Question

Find the area of ∆ABC using Heron’s formula. The vertices of the triangle are A(1,0), B(4,4) and C(7,0).​

Answer 1

Let $AB=a,BC=b,CA=c$
The length of the sides of the triangle are:
$AB=a=\sqrt{(4-1{)}^2+(4-0{)}^2}=5$ units
$BC=b=\sqrt{(7-4{)}^2+(0-4{)}^2}=5$ units
$CA=c=\sqrt{(7-1{)}^2+(0-0{)}^2}=6$ units
Now, semi-perimeter $S=\frac{a+b+c}{2}=\frac{5+5+6}{2}=8$ units
$\therefore$ Area of triangle $ABC=\sqrt{s(s-a)(s-b)(s-c))}$
$=\sqrt{8(8-5)(8-5)(8-6)}$
$=\sqrt{8\times 3\times 3\times 2}$
$=12$ sq. units