(i'm so lost i need with this ! )
the area of a square is calculated with the formula a = s^2, where s is the length of one side of the square.

given that the area of the square shown below is 25cm^2.

answer the following questions:

1. what is the length of any side of the square?

2. what is the length of the diagonal of the square (labeled x in the figure below)? hint: substitute the side length you just found in the pythagorean theorem a^2 + b^2 = c^2.

(be sure to show all your work in complete sentences! )

The side length is 5 cm

The diagonal is 5 sqrt(2) cm

Step-by-step explanation:

A = s^2

We know the area is 25

25 = s^2

Take the square root of each side

sqrt(25)= sqrt(s^2)

5 =s

The side length is 5 cm

2  We use the Pythagorean theorem a^2 + b^2 = c^2.  to find the diagonal

a^2 + b^2 = c^2

5^2 + 5^2 = c^2

25+25 = c^2

50 = c^2

Take the square root of each side.

sqrt(50) = sqrt(c^2)

sqrt(25*2) = sqrt(c^2)

sqrt(25) sqrt(2) = c

5sqrt(2) =c

The diagonal is 5 sqrt(2)

-3

step-by-step explanation:

a drawing really does here:

so i draw the horizontal line y=2 and the vertical line x=-2 at this intersection i put b (this is where my right angle is going to be).   so point b is at (-2,2).   point a is 5 units long and is also on y=2 so this point has to be (3,2) or (-7,2) -doesn't matter i will go with (3,2).

now we know c is at (-2,y) we don't know the y yet bet we know that c is on x=-2.

we have the area of the triangle is .5bh=12.5

we know the base length is 5 (ab)

we have the height is (2-y) if you look at your

plug in

.5(5)(2-y)=12.5

solve for y to get the y-coordinate for c

2.5(2-y)=12.5

divide both sides by 2.5 giving:

2-y=5

now subtract 2 on both sides

-y=3

y=-3

uhhh idk u yolo?