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!

According to the properties of the parallelogram: the diagonals intersect each other.

From the data in the drawing, it follows that diagonal AC and diagonal BD are divided into two equal parts, that is, the diagonals intersect each other:

$AO=OC=8$

$DO=OB=10$

Therefore, the quadrilateral is actually a parallelogram.

Let's review the property: a quadrilateral in which two pairs of opposite angles are equal is a parallelogram.

From the data in the drawing, it follows that:

$D=B=60$

$A=C=120$

Therefore, the quadrilateral is actually a parallelogram.

According to the properties of a parallelogram, any two opposite sides will be equal to each other.

From the data, it can be observed that only one pair of opposite sides are equal and therefore the quadrilateral is not a parallelogram.

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:

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

In a parallelogram, we know that each pair of opposite sides are equal to each other.

The data shows that only one pair of sides are equal to each other:

$AB=DC=8$

Now we try to see that the additional pair of sides are equal to each other.

We replace$x=8$for each of the sides:

$AD=2\times8+9$

$AD=16+9$

$AD=25$

$BC=8+5$

$BC=13$

That is, we find that the pair of opposite sides are not equal to each other:

$25\ne13$

Therefore, the quadrilateral is not a parallelogram.

Remember that in a parallelogram each pair of opposite angles are equal to each other.

The data shows that only one pair of angles are equal to each other:

$D=B=140$

Therefore, we will now find angle C and see if it is equal to angle A, that is, if angle C is equal to 40:

Let's remember that a pair of angles on the same side are equal to 180 degrees, therefore:

$B+C=180$

We replace the existing data:

$140+4x=180$

$4x=180-140$

$4x=40$

Divide by 4:

$\frac{4x}{4}=\frac{40}{4}$

$x=10$

Now we replace X:

$C=4\times10=40$

That is, we found that angles A and C are equal to each other and that the quadrilateral is a parallelogram since each pair of opposite angles are equal to each other.

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!

Let's pay attention to the diagonals, remember that in a parallelogram the diagonals intersect each other.

Therefore, we will find AO, OC, BO, DO and check if they are equal and intersect each other.

We refer to the figure:

$AO=OC$

$9x+1=10x$

We place like terms:

$1=10x-9x$

$1=x$

We replace:

$AO=9\times1+1=10$

$OC=10\times1=10$

Now we know that indeed$AO=OC$

Now we establish that X=1 and see if BO is equal to OD:

$BO=3x-2$

$BO=3\times1-2=$

$BO=3-2=1$

$OD=5x+4$

$OD=5\times1+4$

$OD=5+4=9$

Now we find that: $BO\ne OD$

Since the diagonals do not intersect each other, the quadrilateral is not a parallelogram.

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 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:

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