The Ratio of Similarity: Using Pythagoras' theorem

Let's begin by observing triangle FCE and calculate side FC using the Pythagorean theorem:

$EC^2+FC^2=EF^2$

Let's begin by substituting all the known values into the formula:

$8^2+FC^2=10^2$

$64+FC^2=100$

$FC^2=100-64$

$FC^2=36$

Let's take the square root:

$FC=6$

Since we know that the triangles overlap:

$\frac{AD}{FC}=\frac{DE}{CE}=\frac{AE}{FE}$

Let's again substitute the known values into the formula:

$\frac{AD}{6}=\frac{16}{8}$

$AD=2\times6=12$

Finally let's calculate side CD:

$16+8=24$

Since in a rectangle each pair of opposite sides are equal, we can calculate the perimeter of rectangle ABCD as follows:

$12+24+12+24=24+48=72$

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.

$10\times11\approx110$