Answer
$3y^2\sqrt y$
Work Step by Step
The area of a triangle is $\frac{1}{2}bh$, where b is the base and h is the height.
$a=\frac{1}{2}\times\sqrt{18y}\times\sqrt{2y^4}\longrightarrow$ factor out the perfect squares
$a=\frac{1}{2}\times\sqrt{9\times y\times2}\times\sqrt{2\times y^4}\longrightarrow$ multiplication property of square roots
$a=\frac{1}{2}\times\sqrt9\times\sqrt y\times(\sqrt2\times\sqrt2)\times\sqrt {y^4}\longrightarrow$ simplify
$a=\frac{1}{2}\times3\times\sqrt y\times2\times y^2\longrightarrow$ simplify; combine like terms
$a=3y^2\sqrt y$
