Question

I need help with these questions ASAP!!!!!

Answer 1

Answer:

1.) $x=8\sqrt{2}$

2.) $x=14$

3.) $x=18\sqrt{2}$

4.) $x=9\sqrt{2}$

5.) $x=4\sqrt{2}$

6.) $x=5\sqrt{2}$

7.) $x=12\sqrt{2}$

Step-by-step explanation:

Use the 45°-45°-90° formula:

$hypotenuse=\sqrt{2}*leg$

1.) Insert values:

$x=\sqrt{2}*8$

Simplify:

$x=8\sqrt{2}$

2.) In a 45°-45°-90° angle, the legs have the same value.

$x=14$

3.) x is the hypotenuse. Insert values:

$x=\sqrt{2}*18$

Simplify:

$x=18\sqrt{2}$

4.) Insert values:

$18=\sqrt{2}*x$

Divide $\sqrt{2}$ from both sides:

$\frac{18}{\sqrt{2}}=\frac{\sqrt{2}*x}{\sqrt{2}} \\\\\frac{18}{\sqrt{2}}=x$

Rationalize the left side:

$\frac{\sqrt{2}}{\sqrt{2}}*\frac{18}{\sqrt{2}}=x\\\\\frac{18\sqrt{2}}{\sqrt{4}}\\\\\frac{18\sqrt{2}}{2} \\\\9\sqrt{2}=x$

Simplify:

$x=9\sqrt{2}$

5.) Insert values:

$8=\sqrt{2}*x$

Divide $\sqrt{2}$ from both sides and rationalize:

$\frac{8}{\sqrt{2}}=\frac{\sqrt{2}*x }{\sqrt{2}}\\\\\frac{8}{\sqrt{2}}=x$

$\frac{\sqrt{2} }{\sqrt{2} } *\frac{8}{\sqrt{2}}\\\\\frac{8\sqrt{2}}{\sqrt{4}} \\\\\frac{8\sqrt{2}}{2}\\\\4\sqrt{2}=x$

Simplify:

$x=4\sqrt{2}$

6.) 10 is the hypotenuse. Insert values:

$10=\sqrt{2}*x$

Divide $\sqrt{2}$ from both sides and rationalize:

$\frac{10}{\sqrt{2}}=\frac{\sqrt{2}*x}{\sqrt{2}} \\\\\frac{10}{\sqrt{2}} =x$

$\frac{\sqrt{2}}{\sqrt{2}}*\frac{10}{\sqrt{2}}\\\\\frac{10\sqrt{2}}{\sqrt{4}}\\\\\frac{10\sqrt{2}}{2}\\\\5\sqrt{2}=x$

Simplify:

$x=5\sqrt{2}$

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:

$x=\sqrt{2}*12$

Simplify:

$x=12\sqrt{2}$

Finito.