Perimeter of a Rectangle: Using Pythagoras' theorem

Since in a rectangle every pair of opposite sides are equal to each other, we can claim that:

$AD=BC=11$

We can calculate side FC:

$11-3=FC$

$8=FC$

Let's focus on triangle FCE and calculate side CE using the Pythagorean theorem:

$CF^2+CE^2=FE^2$

Let's substitute the known values into the formula:

$8^2+CE^2=17^2$

$64+CE^2=289$

$CE^2=289-64$

$CE^2=225$

Let's take the square root:

$CE=15$

Since CE equals AB and in a rectangle every pair of opposite sides are equal to each other, we can claim that:

$CE=AB=CD=15$

Now we can calculate the perimeter of the rectangle:

$11+15+11+15=22+30=52$

Since in a rectangle every pair of opposite sides are equal to each other, we can claim that:

$BC=AD=6$

Therefore:

$$$AD-AE=ED$

Let's substitute the known data into the formula:

$6-3=ED$

$3=ED$

Let's focus on triangle EDF and find side DF using the Pythagorean theorem:

$ED^2+DF^2=EF^2$

Let's substitute the known data into the formula:

$3^2+DF^2=5^2$

$9+DF^2=25$

$DF^2=25-9$

$DF^2=16$

Let's find the square root:

$DF=4$

Side DC=8

Since in a rectangle every pair of opposite sides are equal to each other:

$AD=BC=6$

$AB=CD=8$

Now we can calculate the perimeter of the rectangle by adding all sides together:

$6+8+6+8=12+16=28$

Let's focus on triangle BCD in order to find side DC

We'll use the Pythagorean theorem and input the known data:

$BC^2+DC^2=BD^2$

$6^2+DC^2=10^2$

$DC^2=100-36=64$

Let's take the square root:

$DC=8$

Since in a rectangle each pair of opposite sides are equal to each other, we can state that:

$DC=AB=8$

$BC=AD=6$

Now we can calculate the perimeter of the rectangle by adding all sides together:

$8+6+8+6=16+12=28$

Based on the given data, we can claim that:

$OF=OE=3$

$EF=6$

$AB=EF=DC=6$

We'll find side BF using the Pythagorean theorem in triangle BFO:

$OF^2+BF^2=BO^2$

Let's substitute the known values into the formula:

$3^2+BF^2=5^2$

$9+BF^2=25$

$BF^2=25-9$

$BF^2=16$

Let's take the square root:

$BF=4$

Since the triangles overlap:

$BF=DE=4=FC$

From this, we can calculate side BC:

$BC=4+4=8$

Since in a rectangle, each pair of opposite sides are equal to each other, we can claim that AD also equals 8

Now we can calculate the perimeter of rectangle ABCD by adding all sides together:

$6+8+6+8=12+16=28$

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:

$11\times4=44$

Let's first look at triangle BCE and calculate side EC using the Pythagorean theorem:

$BC^2+EC^2=BE^2$

Let's substitute the known values:

$6^2+EC^2=10^2$

$36+EC^2=100$

$EC^2=100-36$

$EC^2=64$

Let's find the square root:

$EC=8$

Let's calculate the area of triangle BCE:

$S=\frac{BC\times EC}{2}$

Let's substitute the known values:

$S=\frac{6\times8}{2}=\frac{48}{2}=24$

According to the given data, the area of triangle BCE is one-third of rectangle ABCD's area, therefore:

$24=\frac{1}{3}$

Let's multiply by 3:

$S=3\times24=72$

The area of the rectangle equals 72

Now let's find side CD

We know that the area of a rectangle equals length times width, meaning:

$S=BC\times DC$

Let's substitute the known values in the formula:

$72=6\times CD$

Let's divide both sides by 6:

$CD=12$

Since in a rectangle opposite sides are equal, AB also equals 12

Now we can calculate the perimeter of rectangle ABCD:

$12+6+12+6=24+12=36$

Let's look at triangle FCE and calculate side FC using the Pythagorean theorem:

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

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

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

$AD=2\times6=12$

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

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

Since the triangles are equal to each other, we can claim that:

$AE=FC$

$AG=CH=8$

$EG=FH=10$

Now let's calculate side AB:

$8+5=13$

Since in a rectangle each pair of opposite sides are equal to each other:

$AB=CD=13$

We can also claim that:

$DH=13-8=5$

Side EF is also equal in length to sides AB and CD which are equal to 13

Now let's calculate side FC using the Pythagorean theorem in triangle FCH:

$HC^2+FC^2=HF^2$

Let's input the known data:

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

$64+FC^2=100$

$FC^2=100-64$

$FC^2=36$

Let's take the square root:

$FC=6$

Now we can calculate the perimeter of rectangle EFCD by adding all sides together:

$13+6+13+6=26+12=38$