Square Geometry Problem: Comparing Diagonal Sums vs Side Lengths in a 4-Unit Square

Look at the square below:

Is the sum of the two diagonals greater than the sum of the 3 sides of the square?

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.

$11.31 < 12$