Area of a triangle

Solution 1. Heron’s Formula

$s = \frac{10+12+18}{2} = 20$

$A = \sqrt{20(20-10)(20-12)(20-18)} = 56.6~m^2$

Solution 2. Cosine Rule or Cosine Law

$10^2 = 12^2+18^2-2(12)(18)cos~A$

$\angle A = 31.586°$

$A = \frac{1}{2}(12)(18)sin~31.586 = 56.6~m^2$

Solution 3. Solve for the height of the triangle by pythagorean theorem

Considering the left side small triangle, we have

$h^2=12^2-(18-x)^2=144-324+36x-x^2=-180+36x-x^2$

Considering the right side small triangle, we have

$h^2=10^2-x^2=100-x^2$

$h^2=h^2$

$-180+36x-x^2=100-x^2$

$36x=280$

$x=\frac{70}{9}$

Solving for $h$ , we have

$h^2=100-(\frac{70}{9})^2$

$h=6.29$

$A=\frac{1}{2}bh=\frac{1}{2}(18)(6.29)=56.6~m^2$