Area of a Triangle – Formulas, Methods & Examples

The area of a closed plane 2D figure is the region of the plane enclosed by the figure’s boundary (perimeter). The area is measured in square units of length such as square centimetre $\left(cm^{2}\right)$ and square metre$\left(m^{2} \right)$.

A triangle is a three-sided plane 2D figure (three-sided polygon) having three vertices (singular vertex) and three angles. Let’s learn what is meant by the area of a triangle and how to find the area of a triangle and what are the different methods of finding it.

Triangle – A 2D Plane Figure

A triangle is a three-sided polygon formed by joining three non-collinear points known as the vertices (singular vertex). The line segments joining the three points (vertices) are called the sides (edges) of a triangle. The region formed between any of the two sides/edges of a triangle is called an angle of a triangle. There are three angles in a triangle whose sum is always $180^{\circ}$.

Triangles are broadly classified under two classifications

• Classification based on sides: On the basis of sides a triangle can be
  • Scalene Triangle: A triangle with no side equal
  • Isosceles Triangle: A triangle with any two sides is equal
  • Equilateral Triangle: A triangle with all three sides are equal
• Classification based on angles
  • Acute Triangle: A triangle where all three angles acute angles $\left(\le 60^{\circ} \right)$
  • Obtuse Triangle: A triangle where one of the angles is an obtuse angle $\left(\gt 90^{\circ} \right)$
  •  Right Triangle: A triangle where one of the angles is a right angle $\left(= 90^{\circ} \right)$

What is the Area of a Triangle?

The area of a triangle is defined as the total space occupied by the three sides of a triangle in a $2$-dimensional plane. The area of a triangle is expressed in square units, like $m^{2}$, $cm^{2}$, $in^{2}$, and so on.

Basic Formula for Area of a Triangle

The basic formula for the area of a triangle is equal to half the product of its base and height, i.e., $A = \frac {1}{2} \times b \times h$. This formula is applicable to all types of triangles, whether it is a scalene triangle, an isosceles triangle, or an equilateral triangle. 

Note: The base and the height of a triangle are perpendicular to each other.

Examples

Ex 1: Find the area of a triangle with a base of $12 mm$ and a height of $5 mm$.

Base of triangle $b = 12 mm$

Height of triangle $h = 5 mm$

Area of triangle = $\frac {1}{2} \times b \times h = \frac {1}{2} \times 12 \times 5 = 30 mm^{2}$

Ex 2: Find the height of a triangle whose area is $75 cm^{2}$ and the base is $25 cm$.

Area of triangle $A = 75 cm^{2}$

Base of triangle $b = 25 cm$

Let the height of the triangle be $h$

$A = \frac {1}{2} \times b \times h => 75 = \frac {1}{2} \times 25 \times h => h = \frac {75 \times 2}{25} = 6 cm$

Therefore, height of triangle = $6 cm$.

Area of Triangle Using Heron’s Formula

When you know the length of all three sides of a triangle, then you use Heon’s formula to find the area of a triangle. To use this formula, you need to find the perimeter and semi-perimeter of a triangle and then apply the formula.

The Heron’s formula is given by $A = \sqrt{s \times \left(s – a \right) \times \left(s – b \right) \times \left(s – c \right)}$, where $a$, $b$ and $c$ are the sides of a triangle

And $s$ is the semi-perimeter of a triangle given by $s = \frac {a + b + c}{2}$

Steps Used to Find the Area of a Triangle Using Heron’s Formula

Heron’s formula is used to find the area of a triangle when the length of the 3 sides of the triangle is known. To use this formula, we need to know the perimeter of the triangle which is the distance covered around the triangle and is calculated by adding the length of all three sides. Heron’s formula has two important steps.

Step 1: Find the semi perimeter (half perimeter) of the given triangle by adding all three sides and dividing it by 2, i.e., semi-perimeter of a triangle $s = \frac {a + b + c}{2}$.

Step 2: Apply the value of the semi-perimeter of the triangle in the main formula called ‘Heron’s Formula’ given by $A = \sqrt{s \times \left(s – a \right) \times \left(s – b \right) \times \left(s – c \right)}$, where $a$, $b$ and $c$ are the sides of a triangle.

Examples

Ex 1: Find the area of a triangle with sides $7 in$, $4 in$, and $6 in$.

The three sides of a triangle are $a = 7 in$, $b = 4 in$ and $c = 6 in$

Semiperimeter $s = \frac {a + b + c}{2} = \frac {7 + 4 + 6}{2} = \frac{17}{2} = 8.5 in $

Area of triangle = $ \sqrt{s \times \left(s – a \right) \times \left(s – b \right) \times \left(s – c \right)} = \sqrt{8.5 \times \left(8.5 – 7 \right) \times \left(8.5 – 4 \right) \times \left(8.5 – 6 \right)}$

$ = \sqrt{8.5 \times 1.5 \times 4.5 \times 2.5} = \sqrt{143.4375} = 11.98 in^{2}$ (Rounded off to $2$ decimal places).

Ex 2: Find the area of an equilateral triangle of side $6 cm$.

The three sides of a triangle are $a = 6 cm$, $b = 6 cm$ and $c = 6 cm$

Semiperimeter $s = \frac {a + b + c}{2} = \frac {6 + 6 + 6}{2} = \frac{18}{2} = 9 cm $

Area of triangle = $ \sqrt{s \times \left(s – a \right) \times \left(s – b \right) \times \left(s – c \right)} = \sqrt{9 \times \left(9 – 6 \right) \times \left(9 – 6 \right) \times \left(9 – 6 \right)}$ 

$= \sqrt{9 \times 3 \times 3 \times 3} = \sqrt{243} = 15.59 cm^{2}$ (Rounded off to $2$ decimal places).

Ex 3: Find the area of an isosceles triangle whose two equal sides are of $7 cm$ and the third side of $4 cm$.

The three sides of a triangle are $a = 7 cm$, $b = 7 cm$ and $c = 4 cm$

Semiperimeter $s = \frac {a + b + c}{2} = \frac {7 + 7 + 4}{2} = \frac{18}{2} = 9 cm $

Area of triangle = $\sqrt{s \times \left(s – a \right) \times \left(s – b \right) \times \left(s – c \right)} = \sqrt{9 \times \left(9 – 7 \right) \times \left(9 – 7 \right) \times \left(9 – 4 \right)} = \sqrt{9 \times 2 \times 2 \times 5} = \sqrt{180} = 13.42 cm^{2}$ (Rounded off to $2$ decimal places).

Area of a Triangle With Two Sides and an Included Angle

When two sides and the included angle of a triangle are given, we use a formula that has three variations according to the given dimensions. 

• When sides $b$, $c$ and included $\angle A$ is known, the area of the triangle is: $\text{Area} \left( \triangle ABC \right) = \frac {1}{2} \times b \times c \times \sin \left(A \right)$
• When sides $c$, $a$ and included $\angle B$ is known, the area of the triangle is: $\text{Area} \left( \triangle ABC \right) = \frac {1}{2} \times c \times a \times \sin \left(B \right)$
• When sides $a$, $b$ and included $\angle C$ is known, the area of the triangle is: $\text{Area} \left( \triangle ABC \right) = \frac {1}{2} \times a \times b \times \sin \left(C \right)$

Ex 1: Find the area of a triangle whose two sides of length $5 cm$ and $8 cm$ and the angle between these two sides is $30^{\circ}$.

Let the two sides of the triangle be $a = 5 cm$ and $b = 8 cm$ and $\angle C = 30^{\circ}$.

Area of triangle = $\text{Area} \left( \triangle ABC \right) = \frac {1}{2} \times a \times b \times \sin \left(C \right) = \frac {1}{2} \times 5 \times 8 \times \sin \left(30^{\circ} \right)$

$ = 20 \times \sin \left(30^{\circ} \right) = 20 \times \frac {1}{2} = 10 cm^{2}$.

Ex 2: The area of a $\triangle XYZ$ is $39 sq m$. If the length of the sides $YZ = 12 m$ and $ZX = 13 m$, find the measure of $angle YZX$.

Area of $\triangle XYZ = 39 sq m$

$YZ = 12 m$ and $ZX = 13 m$

Area of $\triangle XYZ = \frac {1}{2} \times YZ \times ZX \times \sin \left(\angle YZX \right) => 39 = \frac {1}{2} \times 12 \times 13 \times \sin \left(\angle YZX \right)$

$=> \sin \left(\angle YZX \right) = \frac {39 \times 2}{12 \times 13} => \sin \left(\angle YZX \right) = \frac {1}{2} => \angle YZX = 30^{\circ}$.

Area of an Equilateral Triangle

An equilateral triangle is a triangle in which all three sides of a triangle are equal. The formula to calculate the area of an equilateral triangle of side $a$ units is $\frac {\sqrt{3}}{4}a^{2}$.

Derivation of Formula to Find Area of an Equilateral Triangle

Method 1

Consider an equilateral triangle of side $a$ units. From $A$ drop a perpendicular on $BC$ at $D$. 

In $\triangle ADB$, $\angle ADB = 90^{\circ}$ and $AB = a$ and $BD = \frac {a}{2}$

Using Pythagoras theorem, we get $AD = \sqrt{AB^{2} – BD^{2}} = \sqrt{a^{2} – \left(\frac {a}{2} \right)^{2}}$

$ = \sqrt{a^{2} – \frac {a^2}{4}} = \sqrt{\frac {3a^{2}}{4}} = \frac {\sqrt{3}a}{2}$

Now, area of $\triangle ABC = \frac {1}{2} \times BC \times AD = \frac {1}{2} \times a \times \frac {\sqrt{3}a}{2} = \frac {\sqrt{3}a^{2}}{4}$

Method 2

Consider an equilateral triangle of side $a$ units.

By Heron’s formula, area of triangle is $A = \sqrt{s \times \left(s – a \right) \times \left(s – b \right) \times \left(s – c \right)}$, where $a$, $b$ and $c$ are the sides of a triangle where semi-perimeter of a triangle $s = \frac {a + b + c}{2}$.

In an equilateral triangle, the length of all three sides is $a$.

$s = \frac {a + a + a}{2} = \frac {3a}{2}$

Area of triangle = $\sqrt{\frac {3a}{2} \times \left(\frac {3a}{2} – a \right) \times \left(\frac {3a}{2} – a \right) \times \left(\frac {3a}{2} – a \right)} = \sqrt{\frac {3}{2}a \times \frac {1}{2}a \times \frac {1}{2}a \times \frac {1}{2}a} = \sqrt {\frac {3}{16}a^{2}} = \frac {\sqrt{3}a^{2}}{4}$

Examples

Ex 1: Find the area of an equilateral triangle of side $9 cm$.

Length of each side of equilateral triangle $a = 9 cm$

Area of an equilateral triangle = $\frac {\sqrt{3}}{4}a^{2} = \frac {\sqrt{3}}{4}9^{2} = \frac {81 \sqrt{3}}{4} cm^{2}$

Ex 2: Find the side of an equilateral triangle whose area is $4 \sqrt{3} m^{2}$

Area of triangle is $4 \sqrt{3} m^{2}$

Area of an equilateral triangle is $\frac {\sqrt{3}}{4}a^{2}$

Therefore, $\frac {\sqrt{3}}{4}a^{2} = 4 \sqrt{3} => a^{2} = 4 \sqrt{3} \times \frac {4}{\sqrt{3}} = 16$

Length of a side of an equilateral triangle of area $4 \sqrt{3} m^{2}$ is $16 m$

Area of a Right Triangle

A right-angled triangle, also called a right triangle, has one angle equal to $90^{\circ}$. Therefore, the height of the triangle is the length of the perpendicular side.

Area of a Right Triangle = $\frac {1}{2} \times \text{Base} \times \text{Height}$

Examples

Ex 1: Find the area of a right triangle whose legs are of length $3 cm$ and $4 cm$.

Area of a Right Triangle = $\frac {1}{2} \times \text{Base} \times \text{Height}$

Base = $3 cm$ and Height = $4 cm$

Area = $\frac {1}{2} \times 3 \times 4 = 6 cm^{2}$

Ex 2: The area of a right triangle is $120 mm^{2}$ and the perpendicular sides are in the ratio $5:12$. Find the length of each side of the triangle.

Area = $120 mm^{2}$

The ratio of the perpendicular sides is $5:12$

Let the length of the perpendicular sides be $5x$ and $12x$ respectively.

Therefore, $\frac {1}{2} \times 5x \times 12x = 120 => 30x^{2} = 120 => x^{2} = \frac {120}{30} => x^{2} = 4 => x = \sqrt{4} => x = 2$

Length of the two perpendicular sides are $5 \times 2 = 10 mm$ and $12 \times 2 = 24 mm$

And, the length of the hypotenuse is $\sqrt{10^{2} + 24^{2}} = \sqrt{100 + 576} = \sqrt{676} = 26 mm$

In $\triangle ADB$, $\angle ADB = 90^{\circ}$ and $AB = a$ and $BD = \frac {b}{2}$

Using Pythagoras theorem, we get $AD = \sqrt{AB^{2} – BD^{2}} = \sqrt{a^{2} – \left(\frac {b}{2} \right)^{2}} = \sqrt {a^{2} – \frac {b^{2}}{4} }$

Area of an Isosceles Triangle

An isosceles triangle has two of its sides equal and the angles opposite the equal sides are also equal.

Area of an Isosceles Triangle = $\frac {1}{4}b \sqrt{4a^{2} – b^{2}}$  where $b$ is the base and $a$ is the measure of one of the equal sides.

Derivation of Formula of Area of an Isosceles Triangle

Consider an isosceles triangle with base $b$ and length of equal sides as $a$.

From $A$ drop a perpendicular on $BC$ at $D$. 

In $\triangle ADB$, $\angle ADB = 90^{\circ}$ and $AB = a$ and $BD = \frac {b}{2}$

Using Pythagoras theorem, we get $AD = \sqrt{AB^{2} – BD^{2}} = \sqrt{a^{2} – \left(\frac {b}{2} \right)^{2}} = \sqrt{a^{2} – \frac {b^{2}}{4}} = \sqrt {\frac {4a^{2} – b^{2}}{4}} = \frac {\sqrt{4a^{2} – b^{2}}}{2}$

Area of $\triangle ABC = \frac {1}{2} \times \text {BC} \times \text {AD} = \frac {1}{2} \times b \times \frac {\sqrt{4a^{2} – b^{2}}}{2} = \frac {1}{4} \times b \times \sqrt{4a^{2} – b^{2}}$

Examples

Ex 1: Find the area of an isosceles triangle whose length of equal sides is $10 mm$ and that of an unequal side is $7 mm$.

Length of equal sides $a = 10 mm$

Length of unequal side $b = 7 mm$

Area of isosceles triangle = $\frac {1}{4} \times b \times \sqrt{4a^{2} – b^{2}} = \frac {1}{4} \times 7 \times \sqrt{4\times 10^{2} – 7^{2}} = \frac {1}{4} \times 7 \times \sqrt{4\times 100 – 49}$

$ =  \frac {7}{4} \times \sqrt{400 – 49} = \frac {7}{4} \times \sqrt{351} = \frac {7}{4} \times \sqrt{18.735} = 32.79 mm^{2}$

Conclusion

The area of a triangle is the region bounded by its perimeter in the space. The general formula for finding the area of a triangle of base $b$ and height $h$ is $A = \frac {1}{2}bh$. This formula is modified to derive the formulae to find the area of different types of triangles like an equilateral triangle, right triangle, or isosceles triangle.

Practice Problems

1. Find the area of a triangle with
   • base = $14 m$ and height = $8 m$
   • base = $20.54 mm$ and height = $7.4 mm$
2. Find the area of a triangle with a length of three sides as
   • $5 cm$, $9 cm$, and $11 cm$
   • $34 mm$, $27 mm$, and $19 mm$
3. Find the area of a right triangle with sides
   • $3 cm$ and $5 cm$
   • $10 m$ and $8 m$

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• Perimeter of Rectangle – Definition, Formula & Examples
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• Area of Square – Definition, Formula & Examples

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