Special Formula For Finding Area

Solution 1: Heron's Formula

Let the sides of the triangle be $a=12, b=16$ and $c=22$ . Then the semi-perimeter is

$s=\dfrac{a+b+c}{2}=\dfrac{12+16+22}{2}=25$

So the area is

$A=\sqrt{s(s-a)(s-b)(s-c)}=\sqrt{25(25-12)(25-16)(25-22)}\approx \boxed{93.7}$

Solution 2: Cosine Rule

Let the sides of the triangle be $a=12, b=16$ and $c=22$ and let the opposite angle of side $a$ be $A$ , of side $b$ be $B$ and of side $c$ be $C$ . By cosine rule, we have

$c^2=a^2+b^2-2ab \cos C$

$22^2=12^2+16^2-2(12)(16) \cos C$

$C \approx 102.64 ^\circ$

So the area is

$A=\dfrac{1}{2}ab \sin C = \dfrac{1}{2}(12)(16)(\sin 102.64) \approx \boxed{93.7}$