Express the area of each figure as a monomial:
Area of a triangle = 1/2 x base x height
[tex]\begin{gathered} \text{base }=\text{ }4a^2b \\ height=6ab^3 \end{gathered}[/tex]
[tex]A=\frac{1}{2}bh[/tex][tex]\begin{gathered} A=\frac{1}{2}\times4a^2b\times6ab^3 \\ A=2a^2b\times6ab^3 \\ A=12a^3b^4 \end{gathered}[/tex]
Therefore the area of the figure = 12a^3b^4
Hence the correct answer is option C
