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:

$11\times4=44$

First, we must remember the formula for the area of a parallelogram:$\text{Lado }x\text{ Altura}$.

In this case, we will try to find the height CH and the side BC.

We start from the side

First, let's observe the small triangle EBG,

As it is a right triangle, we can use the Pythagorean theorem (

$A^2+B^2=C^2$)

$BG^2+4^2=5^2$

$BG^2+16=25$

$BG^2=9$

$BG=3$

Now, let's start looking for GC.

First, remember that the deltoid has two pairs of equal adjacent sides, therefore:$FC=EC=9$

Now we can also do Pythagoras in the triangle GCE.

$GC^2+4^2=9^2$

$GC^2+16=81$

$GC^2=65$

$GC=\sqrt{65}$

Now we can calculate the side BC:

$BC=BG+GT=3+\sqrt{65}\approx11$

Now, let's observe the triangle BGE and DHC

Angle BGE = 90°
Angle CHD = 90°
Angle CDH=EBG because these are opposite parallel angles.

Therefore, there is a ratio of similarity between the two triangles, so:

$\frac{HD}{BG}=\frac{HC}{GE}$

$\frac{HD}{BG}=\frac{7.5}{3}=2.5$

$\frac{HC}{EG}=\frac{HC}{4}=2.5$

$HC=10$

Now that there is a height and a side, all that remains is to calculate.

$10\times11\approx110$