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?

Video Solution

Solution Steps

00:00 Is the sum of diagonals greater than the sum of 3 sides of the square?

00:03 In a square all sides are equal

00:07 We'll use the Pythagorean theorem in triangle BCD

00:11 We'll substitute appropriate values and solve for BD

00:29 This is the length of diagonal BD

00:32 In a square the diagonals are equal

00:36 Let's calculate the sum of diagonals

00:40 We'll substitute appropriate values and solve for the sum

00:46 Let's calculate the sum of 3 sides of the square

00:54 We'll substitute appropriate values in the equation

01:01 And this is the solution to the question

Step-by-Step Solution

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.

$11.31 < 12$