#21
To find the area of rectangle, you multiple length by width, so that would be [tex](\ \sqrt{2x}\ )(\ \sqrt{6x}\ ) = \sqrt{2x\cdot6x} = \sqrt{12x^2} = x\sqrt{4\cdot3} = \boxed{2x\sqrt3}[/tex]
Since we are talking about area \ length, x need to be positive so we don't need to worry about absolute value now.
#22
The area of triangle is [tex]\dfrac12bh[/tex], so that would be
[tex]\dfrac12\sqrt{18y}\cdot\sqrt{2y^4}\\\ \\=\dfrac{y^2}2\sqrt{18y}\cdot\sqrt{2}\\\ \\=\dfrac{y^2}2\sqrt{18y\cdot2}\\\ \\=\dfrac{y^2}2\sqrt{36y}\\\ \\=\dfrac{y^2}2\sqrt{4\cdot9\cdot y}\\\ \\=\boxed{3y^2\sqrt y}\text{ or }\boxed{3\sqrt{y^5}}[/tex]
Hope this helps.
