You can calculate the area of a triangle if you know the lengths of all three sides, using a formula that has been known for nearly 2000 years.
It is called "Heron's Formula" after Hero of Alexandria (see below)
Just use this two step process:
Step 1: Calculate 's' (half of the triangles perimeter): s=a+b+c2
Step 2: Then calculate the Area : Δ=√s(s−a)(s−b)(s−c)
Derivation of Herons formula:
Let in a ΔABC,
The sides are AB=c,BC=a and AC=b
Draw a AD⊥BC
Let, BC=x
⇒CD=a−x
Now, In ΔADC,∠ADC=90∘
⇒b2=(a−x)2+h2
⇒b2−(a−x)2=h2----(1)
Similarly,
In ΔADB,∠ADB=90∘
⇒c2=x2+h2
⇒c2−x2=h2---(2)
From (1) and (2),
b2−(a−x)2=c2−x2
⇒b2−a2−x2+2ax=c2−x2
⇒b2−a2+2ax=c2
⇒2ax=c2+a2−b2
⇒x=c2+a2−b22a ---(3)
Now, from (2),
c2−x2=h2
⇒c2−(c2+a2−b22a)2=h2
⇒(2ac)2−(c2+a2−b2)2(2a)2=h2
⇒h=√(2ac)2−(c2+a2−b2)2(2a)2
⇒h=√(2ac)2−(c2+a2−b2)22a ---(4)
Now,
Area of ΔABC=12ah
=12a√(2ac)2−(c2+a2−b2)22a
=14√(2ac)2−(c2+a2−b2)2
=14√(2ac+c2+a2−b2)(2ac−(c2+a2−b2)
=14√(2ac+c2+a2−b2)(b2−(c2+a2−2ac))
=14√((a+c)2−b2)(b2−(a−c)2)
=14√(a+b+c)(a−b+c)(−a+b+c)(a+b−c)----(5)
Let, Perimeter of triangle ABC is,
2s=a+b+c ---(i)
⇒s=a+b+c2 ---(ii)
2s=a+b+c
⇒2s−2a=a+b+c−2a=−a+b+c---(iii)
2s−2b=a+b+c−2b=a−b+c ----(iv)
2s−2c=a+b+c−2c=a+b−c -----(v)
Now, from (5)
Area of ΔABC=14√(a+b+c)(a−b+c)(−a+b+c)(a+b−c)
=14√(2s)(2s−2b)(2s−2a)(2s−2c)
=14√2.2.2.2.(s)(s−b)(s−a)(s−c)
=14×4√s(s−b)(s−a)(s−c)
=√s(s−b)(s−a)(s−c)
Therefore, Area of ΔABC=√s(s−b)(s−a)(s−c)
i.e., Δ=√s(s−b)(s−a)(s−c)
Hence, proved