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Among the variety of geometric shapes, a quadrilateral such as a rhombus stands out noticeably. Even its name itself is not typical for the designation of quadrangles. And although in geometry it is found much less frequently than such simple figures as a circle, triangle, square or rectangle, it also cannot be ignored.
Below are the definition, properties and characteristics of rhombuses.
Definition
A rhombus is a parallelogram with equal sides. A rhombus is called a square if all its angles are right angles. Most a shining example The diamond is the image of the diamond suit on a playing card. In addition, the rhombus was often depicted on various coats of arms. An example of a diamond in everyday life is a basketball court.
Properties
- The opposite sides of a rhombus lie on parallel lines and have the same length.
- The intersection of the diagonals of a rhombus occurs at an angle of 90° at one point, which is their midpoint.
- The diagonals of a rhombus bisect the angle from which they originated.
- Based on the properties of a parallelogram, we can derive the sum of the squares of the diagonals. According to the formula, it is equal to the side raised to a quadratic power and multiplied by four.
Signs
We must clearly understand that any rhombus is a parallelogram, but at the same time, not every parallelogram has all the indicators of a rhombus. To distinguish these two geometric shapes, you need to know the characteristics of a rhombus. The following are the characteristic features of this geometric figure:
- Any two sides with a common vertex are equal.
- The diagonals intersect at an angle of 90°C.
- At least one diagonal divides the angles from whose vertex points it emerges in half.
Area formulas
Basic formula:
- S = (AC*BD)/2
Based on the properties of a parallelogram:
- S = (AB*H AB)
Based on the size of the angle between two adjacent sides of the rhombus:
- S = AB2*sinα
If we know the length of the radius of a circle inscribed in a rhombus:
- S = 4r 2 /(sinα), where:
- S - area;
- AB, AC, BD - designation of sides;
- H - height;
- r - radius of the circle;
- sinα - sine alpha.
Perimeter
To calculate the perimeter of a rhombus, you just need to multiply the length of any of its sides by four.
Construction of the drawing
Some people have difficulty constructing a diamond pattern. Even if you have already figured out what a rhombus is, it is not always clear how to construct its drawing accurately and in compliance with the necessary proportions.
There are two ways to construct a diamond pattern:
- First construct one diagonal, then a second diagonal perpendicular to it, and then connect the ends of the segments of adjacent pairs of parallel sides of the rhombus.
- First set aside one side of the rhombus, then construct a segment equal in length parallel to it, and connect the ends of these segments also in pairs in parallel.
Be careful when constructing - if in the drawing you make the length of all sides of the rhombus the same, you will not get a rhombus, but a square.
In Figure 1, $ABCD$ is a rhombus, $A B=B C=C D=A D$. Since a rhombus is a parallelogram, it has all the properties of a parallelogram, but there are also properties inherent only to a rhombus.
You can fit a circle into any rhombus. The center of a circle inscribed in a rhombus is the intersection point of its diagonals. The radius of the circle is equal to half the height of the rhombus $r=\frac(A H)(2)$ (Fig. 1)
Properties of a rhombus
- The diagonals of a rhombus are perpendicular;
- The diagonals of a rhombus are the bisectors of its angles.
Signs of a diamond
- A parallelogram whose diagonals intersect at right angles is a rhombus;
- A parallelogram whose diagonals are the bisectors of its angles is a rhombus.
Examples of problem solving
Example
Exercise. The diagonals of the rhombus $ABCD$ are 6 and 8 cm. Find the side of the rhombus.
Solution. Let's make a drawing (Fig. 1). Let, for definiteness, $A C=6$ cm, $B D=8$ cm. By the property of a rhombus, its diagonals intersect at right angles. At the point of intersection, the diagonals are divided in half (a property of a parallelogram, and a rhombus is a special case of a parallelogram).
Consider the triangle $A O B$. It is rectangular ($\angle O=90^(\circ)$), $A O=\frac(A C)(2)=\frac(6)(2)=3$ cm, $B O=\frac(B D) (2)=\frac(8)(2)=4$ cm. Let us write the Pythagorean theorem for this triangle:
$$A B^(2)=A O^(2)+B O^(2)$$
Let's substitute the found values of $AO$ and $BO$,
$A B^(2)=3^(2)+4^(2)$
Answer. The side of a rhombus is 5 cm.
Example
Exercise. In a rhombus with a side of 4 cm, one of the angles is equal to $60^(\circ)$. Find the diagonals of the rhombus.
Solution. Let's make a drawing (Fig. 2).
Let $\angle B=60^(\circ)$ for definiteness. Then, by the property of a rhombus, the diagonal $BD$ is the bisector of the angle $B$, $\angle A B O=\angle O B C=\frac(\angle B)(2)=30^(\circ)$. Consider $\Delta O B C$, it is rectangular ($\angle B O C=90^(\circ)$) because the diagonals of a rhombus intersect at right angles. Since $\angle O B C=30^(\circ), O C=\frac(B C)(2)=2$ dm is the leg lying opposite the angle of $30^(\circ)$. Using the Pythagorean theorem we find $B O$:
$$B O=\sqrt(B C^(2)-O C^(2))$$
$$B O=\sqrt(4^(2)-2^(2))$$
$$B O=\sqrt(12)$$
$$B O=2 \sqrt(3)$$
The diagonals of a rhombus are divided in half at the point of intersection, so
$B D=2 \cdot B O=2 \cdot 2 \sqrt(3)=4 \sqrt(3)$ (dm)
$A C=2 \cdot O C=2 \cdot 2=4$ (dm)
Answer.$B D=4 \sqrt(3)$ dm, $A C=4$ dm
Example
Exercise. In a rhombus, the angle formed by one of the diagonals and the side of the rhombus is equal to $27^(\circ)$. Find the angles of the rhombus.
Solution. Let's make a drawing (Fig. 3)
To be specific, $\angle K L O=27^(\circ)$. The diagonals in a rhombus are the bisectors of its angles, so $\angle L=2 \cdot \angle K L O=2 \cdot 27^(\circ)=54^(\circ)$. Since a rhombus is a parallelogram, the following properties apply to it: the sum of the angles adjacent to one side is equal to $180^(\circ)$ and the opposite angles are equal. That's why,
$\angle M=\angle K=180^(\circ)-\angle L=180^(\circ)-54^(\circ)=126^(\circ)$
Answer.$\angle N=\angle L=54^(\circ)$
$\angle M=\angle K=126^(\circ)$
with equal sides. A rhombus with right angles is square .
A rhombus is considered as a type of parallelogram, with two adjacent equal sides either with mutually perpendicular diagonals, or with diagonals dividing the angle into 2 equal parts.
Properties of a rhombus.
1. Rhombus is a parallelogram, so opposite sides have the same length and are parallel in pairs, AB || CD, AD || Sun.
2. Angle of intersection of diagonals rhombus is straight (AC⊥ BD) and the intersection point are divided into two identical parts. That is, the diagonals divide the rhombus into 4 rectangular triangles.
3. Diagonals of a rhombus are the bisectors of its angles (∠ DCA =∠ B.C.A.∠ ABD =∠ CBD etc. ).
4. Sum of squares of diagonals equals the square of the side multiplied by four (derived from the parallelogram identity).
Signs of a diamond.
Parallelogram ABCD will be called a rhombus only if at least one of the conditions is met:
1. Its 2 adjacent sides have the same length (that is, all sides of a rhombus are equal, AB=BC=CD=AD).
2. The angle of intersection of the diagonals of a straight line ( A.C.⊥ BD).
3. 1 of the diagonals divides the angles that contain it in half.
We may not know in advance that the quadrilateral turns out to be a parallelogram, but we know that all its sides are equal. So this quadrilateral is a rhombus.
Symmetry of a rhombus.
The rhombus is symmetrical relative to all its diagonals, it is often used in ornaments and parquet floors.
Perimeter of a rhombus.
Perimeter of a geometric figure- the total length of the boundaries of a flat geometric figure. The perimeter has the same dimension as the length.
AB \parallel CD,\;BC \parallel AD
AB = CD,\;BC = AD
2. The diagonals of a rhombus are perpendicular.
AC\perp BD
Proof
Since a rhombus is a parallelogram, its diagonals are divided in half.
This means that \triangle BOC = \triangle DOC on three sides (BO = OD, OC - joint, BC = CD). We get that \angle BOC = \angle COD and they are adjacent.
\Rightarrow \angle BOC = 90^(\circ) and \angle COD = 90^(\circ) .
3. The intersection point of the diagonals divides them in half.
AC=2\cdot AO=2\cdot CO
BD=2\cdot BO=2\cdot DO
4. The diagonals of a rhombus are the bisectors of its angles.
\angle 1 = \angle 2; \; \angle 5 = \angle 6;
\angle 3 = \angle 4; \; \angle 7 = \angle 8.
Proof
Due to the fact that the diagonals are divided in half by the intersection point, and all sides of the rhombus are equal to each other, the entire figure is divided by the diagonals into 4 equal triangles:
\triangle BOC,\; \triangle BOA,\; \triangle AOD,\; \triangle COD.
This means that BD, AC are bisectors.
5. Diagonals form 4 right triangles from a rhombus.
6. Any rhombus can contain a circle with its center at the point of intersection of its diagonals.
7. The sum of the squares of the diagonals is equal to the square of one of the sides of the rhombus multiplied by four
AC^2 + BD^2 = 4\cdot AB^2
Signs of a diamond
1. A parallelogram with perpendicular diagonals is a rhombus.
\begin(cases) AC \perp BD \\ ABCD \end(cases)- parallelogram, \Rightarrow ABCD - rhombus.
Proof
ABCD is a parallelogram \Rightarrow AO = CO ; BO = OD. It is also stated that AC \perp BD \Rightarrow \triangle AOB = \triangle BOC = \triangle COD = \triangle AOD- on 2 legs.
It turns out that AB = BC = CD = AD.
Proven!
2. When in a parallelogram at least one of the diagonals divides both angles (through which it passes) in half, then this figure will be a rhombus.
Proof
On a note: not every figure (quadrangle) with perpendicular diagonals will be a rhombus.
Eg:
This is no longer a rhombus, despite the perpendicularity of the diagonals.
To differentiate, it is worth remembering that first the quadrilateral must be a parallelogram and have