}}.$

Altitudes of an obtuse triangle

Let the angle at $C$ be obtuse. The altitude from $C$ lies inside the triangle, but the altitudes from $A$ and $B$ lie outside it: their feet $A_1$ and $B_1$ lie on the extensions of the sides $BC$ and $AC$ beyond $C.$ The altitudes themselves do not meet. Their lines meet at the orthocenter $H,$ which lies outside the triangle, beyond the vertex of the obtuse angle.

A nice fact: in this figure $C$ is the orthocenter of triangle $ABH.$ Of the four points $A,$ $B,$ $C$ and $H,$ each one is the orthocenter of the triangle formed by the other three.

Formulas for the Altitudes

We use the standard notation: the sides $a,$ $b,$ $c,$ the angles $\alpha,$ $\beta,$ $\gamma,$ the area $S,$ the semiperimeter ${p=\frac{a+b+c}{2}},$ the radius $R$ of the circumscribed circle and the radius $r$ of the inscribed circle.

Known	Formula	Where it comes from
The area and the side	${h_a=\frac{2S}{a}}$	${S=\frac12\,a\,h_a}$
The three sides	${h_a}={\frac{2\sqrt{p(p-a)(p-b)(p-c)}}{a}}$	Heron's formula for the area
A side and an angle	${h_a=b\sin\gamma=c\sin\beta}$	the right triangles $AA_1C$ and $AA_1B$
Two sides and $R$	${h_a=\frac{bc}{2R}}$	${S=\frac{abc}{4R}}$

The formulas for $h_b$ and $h_c$ are the same with the letters changed, for example ${h_c}={\frac{2S}{c}}={a\sin\beta}={b\sin\alpha}={\frac{ab}{2R}}.$

Altitudes and sides

One more relation connects the altitudes with the radius of the inscribed circle:

${\frac{1}{h_a}+\frac{1}{h_b}+\frac{1}{h_c}=\frac1r}.$

Indeed, ${\frac{1}{h_a}=\frac{a}{2S}},$ so the sum equals ${\frac{a+b+c}{2S}=\frac{2p}{2S}=\frac{p}{S}=\frac1r},$ because ${S=p\,r}.$

Altitudes of special triangles

• Equilateral triangle with side $a.$ All three altitudes are equal, ${h=\frac{a\sqrt3}{2}}.$
• Isosceles triangle with legs $b$ and base $c.$ The altitude to the base is ${h_c=\sqrt{b^2-\frac{c^2}{4}}};$ it is also the median and the angle bisector. The altitudes to the two legs are equal.
• Right triangle with legs $a,$ $b$ and hypotenuse $c.$ Two of the altitudes are the legs, and the altitude to the hypotenuse is ${h_c=\frac{ab}{c}}.$ It divides the hypotenuse into the segments ${m=AC_1}$ and ${n=C_1B},$ and ${h_c^2=m\cdot n}$ (on the page Triangles the foot $C_1$ is called $H$).

Solved Problems

Problem 1. Find the three altitudes of the triangle with sides 13, 14 and 15.

Solution. ${p=21}$ and, by Heron's formula, ${S=\sqrt{21\cdot 8\cdot 7\cdot 6}=84}.$ So

${h_{13}=\frac{168}{13}\approx 12.92},$ ${h_{14}=\frac{168}{14}=12},$ ${h_{15}=\frac{168}{15}=11.2}.$

As expected, the longest altitude goes to the shortest side.

Problem 2. The legs of a right triangle are 6 cm and 8 cm. Find the altitude to the hypotenuse.

Solution. ${c=\sqrt{36+64}=10}$ cm, so ${h_c=\frac{6\cdot 8}{10}=4.8}$ cm.

Problem 3. The legs of an isosceles triangle are 10 cm and its base is 12 cm. Find the altitude to the base and the altitude to a leg.

Solution. The altitude to the base is ${\sqrt{10^2-6^2}=8}$ cm, so the area is ${\frac12\cdot 12\cdot 8=48}$ cm². The altitude to a leg is ${\frac{2S}{10}=\frac{96}{10}=9.6}$ cm.

Problem 4. In an acute triangle ${\alpha=70^\circ}$ and ${\beta=60^\circ}.$ The altitudes $AA_1$ and $BB_1$ meet at $H.$ Find ${\angle AHB}$ and ${\angle BAA_1}.$

Solution. ${\gamma}={180^\circ-70^\circ-60^\circ}={50^\circ},$ so ${\angle AHB}={180^\circ-50^\circ}={130^\circ}.$ In the right triangle $ABA_1$ we get ${\angle BAA_1}={90^\circ-60^\circ}={30^\circ}.$

Problem 5. Two sides of a triangle are 8 cm and 12 cm, and the altitude to the side 8 cm is 6 cm. Find the altitude to the side 12 cm.

Solution. ${2S=8\cdot 6=48},$ so the altitude to the side 12 cm is ${\frac{48}{12}=4}$ cm.

Problem 6. In triangle $ABC$ the side ${b=10}$ cm and ${\gamma=30^\circ}.$ Find the altitude $h_a.$

Solution. ${h_a=b\sin\gamma=10\cdot\frac12=5}$ cm.

Problem 7. The sides of a triangle are in the ratio ${3:4:6}.$ Find the ratio of its altitudes.

Solution. ${h_a:h_b:h_c=\frac13:\frac14:\frac16}.$ Multiplying by 12 we get ${4:3:2}.$

Common Mistakes

Mistake	Wrong	Right
An altitude is always inside the triangle	The foot of every altitude lies on a side	In an obtuse triangle two feet lie on the extensions of the sides
Altitude and median mixed up	The altitude goes to the midpoint of the opposite side	That is the median. They coincide only for the altitude to the base of an isosceles triangle
The longest altitude to the longest side	Sides 5, 12, 13: the longest altitude is the one to 13	${a\,h_a=2S}$ is the same for all sides, so the longest altitude goes to the shortest side, 5
The altitude to the hypotenuse taken as half the hypotenuse	Legs 6 and 8, so ${h_c=\frac{10}{2}=5}$	Half the hypotenuse is the median. The altitude is ${h_c=\frac{6\cdot 8}{10}=4.8}$
The orthocenter of a right triangle	It is the midpoint of the hypotenuse	It is the vertex of the right angle. The midpoint of the hypotenuse is the center of the circumscribed circle

Practice Problems

Try them yourself, then open the answer.

1. A side of a triangle is 10 cm and the altitude to it is 7 cm. Find the area.

${S=\frac12\cdot 10\cdot 7=35}$ cm²

2. The area of a triangle is 24 cm² and one side is 8 cm. Find the altitude to this side.

${h=\frac{2\cdot 24}{8}=6}$ cm

3. Find the altitude of an equilateral triangle with side 4 cm.

${h=\frac{4\sqrt3}{2}=2\sqrt3\approx 3.46}$ cm

4. The legs of a right triangle are 5 and 12. Find the altitude to the hypotenuse.

${c=13},$ so ${h=\frac{5\cdot 12}{13}=\frac{60}{13}\approx 4.62}.$

5. The legs of an isosceles triangle are 5 cm and its base is 8 cm. Find the altitude to the base and the altitude to a leg.

The altitude to the base is ${\sqrt{25-16}=3}$ cm, and ${S=\frac12\cdot 8\cdot 3=12}$ cm². The altitude to a leg is ${\frac{2\cdot 12}{5}=4.8}$ cm.

6. Where is the orthocenter of an obtuse triangle?

Outside the triangle, beyond the vertex of the obtuse angle.

7. In an acute triangle ${\angle AHB=110^\circ},$ where $H$ is the orthocenter. Find the angle $\gamma.$

${\angle AHB=180^\circ-\gamma},$ so ${\gamma=70^\circ}.$

8. Two sides of a triangle are 6 and 8, and the altitude to the side 6 is 4. Find the altitude to the side 8.

${2S=6\cdot 4=24},$ so the altitude is ${\frac{24}{8}=3}.$

9. Find the shortest altitude of the triangle with sides 7, 8 and 9.

${p=12}$ and ${S}={\sqrt{12\cdot 5\cdot 4\cdot 3}}={12\sqrt5}.$ The shortest altitude goes to the longest side: ${h=\frac{24\sqrt5}{9}=\frac{8\sqrt5}{3}\approx 5.96}.$

10. The altitude to the hypotenuse divides it into segments of 4 and 9. Find the altitude.

${h^2=4\cdot 9=36},$ so ${h=6}.$

More about triangles: Triangles · Classification · Median · Area formulas