Rhombus, kite, or diamond?

A rhombus is a polygon with four sides of equal length. If we talk about the "extended family," we can admit that a rhombus is, in fact, a particular case of the figures kite and parallelogram. On the other hand, if a certain rhombus is also characterized by having four equivalent angles (that is, each of 90º) the rhombus becomes a square.

Below are the main characteristics that describe a rhombus. To explain them clearly, we will use the following illustration:

• All sides of the rhombus have the same length. That is:$BC=CD=DE=EB$
• The opposite sides of the rhombus are parallel. That is: $BC||DE ,EB ||CD$
• The opposite angles of the rhombus have the same amplitude. That is: $\sphericalangle C=\sphericalangle E,\sphericalangle D=B$
• If a height is measured from each vertex, all four will have the same length. In the illustration, a single height $H$ is seen descending from the vertex $B$ to the side $DE$.
• A circle can be inscribed in any rhombus.
• In every rhombus, there are two perpendicular diagonals. That is: $BD$ is perpendicular to $CE$.
• The meeting point of the two diagonals divides each diagonal into two equal parts. That is: $BK=KD,=KC=KE$
• Each diagonal divides the angles of the rhombus into two equal angles. That is:  

$\sphericalangle B1=\sphericalangle B2,\sphericalangle C1=\sphericalangle C2,\sphericalangle D1=\sphericalangle D2,\sphericalangle E1=\sphericalangle E2$

To demonstrate that a quadrilateral is a rhombus we can use the direct or indirect method. If we opt for the direct method, we must prove that the four sides of said quadrilateral have the same length. Using the indirect method, we must first demonstrate that the polygon is a parallelogram. After having managed to prove that a certain quadrilateral is a parallelogram, we are left with three options:

• Prove that the diagonals of the parallelogram act as bisectors
• Prove that the diagonals of the parallelogram are perpendicular
• Prove that the two adjacent sides of the parallelogram are equivalent

Depending on the data available in the exercise, we can choose the most appropriate method to act and thus demonstrate that a certain parallelogram is, in fact, a rhombus.

There are several ways to calculate the area of a rhombus. In this section, we will briefly mention two formulas to calculate the area of the rhombus. If you want to expand your knowledge on the subject, we invite you to read the full article about the area of the rhombus.

To see it in a simple way, also in this case, we will rely on the following illustration:

The lengths of the diagonals are multiplied and divided by $2$.

That is:

$A=\frac{\left(CL\times KM\right)}{2}$

One of the sides is multiplied by the height.

That is to say:

$A=CT\times ML$

We have reached the simplest part of all, which is the calculation of the perimeter of a rhombus. Let's remember that, the perimeter of a rhombus consists of the sum of the lengths of its sides.

Because the four sides of the rhombus have the same length, knowing just one is enough. Next, we will multiply the length of one side by $4$ and obtain the perimeter of the rhombus.

Given the rhombus $\text{QRST}$ .  

One of the angles is given. Using the data from the illustration, find the remaining three angles of the rhombus.

Solution: 

Let's focus on the given data. We know the vertex angle $R$ which is equal to $70^o$

According to what we know about the properties of the rhombus, the opposite angles have the same amplitude, therefore, the vertex angle $T$ will also be equal to $70^o$ .

The rhombus is a quadrilateral, therefore, the sum of all its angles is $360^o$ degrees, we will obtain that the sum of the angles $S$ and $Q$ is: $220=140-360=360-70\times2$ .

Let's remember again that, the opposite angles of the rhombus have the same amplitude. From this it follows that, each of the remaining vertex angles, $Q$ and $S$ , will measure $110^o$ .

Answer: the angles of the rhombus are $110^o,110^o,70^o,70^o$

Given the rhombus $\text{ABCD}$.

According to the data in the illustration, find the perimeter and the area of the rhombus.

Solution: 

First, we will observe the scheme. We know one of the sides of the rhombus. According to the properties we have learned, the four angles of the rhombus are equivalent.

Therefore, it follows: $AB=BC=CD=DA=5$.

The perimeter of the rhombus is the sum of its four sides, therefore, we will multiply $5$ by $4$ and it will give us $20$ cm.

Now let's move on to calculating the area of the rhombus. To do this, we will choose the first option we studied, that is, we will multiply the diagonals and divide by $2$.

In the illustration, it can be seen that $4=AK$. According to the properties of the rhombus, the intersection point of the diagonals of the rhombus crosses them, which leads to $AC=8$. We just need to find the second diagonal.

Let's remember another characteristic of the rhombus: the diagonals of the rhombus are perpendicular. That is, the triangle $ABK$ is a right triangle to which the Pythagorean theorem is applicable. Based on the Pythagorean theorem, we will calculate $BK$.

It follows:

$BK=\sqrt{25-16}=\sqrt{9}=3$

Again, the diagonals intersect and, therefore: $BD=6$.

That is, it follows:

$A=\frac{\left(AC\times BD\right)}{2}=\frac{\left(8\times6\right)}{2}=\frac{48}{2}=24$

Answer:

The perimeter of the rhombus is $20$ cm.

The area of the rhombus is $24$ cm.