Find the area of \u200b\u200bthe parallelogram on the sides. Square Pollogram
What is a parallelogram? The parallelogram is called a quadrilateral who has opposite sides pairly parallel.
1. The area of \u200b\u200bthe parallelogram is calculated by the formula:
\\ [\\ LARGE S \u003d A \\ CDOT H_ (A) \\]
where:
A - side of the parallelogram,
H a is the height carried out to this side.
2. If the length of two adjacent sides of the parallelogram and the angle between them are known, then the parallelogram area is calculated by the formula:
\\ [\\ Large s \u003d a \\ cdot b \\ cdot sin (\\ alpha) \\]
3. If the diagonal parallelogram is set and the angle is known between them, the parallelogram area is calculated by the formula:
\\ [\\ LARGE S \u003d \\ FRAC (1) (2) \\ CDOT D_ (1) \\ CDOT D_ (2) \\ CDOT SIN (\\ ALPHA) \\]
Properties of parallelogram
In the parallelogram, the opposite directions are equal: \\ (ab \u003d cd \\), \\ (bc \u003d ad \\)
In the parallelogram, opposite angles are equal: \\ (\\ angle a \u003d \\ angle c \\), \\ (\\ angle b \u003d \\ angle d \\)
The diagonal of the parallelogram at the intersection point is divided by half \\ (AO \u003d OC \\), \\ (BO \u003d OD \\)
The diagonal of the parallelogram divides it into two equal triangles.
The sum of the angles of the parallelogram, adjacent to one side equal to 180 o:
\\ (\\ angle a + \\ angle b \u003d 180 ^ (O) \\), \\ (\\ ANGLE B + \\ ANLE C \u003d 180 ^ (O) \\)
\\ (\\ ANGLE C + \\ ANGLE D \u003d 180 ^ (O) \\), \\ (\\ ANGLE D + \\ ANGLE A \u003d 180 ^ (O) \\)
The diagonals and side of the parallelogram are associated with the following ratio:
\\ (d_ (1) ^ (2) + d_ (2) ^ 2 \u003d 2a ^ (2) + 2b ^ (2) \\)
In a parallelogram, the angle between heights is equal to its acute corner: \\ (\\ angle k b h \u003d \\ angle a \\).
The bisector of the angles adjacent to one side of the parallelogram are mutually perpendicular.
Bissectrix of two opposite corners of the parallelogram are parallel.
Signs of parallelogram
The quadrilateral will be a parallelogram if:
\\ (AB \u003d CD \\) and \\ (ab || CD \\)
\\ (AB \u003d CD \\) and \\ (BC \u003d AD \\)
\\ (AO \u003d OC \\) and \\ (BO \u003d OD \\)
\\ (\\ angle a \u003d \\ angle c \\) and \\ (\\ angle b \u003d \\ angle d \\)
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The output of the area of \u200b\u200bthe area of \u200b\u200bthe parallelogram is reduced to the construction of a rectangle equal to this parallelogram in the area. We take one side of the parallelogram for the base, and the perpendicular, carried out from any point of the opposite side to a straight line, which contain the base will be called a parallelogram height. Then the area of \u200b\u200bthe parallelogram will be equal to the product of its base to height.
Theorem.The area of \u200b\u200bthe parallelogram is equal to the product of its base to height.
Evidence. Consider parallelograms with an area. Let's face the base and carry out the height (Figure 2.3.1). It is required to prove that.
Figure 2.3.1
We first prove that the area of \u200b\u200bthe rectangle is also equal. The trapezion is made of a triangle parallelogram. On the other hand, it is composed of a rectangle of the NVCC and a triangle. But rectangular triangles are equal to hypotenuse and acute angle (their hypothenuisaes, as opposite sides, the parallelogram, and the angles 1 and 2 are equal to both respective angles when crossing parallel direct), so they are equal. Consequently, the area of \u200b\u200bthe parallelogram of the rectangle is equal, that is, the area is rectangle. By the rectangle area theorem, but since, then.
Theorem is proved.
Example 2.3.1.
In a rhombus with a party and a sharp corner, a circle is inscribed. Determine the area of \u200b\u200bthe quadriller, whose vertices are the point of touching the circle with the sides of the rhombus.
Decision:
The radius inscribed in the rhombus of the circle (Figure 2.3.2), since the quadricular rectangle, since its angles are based on the diameter of the circle. His area, where (catat lying against the angle) ,. 
Figure 2.3.2
So, 
Answer:
Example 2.3.2.
Danzh, the diagonal of which is 3 cm and 4 cm. From the top of the stupid angle, a tranny area was taken
Decision:
Roma Area (Figure 2.3.3).
So, 
Answer:
Example 2.3.3.
The area of \u200b\u200bthe quadril is equal to find the area of \u200b\u200bthe parallelogram, the sides of which are equal and parallel to the diagonals of the quadril.
Decision:
Since both (Figure 2.3.4), then parallelograms and, it means.
Figure 2.3.4.
Similarly, we get from where it follows that.
Answer:.
2.4 Triangle Square
There are several formulas for calculating the triangle area. Consider those studied at school.
The first formula flows out of the formula of the Pollogram area and is offered by students in the form of the theorem.
Theorem. The area of \u200b\u200bthe triangle is equal to half the work of its base to height.
Evidence. Let - the area of \u200b\u200bthe triangle. Let's face the bottom of the triangle and spend the height. We prove that:
Figure 2.4.1
Thunderstand the triangle to the parallelogram, as shown in the figure. Triangles on three sides (- their common party, and the opposite sides of parallel gram), therefore their square is equal. Consequently, the area S is the ABS triangle is equal to half the area of \u200b\u200bthe parallelogram, i.e. 
Theorem is proved.
It is important to draw students the attention of two consequences arising from this theorem. Namely:
area rectangular triangle It is half the work of its cathets.
if the height of two triangles are equal, then their areas belong as a basis.
These two consequences play important role In solving a different kind of tasks. With a support for this, another theorem is proved, which has widespread use when solving problems.
Theorem. If the angle of one triangle is equal to the angle of another triangle, then their areas are related as works of parties to equal angles.
Evidence. Let the objects of the triangles that are coaled.
Figure 2.4.2
We prove that: 
.
Take a triangle. On triangular to peak up with the top, and parties, respectively, on Lucia.
Figure 2.4.3.
Triangles have a total height, therefore,. Triangles have a total height -, therefore,. Multiplying the equality obtained, we get 
.
Theorem is proved.
Second formula.The triangle area is equal to half the work of two sides on the sine of the corner between them. There are several ways to proof this formula, and I will accustomed one of them.
Evidence.From the geometry known the theorem that the area of \u200b\u200bthe triangle is equal to half the product of the base for height, lowered to this base:
In the case of an acute triangle. In case of dull angle. Ho, and therefore 
. So, in both cases. Substituting a triangle square in the geometric formula, we obtain the trigonometric formula of the triangle area:
Theorem is proved.
Third formula For the triangle area, the formula of Geron, is named after the ancient Greek scientist Gerona Alexandrian, who lived in the first century of our era. This formula allows you to find the area of \u200b\u200bthe triangle, knowing it. It is convenient because it allows you to make any additional constructions and not measure the corners. Its conclusion is based on the second one we considered the formulas of the area of \u200b\u200bthe triangle and the cosine theorem: and.
Before proceeding to the implementation of this plan, we note that
Similarly, we have:
Now we will express cosine through and:
Since any angle in the triangle is greater and less, then. It means 
.
Now we transform each of the factors in the guoked expression. We have:
Substituting this expression in the formula for the area, we get:
The topic "The Square of the Triangle" is of great importance in the school course of mathematics. The triangle is the simplest of geometric shapes. It is a "structural element" of school geometry. The overwhelming majority of geometric tasks are reduced to solving triangles. Not an exception and task of finding the area of \u200b\u200bthe right and arbitrary N-parliament.
Example 2.4.1.
What is an equifiable triangle area, if its base, and the side side?
Decision:
-isosceles,
Figure 2.4.4.
We carry out the property of an equilibrium triangle - median and height. Then 
In Pythagore's theorem:
We find the area of \u200b\u200bthe triangle:
Answer:
Example 2.4.2.
In the rectangular triangle of the bisector of an acute angle divides the opposite catt on the segments of 4 and 5 cm. Determine the area of \u200b\u200bthe triangle.
Decision:
Let (Figure 2.4.5). Then (since bd - bisector). From here you have 
, i.e. It means
Figure 2.4.5.
Answer: 
Example 2.4.3.
Find an equifiable triangle area if its base is equal, and the length of the height conducted to the base is equal to the length of the segment connecting the middle of the base and the side.
Decision:
By condition, the middle line (Figure 2.4.6). So what you like:
or 
AT SUCCESSORY, 
Before you know how to find a parallelogram area, we need to remember what a parallelogram is and what is called it high. The parallelogram is a quadrangle, the opposite sides of which are parallel parallel (lie on parallel straight lines). Perpendicular, conducted from an arbitrary point of the opposite side to a direct, containing this side called a parallelogram height.
Square, rectangle and rhombus are particular cases of a parallelogram.
The area of \u200b\u200bthe parallelogram is indicated as (s).
Formulas finding the area of \u200b\u200bthe parallelogram
S \u003d A * H, where A is the base, H is the height that is conducted to the base.
S \u003d A * B * SINα, where a and b is the base, and α is the angle between the bases a and b.
S \u003d P * R, where p is a half-meter, R is the radius of the circle that is written in the parallelogram.
The area of \u200b\u200bthe parallelogram, which is formed by vectors A and B is equal to the module of the product of the specified vectors, namely:
Consider Example # 1: Dan Pollogram, the side of which is 7 cm, and the height is 3 cm. How to find a parallelogram area, the formula for solving we need.
Thus, S \u003d 7x3. S \u003d 21. Answer: 21 cm 2.
Consider Example # 2: The bases of 6 and 7 cm are given, and the angle between the bases of 60 degrees is given. How to find a parallelogram area? The formula used to solve:
Thus, we first find sinus angle. Sinus 60 \u003d 0.5, respectively, S \u003d 6 * 7 * 0.5 \u003d 21 Answer: 21 cm 2.
I hope these examples will help you when solving tasks. And remember, the main thing is knowledge of formulas and attentiveness.
When solving tasks on this topic except basic properties parallelogram and the corresponding formulas can be remembered and applied as follows:
- Bissectrice of the inner corner The parallelogram cuts off from it an equifiable triangle
- Inner angles bisectors adjacent to one side parallelogram mutually perpendicular
- Bissectrix, emerging from opposite internal angles, parallelogram, parallel to each other or lie on one straight line
- The sum of the squares of the diagonals of the parallelogram is equal to the sum of the squares of its sides
- The area of \u200b\u200bthe parallelogram is equal to half the work of diagonals on the sine corner between them
Consider the tasks when solving these properties.
Task 1.
The bisector of the angle with the parallelogram of the AVD crosses the side of the AD at the point M and the continuation of the side of the AV per point A at the point E. Find the perimeter of the parallelogram if Ae \u003d 4, DM \u003d 3.
Decision.
1. The triangle shade is a chaired. (Property 1). Therefore, Cd \u003d Md \u003d 3 cm.
2. Triangle EAM is a preceded.
Consequently, AE \u003d AM \u003d 4 cm.
3. AD \u003d AM + MD \u003d 7 cm.
4. Perimeter ABSD \u003d 20 cm.
Answer. 20 cm.
Task 2.
In the convex four-trigger AVD diagonal was carried out. It is known that the Square of the triangles of the AVD, ACD, ADD is equal. Prove that this quadril is a parallelogram.
Decision.
1. Let be - the height of the AVD triangle, CF is the height of the ACD triangle. Since, according to the condition of the task of the area of \u200b\u200btriangles, they also have a common base of AD, then the height of these triangles is equal. Ve \u003d CF.
2. ve, CF perpendicular to AD. Points in and from are located on one side relative to the direct AD. Ve \u003d CF. Consequently, the direct sun || AD. (*)
3. Let AL - the height of the ACD triangle, Bk - the height of the BCD triangle. Since, according to the condition of the task of the area of \u200b\u200btriangles, they also have a general base of CD, then the height of these triangles is equal. Al \u003d bk.
4. Al and BK perpendicular to CD. Points in and a are located on one side relative to the straight CD. Al \u003d bk. Consequently, direct Av || CD (**)
5. From the conditions (*), (**) flows - AVD parallelograms.
Answer. Proved. AVD - parallelogram.
Task 3.
On the sides of the aircraft and the CD, the parallelogram of the ABSDs are noted points M and H, respectively, so that the segments of VM and HD intersect at the point O;<ВМD = 95 о,
Decision.
1. In triangle Dom<МОD = 25 о (Он смежный с <ВОD = 155 о); <ОМD = 95 о. Тогда <ОDМ = 60 о.
2. In the rectangular triangle DNS Then<НСD = 30 о. СD: НD = 2: 1 But cd \u003d av. Then av: nd \u003d 2: 1. 3. <С = 30 о, 4. <А = <С = 30 о, <В = Answer: AV: HD \u003d 2: 1,<А = <С = 30 о, <В = Task 4. One of the diagonals of the parallelogram with length of 4√6 is based on an angle of 60 o, and the second diagonal is with the same base angle 45 o. Find the second diagonal. Decision.
1. AO \u003d 2√6. 2. To the triangle AOD apply the theorem of the sinuses. JSC / SIN D \u003d OD / SIN A. 2√6 / SIN 45 O \u003d OD / SIN 60 O. OD \u003d (2√6Sin 60 O) / SIN 45 O \u003d (2√6 · √3 / 2) / (√2 / 2) \u003d 2√18 / √2 \u003d 6. Answer: 12.
Task 5. The parallelogram with the parties 5√2 and 7√2, the smaller angle between the diagonals is equal to a smaller corner of the parallelogram. Find the sum of the lengths of diagonals. Decision.
Let D 1, D 2 - diagonally the parallelogram, and the angle between the diagonals and the smaller angle of the parallelogram is equal to f. 1. Count two different S abcd \u003d ab · ad · sin a \u003d 5√2 · 7√2 · sin f, S abcd \u003d 1/2 AS · cd · sin AOs \u003d 1/2 · d 1 d 2 sin f. We obtain equality 5√2 · 7√2 · sin f \u003d 1 / 2D 1 d 2 sin F or 2 · 5√2 · 7√2 \u003d d 1 d 2; 2. Using the ratio between the parties and diagonals of the parallelogram will install equality (AB 2 + AD 2) · 2 \u003d AC 2 + CD 2. ((5√2) 2 + (7√2) 2) · 2 \u003d d 1 2 + d 2 2. d 1 2 + d 2 2 \u003d 296. 3. Make a system: (D 1 2 + d 2 2 \u003d 296, Multiply the second equation of the system on 2 and fold with the first. We obtain (D 1 + d 2) 2 \u003d 576. Hence the ID 1 + D 2 i \u003d 24. Since d 1, d 2 - the length of the diagonals of the parallelogram, then D 1 + d 2 \u003d 24. Answer: 24.
Task 6. Sides parallelogram 4 and 6. The sharp corner between the diagonals is 45 o. Find the Pollogram area. Decision.
1. From the triangle AOs, using the cosine theorem, we write the ratio between the side of the parallelogram and the diagonals. AB 2 \u003d AO 2 + in 2 2 · JSC · COS AOs. 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 \u003d 16. d 1 2 + d 2 2 - d 1 · d 2 √2 \u003d 64. 2. Similarly, write down the ratio for the AOD triangle. We take into account what<АОD = 135 о и cos 135 о = -cos 45 о = -√2/2. We obtain the equation d 1 2 + d 2 2 + d 1 · d 2 √2 \u003d 144. 3. We have the system Survived from the second equation first, we obtain 2D 1 · d 2 √2 \u003d 80 or d 1 · d 2 \u003d 80 / (2√2) \u003d 20√2 4. S ABCD \u003d 1/2 AS · cd · Sin AOs \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 a fully system, anticipating that we need a product of diagonals in this task to calculate the area. Answer: 10. Task 7. The area of \u200b\u200bthe parallelogram is equal to 96, and its parties are 8 and 15. Find the square of the smallest diagonal. Decision.
1. S ABCD \u003d AV · AD · SIN Vad. Make a substitution in the formula. We obtain 96 \u003d 8 · 15 · SIN VAD. Hence SIN Vad \u003d 4/5. 2. Find COS WD. SIN 2 VAD + COS 2 WD \u003d 1. (4/5) 2 + COS 2 WD \u003d 1. COS 2 WD \u003d 9/25. By the condition of the problem, we find the length of a smaller diagonal. The BD diagonal will be smaller if the angle is sharp. Then cos wad \u003d 3/5. 3. From the AVD triangle on the cosine theorem will find the square of the VD diagonal. Cd 2 \u003d AB 2 + AD 2 - 2 · AV · cd · cos wad. Cd 2 \u003d 8 2 + 15 2 - 2 · 8 · 15 · 3/5 \u003d 145. Answer: 145.
Have questions? Do not know how to solve the geometric problem? the site, with full or partial copying of the material reference to the original source is required. Formula for square parallelogram The area of \u200b\u200bthe parallelogram is equal to the product of its side to the height, lowered on this side. Evidence If the parallelogram is a rectangle, then the equality is made by the rectangle area theorem. Next, we believe that the corners of the parallelogram are not direct. Let $ angle $ \\ angle Bad $ acute and $ AD\u003e AB $ be in the parallelogram of $ ABCD $. Otherwise we rename the vertices. Then the height of $ BH $ from the top $ b $ to direct $ ad $ falls to the side of $ ad $, as the cattat $ ah $ shorter hypotenuse $ AB $, and $ AB< AD$. Основание $K$ высоты $CK$ из точки $C$ на прямую $AB$ лежит на продолжении отрезка $AD$ за точку $D$, так как угол $\angle BAD$ острый, а значит $\angle CDA$ тупой. Вследствие параллельности прямых $BA$ и $CD$ $\angle BAH = \angle CDK$. В параллелограмме противоположные стороны равны, следовательно, по стороне и двум углам, треугольники $\triangle ABH = \triangle DCK$ равны. Compare the area of \u200b\u200bthe $ ABCD $ parallelogram and the $ HBCK $ rectangle area. The parallelogram area is larger on the area of \u200b\u200b$ \\ triangle ABH $, but less on the area of \u200b\u200b$ \\ triangle dck $. Since these triangles are equal, then their square is equal. So, the area of \u200b\u200bthe parallelogram is equal to the square of the rectangle with the side of the side to the side and the height of the parallelogram. Formula for square parallelogram through side and sinus The area of \u200b\u200bthe parallelogram is equal to the product of neighboring sides to the corner sinus between them. Evidence $ ABCD $ parallelogram height, lowered to $ AB $ side is equal to a piece of $ BC $ segment on a $ \\ angle Angc $ angle. It remains to apply the previous statement. Formula for square parallelogram via diagonal The area of \u200b\u200bthe parallelogram is equal to half the work of diagonals on the sine corner between them. Evidence Let the diagonal of the $ ABCD $ parallelogram intersect at $ o $'s point at $ \\ alpha $. Then $ AO \u003d OC $ and $ BO \u003d OD $ for the property of the parallelogram. Sinuses of the corners, in the amount of $ 180 ^ \\ Circ $ are equal, $ \\ angle aob \u003d \\ angle Cod \u003d 180 ^ \\ CIRC - \\ ANGLE BOC \u003d 180 ^ \\ CIRC - \\ Angle AOD $. So, the sines of angles with the intersection of diagonals are equal to $ \\ sin \\ alpha $. $ S_ (abcd) \u003d s _ (\\ triangle aob) + s _ (\\ triangle boc) + s _ (\\ triangle cod) + s _ (\\ triangle AOD) $ according to the axiom of measuring area. Apply the formula of the triangle area $ s_ (ABC) \u003d \\ DFRAC (1) (2) \\ CDOT AB \\ CDOT BC \\ SIN \\ ANGC $ for these triangles and angles when crossing diagonals. The sides of each are equal to half the diagonals, sines are also equal. Consequently, the area of \u200b\u200ball four triangles is equal to $ s \u003d \\ dFrac (1) (2) \\ CDOT \\ DFRAC (AC) (2) \\ CDOT \\ DFRAC (BD) (2) \\ CDOT \\ SIN \\ ALPHA \u003d \\ DFRAC (AC \\ Summing all the above, we get $ S_ (abcd) \u003d 4S \u003d 4 \\ CDOT \\ DFRAC (AC \\ CDOT BD) (8) \\ Sin \\ Alpha \u003d \\ DFRAC (AC \\ CDOT BD \\ CDOT \\ SIN \\ ALPHA) (2) $
(
(Since in a rectangular triangle catat, which lies against an angle of 30 o, equal to half of the hypotenuse).
ways to its area.
(D 1 + d 2 \u003d 140.
(D 1 2 + d 2 2 - d 1 · d 2 √2 \u003d 64,
(D 1 2 + d 2 2 + d 1 · d 2 √2 \u003d 144.
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