Heron's Formula

Heron’s formula is one of the most important concepts used to find the area of a triangle when all the sides are known. Hero of Alexandra was a great mathematician who derived the formula for the calculation of the area of a triangle using the length of all three sides. This derived formula is called Heron’s formula. He also extended this idea to find the area of quadrilateral and also higher order polygons. This is one of the important topics in class 9 chapter 12.

Formula to Find the Area of a Triangle

So, the area of a triangle can be found using the following heron’s formula:

Herons formula, where S is semi-perimeter = (a+b+c) / 2

a, b, c are the three sides of the triangle.

Derivation of Heron’s Formula:

METHOD 1: Let us prove the result using the law of cosines:

Let a, b, c be the sides of the triangle and α, β, γ are opposite angles to the sides.

We know that,  the law of cosines is 

\( Cos \gamma  =\frac{a^2+b^2-c^2}{2ab}\)

Again, using trig identity, we have

\(Sin \gamma = \sqrt{1-cos^2γ} \)

= \(\frac{4a^2b^2-(a^2+b^2-c^2)^2}{2ab}\)

Here, Base of triangle = a

Altitude = b sinγ


Heron’s Formula

METHOD 2:  Area of a Triangle with 3 Sides

Area of ∆ABC is given by

Heron's Formula Derivation

\(A~=~\frac{1}{2} bh\) _ _ _ _ (i)

Draw a perpendicular BD on AC

Consider a ∆ADB

\(x^2~+~ h^2~ = ~c^2\)

\(x^2~ =~ c^2~ -~ h^2\)—(ii)

\(⇒x~ =~\sqrt{c^2 -h^2}\)—(iii)

Consider a ∆CDB

\((b-x)^2~ + ~h^2~ =~a^2\)

\((b-x)^2~ =~a^2~ -~h^2\)

\(b^2~ -~2bx~+~x^2~=~a^2 – h^2\)

Substituting the value of \(x\) and \(x^2\) from equation (ii) and (iii), we get

\(b^2~ – ~2b\sqrt{c^2 -h^2}~+~ c^2 -h^2~ = ~a^2~ -~ h^2\)

\(b^2~ + ~c^2~ -~a^2~ =~ 2b\sqrt{c^2 ~-~h^2}\)

Squaring on both sides

\((b^2 + c^2 – a^2 )^2~ =~ 4b^2 (c^2 -h^2)\)

\(\frac{(b^2 + c^2 -a^2 )^2}{4b^2} = c^2 – h^2\)

\(h^2~ = ~c^2 ~-~\frac{(b^2 +c^2 -a^2 )^2}{4b^2}\)

\(h^2~ =~\frac{4b^2 c^2 – (b^2 +c^2 -a^2 )^2 }{4b^2}\)

\(h^2~=~\frac{(2bc)^2 -(b^2+c^2- a^2 )^2}{4b^2}\)

\(h^2~=~\frac{[2bc+(b^2 +c^2 -a^2 )][2bc-(b^2+c^2-a^2)]}

\(h^2~ =~\frac{[(b^2+2bc+c^2)-a^2][a^2-(b^2- 2bc+c^2)]}{4b^2}\)

\(h^2~=~\frac{[(b+c)^2 – a^2 ].[a^2 -(b-c)^2 ]}{4b^2}\)



The perimeter of a ∆ABC is

\(P~=~ a+b+c\)

\(⇒~h^2~ =~\frac{P(P – 2a)(P – 2b)(P -2c)}{4b^2}\)

\(⇒~h~ = \sqrt{P(P – 2a)(P – 2b)(P -2c)}{2b}\)

Substituting the value of h in equation (i)

\(A~=~\frac{1}{2} b \frac{\sqrt{P(P – 2a)(P – 2b)(P -2c)}}{2b}\)

\(A~=~\frac{1}{4}\sqrt{(P(P – 2a)(P – 2b)(P -2c)}\)

\(A~=~\sqrt{\frac{1}{16} P(P – 2a)(P – 2b)(P -2c)}\)

\(A~=~\sqrt{\frac{P}{2}\left(\frac{P – 2a}{2}\right)\left(\frac{P – 2b}{2}\right)\left(\frac{P -2c}{2}\right)}\)

Semi perimeter(S) = \(\frac{perimeter}{2}~=~\frac{P}{2}\)

\(⇒~A~=~\sqrt{S(S – a)(S – b)(S – c)}\)

Note: Heron’s formula is applicable to all type of triangles and the formula can also be derived using the law of cosines and the law of Cotangents.


Let us now look into some example to have a brief insight into the topic:

Example 1: Find the area of a trapezium, length of whose parallel sides is given as 22cm and 12 cm and the length of other sides are 14 cm each.

Solution: Let PQRS be the given trapezium in which PQ= 22cm, SR= 12cm,


Constructions: Draw OR||PS

Now, PORS is a parallelogram in which PS||OR and PO||SR

Therefore, PO=SR=12cm

⇒OQ = PQ-PO = 22 -12 = 10cm

In ∆OQR , we have

\(S~ = ~\frac{14+14+10}{2}~ =~\frac{38}{2}~=~19\)

Area of ∆OQR = \(\sqrt{S(S – a)(S – b)(S –C)}\)

= \( \sqrt{ (19(19 – 14)(19 – 14)(19 –10))}\)

= \( \sqrt{4275}\)

= \(15 \sqrt{19} cm^2\) ………(i)

We know that Area = \(\frac{1}{2} \times b \times h\)

\(\Rightarrow 15\sqrt{19} = \frac{1}{2} \times 10 \times h\)

\(\Rightarrow h = 5\sqrt{19}\) …………..(ii)

Area of trapezium = \(\frac{1}{2}~ (PQ+SR) × h \)

= \(\frac{1}{2}(22+12) × 3 \sqrt{19} \)

= \(51 \sqrt{19} cm^2\)

Example 2: Find the area of the triangle whose sides measure 10 cm, 17 cm and 21 cm.  Also determine the length of the altitude on the side which measures 17 cm? 

Solution: S = \(\frac{a+b+c}{2}\) = \(\frac{10+17+21}{2}\) = 24

Area of Triangle= \(\sqrt{s(s-a)(s-b)(s-c)}\)

\(\sqrt{24 \times 14 \times 7 \times 3}\)

\(\sqrt{7056}\) = 84 square cm

Taking 17 cm as the base length we need to find the height

Area, A = 1/2 x base x height

1/2  x 17 x h = 84   or h = 168/17 = 9.88 cm   (Rounded to the nearest hundredth).

To solve more problems based on Heron’s Formula, Register with BYJU’S today and clear your all doubts on maths concepts.  We provide detailed and step by step solutions to all questions asked by you. Also, take free tests to practice for exams.

1 Comment

  1. very nice

Leave a Comment

Your email address will not be published. Required fields are marked *