# Chapter 10 - Radical Expressions and Equations - Mid-Chapter Quiz - Page 619: 22

$3y^2\sqrt y$

#### 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 $a=3y^2\sqrt y$

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