We calculate the sum of angles and the area of a parallelogram: properties and signs. How to find the area of a parallelogram? How to calculate the area of a parallelogram
Geometric area- a numerical characteristic of a geometric figure showing the size of this figure (part of the surface bounded by a closed contour of this figure). The size of the area is expressed by the number of square units contained in it.
Triangle area formulas
- Triangle area formula for side and height
Area of a triangle equal to half the product of the length of a side of a triangle and the length of the altitude drawn to this side - The formula for the area of a triangle given three sides and the radius of the circumscribed circle
- The formula for the area of a triangle given three sides and the radius of an inscribed circle
Area of a triangle is equal to the product of the half-perimeter of the triangle and the radius of the inscribed circle. where S is the area of the triangle,
- the lengths of the sides of the triangle,
- the height of the triangle,
- the angle between the sides and,
- radius of the inscribed circle,
R - radius of the circumscribed circle,
Square area formulas
- The formula for the area of a square given the length of a side
square area is equal to the square of its side length. - The formula for the area of a square given the length of the diagonal
square area equal to half the square of the length of its diagonal.S= 1 2 2 where S is the area of the square,
is the length of the side of the square,
is the length of the diagonal of the square.
Rectangle area formula
- Rectangle area is equal to the product of the lengths of its two adjacent sides
where S is the area of the rectangle,
are the lengths of the sides of the rectangle.
Formulas for the area of a parallelogram
- Parallelogram area formula for side length and height
Parallelogram area - The formula for the area of a parallelogram given two sides and the angle between them
Parallelogram area is equal to the product of the lengths of its sides multiplied by the sine of the angle between them.a b sinα
where S is the area of the parallelogram,
are the lengths of the sides of the parallelogram,
is the height of the parallelogram,
is the angle between the sides of the parallelogram.
Formulas for the area of a rhombus
- Rhombus area formula given side length and height
Rhombus area is equal to the product of the length of its side and the length of the height lowered to this side. - The formula for the area of a rhombus given the length of the side and the angle
Rhombus area is equal to the product of the square of the length of its side and the sine of the angle between the sides of the rhombus. - The formula for the area of a rhombus from the lengths of its diagonals
Rhombus area is equal to half the product of the lengths of its diagonals. where S is the area of the rhombus,
- length of the side of the rhombus,
- the length of the height of the rhombus,
- the angle between the sides of the rhombus,
1, 2 - the lengths of the diagonals.
Trapezium area formulas
- Heron's formula for a trapezoid
Where S is the area of the trapezoid,
- the length of the bases of the trapezoid,
- the length of the sides of the trapezoid,
As in Euclidean geometry, the point and the straight line are the main elements of the theory of planes, so the 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 a 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 a quadrilateral ABCD. The sides are called the bases (AB, BC, CD and AD), the perpendicular drawn from any vertex to the opposite side of this vertex is called the height (BE and BF), the lines AC and BD are the diagonals.
Attention! Square, rhombus and rectangle are special cases of parallelogram.
Sides and angles: ratio features
Key properties, by and large, predetermined by the designation itself, they are proved by the theorem. These characteristics are as follows:
- Sides that are opposite are identical in pairs.
- Angles that are opposite to each other are equal in pairs.
Proof: consider ∆ABC and ∆ADC, which are obtained by dividing quadrilateral ABCD by line AC. ∠BCA=∠CAD and ∠BAC=∠ACD, since AC is common to them (vertical angles for BC||AD and AB||CD, respectively). It follows from this: ∆ABC = ∆ADC (the second criterion for the equality of triangles).
Segments AB and BC in ∆ABC correspond in pairs to 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 identical in pairs, then ∠A = ∠C. The property has been proven.
Characteristics of the figure's diagonals
Main feature these parallelogram lines: the point of intersection bisects them.
Proof: let m. E be the intersection point of the diagonals AC and BD of the figure ABCD. They form two commensurate triangles - ∆ABE and ∆CDE.
AB=CD since they are opposite. According to lines and secants, ∠ABE = ∠CDE and ∠BAE = ∠DCE.
According to the second sign of equality, ∆ABE = ∆CDE. This means that the elements ∆ABE and ∆CDE are: AE = CE, BE = DE and, moreover, they are commensurate parts of AC and BD. The property has been proven.
Features of adjacent corners
At adjacent sides, the sum of the angles is 180°, since they lie on the same side of the parallel lines and the secant. For quadrilateral ABCD:
∠A+∠B=∠C+∠D=∠A+∠D=∠B+∠C=180º
Bisector properties:
- , dropped to one side, are perpendicular;
- opposite vertices have parallel bisectors;
- the triangle obtained by drawing the bisector will be isosceles.
Determining the characteristic features of a parallelogram by the theorem
The features of this figure follow from its main theorem, which reads as follows: quadrilateral is considered a parallelogram in the event that its diagonals intersect, and this point divides them into equal segments.
Proof: Let lines AC and BD of quadrilateral ABCD intersect in t. E. Since ∠AED = ∠BEC, and AE+CE=AC BE+DE=BD, then ∆AED = ∆BEC (by the first sign of equality of triangles). That is, ∠EAD = ∠ECB. They are also the interior crossing angles of the secant AC for lines AD and BC. Thus, by definition of parallelism - AD || BC. A similar property of the lines BC and CD is also derived. The theorem has been proven.
Calculating the area of a figure
The area of this figure found in several ways 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 to the rectangle EBCF, since they also consist of proportionate figures: S ABE and S EBCD, as well as S DCF and S EBCD. It follows 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, we 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 - 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, i.e. . Transforming the ratio, 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 an angle, which they create when they intersect, you can also find the area.
Proof: AC and BD intersecting 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 from the expression , where a=BE, b=AE, ∠γ =∠AEB. Since , then a single value of the sine is used in the calculations. 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 quadrangle 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 - that is. - we build vectors and . Next, we build 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 sides | |
| along the diagonals and the cosine of the angle between them | 
|
| diagonally and sideways | 
|
| through height and opposite vertex | |
| Finding the length of the diagonals | |
| on the sides and the size of the top between them | |
| along the sides and one of the diagonals | 
ConclusionThe parallelogram, as one of the key figures of geometry, is used in life, for example, in construction when calculating the area of \u200b\u200bthe site or other measurements. Therefore, knowledge about the distinguishing features and methods for calculating its various parameters can be useful at any time in life. |
Before we learn how to find the area of a parallelogram, we need to remember what a parallelogram is and what is called its height. A parallelogram is a quadrilateral whose opposite sides are pairwise parallel (lie on parallel lines). The perpendicular drawn from an arbitrary point on the opposite side to the line containing this side is called the height of the parallelogram.
Square, rectangle and rhombus are special cases of parallelogram.
The area of a parallelogram is denoted as (S).
Formulas for finding the area of a parallelogram
S=a*h, where a is the base, h is the height that is drawn to the base.
S=a*b*sinα, where a and b are the bases, and α is the angle between the bases a and b.
S \u003d p * r, where p is the semi-perimeter, r is the radius of the circle that is inscribed in the parallelogram.
The area of the parallelogram formed by the vectors a and b is equal to the modulus of the product of the given vectors, namely:
Consider example No. 1: A parallelogram is given, the side of which is 7 cm, and the height is 3 cm. How to find the area of \u200b\u200bthe parallelogram, we need a formula for solving.
So S= 7x3. S=21. Answer: 21 cm 2.
Consider example No. 2: The bases are 6 and 7 cm, and the angle between the bases is 60 degrees. How to find the area of a parallelogram? Formula used to solve:
Thus, first we find the sine of the angle. Sine 60 \u003d 0.5, respectively S \u003d 6 * 7 * 0.5 \u003d 21 Answer: 21 cm 2.
I hope that these examples will help you in solving problems. And remember, the main thing is knowledge of formulas and attentiveness
What is a parallelogram? A parallelogram is a quadrilateral whose opposite sides are pairwise parallel.
1. The area of a parallelogram is calculated by the formula:
\[ \LARGE S = a \cdot h_(a)\]
Where:
a is the side of the parallelogram,
h a is the height drawn to this side.
2. If the lengths of two adjacent sides of the parallelogram and the angle between them are known, then the area of the parallelogram is calculated by the formula:
\[ \LARGE S = a \cdot b \cdot sin(\alpha) \]
3. If the diagonals of the parallelogram are given and the angle between them is known, then the area of the parallelogram is calculated by the formula:
\[ \LARGE S = \frac(1)(2) \cdot d_(1) \cdot d_(2) \cdot sin(\alpha) \]
Parallelogram Properties
In a parallelogram, opposite sides are equal: \(AB = CD \) , \(BC = AD \)
In a parallelogram, opposite angles are: \(\angle A = \angle C \) , \(\angle B = \angle D \)
The diagonals of the parallelogram at the point of intersection are bisected \(AO = OC \) , \(BO = OD \)
The diagonal of a parallelogram divides it into two equal triangles.
The sum of the angles of a parallelogram adjacent to one side is 180 o:
\(\angle A + \angle B = 180^(o) \), \(\angle B + \angle C = 180^(o)\)
\(\angle C + \angle D = 180^(o) \), \(\angle D + \angle A = 180^(o)\)
The diagonals and sides of a parallelogram are related by the following relationship:
\(d_(1)^(2) + d_(2)^2 = 2a^(2) + 2b^(2) \)
In a parallelogram, the angle between the heights is equal to its acute angle: \(\angle K B H =\angle A \) .
Bisectors of angles adjacent to one side of a parallelogram are mutually perpendicular.
Bisectors of two opposite angles of a parallelogram are parallel.
Parallelogram features
A quadrilateral is a parallelogram if:
\(AB = CD \) and \(AB || CD \)
\(AB = CD \) and \(BC = AD \)
\(AO = OC \) and \(BO = OD \)
\(\angle A = \angle C \) and \(\angle B = \angle D \)
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When solving problems on this topic, in addition to basic properties parallelogram and the corresponding formulas, you can remember and apply the following:
- The bisector of the interior angle of a parallelogram cuts off an isosceles triangle from it
- Bisectors of internal angles adjacent to one of the sides of a parallelogram are mutually perpendicular
- Bisectors coming from opposite internal angles of a parallelogram, parallel to each other or lie on one straight line
- The sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides
- The area of a parallelogram is half the product of the diagonals times the sine of the angle between them.
Let's consider the tasks in the solution of which these properties are used.
Task 1.
The bisector of angle C of parallelogram ABCD intersects side AD at point M and the continuation of side AB beyond point A at point E. Find the perimeter of the parallelogram if AE \u003d 4, DM \u003d 3.
Solution.
1. Triangle CMD isosceles. (Property 1). Therefore, CD = MD = 3 cm.
2. Triangle EAM is isosceles.
Therefore, AE = AM = 4 cm.
3. AD = AM + MD = 7 cm.
4. Perimeter ABCD = 20 cm.
Answer. 20 cm
Task 2.
Diagonals are drawn in a convex quadrilateral ABCD. It is known that the areas of triangles ABD, ACD, BCD are equal. Prove that the given quadrilateral is a parallelogram.
Solution.
1. Let BE be the height of triangle ABD, CF be the height of triangle ACD. Since, according to the condition of the problem, the areas of the triangles are equal and they have a common base AD, then the heights of these triangles are equal. BE = CF.
2. BE, CF are perpendicular to AD. Points B and C are located on the same side of the line AD. BE = CF. Therefore, the line BC || AD. (*)
3. Let AL be the altitude of triangle ACD, BK the altitude of triangle BCD. Since, according to the condition of the problem, the areas of the triangles are equal and they have a common base CD, then the heights of these triangles are equal. AL = BK.
4. AL and BK are perpendicular to CD. Points B and A are located on the same side of the straight line CD. AL = BK. Therefore, the line AB || CD (**)
5. Conditions (*), (**) imply that ABCD is a parallelogram.
Answer. Proven. ABCD is a parallelogram.
Task 3.
On the sides BC and CD of the parallelogram ABCD, the points M and H are marked, respectively, so that the segments BM and HD intersect at the point O;<ВМD = 95 о,
Solution.
1. In the triangle DOM<МОD = 25 о (Он смежный с <ВОD = 155 о); <ОМD = 95 о. Тогда <ОDМ = 60 о.
2. In a right triangle DHC Then<НСD = 30 о. СD: НD = 2: 1 But CD = AB. Then AB: HD = 2: 1. 3. <С = 30 о, 4. <А = <С = 30 о, <В = Answer: AB: HD = 2: 1,<А = <С = 30 о, <В = Task 4. One of the diagonals of a parallelogram of length 4√6 makes an angle of 60° with the base, and the second diagonal makes an angle of 45° with the same base. Find the second diagonal. Solution.
1. AO = 2√6. 2. Apply the sine theorem to the triangle AOD. AO/sin D = OD/sin A. 2√6/sin 45 o = OD/sin 60 o. OD = (2√6sin 60 o) / sin 45 o = (2√6 √3/2) / (√2/2) = 2√18/√2 = 6. Answer: 12.
Task 5. For a parallelogram with sides 5√2 and 7√2, the smaller angle between the diagonals is equal to the smaller angle of the parallelogram. Find the sum of the lengths of the diagonals. Solution.
Let d 1, d 2 be the diagonals of the parallelogram, and the angle between the diagonals and the smaller angle of the parallelogram be φ. 1. Let's count two different S ABCD \u003d AB AD sin A \u003d 5√2 7√2 sin f, S ABCD \u003d 1/2 AC BD sin AOB \u003d 1/2 d 1 d 2 sin f. We obtain the equality 5√2 7√2 sin f = 1/2d 1 d 2 sin f or 2 5√2 7√2 = d 1 d 2 ; 2. Using the ratio between the sides and diagonals of the parallelogram, we write the equality (AB 2 + AD 2) 2 = AC 2 + BD 2. ((5√2) 2 + (7√2) 2) 2 = d 1 2 + d 2 2 . d 1 2 + d 2 2 = 296. 3. Let's make a system: (d 1 2 + d 2 2 = 296, Multiply the second equation of the system by 2 and add it to the first. We get (d 1 + d 2) 2 = 576. Hence Id 1 + d 2 I = 24. Since d 1, d 2 are the lengths of the diagonals of the parallelogram, then d 1 + d 2 = 24. Answer: 24.
Task 6. The sides of the parallelogram are 4 and 6. The acute angle between the diagonals is 45 o. Find the area of the parallelogram. Solution.
1. From the triangle AOB, using the cosine theorem, we write the relationship between the side of the parallelogram and the diagonals. AB 2 \u003d AO 2 + VO 2 2 AO VO cos AOB. 4 2 \u003d (d 1 / 2) 2 + (d 2 / 2) 2 - 2 (d 1 / 2) (d 2 / 2) cos 45 o; d 1 2/4 + d 2 2/4 - 2 (d 1/2) (d 2/2)√2/2 = 16. d 1 2 + d 2 2 - d 1 d 2 √2 = 64. 2. Similarly, we write the relation for the triangle AOD. We take into account that<АОD = 135 о и cos 135 о = -cos 45 о = -√2/2. We get the equation d 1 2 + d 2 2 + d 1 d 2 √2 = 144. 3. We have a system Subtracting the first from the second equation, we get 2d 1 d 2 √2 = 80 or d 1 d 2 = 80/(2√2) = 20√2 4. S ABCD \u003d 1/2 AC BD sin AOB \u003d 1/2 d 1 d 2 sin α \u003d 1/2 20√2 √2/2 \u003d 10. Note: In this and in the previous problem, there is no need to solve the system completely, foreseeing that in this problem we need the product of diagonals to calculate the area. Answer: 10. Task 7. The area of the parallelogram is 96 and its sides are 8 and 15. Find the square of the smaller diagonal. Solution.
1. S ABCD \u003d AB AD sin VAD. Let's do a substitution in the formula. We get 96 = 8 15 sin VAD. Hence sin VAD = 4/5. 2. Find cos BAD. sin 2 VAD + cos 2 VAD = 1. (4/5) 2 + cos 2 BAD = 1. cos 2 BAD = 9/25. According to the condition of the problem, we find the length of the smaller diagonal. Diagonal BD will be smaller if angle BAD is acute. Then cos BAD = 3 / 5. 3. From the triangle ABD, using the cosine theorem, we find the square of the diagonal BD. BD 2 \u003d AB 2 + AD 2 - 2 AB BD cos BAD. ВD 2 \u003d 8 2 + 15 2 - 2 8 15 3 / 5 \u003d 145. Answer: 145.
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(Since in a right triangle, the leg that lies opposite an angle of 30 o is equal to half the hypotenuse).
ways of its area.
(d 1 + d 2 = 140.
(d 1 2 + d 2 2 - d 1 d 2 √2 = 64,
(d 1 2 + d 2 2 + d 1 d 2 √2 = 144.
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