Answer:
[tex]5x^2 y^2[/tex]
Step-by-step explanation:
We need to use the properties shown below to solve this:
1. [tex]\sqrt[n]{x^a} =x^{\frac{a}{n}}[/tex]
2. [tex]\sqrt{x}\sqrt{x} =x[/tex]
3. [tex]\sqrt{x} \sqrt{y}=\sqrt{x*y}[/tex]
Area of a triangle is given by 1/2 * base * height, so we do that and simplify:
[tex]A=\frac{1}{2}(\sqrt{5x^3} )(2\sqrt{5xy^4} )\\A=\frac{1}{2}(5x^3)^{\frac{1}{2}}*2*(5xy^4)^{\frac{1}{2}}\\A=\sqrt{5}x^{\frac{3}{2}}*\sqrt{5}\sqrt{x} } y^2\\A=\sqrt{5} \sqrt{5}x^{\frac{3}{2}} x^{\frac{1}{2}}y^2\\A=5*x^2y^2\\A=5x^2 y^2[/tex]
