A parallelogram is a quadrilateral whose opposite sides are parallel in pairs.
A parallelogram is a quadrilateral whose opposite sides are parallel in pairs. The area of a parallelogram is equal to the product of its base (a) and height (h). You can also find its area through two sides and an angle and through diagonals.
Properties of a parallelogram
1. Opposite sides are identical
First of all, let's draw the diagonal \(AC\) . We get two triangles: \(ABC\) and \(ADC\).
Since \(ABCD\) is a parallelogram, the following is true:
\(AD || BC \Rightarrow \angle 1 = \angle 2\) like lying crosswise.
\(AB || CD \Rightarrow \angle3 = \angle 4\) like lying crosswise.
Therefore, (according to the second criterion: and \(AC\) is common).
And that means \(\triangle ABC = \triangle ADC\), then \(AB = CD\) and \(AD = BC\) .
2. Opposite angles are identical
According to the proof properties 1 We know that \(\angle 1 = \angle 2, \angle 3 = \angle 4\). Thus the sum of opposite angles is: \(\angle 1 + \angle 3 = \angle 2 + \angle 4\). Considering that \(\triangle ABC = \triangle ADC\) we get \(\angle A = \angle C \) , \(\angle B = \angle D \) .
3. Diagonals are divided in half by the intersection point
By property 1 we know that opposite sides are identical: \(AB = CD\) . Once again, note the crosswise lying equal angles.
Thus it is clear that \(\triangle AOB = \triangle COD\) according to the second sign of equality of triangles (two angles and the side between them). That is, \(BO = OD\) (opposite the angles \(\angle 2\) and \(\angle 1\) ) and \(AO = OC\) (opposite the angles \(\angle 3\) and \( \angle 4\) respectively).
Signs of a parallelogram
If only one feature is present in your problem, then the figure is a parallelogram and you can use all the properties of this figure.
For better memorization, note that the parallelogram sign will answer the following question - "how to find out?". That is, how to find out that a given figure is a parallelogram.
1. A parallelogram is a quadrilateral whose two sides are equal and parallel
\(AB = CD\) ; \(AB || CD \Rightarrow ABCD\)- parallelogram.
Let's take a closer look. Why \(AD || BC \) ?
\(\triangle ABC = \triangle ADC\) By property 1: \(AB = CD \) , \(\angle 1 = \angle 2 \) lying crosswise when \(AB \) and \(CD \) and the secant \(AC \) are parallel.
But if \(\triangle ABC = \triangle ADC\), then \(\angle 3 = \angle 4 \) (lie opposite \(AD || BC \) (\(\angle 3 \) and \(\angle 4 \) - those lying crosswise are also equal).
The first sign is correct.
2. A parallelogram is a quadrilateral whose opposite sides are equal
\(AB = CD \) , \(AD = BC \Rightarrow ABCD \) is a parallelogram.
Let's consider this sign. Let's draw the diagonal \(AC\) again.
By property 1\(\triangle ABC = \triangle ACD\).
It follows that: \(\angle 1 = \angle 2 \Rightarrow AD || BC \) And \(\angle 3 = \angle 4 \Rightarrow AB || CD \), that is, \(ABCD\) is a parallelogram.
The second sign is correct.
3. A parallelogram is a quadrilateral whose opposite angles are equal
\(\angle A = \angle C\) , \(\angle B = \angle D \Rightarrow ABCD\)- parallelogram.
\(2 \alpha + 2 \beta = 360^(\circ) \)(since \(\angle A = \angle C\) , \(\angle B = \angle D\) by condition).
It turns out, \(\alpha + \beta = 180^(\circ) \). But \(\alpha \) and \(\beta \) are internal one-sided at the secant \(AB \) .
Just as in Euclidean geometry, a point and a straight line are the main elements of the theory of planes, so a parallelogram is one of the key figures of convex quadrilaterals. From it, like threads from a ball, flow the concepts of “rectangle”, “square”, “rhombus” and other geometric quantities.
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Definition of parallelogram
convex quadrilateral, consisting of segments, each pair of which is parallel, is known in geometry as a parallelogram.
What a classic parallelogram looks like is depicted by a quadrilateral ABCD. The sides are called bases (AB, BC, CD and AD), the perpendicular drawn from any vertex to the side opposite to this vertex is called height (BE and BF), lines AC and BD are called diagonals.
Attention! Square, rhombus and rectangle are special cases of parallelogram.
Sides and angles: features of the relationship
Key properties, by and large, predetermined by the designation itself, they are proved by the theorem. These characteristics are as follows:
- The sides that are opposite are identical in pairs.
- Angles opposite each other are equal in pairs.
Proof: Consider ∆ABC and ∆ADC, which are obtained by dividing the quadrilateral ABCD with the straight line AC. ∠BCA=∠CAD and ∠BAC=∠ACD, since AC is common for them (vertical angles for BC||AD and AB||CD, respectively). It follows from this: ∆ABC = ∆ADC (the second sign of equality of triangles).
The segments AB and BC in ∆ABC correspond in pairs to the lines CD and AD in ∆ADC, which means that they are identical: AB = CD, BC = AD. Thus, ∠B corresponds to ∠D and they are equal. Since ∠A=∠BAC+∠CAD, ∠C=∠BCA+∠ACD, which are also pairwise identical, then ∠A = ∠C. The property has been proven.
Characteristics of the diagonals of a figure
Main feature of these lines of a parallelogram: the point of intersection divides them in half.
Proof: Let i.e. be the intersection point of diagonals AC and BD of figure ABCD. They form two commensurate triangles - ∆ABE and ∆CDE.
AB=CD since they are opposites. According to lines and secants, ∠ABE = ∠CDE and ∠BAE = ∠DCE.
By the second criterion of equality, ∆ABE = ∆CDE. This means that the elements ∆ABE and ∆CDE: AE = CE, BE = DE and at the same time they are proportional parts of AC and BD. The property has been proven.
Features of adjacent corners
Adjacent sides have a sum of angles equal to 180°, since they lie on the same side of parallel lines and a transversal. For quadrilateral ABCD:
∠A+∠B=∠C+∠D=∠A+∠D=∠B+∠C=180º
Properties of the bisector:
- , lowered to one side, are perpendicular;
- opposite vertices have parallel bisectors;
- the triangle obtained by drawing a bisector will be isosceles.
Determination of the characteristic features of a parallelogram using the theorem
The characteristics of this figure follow from its main theorem, which states the following: a quadrilateral is considered a parallelogram in the event that its diagonals intersect, and this point divides them into equal segments.
Proof: let the lines AC and BD of the quadrilateral ABCD intersect at i.e. Since ∠AED = ∠BEC, and AE+CE=AC BE+DE=BD, then ∆AED = ∆BEC (by the first criterion for the equality of triangles). That is, ∠EAD = ∠ECB. They are also the internal cross angles of the secant AC for lines AD and BC. Thus, by definition of parallelism - AD || B.C. A similar property of lines BC and CD is also derived. The theorem is proven.
Calculating the area of a figure
Area of this figure found by several methods one of the simplest: multiplying the height and the base to which it is drawn.
Proof: draw perpendiculars BE and CF from vertices B and C. ∆ABE and ∆DCF are equal, since AB = CD and BE = CF. ABCD is equal in size to rectangle EBCF, since they consist of commensurate figures: S ABE and S EBCD, as well as S DCF and S EBCD. It follows from this that the area of this geometric figure is the same as that of a rectangle:
S ABCD = S EBCF = BE×BC=BE×AD.
To determine the general formula for the area of a parallelogram, let us denote the height as hb, and the side - b. Respectively:
Other ways to find area
Area calculations through the sides of the parallelogram and the angle, which they form, is the second known method.
,
Spr-ma - area;
a and b are its sides
α is the angle between segments a and b.
This method is practically based on the first, but in case it is unknown. always cuts off a right triangle whose parameters are found by trigonometric identities, that is. Transforming the relation, we get . In the equation of the first method, we replace the height with this product and obtain a proof of the validity of this formula.
Through the diagonals of a parallelogram and the angle, which they create when they intersect, you can also find the area.
Proof: AC and BD intersect to form four triangles: ABE, BEC, CDE and AED. Their sum is equal to the area of this quadrilateral.
The area of each of these ∆ can be found by the expression , where a=BE, b=AE, ∠γ =∠AEB. Since , the calculations use a single sine value. That is . Since AE+CE=AC= d 1 and BE+DE=BD= d 2, the area formula reduces to:
.
Application in vector algebra
The features of the constituent parts of this quadrilateral have found application in vector algebra, namely the addition of two vectors. The parallelogram rule states that if given vectorsAndNotare collinear, then their sum will be equal to the diagonal of this figure, the bases of which correspond to these vectors.
Proof: from an arbitrarily chosen beginning - i.e. - construct vectors and . Next, we construct a parallelogram OASV, where the segments OA and OB are sides. Thus, the OS lies on the vector or sum.
Formulas for calculating the parameters of a parallelogram
The identities are given under the following conditions:
- a and b, α - sides and the angle between them;
- d 1 and d 2, γ - diagonals and at the point of their intersection;
- h a and h b - heights lowered to sides a and b;
| Parameter | Formula |
| Finding the sides | |
| along the diagonals and the cosine of the angle between them | 
|
| along diagonals and sides | 
|
| through the height and the opposite vertex | |
| Finding the length of diagonals | |
| on the sides and the size of the apex between them | |
A parallelogram is a quadrilateral whose opposite sides are parallel in pairs. The following figure shows parallelogram ABCD. It has side AB parallel to side CD and side BC parallel to side AD.
As you may have guessed, a parallelogram is a convex quadrilateral. Let's consider the basic properties of a parallelogram.
Properties of a parallelogram
1. In a parallelogram, opposite angles and opposite sides are equal. Let's prove this property - consider the parallelogram presented in the following figure.
Diagonal BD divides it into two equal triangles: ABD and CBD. They are equal along the side BD and the two angles adjacent to it, since the angles lying crosswise at the secant BD of parallel lines BC and AD and AB and CD, respectively. Therefore AB = CD and
BC = AD. And from the equality of angles 1, 2, 3 and 4 it follows that angle A = angle1 + angle3 = angle2 + angle4 = angle C.
2. The diagonals of a parallelogram are divided in half by the point of intersection. Let point O be the intersection point of diagonals AC and BD of parallelogram ABCD.
Then triangle AOB and triangle COD are equal to each other, along the side and two adjacent angles. (AB = CD since these are opposite sides of the parallelogram. And angle1 = angle2 and angle3 = angle4 are like crosswise angles when the lines AB and CD intersect with the secants AC and BD, respectively.) From this it follows that AO = OC and OB = OD, which and needed to be proven.
All main properties are illustrated in the following three figures.
In order to determine whether a given figure is a parallelogram, there are a number of signs. Let's look at the three main features of a parallelogram.
1 parallelogram sign
If two sides of a quadrilateral are equal and parallel, then this quadrilateral will be a parallelogram.
Proof:
Consider the quadrilateral ABCD. Let the sides AB and CD be parallel. And let AB=CD. Let's draw the diagonal BD in it. It will divide this quadrilateral into two equal triangles: ABD and CBD.
These triangles are equal to each other along two sides and the angle between them (BD is the common side, AB = CD by condition, angle1 = angle2 as crosswise angles with the transversal BD of parallel lines AB and CD.), and therefore angle3 = angle4.
And these angles will lie crosswise when the lines BC and AD intersect with the secant BD. It follows from this that BC and AD are parallel to each other. We have that in the quadrilateral ABCD the opposite sides are pairwise parallel, and therefore the quadrilateral ABCD is a parallelogram.
Parallelogram sign 2
If in a quadrilateral the opposite sides are equal in pairs, then this quadrilateral will be a parallelogram.
Proof:
Consider the quadrilateral ABCD. Let's draw the diagonal BD in it. It will divide this quadrilateral into two equal triangles: ABD and CBD.
These two triangles will be equal to each other on three sides (BD is the common side, AB = CD and BC = AD by condition). From this we can conclude that angle1 = angle2. It follows that AB is parallel to CD. And since AB = CD and AB is parallel to CD, then according to the first criterion of a parallelogram, the quadrilateral ABCD will be a parallelogram.
3 parallelogram sign
If the diagonals of a quadrilateral intersect and are bisected by the point of intersection, then this quadrilateral will be a parallelogram.
Consider the quadrilateral ABCD. Let us draw two diagonals AC and BD in it, which will intersect at point O and are bisected by this point.
Triangles AOB and COD will be equal to each other, according to the first sign of equality of triangles. (AO = OC, BO = OD by condition, angle AOB = angle COD as vertical angles.) Therefore, AB = CD and angle1 = angle 2. From the equality of angles 1 and 2, we have that AB is parallel to CD. Then we have that in the quadrilateral ABCD the sides AB are equal to CD and parallel, and according to the first criterion of a parallelogram, the quadrilateral ABCD will be a parallelogram.
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