Area of Triangle With 3 Sides
We know that the area of a triangle is ½ × Base × Height where "Base" is any side of the triangle and "Height" is the length of the altitude drawn to the "Base" from its opposite vertex. But what if we are not aware of the altitude of any side? What if we need to find the area of a triangle with its 3 sides given? The formula used in such instances is Heron's Formula.
For example, to find the area of a triangle with 3 sides given as: 3, 6, and 7, we assume that a = 3, b = 6, and c = 7. Then the semiperimeter is, s = (a + b + c)/2 = (3 + 6 + 7)/2 = 8. We will find the area of the triangle using the Heron's formula.
A = √[s(sa)(sb)(sc)]
= √[8(83)(86)(87)]
= √[8 × 5 × 2 × 1]
= √(80)
≈ 8.94
This formula was derived by a mathematician known as the Hero of Alexandria. Let us see what are the alternative formulas to find the area of a triangle with 3 sides and let us see how all these formulas are derived.
1.  Area of Triangle with 3 Sides Formula 
2.  Proof of Area of Triangle with 3 Sides Formulas 
3.  FAQs on Area of Triangle with 3 Sides 
Area of Triangle With 3 Sides Formula
We already have seen Heron's formula to find the area of a triangle with 3 sides which says if a, b, and c are the three sides of a triangle, then its area is,
Area = √[s(sa)(sb)(sc)]
Here, "s" is the semiperimeter of the triangle. i.e., s = (a + b + c)/2. But there are 3 alternate formulas to find the same. You can find them in the below image.
Let us solve the above example (of finding the area of a triangle with 3 sides 3, 6, and 7) using one of the above formulas. As earlier, we will assume that a = 3, b = 6, and c = 7.
\( \begin{align} \text{Area }&= \dfrac{\sqrt{4a^2b^2(a^2+b^2c^2)^2}}{4} \\[0.2cm] &=\dfrac{ \sqrt{4(3^2)(6^2)(3^2+6^27^2)^2}}{4} \\[0.2cm] & \approx 8.94 \text{ square units} \end{align}\)
Proof of Area of Triangle With 3 Sides Formulas
We have seen multiple formulas to find the area of a triangle with 3 sides. But how can we derive these formulas? These formulas are connected. Heron's formula is derived from one of the above 3 formulas. So if we can derive one of the above 3 formulas, deriving Heron's formula is easy.
Consider the above triangle.
Using law of cosines, cos A = (b^{2} + c^{2}  a^{2}) / 2bc.
Using one of the Pythagorean identities,
sin^{2} A = 1  cos^{2} A
\( \begin{align}\sin A &= \sqrt{1\cos^2A}\\[0.2cm]\sin A &= \sqrt{1  \dfrac{(b^2+c^2a^2)^2}{4b^2c^2}} \\[0.2cm] \sin A &= \dfrac{\sqrt{4b^2c^2  (b^2+c^2a^2)^2}}{2bc} \\[0.2cm] \dfrac{1}{2} bc \sin A& = \dfrac{\sqrt{4b^2c^2  (b^2+c^2a^2)^2}}{4} \end{align}\)
We know that one of the formulas of the area of a triangle is ½ bc sin A. Thus, the area of triangle = \(\dfrac{\sqrt{4b^2c^2  (b^2+c^2a^2)}}{4}\).
We derived one of the above 3 formulas. Note that we can derive the other two formulas also in the same way. Can you give it a try?
Now, we will derive Heron's formula using the above formula just by applying some algebraic techniques. The above formula can be written as:
\( \begin{align} &\text{Area }\\[0.2cm] &= \dfrac{\sqrt{(2bc)^2 (b^2+c^2a^2)^2}}{4}\\[0.2cm] &= \dfrac{1}{4} \sqrt{[ (b^2+c^2+2bc)  a^2] [ a^2  (b^2+c^22bc)}\\[0.2cm] &= \dfrac{1}{4} \sqrt{[ (b+c)^2a^2] [a^2(bc)^2]}\\[0.2cm] &= \dfrac{1}{4} \sqrt{(b+c+a)(b+ca)(a+bc)(ab+c)}\\[0.2cm] \end{align} \)
We know that the semiperimeter of a triangle is, s = (a + b + c)/2. From this,
a + b + c = 2s
b + c  a = 2s  2a
a + b  c = 2s  2c
a  b + c = 2s  2b
Substituting all these values in the last step,
\( \begin{align}& \text{Area } \\[0.2cm]&= \dfrac{1}{4} \sqrt{2s (2s2a)(2s2c)(2s2b)}\\[0.2cm] &= \dfrac{4}{4} \sqrt{s(sa)(sc)(sb)}\\[0.2cm] &= \sqrt{s(sa)(sb)(sc)} \end{align}\)
Hence, we proved the Heron's formula as well.
Solved Examples on Area of a Triangle With 3 Sides

Example 1
Two sides of a given triangle are 8 units and 11 units. Its semiperimeter is 16 units. What is its area?
Solution
The two sides of the given triangle are,
a = 8; b = 11.
Its semiperimeter is, s = 16.
We know that
2s = a + b + c
2(16) = 8 + 11 + c
c = 13
We will calculate the area using Heron's formula.
A =√[s(sa)(sb)(sc)]
= √[16(168)(1611)(1613)]
= √[16 × 8 × 5 × 3]
= √16 × √8 × √5 × √3
= 4 × 2√2 × √5 ×√3
= 8 √30
∴ Area of the given triangle = 8√30 square units.

Example 2
The perimeter of a triangle is 18 units whose sides are in the ratio 2:3:4. Find its area.
Solution
Let us assume that the sides of the given triangle are
a = 2x; b = 3x; c = 4x
Then its perimeter is
2x + 3x + 4x = 18
9x = 18
x = 2
So the sides of the given triangle are
a = 2x = 2(2) = 4
b = 3x = 3(2) = 6
c = 4x = 4(2) =8
The semiperimeter of the triangle is,
s = (a + b + c)/2 = (4 + 6 + 8)/2 = 18/2 =9.
Now, we will find the area of the triangle using Heron's formula:
A =√[s(sa)(sb)(sc)]
= √[9(94)(96)(98)]
= √[9 × 5 × 3 × 1]
= 3 √15
∴ The area of the triangle = 3√15 square units.
FAQs on Area of Triangle with 3 Sides
1. How Do You Find the Area of a Triangle With 3 Sides?
We use one of the formulas we discussed on this page to find the area of a triangle with 3 sides. The most popular formula is Heron's formula which says, the area of a triangle whose sides are a, b, and c is √[s(sa)(sb)(sc)].
2. What Is an Irregular Triangle?
An irregular triangle is a triangle whose all three sides are not of the same lengths. It is also known as a scalene triangle (if all three sides are different).
3. How Do You Find the Area of an Irregular Triangle?
We use one of the formulas we discussed on this page to find the area of an irregular triangle with 3 sides. The most popular formula is Heron's formula which says, the area of a triangle whose sides are a, b, and c is √[s(sa)(sb)(sc)].
4. How Do You Find the Angle of a Triangle With Three Sides?
We can use the law of cosines to find the angles of a triangle with three sides. If a, b, and c are three sides of a triangle, then using the law of cosines,
a^{2} = b^{2} + c^{2}  2bc cos A
b^{2} = c^{2} + a^{2}  2ca cos B
c^{2} = a^{2} + b^{2}  2ab cos C
5. How To Find the Lengths of the Sides of a Triangle Given 3 Angles Only?
Let us recall the fact that two similar triangles have the same angles but different sides (sides are in proportion). Thus, there are an infinite number of triangles with the same set of any three given angles. So it is not possible to find the sides of a triangle if we just know all 3 angles, we need to know at least one side to determine the other two sides.
6. What Is Heron’s Formula Used For?
Heron's formula is used to find the area of a triangle given its 3 sides. It says the area of a triangle whose sides are a, b, and c is √[s(sa)(sb)(sc)].
7. Who Is Heron’s Formula Named After?
Heron's formula is derived by a mathematician known as the Hero of Alexandria.