If a room is 10x10 what is lenght in the middle?

You are going to have to use the Pythagorean theorem

a^2 + b^2 = c^2

Where c is the hypotenuse

so

c^2 = 10^2 + 10^2

c = sqrt( 100+100)

if a room is 10x10. The lenght from the far left angle of the room to the close right angle of the room is given by 10 sqrt(2) = 14.14

You mean the diagonal?

The diagonal and the two walls that meet at a corner form a triangle with a right angle. That means you can use the Pythagorean theorem to find any length you need.

a² + b² = c²

with c being the Hypotenuse (the longest leg of the triangle, in this case that's your diagonal), and a,b being the other two legs.

Now just plug in the values:

10² + 10² = c²

c² = 100 + 100

c = √200 = approx. 14.14 units.

In general, the diagonal of a square (as it is here with 10x10) is always √2 * a.

√2 * 10 = 14.14

Think about it like it would be a square 10 by 10.

You want to find the diagonal of the square.

The special right triangle- 45-45-90 tells you that three sides are x, x, and x root 2.

So the sides of the room is 10, while the diagonal, which is what you need to find, is 10 root 2, which is about 14.14.

It would be the sqrt of 200, or about 14.14.

This is from the formula a^2+b^2=c^2.

You would do 10^2+10^2 or 100+100 or 200.

This number is equal to c^2. Next 200=c^2, so c=sqrt of 200.

14.14

I think Pythagorean theorem would work here. A^2 + B^2 = C^2

10^2 + 10^2= 200

square root of 200 is your answer

10x10 Room

I used A^2 + B^2 = C^2 and got 14.14