Area of the Square: True / false

Let's remember that the area of the square is equal to the side of the square raised to the second power.

Also, the perimeter of the square is equal to the side multiplied by 4.

We calculate the area of the square:
$A=3^2=9$

We calculate the perimeter of the square:

$3\times4=12$

Therefore, the perimeter is greater than the area of the square.

Remember that the area of the square is equal to the side of the square raised to the 2nd power.

Remember that the perimeter of the square is equal to the side of the square multiplied by 4.

We calculate the area of the square:

$A=2^2=4$

We calculate the perimeter of the square:

$2\times4=8$

Therefore, the perimeter is greater than the area.

Remember that the area of the square is equal to the side of the square raised to the 2nd power.

Remember that the perimeter of the square is equal to the side of the square multiplied by 4.

We calculate the area of the square:

$A=4.5^2=20.25$

We calculate the perimeter of the square:

$4.5\times4=18$

Therefore, the perimeter is not greater than the area.

Let's remember that the area of the square is equal to the side of the square raised to the second power.

Let's remember that the perimeter of the square is equal to the side multiplied by 4.

We calculate the area of the square:

$A=4^2=16$

We calculate the perimeter of the square:

$4\times4=16$

Therefore, the perimeter is not greater than the area (they are equal).

Remember that the area of the square is equal to the side of the square raised to the 2nd power.

Remember that the perimeter of the square is equal to the side of the square multiplied by 4.

We calculate the area of the square:

$A=10^2=100$

We calculate the perimeter of the square:

$10\times4=40$

Therefore, the perimeter is not greater than the area.

Given that we have one side equal to 6, we can multiply and calculate the area:

$6^2=36$

The perimeter can also be calculated:

$6\times4=24$

From this we can conclude that the area of the square is greater than its perimeter: 36 > 24

Remember that the area of the square is equal to the side of the square raised to the 2nd power.

Remember that the perimeter of the square is equal to the side of the square multiplied by 4.

We calculate the area of the square:

$A=9^2=81$

We calculate the perimeter of the square:$9\times4=36$

Therefore, the area is greater than the perimeter.

Let's remember that the area of the square is equal to the side of the square raised to the second power.

Keep in mind that the circumference of the square is equal to the side of the square times 2.

Let's remember that the perimeter of the square is equal to the side multiplied by 4.

Calculate the area of the square:

$A=5^2=25$

Then calculate the perimeter of the square:

$5\times4=20$

Therefore, the perimeter is not greater than the area.

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