We are given that the legs of a triangle to be:
9mm, 6mm, and 12mm.
Solution
We can find the area of a triangle using Heron's formula:
[tex]\begin{gathered} \text{Area = }\sqrt[]{s(s-a)(s-b)(s-c)} \\ \text{where, s = }\frac{a+b+c}{2} \end{gathered}[/tex]s is called the semi-perimeter, a, b and c are the sides of the triangle.
The semi-perimeter(s) for the given triangle would be:
[tex]\begin{gathered} s\text{ = }\frac{9+6+12}{2} \\ s\text{ = }\frac{27}{2} \\ =\text{ 13.5} \end{gathered}[/tex]Hence, the area (A) :
[tex]\begin{gathered} A\text{ = }\sqrt[]{13.5(13.5-9)(13.5-6)(13.5-12)} \\ =\text{ }\sqrt[]{13.5\times4.5\times7.5\times1.5} \\ =\text{ 26.1426} \\ =26.1mm^2 \end{gathered}[/tex]Answer: Option C
