Question

I really need help guys can some one plz help me it's for 10 points

Answer 1

Answer:

$\sqrt{8}----- 2units,2units$

$\sqrt{7} ------\sqrt{5}units,\sqrt{2}units$

$\sqrt{5}-------1unit,2units$

$3------2units, \sqrt{5}units$

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.

$Formula: c^2=a^2+b^2$

1. $a=\sqrt{5}, b=\sqrt{2}$

Plug this into the formula.

$c=\sqrt{(\sqrt{5} )^2+(\sqrt{2})^2 }$

The square here eliminates the roots.

$c=\sqrt{5+2}\\ c=\sqrt{7}$

2. $a=\sqrt{3}, b=4$

$c=\sqrt{(\sqrt{3} )^2+(4)^2}\\ c=\sqrt{3+16}\\ c=\sqrt{19}$

3. $a=2, b=\sqrt{5}$

$c=\sqrt{(2)^2+(\sqrt{5})^2 }\\ c=\sqrt{4+5}\\ c=\sqrt{9}\\ c=3$

4. $a=2, b=2$

$c=\sqrt{(2)^2+(2)^2}\\ c=\sqrt{4+4}\\ c=\sqrt{8}$

5. $a=1, b=2$

$c=\sqrt{(1)^2+(2)^2}\\ c=\sqrt{1+4}\\ c=\sqrt{5}$