Answer:
[tex]\sf 2(3\sqrt{8}+4\sqrt{20}+2\sqrt{24})[/tex]
[tex]\sf 2(3\sqrt{8})+2(4\sqrt{20}+2\sqrt{24})[/tex]
[tex]\sf 12 \sqrt{2}+16 \sqrt{5}+8 \sqrt{6}[/tex]
[tex]\sf 6 \cdot 8^{\frac{1}{2}}+8 \cdot 20^{\frac{1}{2}}+4 \cdot 24^{\frac{1}{2}}[/tex]
[tex]\sf 12 \cdot 2^{\frac{1}{2}}+16 \cdot 5^{\frac{1}{2}}+8 \cdot 6^{\frac{1}{2}}[/tex]
Step-by-step explanation:
Perimeter of a rectangle
Perimeter = 2(width + length)
Given information:
[tex]\textsf{width} = 3\sqrt{8}[/tex]
[tex]\textsf{length} = 4\sqrt{20}+2\sqrt{24}[/tex]
Equivalent Expression 1
Substitute the given information into the formula:
[tex]\sf Perimeter = 2(3\sqrt{8}+4\sqrt{20}+2\sqrt{24})[/tex]
Equivalent Expression 2
Using the distributive property law, this can also be written as:
[tex]\sf Perimeter = 2(3\sqrt{8})+2(4\sqrt{20}+2\sqrt{24})[/tex]
Equivalent Expression 3
Distribute the parentheses and simplify the radicals:
[tex]\begin{aligned}\sf Perimeter & = \sf 2(3\sqrt{8}+4\sqrt{20}+2\sqrt{24})\\& = \sf 6\sqrt{8}+8\sqrt{20}+4\sqrt{24}\\& = \sf 6\sqrt{4 \cdot 2}+8\sqrt{4 \cdot 5}+4\sqrt{4 \cdot 6}\\& = \sf 6\sqrt{4}\sqrt{2}+8\sqrt{4}\sqrt{5}+4\sqrt{4}\sqrt{6}\\& = \sf 6 \cdot 2 \sqrt{2}+8 \cdot 2 \sqrt{5}+4 \cdot 2 \sqrt{6}\\& = \sf 12 \sqrt{2}+16 \sqrt{5}+8 \sqrt{6}\end{aligned}[/tex]
Equivalent Expression 4
Distribute the parentheses and rewrite the square roots as [tex]\sf \sqrt{a}=a^{\frac{1}{2}}[/tex] :
[tex]\begin{aligned}\sf Perimeter & = \sf 2(3\sqrt{8}+4\sqrt{20}+2\sqrt{24})\\& = \sf 6\sqrt{8}+8\sqrt{20}+4\sqrt{24}\\ & = \sf 6 \cdot 8^{\frac{1}{2}}+8 \cdot 20^{\frac{1}{2}}+4 \cdot 24^{\frac{1}{2}}\end{aligned}[/tex]
Equivalent Expression 5
Rewrite the square roots as [tex]\sf \sqrt{a}=a^{\frac{1}{2}}[/tex] :
[tex]\sf Perimeter=12 \cdot 2^{\frac{1}{2}}+16 \cdot 5^{\frac{1}{2}}+8 \cdot 6^{\frac{1}{2}}[/tex]
