Altitude of a Triangle with Examples and FAQs

What is the Altitude of a triangle?

The altitude or height of a triangle is the line segment that joins a vertex to its opposite side and is perpendicular to it. As a triangle has three sides vis-à-vis it has three altitudes.

The main importance of altitude is in the calculation of the area of a triangle.

Area of Triangle = \frac{1}{2} × Base × Height

Here, height is the altitude of the triangle.

The point at which the three altitudes of a triangle meet(intersect) is the orthocenter of that triangle.

In ∆ABC the perpendicular drawn from vertex A to the side BC is the altitude of the triangle. Where BC is the base with respect to the altitude AD. D is the point of intersection of base and altitude.

Properties of Altitude of a triangle

The various properties of altitudes are:

• Any given triangle has only three altitudes.
• The position of the altitude can be either inside the triangle, outside the triangle, or it can even be the side of the triangle.
• It makes a 90° angle with the side opposite to it.
• The point of intersection of three altitudes of a triangle is its orthocenter, which can lie inside or outside the triangle.

Altitudes of different types of triangle

Equilateral Triangle

In an equilateral triangle, the altitude bisects its base and the angle at its respective vertex.

The image shows an equilateral triangle ∆XYZ.

We know, for an equilateral triangle all the angles are equal to 60°and let all sides of the triangle be ‘A’. Let ‘H’ be the altitude. In the triangle above we have:

Therefore, for an equilateral triangle, its Altitude is \frac{\sqrt{3}}{2} \times(\text { side }).

Isosceles Triangle

We know, for an isosceles triangle two sides have equal dimensions. 

∆PQR is an isosceles triangle, PQ = PR = x, QR = y and let PO = h.

The altitude PO of the triangle bisects the side QR ⇒ QO = OR  = y/2

∆POQ is a right-angle triangle, hence, by Pythagoras Theorem,

Substituting the values of PO, QO and PQ we get, 

\therefore h=\sqrt{x^{2}-\frac{y^{2}}{4}}

Right Angled Triangle

The above triangle represents a right-angled triangle which is right-angled at A. A perpendicular is dropped from vertex A onto the hypotenuse BC.

AD is thus the altitude of the triangle. Let its length be h.

By the right triangle altitude theorem this altitude divides the triangle, ∆ABC into two similar triangles, i.e., ∆BDA 〜∆ACD.

As  ∆BDA 〜 ∆ACD, we have 

\frac{B D}{A D}=\frac{A D}{D C}

\Rightarrow A D^{2}=B D \times D C

\therefore h^{2}=p \times q

h=\sqrt{p q}

AB and AC are also altitudes of the triangle shown above and as they are perpendicular to each other when one is considered to be the altitude the other will be the base.

Obtuse Triangle

We know, that in a triangle when one angle is greater than 90°, it is termed as an obtuse triangle. In the case of an obtuse triangle, the altitude lies outside the triangle and is constructed by extending the base as seen in the figure below.

Examples

Example 1: Calculate the length of altitude of an equilateral triangle in which each side is 12 units in length. Also, determine the area of the same triangle.

Solution:

We know, that for an equilateral triangle its Altitude is \frac{\sqrt{3}}{2} \times(\text { side }).

Hence, the Altitude of the given triangle =\frac{\sqrt{3}}{2} \times 12 \text { units }=6 \sqrt{3} \text { units. }

∴ The altitude of the given triangle is 6 \sqrt{3} units.

The formula for the area of a triangle =\frac{1}{2}\text{ base }\times \text { height }

Base = side length = 12 units and height = altitude = 6 \sqrt{3} \text { units }

\therefore \text { Area }=\frac{1}{2} \times 12 \times 6 \sqrt{3}=36 \sqrt{3} \text { unit }^{2} \text {. }

Hence, the area of the triangle is 36 \sqrt{3} \text { unit }^{2}.

Example 2: The area of a triangle is 30 sq units. Determine its altitude if the base is 12 units.

Solution:

We know the formula for the area of a triangle, i.e.,

\text { Area }=\frac{1}{2} \times \text { Base } \times \text { Height }

\Rightarrow \text { Height }=\frac{2 \times \text { Area }}{\text { Base }}=\frac{2 \times 30}{12}=5 \text { units }

As height and altitude are the same. Hence, the altitude of the given triangle is 5 units in length.