Area of a Parallelogram: Using additional geometric shapes

Let's first calculate the sides of the rectangle:

$AEDF=AE\times ED$

Let's input the known data:

$35=7\times ED$

Let's divide the two legs by 7:

$ED=5$

Since AEDF is a rectangle, we can claim that:

ED=FD=7

Let's calculate side CD:

$2+7=9$

Let's calculate the area of parallelogram ABCD:

$ABCD=CD\times ED$

Let's input the known data:

$ABCD=9\times5=45$

First, we add letters as reference points:

Let's observe points A and B.

We know that two tangent lines to a circle that start from the same point are parallel to each other.

Therefore:

$AE=AF=3$
$BG=BF=6$

And from here we can calculate:

$AB=AF+FB=3+6=9$

Now we need the height of the parallelogram.

We know that F is tangent to the circle, so the diameter that comes out of point F will also be the height of the parallelogram.

It is also known that the diameter is equal to two radii.

Since the circumference is 25.13.

Circumference formula:$2\pi R$
We replace and solve:

$2\pi R=25.13$
$\pi R=12.565$
$R\approx4$

The height of the parallelogram is equal to two radii, that is, 8.

And from here you can calculate with a parallelogram area formula:

$AlturaXLado$

$9\times8\approx72$

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.

$10\times11\approx110$

First, we add letters as reference points:

Let's observe points A and B.

We know that two tangent lines to a circle that start from the same point are parallel to each other.

Therefore:

$AE=AF=3$
$BG=BF=6$

From here we can calculate:

$AB=AF+FB=3+6=9$

Now we need the height of the parallelogram.

We know that F is tangent to the circle, so the diameter that comes out of point F will also be the height of the parallelogram.

It is also known that the diameter is equal to two radii.

It is known that the circumference of the circle is 25.13.

Formula of the circumference:$2\pi R$
We replace and solve:

$2\pi R=25.13$
$\pi R=12.565$
$R\approx4$

The height of the parallelogram is equal to two radii, that is, 8.

And from here it is possible to calculate the area of the parallelogram:

$\text{Lado }x\text{ Altura}$$9\times8\approx72$

Now, we calculate the area of the circle according to the formula:$\pi R^2$

$\pi4^2=50.26$

Now, subtract the area of the circle from the surface of the trapezoid to get the answer:

$72-56.24\approx21.73$

Let's try to calculate the area in two ways.

In the first method, we will try to use the rhombus ECFD:

Let's try to calculate according to the formula $area=DC\times h_{DC}$

We will lower a height to DC and see that we do not have enough data to calculate, so we will not be able to calculate the area of the parallelogram using the rhombus.

In the second method , we will try to use the circle:

$area=BC\times h_{BC}$We will lower a height to BC and see that we do not have enough data to calculate, so we will not be able to calculate the area of the parallelogram using the circle.

From this it follows that we do not have enough data to calculate the area of parallelogram ABCD and therefore the exercise cannot be solved.