Using the Pythagorean Theorem: 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

$\sqrt{21}\times9=41.24$

DE crosses AB and AC, that is to say:

$AD=DB=\frac{1}{2}AB=\frac{1}{2}\times6=3$

$AE=EC=\frac{1}{2}AC=\frac{1}{2}\times10=5$

Now let's look at triangle ADE, two sides of which we have already calculated.

Now we can find the third side DE using the Pythagorean theorem:

$AD^2+DE^2=AE^2$

We substitute our values into the formula:

$3^2+DE^2=5^2$

$9+DE^2=25$

$DE^2=25-9$

$DE^2=16$

We extract the root:

$DE=\sqrt{16}=4$

Now let's look at triangle ABC, two sides of which we have already calculated.

Now we can find the third side (BC) using the Pythagorean theorem:

$AB^2+BC^2=AC^2$

We substitute our values into the formula:

$6^2+BC^2=10^2$

$36+BC^2=100$

$BC^2=100-36$

$BC^2=64$

We extract the root:

$BC=\sqrt{64}=8$

Now we have all the data needed to calculate the area of the trapezoid DECB using the formula:

(base + base) multiplied by the height divided by 2:

Keep in mind that the height in the trapezoid is DB.

$S=\frac{(4+8)}{2}\times3$

$S=\frac{12\times3}{2}=\frac{36}{2}=18$

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:

$11\times4=44$

To find the area of the trapezoid, you must remember its formula:$\frac{(base+base)}{2}+\text{altura}$We will focus on finding the bases.

To find GF we use the Pythagorean theorem: $A^2+B^2=C^2$ In triangle AFG

We replace:

$3^2+GF^2=5^2$

We isolate GF and solve:

$9+GF^2=25$

$GF^2=25-9=16$

$GF=4$

We will do the same process with side DB in triangle ABD:

$8^2+DB^2=17^2$

$64+DB^2=289$

$DB^2=289-64=225$

$DB=15$

From here there are two ways to finish the exercise:

1. Calculate the area of the trapezoid GFBD, prove that it is equal to the trapezoid EGDC and add them up.

2. Use the data we have revealed so far to find the parts of the trapezoid EFBC and solve.

Let's start by finding the height of GD:

$GD=AD-AG=8-3=5$

Now we reveal that EF and CB:

$GF=GE=4$

$DB=DC=15$

This is because in an isosceles triangle, the height divides the base into two equal parts then:

$EF=GF\times2=4\times2=8$

$CB=DB\times2=15\times2=30$

We replace the data in the trapezoid formula:

$\frac{8+30}{2}\times5=\frac{38}{2}\times5=19\times5=95$

To find the perimeter of the trapezoid, all its sides must be added:

We will focus on finding the bases.

To find GF we use the Pythagorean theorem: $A^2+B^2=C^2$in the triangle AFG

We replace

$3^2+GF^2=5^2$

We isolate GF and solve:

$9+GF^2=25$

$GF^2=25-9=16$

$GF=4$

We perform the same process with the side DB of the triangle ABD:

$8^2+DB^2=17^2$

$64+DB^2=289$

$DB^2=289-64=225$

$DB=15$

We start by finding FB:

$FB=AB-AF=17-5=12$

Now we reveal EF and CB:

$GF=GE=4$

$DB=DC=15$

This is because in an isosceles triangle, the height divides the base into two equal parts so:

$EF=GF\times2=4\times2=8$

$CB=DB\times2=15\times2=30$

All that's left is to calculate:

$30+8+12\times2=30+8+24=62$

Let's look at triangle ADK to calculate side AD:

$AD^2+DK^2=AK^2$

Let's input the given data:

$AD^2+4^2=5^2$

$AD^2+16=25$

We'll move 16 to the other side and change the appropriate sign:

$AD^2=25-16$

$AD^2=9$

We'll take the square root and get:

$AD=3$

Since AD is a side of rectangle ABCD, we can calculate side AB as follows:

$S=AB\times AD$

Let's input the given data:

$24=3\times AB$

We'll divide both sides by 3:

$AB=8$

There are two ways to solve the exercise:

It is possible to drop a height from one of the vertices, as we know

In an equilateral triangle, the height intersects the base,

This creates a right triangle whose two sides are 6 and 3,

Using the Pythagorean theorem$A^2+B^2=C^2$

We can find the length of the missing side.

$3^2+X^2=6^2$

We convert the formula

$6^2-3^2=X^2$

$36-9=27$

Therefore, the height of the triangle is equal to:$\sqrt{27}$

From here we calculate with the usual formula for the area of a triangle.

$\frac{6\times\sqrt{27}}{2}=15.588$

Option B for the solution is through the formula for the area of an equilateral triangle:

$S=\frac{\sqrt{3}\times X^2}{4}$

Where X is one of the sides.

$\frac{\sqrt{3}\times6^2}{4}=\frac{62.353}{4}=15.588$

To find the missing side, we use the Pythagorean theorem in the upper triangle.

Since the triangle is isosceles, we know that the length of both sides is 7.

Therefore, we apply Pythagoras$A^2+B^2=C^2$$7^2+7^2=49+49=98$

Therefore, the area of the missing side is:$\sqrt{98}$

The area of a rectangle is the multiplication of the sides, therefore:

$\sqrt{98}\times10=98.99\approx99$