**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:**

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

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 = \(\frac{1}{2}~ (PQ+SR) × h \)

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

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

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