Area of a Parallelogram: Using Pythagoras' theorem

To find the area,

first, the height of the parallelogram must be found.

To conclude, let's take a look at triangle EBC.

Since we know it is a right triangle (since it is the height of the parallelogram)

the Pythagorean theorem can be used:

$a^2+b^2=c^2$

In this case: $EB^2+EC^2=BC^2$

We place the given information: $2^2+EC^2=5^2$

We isolate the variable:$EC^2=5^2+2^2$

We solve:$EC^2=25-4=21$

$EC=\sqrt{21}$

Now all that remains is to calculate the area.

It is important to remember that for this, the length of each side must be used.
That is, AE+EB=2+7=9

$\sqrt{21}\times9=41.24$

In the first step, we must find the length of EC, which we will identify with an X.

We know that the perimeter of a rectangle is the sum of all its sides (AE+EC+CF+FA),

Since in a rectangle the opposite sides are equal, the formula can also be written like this: 2AE=2EC.

We replace the known data:

$2\times8+2X=24$

$16+2X=24$

We isolate X:

$2X=8$

and divide by 2:

$X=4$

Now we can use the Pythagorean theorem to find EB.

(Pythagoras: $A^2+B^2=C^2$)

$EB^2+4^2=5^2$

$EB^2+16=25$

We isolate the variable

$EB^2=9$

We take the square root of the equation.

$EB=3$

The area of a parallelogram is the height multiplied by the side to which the height descends, that is$AB\times EC$.

$AB=\text{ AE}+EB$

$AB=8+3=11$

And therefore we will apply the area formula:

$11\times4=44$