How to Calculate the Area of a Triangle

A triangle is a fundamental planar geometric figure determined by three non-collinear points, known as vertices, connected by three straight line segments, known as sides. Determining the area enclosed by a triangle is essential in geometric computations, coordinate geometry, and trigonometry. The area can be determined by various formulae depending on the given data, such as side lengths, altitude, or included angles.

Canonical Formula for Area of a Triangle Using Base and Altitude

Let $ABC$ be a triangle with base $BC = b$ and corresponding altitude (height) from vertex $A$ to side $BC$ as $h$. The area, denoted by $A_\triangle$, is given by the formula:

$A_\triangle = \dfrac{1}{2} \times b \times h$

This formula holds for any triangle, regardless of type, provided the length of the base and its corresponding altitude are known. Both $b$ and $h$ must be in the same unit, and the area is expressed in square units. For comprehensive reading on polygonal area formulae, refer to the Area Of Triangle Formula page.

Area Formulae for Special Classes of Triangles

An equilateral triangle is defined by all sides being equal in length, denoted $a$. The formula for its area is derived by constructing an altitude, using the Pythagorean theorem, and is given by:

$A_{\text{equilateral}} = \dfrac{\sqrt{3}}{4} a^2$

A right-angled triangle is characterized by one angle being exactly $90^\circ$. Let the perpendicular sides be $b$ and $h$. The area is:

$A_{\text{right triangle}} = \dfrac{1}{2} b \, h$

An isosceles triangle has two equal sides, denoted $a$, and a base $b$. Dropping a perpendicular from the vertex opposite the base bisects the base. If $h$ is the altitude to $b$, then using the Pythagorean theorem:

Let $h = \sqrt{a^2 - \left(\frac{b}{2}\right)^2}$ so the area is

$A_{\text{isosceles}} = \dfrac{1}{2} b h = \dfrac{1}{2} b \sqrt{a^2 - \left( \frac{b}{2}\right)^2} = \dfrac{b}{4} \sqrt{4a^2 - b^2}$

Derivation of Area by Heron's Formula (Three Sides Known)

When the lengths of all three sides $a$, $b$, and $c$ of a triangle are known and no altitude measure is given, the area can be obtained by Heron's formula. The semi-perimeter $s$ is defined as $s = \dfrac{a+b+c}{2}$. The area is then:

$A_\triangle = \sqrt{ s(s-a)(s-b)(s-c) }$

The steps for the derivation are as follows.

Let the triangle have sides $a$, $b$, and $c$ opposite vertices $A$, $B$, and $C$ respectively. Draw the triangle with base $c$, and let $h$ be the altitude from vertex $A$ to base $c$. The area is $A_\triangle = \dfrac{1}{2} c h$.

Let $b$ and $a$ be the other two sides. Drop perpendicular $AD$ from $A$ to $BC$, such that $BD = x$ and $DC = c - x$. By geometry, $AB = b$; $AC = a$.

From the right triangle $ABD$: $AB^2 = h^2 + x^2 \implies h^2 = b^2 - x^2$

From the right triangle $ADC$: $AC^2 = h^2 + (c-x)^2 \implies h^2 = a^2 - (c-x)^2$

Set the two expressions for $h^2$ equal: $b^2 - x^2 = a^2 - (c-x)^2$

Expand $(c-x)^2$: $= c^2 - 2c x + x^2$

Thus $b^2 - x^2 = a^2 - c^2 + 2c x - x^2$

Bring $x^2$ terms to one side: $b^2 = a^2 - c^2 + 2c x$

$2c x = b^2 - a^2 + c^2$ so $x = \dfrac{b^2 - a^2 + c^2}{2c}$

Recall $h^2 = b^2 - x^2$, substitute the value of $x$:

$h^2 = b^2 - \left( \dfrac{b^2 - a^2 + c^2}{2c} \right)^2$

So, $A_\triangle = \dfrac{1}{2} c h$

After algebraic expansion and simplification (see standard textbooks for full expansion), we obtain:

$A_\triangle = \sqrt{ s(s-a)(s-b)(s-c) }$

Heron's formula allows computation of the area without knowledge of altitudes or angles. For formulas related to four-sided figures, see Area Formula For Quadrilateral.

Area When Two Sides and the Included Angle Are Known (Sine Formula)

Consider a triangle where two sides, $b$ and $c$, and the included angle $A$ are given. The area is given by the trigonometric (sine) formula:

$A_\triangle = \dfrac{1}{2} b c \sin A$

More generally, for sides $a$ and $c$ and included angle $B$, $A_\triangle = \dfrac{1}{2} a c \sin B$, or for sides $a$ and $b$ with included angle $C$, $A_\triangle = \dfrac{1}{2} a b \sin C$.

This formula follows from interpreting one side as the base and constructing the altitude using trigonometric ratios. For a review of area formulae using trigonometric identities, refer to the Area Of Triangle Formula resource.

Stepwise Example: Area Calculation Using Different Formulae

Given: Triangle with base $b = 8$ cm and height $h = 6$ cm.

Substitution: $A_\triangle = \dfrac{1}{2} \times 8 \times 6$

Simplification: $A_\triangle = \dfrac{1}{2} \times 48 = 24$

Final result: $A_\triangle = 24$ cm$^2$

Given: Triangle with sides $a=7$ cm, $b=8$ cm, $c=9$ cm.

Step 1: Compute semi-perimeter $s = \dfrac{7+8+9}{2} = 12$ cm.

Step 2: Compute area: $A_\triangle = \sqrt{12(12-7)(12-8)(12-9)}$

Simplification: $A_\triangle = \sqrt{12 \times 5 \times 4 \times 3}$

$= \sqrt{720} = 12 \sqrt{5}$

Final result: $A_\triangle = 12\sqrt{5}$ cm$^2$

Given: Triangle with $b = 11$ units, $c = 14$ units, included angle $A = 30^\circ$.

Substitution: $A_\triangle = \dfrac{1}{2} \times 11 \times 14 \times \sin 30^\circ$

Simplification: $\sin 30^\circ = \dfrac{1}{2}$, so $A_\triangle = \dfrac{1}{2} \times 11 \times 14 \times \dfrac{1}{2}$

$= \dfrac{1}{2} \times 11 \times 7 = \dfrac{1}{2} \times 77 = 38.5$

Final result: $A_\triangle = 38.5$ square units

Area Formula Manipulations — Finding Missing Quantities

Given the area $A$ and the base $b$ of a triangle, the corresponding altitude $h$ can be computed as $h = \dfrac{2A}{b}$. Similarly, if the area $A$ and the altitude $h$ are given, the base $b = \dfrac{2A}{h}$.

Determining the area of any polygon by triangle decomposition is often leveraged in higher geometry. For the area of sectors of circles, see Area Of A Sector Of A Circle Formula.

Summary of Area Formulae Relevant to the Triangle

The main formulae for the area of a triangle include:

$1.\ $ Given base and altitude: $A_\triangle = \dfrac{1}{2} b h$

$2.\ $ Given all three sides: $A_\triangle = \sqrt{ s(s-a)(s-b)(s-c) }$

$3.\ $ Given two sides and included angle: $A_\triangle = \dfrac{1}{2} ab \sin C$

$4.\ $ Equilateral: $A_\triangle = \dfrac{\sqrt{3}}{4} a^2$

$5.\ $ Isosceles (sides $a$, base $b$): $A_\triangle = \dfrac{b}{4} \sqrt{4a^2-b^2}$

Comprehensive mastery of these formulae is critical for applications in geometry, coordinate geometry, trigonometry, and for mathematical examinations such as JEE Main. For equilateral triangles, review Area Of Equilateral Triangle Formula.