Area of a Parallelogram - Examples, Exercises and Solutions

To solve the exercise, we need to remember the formula for the area of a parallelogram:

Side * Height perpendicular to the side

We can identify that in the diagram, although it's not presented in the way we're familiar with, we are given the two essential pieces of information -

Side = 6

Height = 5

Let's substitute into the formula and calculate:

6*5=30

And that's the solution!

Since we know that ABCD is a parallelogram, according to the properties of parallelograms, each pair of opposite sides are equal and parallel.

Therefore $CD=AB=10$

We will calculate the area of the parallelogram using the formula of side multiplied by the height drawn from that side, so the area of the parallelogram is equal to:

$S_{ABCD}=10\times7=70cm^2$

The formula for the area of a parallelogram:

Height * The side to which the height descends.

We replace in the formula all the known data, including the area:

5*DC = 70

We divide by 5:

DC = 70/5 = 14

And that's how we reveal the unknown!

First, let's remember that the perimeter of a parallelogram is the sum of its sides,

which is

AB+BC+CD+DA

We recall that in a parallelogram, opposite sides are equal, so
BC=AD=6

Let's substitute in the formula:

2AB+12=47

2AB=35

AB=17.5

Now, after finding the missing sides, we can continue to calculate the area.

Remember, the area of a parallelogram is side*height to the side.

17.5*8= 140

We are told that ABCD is a parallelogram,$AB=CD=8$According to the properties of a parallelogram, each pair of opposite sides are equal and parallel.

Hence to find AE we will need to use the area given to us in the formula in order to determine the area of the parallelogram:

$S=DC\times AE$

$40=8\times AE$

We divide both sides of the equation by 8:

$8AE:8=40:8$

$AE=5$

In this exercise, we are given two heights and two sides.

It is important to keep in mind: The external height can also be used to calculate the area

Therefore, we can perform the operation of the following exercise:

The height BF * the side AD

8*6

The height BE the side DC
4*12

 The solution of these two exercises is 48, which is the area of the parallelogram.

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

$\sqrt{21}\times9=41.24$

Keep in mind that AB is 3 cm greater than AE, so we must pay attention to the data when we put the formula to calculate the parallelogram:

Height multiplied by the side of the height:

$AB\times AE=S$

We will mark AE with the letter a and therefore AB will be a+3:

$a\times(a+3)=32$

We open the parentheses:

$a^2+3a=32$

We use the trinomial/roots formula:

$a^2+3a-32=0$$(a+8)(a-5)=0$

That means we have two options:

$a=-8,a=5$

Since it is not possible to place a negative side in the formula to calculate the area$a=5$

Now we can calculate the sides:

$AE=5$

$AB=5+3=8$

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:

$ABCD=9\times5=45$

The perimeter of the parallelogram is calculated as follows:

$S_{ABCD}=AB+BC+CD+DA$ Since ABCD is a parallelogram, each pair of opposite sides is equal, and therefore, AB=DC and AD=BC

According to the figure that the side of the parallelogram is 2 times larger than the side adjacent to it, it can be argued that$AB=DC=2BC$

We inut the data we know in the formula to calculate the perimeter:

$P_{ABCD}=2BC+BC+2BC+BC$

We replace the given perimeter in the formula and add up all the BC coefficients accordingly:

$24=6BC$

We divide the two sections by 6

$24:6=6BC:6$

$BC=4$

We know that$AB=DC=2BC$We replace the data we obtained (BC=4)

$AB=DC=2\times4=8$

As ABCD is a parallelogram, then all pairs of opposite sides are equal, therefore BC=AD=4

To find EC we use the formula:$A_{ABCD}=AB\times EC$

We replace the existing data:

$24=8\times EC$

We divide the two sections by 8$24:8=8EC:8$

$3=EC$

Let's call CE as X

According to the data

$AB=x+2,AD=x-3$

The perimeter of the parallelogram:

$2(AB+AD)$

$38=2(x+2+x-3)$

$38=2(2x-1)$

$38=4x-2$

$38+2=4x$

$40=4x$

$x=10$

Now it can be argued:

$AD=10-3=7,CE=10$

The area of the parallelogram:

$CE\times AD=10\times7=70$

To calculate the ratio between the sides we will use the existing figure:

$\frac{A_{AED}}{A_{ABCE}}=\frac{1}{3}$

We calculate the ratio between the sides according to the formula to find the area and then replace the data.

We know that the area of triangle ADE is equal to:

$A_{ADE}=\frac{h\times DE}{2}$

We know that the area of the parallelogram is equal to:

$A_{ABCD}=h\times EC$

We replace the data in the formula given by the ratio between the areas:

$\frac{\frac{1}{2}h\times DE}{h\times EC}=\frac{1}{3}$

We solve by cross multiplying and obtain the formula:

$h\times EC=3(\frac{1}{2}h\times DE)$

We open the parentheses accordingly:

$h\times EC=1.5h\times DE$

We divide both sides by h:

$EC=\frac{1.5h\times DE}{h}$

We simplify to h:

$EC=1.5DE$

Therefore, the ratio between is: $\frac{EC}{DE}=\frac{1}{1.5}$

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:

$11\times4=44$

Given that angle ACB is equal to angle CBE, it follows that AC is parallel to BE

since alternate angles between parallel lines are equal.

As we know that ABCD is a parallelogram, AB is parallel to DC and therefore AB is also parallel to CE since it is a line that continues DC.

Given that AC is parallel to BE and, in addition, AB is parallel to CE, it can be argued that ABCE is a parallelogram and, therefore, each pair of opposite sides in a parallelogram are parallel and equal.

From this it is concluded that AB=CE=9

Now we calculate the area of the parallelogram ABCD according to the data.

$S_{ABCD}=AB\times BF$

We replace the data accordingly:

$S_{ABCD}=9\times6=54$

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:

$AlturaXLado$

$9\times8\approx72$