Altitude of a triangle is the side that is perpendicular to the base. A triangle has three sides altitude, base and hypotenuse. The altitude of the triangle is the perpendicular drawn from the vertex of the triangle to the opposite side. The altitude is also referred to as the height or perpendicular of the triangle. It is important for students to have appropriate knowledge of all the properties of triangles as it will help them solve sums related to triangles without experiencing any challenges.

It is necessary for students to understand the basic concepts associated with triangles to be able to attend test papers related to the same. Embibe offers MCQ mock tests, previous year question papers and samples test papers. Students can practice these test papers for and can download the NCERT books and solution sets for.

Triangle: Definition

A triangle is a three-sided polygon. It includes three sides, three vertices and three angles. Three sides of a triangle are referred to as base, hypotenuse and height. It is important for students to be able to identify these sides independently to be able to apply

Triangles: Definition of Altitude

The perpendicular drawn from any vertex to the opposite side is called the altitude of the triangle from that vertex.

In the above figure, a perpendicular \(AD\) is drawn from the vertex \(A\) to the side \(BC.\)
So, \(AD\) is called the altitude of the triangle \(ABC.\)

The perpendicular doesn’t need to be drawn from the triangle’s top vertex to the opposite side to get altitude. We can draw a perpendicular from any vertex of the triangle to the opposite sides to get altitude, as shown in the figure above.

In the above figure, perpendiculars \(AD, BE,\) and \(CF\) are drawn from the vertices \(A, B\) and \(C\) on the opposite sides \(BC, CA\) and \(AB,\) respectively. In this case, \(AD\) is considered the altitude of the triangle from vertex \(A\) concerning base \(BC.\) Similarly, \(BE\) and \(CF\) are considered altitudes of the triangle from vertex \(B\) and \(C\) concerning bases \(CA\) and \(AB,\) respectively.

Altitudes of Different Triangles

We can classify the triangles concerning their sides and the angles. Then, we will explain the different types of altitude of different kinds of triangles.

Altitude of an Obtuse Triangle

If one angle in a triangle is an obtuse-angle, then the triangle is called an obtuse-angled triangle. An angle whose measure is more than \({90^{\rm{o}}}\) and less than \({180^{\rm{o}}}\) is called an obtuse-angled triangle. The other two angles are acute.

The altitude is outside the triangle for an obtuse-angled triangle. For these triangles, the base is stretched, and then a perpendicular is constructed from the opposite vertex to the base. The altitude of an obtuse triangle is shown in the triangle following.

Altitude of an Acute Triangle

If all the three angles in a triangle are acute, then the triangle is called an acute-angled triangle. An angle measuring more than \({0^{\rm{o}}}\) but less than \({90^{\rm{o}}}\) is called an acute angle. For an acute-angled triangle, the altitudes can be drawn inside the triangle.

Altitude of a Right Triangle

In a right-angled triangle, the perpendicular side and the base can be considered as altitudes of it.
For a right-angled triangle, the altitude from the vertex to the hypotenuse divides the triangle into two similar triangles.

Here, \(△ADC\), \(△BCD\) are similar triangles according to the \(AA\) similarity.

Altitude of an Equilateral Triangle

If all the three sides of a triangle are the same in length, then the triangle is referred to as equilateral triangle. The altitude of an equilateral triangle bisects its base and the opposite angle of the base.

Consider an equilateral \(△ABC\) where \(BD\) is the altitude \((h).\)

\(AB = BC = AC\)
\(\angle ABD = \angle CBD\)
\(AD = CD\)

We can say the altitude of an equilateral triangle divide it into two congruent triangles using \(SSS\) congruency. Three altitudes are there for an equilateral triangle.

Altitude: Isosceles Triangle

If any two of the three sides of a triangle are equal to each other, then the triangle is called isosceles triangle.

The altitude of an isosceles triangle bisects the angle of the vertex and also bisects the base. Thus, the altitude of the isosceles triangle divides the triangle into two congruent triangles using \(SSS\) congruency.

Altitude: Scalene Triangle

If all the three sides of a triangle are different in length or if none of the sides of the triangle is equal to each other, then the triangle is called a scalene triangle.

If we know the length of all the sides of a triangle, we can easily find the length of its height.
Area of a scalene triangle\( = \sqrt {s(s – a)(s – b)(s – c)} \) where \(a,b,c\) are the sides of the triangle, and \(s\) is the semi-perimeter.

Area of a triangle\({\rm{ = }}\frac{{\rm{1}}}{{\rm{2}}}{\rm{ \times base \times altitude}}\)
Since \(\sqrt {s(s – a)(s – b)(s – c)} = \frac{{\rm{1}}}{{\rm{2}}}{\rm{ \times base \times altitude}}\)
\( \Rightarrow {\rm{altitude}} = \frac{{2\sqrt {s(s – a)(s – b)(s – c)} }}{{{\rm{ base }}}}\)

Altitude of a Triangle: Formulas

Triangle Type	Altitude Formula
Equilateral Triangle	\(h = \frac{{\sqrt 3 }}{2}a\) where the length of each side is \(a.\)
Isosceles Triangle	\(h = \sqrt {{a^2} – \frac{{{b^2}}}{4}} \) where \(a\) is the length of the congruent sides and \(b\) is the base.
Right Triangle	\(h = \sqrt {xy} \)  where \(x\) and \(y\) are the line segments formed by the altitude drawn on the hypotenuse
Scalene Triangle	\(h = \frac{{2\sqrt {s(s – a)(s – b)(s – c)} }}{{{\rm{base}}}}\)where \(a, b, c\) are the sides of the triangle, and \(s\) is the semi perimeter.

Altitudes of Triangles: Applications

The orthocentre has a significant role in the study of triangles. Therefore, knowing about the orthocentre, the study of the altitudes is important. Altitude is one of the most important parts of the triangles as we use it to find the area of the triangle, congruency proofs, similarity proof, etc. So, let us study the application of altitudes in geometry.

Application of Altitudes to Find the Orthocentre

Definition of Orthocentre: A triangle has three altitudes. These three altitudes always meet at a point. The point of intersection of the altitudes is called the orthocentre of the triangle.

In the above triangle, \(ABC, O\) is the orthocenter.

Orthocentre in an Acute-Angled Triangle

In an acute-angled triangle, the orthocenter lies inside the triangle.

Orthocentre in an Obtuse-Angled Triangle

In obtuse-angled triangles, the orthocenter lies outside the triangle.

In an obtuse-angled \(△ABC\) the altitudes are drawn from the vertices \(A, B\) and \(C\) on their corresponding opposite sides \(BC, AC\) and \(AB\) have to be extended to meet at an external point of the \(△ABC\).

Orthocentre in a Right-Angled Triangle

In a right-angled triangle, the orthocenter lies on the vertex forming the right angle. Look at the figure below.

In \(△ABC ,AC\) is the side, as well as altitude from the vertex \(A. BC\), is the side as well as altitude from the vertex \(B. AC\) and \(BC\) meet at \(C.\) The altitude from the opposite side  (hypotenuse) \(AB\) will pass through \(C.\)

Hence, \(C\) is the vertex containing the right-angle and the orthocentre as well.

Application of Altitudes to Find the Area of a Triangle

Area of triangle \({\rm{ = }}\frac{{\rm{1}}}{{\rm{2}}}{\rm{ \times base \times height}}{\rm{.}}\)

The above formula is used when the length of any side and the corresponding height is known or given. For the above figure, the area of the triangle \( = \frac{1}{2} \times {\rm{base \times height}} = \frac{1}{2} \times BC \times AD.\)

Altitude of a Triangle: Solved Examples

Q.1. The area of a right-angled triangle is \({\rm{81}}\,{\rm{sq}}{\rm{.}}\,{\rm{units}}{\rm{.}}\) Find the length of the altitude if the length of the base is \({\rm{9}}\,{\rm{units}}{\rm{.}}\)
Ans: Given, the area of the triangle is \({\rm{81}}\,{\rm{sq}}{\rm{.}}\,{\rm{units}}{\rm{.}}\), and the base is \({\rm{9}}\,{\rm{units}}{\rm{.}}\)
We know that the area of the right-angles triangle is \({\rm{ = }}\frac{{\rm{1}}}{{\rm{2}}}{\rm{ \times base \times height}}{\rm{.}}\)
Thus, \( \Rightarrow {\rm{height = }}\frac{{{\rm{81 \times 2}}}}{{\rm{9}}}{\rm{ = 18}}\,{\rm{units}}{\rm{.}}\)
Hence, the length of the altitude is \({\rm{18}}\,{\rm{units}}{\rm{.}}\)

Q.2. In \(△ABC ,CD\) is perpendicular to \(AB\) and \(AD = 3\,{\rm{cm}}\) and \(BD = 3\,{\rm{cm}}\) then, find the value of \(CD\).
Ans:

The right triangle altitude theorem states that the height drawn on the hypotenuse is equal to the geometric mean of line segments made by the height on the hypotenuse. We know that, \(h = \sqrt {xy} \) 
Thus, \(h = \sqrt {3 \times 6} = 3\sqrt 2 \;{\rm{cm}}.\)

Q.3. Calculate the length of the altitude of an isosceles triangle whose base is \({\rm{3}}\,{\rm{cm}}\) and congruent sides are \({\rm{5}}\,{\rm{cm}}\).
Ans: Given, congruent sides are \({\rm{5}}\,{\rm{cm,}}\) and the base is \({\rm{3}}\,{\rm{cm}}\).
Let us say the congruent sides are \(a\) and the base is \(b\).
Now, the formula of the altitude of the isosceles triangle\( = \sqrt {{a^2} – \frac{{{b^2}}}{4}} \)
\( = \sqrt {{5^2} – \frac{{{3^2}}}{4}} \)
\( = \frac{{\sqrt {91} }}{2}\,{\rm{cm}}\)
Hence, the length of the altitude of the triangle is \(\frac{{\sqrt {91} }}{2}\,{\rm{cm}}{\rm{.}}\)

Q.4. An equilateral triangle has each side of length \({\rm{32}}\,{\rm{inches}}{\rm{.}}\) Find the altitude of the triangle.
Ans: Given, the side of the equilateral triangle is \({\rm{32}}\,{\rm{inches}}{\rm{.}}\)
We know that the formula of the altitude of an equilateral triangle\({\rm{ = }}\frac{{\sqrt {\rm{3}} }}{{\rm{2}}}{\rm{ \times side = }}\frac{{\sqrt {\rm{3}} }}{{\rm{2}}}{\rm{ \times 32 = 16}}\sqrt {\rm{3}} \,{\rm{inches}}{\rm{.}}\)
Hence, the altitude is \({\rm{16}}\sqrt {\rm{3}} \,{\rm{inches}}{\rm{.}}\)

Question-5:  \(△ABC\) is a right triangle with \(AC=3, AB=5, BC=4.\) What is the length of its height \(CD\)? 
Answer: The height \(CD,\) divides the \(△ABC\), in two triangles, \(△ADC\) and \(△CDB\) with the same proportions as the original triangle, \(△ABC.\)
So, \(\frac{{CD}}{{BC}} = \frac{{AC}}{{AB}}\)
\( \Rightarrow CD = \frac{{BC \times AC}}{{AB}}\)
\( \Rightarrow \frac{{3 \times 4}}{5} = \frac{{12}}{5}\,{\rm{units}}\)
Therefore, the length of the altitude is \(\frac{{12}}{5}\,{\rm{units}}{\rm{.}}\)

Summary

In this article, we have discussed the definition of altitude of a triangle, properties and formulas of altitudes of different triangles, and the application of altitude of a triangle in mathematics. We hope it will help you to understand the concept.

Frequently Asked Questions (FAQ) – Altitude of a Triangle

Frequently asked questions related to altitude of a triangle is listed as follows:

Q.1. Are there any differences between the altitude and the height of a triangle?
Ans: No, the altitude and the height of a triangle are not different. They are the same.

Q.2. What are the formulas of altitudes of the triangles?
Ans: There are different formulas of altitude for different types of triangles.
The formula of the altitude of an equilateral triangle, \(h = \frac{{\sqrt 3 }}{2}a,\) where the length of each side is \(a.\)
The formula of the altitude of an isosceles triangle, \(h = \sqrt {{a^2} – \frac{{{b^2}}}{4}} \) where \(a\) is the length of the congruent sides and \(b\) is the base.
The formula of the altitude of a right triangle \(h = \sqrt {xy} \)  where \(x\) and \(y\) are the line segments formed by the altitude drawn on the hypotenuse.

Q.3. What do you understand by the altitude and the median of a triangle?
Ans: The perpendicular drawn from any vertex to the side opposite to the vertex is called the altitude of the triangle from that vertex.
The median of a triangle is the line segment drawn from the vertex to the opposite side, and it joins the midpoint of the opposite side.

Q.4. How many altitudes are possible for a triangle?
Ans: Maximum of three altitudes can be drawn in a triangle.

Q.5. Is the altitude of a triangle always \({90^{\rm{o}}}\)?
Ans: The perpendicular drawn from any vertex to the side opposite to the vertex is called the altitude of the triangle from that vertex.
Hence, an altitude is always \({90^{\rm{o}}}\) as it is always perpendicular to the side opposite to the vertex from where it is drawn.

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