Heron's Formula

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 as Heron’s formula. He also extended this idea to find the area of quadrilaterals and also higher order polygons.

Derivation of Heron’s formula:

Heron’s Formula

Area of ∆ABC is given by

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

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

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

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

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

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

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

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

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

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

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

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 about the topic:

Let’s Work Out-

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

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

PS=QR=14cm.

Constructions: Draw OR||PS

Heron’s Formula

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 = \(\frac{1}{2}~ (PQ+SR) × h \)

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

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

To solve more problems based on Heron’s Formula NCERT solution, visit BYJU’S which provides detailed and step by step solutions to all questions in an NCERT Books. Also, take free tests to practice for exams.

 


Practise This Question

Three prizes are to be distributed in a quiz contest. The value of second prize is 5/6 of the value of first prize and the value of third prize is 4/5 of second prize. If the total value of three prizes is Rs.1500. Find the value of each prize. (in order of 1st, 2nd & 3rd)