Calculate Square Perimeter Using Diagonal Expression: 3√2×(3²-2³)-2√2

ABCD is a square.

The length of the diagonal:
$3\sqrt{2}\times\left(3^2-2^3\right)-2\sqrt{2}$

What is the perimeter of the square ABCD?

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.