Respuesta :
Let's see
Area
[tex]\\ \rm\rightarrowtail \dfrac{1}{2}\times Base\times Height [/tex]
[tex]\\ \rm\rightarrowtail \dfrac{1}{2}(6xy^5)(4x^2y)[/tex]
[tex]\\ \rm\rightarrowtail 3xy^5(4x^2y)[/tex]
[tex]\\ \rm\rightarrowtail 12x^3y^6[/tex]
[tex]\text{Area of triangle} = \dfrac 12 \times \text{Height} \times \text{Base} \\\\~~~~~~~~~~~~~~~~~~~~~~=\dfrac 12 \cdot 6xy^5\cdot 4x^2y\\\\~~~~~~~~~~~~~~~~~~~~~~=12x^3y^6 ~ \text{units}^2[/tex]