Area of the Square: Using Pythagoras' theorem

Let's look at triangle BCD, let's calculate the diagonal by the Pythagorean theorem:

$DC^2+BC^2=BD^2$

As we are given one side, we know that the other sides are equal to 4, so we will replace accordingly in the formula:

$4^2+4^2=BD^2$

$16+16=BD^2$

$32=BD^2$

We extract the root:$BD=AC=\sqrt{32}$

Now we calculate the sum of the diagonals:

$2\times\sqrt{32}=11.31$

Now we calculate the sum of the 3 sides of the square:

$4\times3=12$

And we reveal that the sum of the two diagonals is less than the sum of the 3 sides of the square.

$11.31 < 12$

The problem involves the square ABCD, and we need to determine its perimeter, given the expression for the length of its diagonal. Here's the step-by-step solution:

Let's denote the side of the square ABCD as $s$. The diagonal of a square can be calculated using Pythagoras' theorem as:

• $d = s\sqrt{2}$

The problem provides an expression for the length of the diagonal:

• $3\sqrt{2}\times(3^2-2^3)-2\sqrt{2}$

Let's simplify this expression step by step.

First, calculate the powers:

• $3^2 = 9$

• $2^3 = 8$

Subtract these values:

• $3^2 - 2^3 = 9 - 8 = 1$

Substitute back into the expression for the diagonal:

• $3\sqrt{2} \times 1 - 2\sqrt{2}$

This simplifies to:

• $3\sqrt{2} - 2\sqrt{2}$

• $(3 - 2)\sqrt{2} = 1\sqrt{2} = \sqrt{2}$

So, the length of the diagonal is $\sqrt{2}$.

We know from the formula for the diagonal of a square that $d = s\sqrt{2}$. Given $d = \sqrt{2}$, we can equate:

• $s\sqrt{2} = \sqrt{2}$

Thus:

• $s = 1$

Therefore, the perimeter of the square ABCD is:

• $4 \times s = 4 \times 1 = 4$

Hence, the perimeter of the square ABCD is 4.