Answer:
1.) [tex]x=8\sqrt{2}[/tex]
2.) [tex]x=14[/tex]
3.) [tex]x=18\sqrt{2}[/tex]
4.) [tex]x=9\sqrt{2}[/tex]
5.) [tex]x=4\sqrt{2}[/tex]
6.) [tex]x=5\sqrt{2}[/tex]
7.) [tex]x=12\sqrt{2}[/tex]
Step-by-step explanation:
Use the 45°-45°-90° formula:
[tex]hypotenuse=\sqrt{2}*leg[/tex]
1.) Insert values:
[tex]x=\sqrt{2}*8[/tex]
Simplify:
[tex]x=8\sqrt{2}[/tex]
2.) In a 45°-45°-90° angle, the legs have the same value.
[tex]x=14[/tex]
3.) x is the hypotenuse. Insert values:
[tex]x=\sqrt{2}*18[/tex]
Simplify:
[tex]x=18\sqrt{2}[/tex]
4.) Insert values:
[tex]18=\sqrt{2}*x[/tex]
Divide [tex]\sqrt{2}[/tex] from both sides:
[tex]\frac{18}{\sqrt{2}}=\frac{\sqrt{2}*x}{\sqrt{2}} \\\\\frac{18}{\sqrt{2}}=x[/tex]
Rationalize the left side:
[tex]\frac{\sqrt{2}}{\sqrt{2}}*\frac{18}{\sqrt{2}}=x\\\\\frac{18\sqrt{2}}{\sqrt{4}}\\\\\frac{18\sqrt{2}}{2} \\\\9\sqrt{2}=x[/tex]
Simplify:
[tex]x=9\sqrt{2}[/tex]
5.) Insert values:
[tex]8=\sqrt{2}*x[/tex]
Divide [tex]\sqrt{2}[/tex] from both sides and rationalize:
[tex]\frac{8}{\sqrt{2}}=\frac{\sqrt{2}*x }{\sqrt{2}}\\\\\frac{8}{\sqrt{2}}=x[/tex]
[tex]\frac{\sqrt{2} }{\sqrt{2} } *\frac{8}{\sqrt{2}}\\\\\frac{8\sqrt{2}}{\sqrt{4}} \\\\\frac{8\sqrt{2}}{2}\\\\4\sqrt{2}=x[/tex]
Simplify:
[tex]x=4\sqrt{2}[/tex]
6.) 10 is the hypotenuse. Insert values:
[tex]10=\sqrt{2}*x[/tex]
Divide [tex]\sqrt{2}[/tex] from both sides and rationalize:
[tex]\frac{10}{\sqrt{2}}=\frac{\sqrt{2}*x}{\sqrt{2}} \\\\\frac{10}{\sqrt{2}} =x[/tex]
[tex]\frac{\sqrt{2}}{\sqrt{2}}*\frac{10}{\sqrt{2}}\\\\\frac{10\sqrt{2}}{\sqrt{4}}\\\\\frac{10\sqrt{2}}{2}\\\\5\sqrt{2}=x[/tex]
Simplify:
[tex]x=5\sqrt{2}[/tex]
7.) Draw the figure like the squares in problems 3 and 10. The problem says that the perimeter is 48, so divide 48 by 4, which is 12. A side is 12 meters (or a leg). The diagonal is the hypotenuse of a triangle. Insert values:
[tex]x=\sqrt{2}*12[/tex]
Simplify:
[tex]x=12\sqrt{2}[/tex]
Finito.
