Answer:
Triangle has the greatest perimeter
Step-by-step explanation:
Triangle:
[tex]2\sqrt{48}=2\sqrt{2*2*2*2*3}=2*2*2\sqrt{3}=8\sqrt{3}\\\\\\3\sqrt{75}=3*\sqrt{3*5*5}=3*5*\sqrt{3}=15\sqrt{3}\\\\\\6\sqrt{12}=6*\sqrt{2*2*3}=6*2*\sqrt{3}=12\sqrt{3}\\\\\\2\sqrt{108}=2*\sqrt{2*2*3*3*3}=2*2*3*\sqrt{3}=12\sqrt{3}\\\\\\3\sqrt{147}=3*\sqrt{3*7*7}=3*7\sqrt{3}=21\sqrt{3}\\\\Perimeter=3\sqrt{147}+2\sqrt{108}+6\sqrt{12}+2\sqrt{48}+3\sqrt{75}\\\\ =21\sqrt{3}+12\sqrt{3}+12\sqrt{3}+8\sqrt{3}+15\sqrt{3}\\\\=(21+12+12+8+15)\sqrt{3}\\\\=68\sqrt{3}\\[/tex]
= 68 * 1.732 = 117.78
Square:
[tex]\frac{1}{3}\sqrt{63}=\frac{1}{3}\sqrt{3*3*7}=\frac{1}{3}*3*\sqrt{7}=\sqrt{7}\\\\Perimeter=4*side=4*\sqrt{7}=4\sqrt{7} = 4* 2.6458 = 10.58[/tex]
Rectangle:
[tex]\frac{1}{4}\sqrt{100}=\frac{1}{4}\sqrt{10*10}=\frac{1}{4}*10=\frac{5}{2}\\\\\\Perimeter=\frac{1}{4}\sqrt{100}+2\sqrt{48}+4\sqrt{75}+\frac{1}{4}\sqrt{100}+2\sqrt{48}+4\sqrt{75}\\\\=\frac{5}{2}+8\sqrt{3}+20\sqrt{3}+\frac{5}{2}+8\sqrt{3}+20\sqrt{3}\\\\=\frac{5+5}{2}+(8+8+20+20)\sqrt{3} \\\\=\frac{10}{2}+56\sqrt{3}+\\\\=5+56\sqrt{3}\\\\=5+56*1.732=5+96.99=102[/tex]
