Answer:
[tex]\sqrt{8}----- 2units,2units[/tex]
[tex]\sqrt{7} ------\sqrt{5}units,\sqrt{2}units[/tex]
[tex]\sqrt{5}-------1unit,2units[/tex]
[tex]3------2units, \sqrt{5}units[/tex]
Step-by-step explanation:
Use pythagorean's theorem to solve each pair individually.
The instructions say that you have the two sides (a and b) and you have to match it with their hypothenuse.
[tex]Formula: c^2=a^2+b^2[/tex]
1. [tex]a=\sqrt{5}, b=\sqrt{2}[/tex]
Plug this into the formula.
[tex]c=\sqrt{(\sqrt{5} )^2+(\sqrt{2})^2 }[/tex]
The square here eliminates the roots.
[tex]c=\sqrt{5+2}\\ c=\sqrt{7}[/tex]
2. [tex]a=\sqrt{3}, b=4[/tex]
[tex]c=\sqrt{(\sqrt{3} )^2+(4)^2}\\ c=\sqrt{3+16}\\ c=\sqrt{19}[/tex]
3. [tex]a=2, b=\sqrt{5}[/tex]
[tex]c=\sqrt{(2)^2+(\sqrt{5})^2 }\\ c=\sqrt{4+5}\\ c=\sqrt{9}\\ c=3[/tex]
4. [tex]a=2, b=2[/tex]
[tex]c=\sqrt{(2)^2+(2)^2}\\ c=\sqrt{4+4}\\ c=\sqrt{8}[/tex]
5. [tex]a=1, b=2[/tex]
[tex]c=\sqrt{(1)^2+(2)^2}\\ c=\sqrt{1+4}\\ c=\sqrt{5}[/tex]
