How to find the area of ​​a parallelogram, triangle, trapezoid. How to find the area of ​​a parallelogram? Area of ​​a parallelogram with two heights

Task 4.

One of the diagonals of a parallelogram with a length of 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. We apply the sine theorem to triangle AOD.

AO/sin D = OD/sin A.

2√6/sin 45 o = OD/sin 60 o.

ОD = (2√6sin 60 о) / sin 45 о = (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 is equal to φ.

1. Let's count two different
ways its area.

S ABCD = AB AD sin A = 5√2 7√2 sin f,

S ABCD = 1/2 AC ВD sin AOB = 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 relationship 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 create a system:

(d 1 2 + d 2 2 = 296,
(d 1 + d 2 = 140.

Let's 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 degrees. Find the area of ​​the parallelogram.

Solution.

1. From triangle AOB, using the cosine theorem, we write the relationship between the side of the parallelogram and the diagonals.

AB 2 = AO 2 + VO 2 2 · AO · VO · cos AOB.

4 2 = (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.

Let's 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
(d 1 2 + d 2 2 – d 1 · d 2 √2 = 64,
(d 1 2 + d 2 2 + d 1 · d 2 √2 = 144.

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 = 1/2 AC ВD sin AOB = 1/2 d 1 d 2 sin α = 1/2 20√2 √2/2 = 10.

Note: In this and the previous problem there is no need to solve the system completely, anticipating 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 = AB · AD · sin ВAD. Let's make a substitution in the formula.

We get 96 = 8 · 15 · sin ВAD. Hence sin ВAD = 4/5.

2. Let's find cos VAD. sin 2 VAD + cos 2 VAD = 1.

(4 / 5) 2 + cos 2 VAD = 1. cos 2 VAD = 9 / 25.

According to the conditions of the problem, we find the length of the smaller diagonal. The diagonal ВD will be smaller if the angle ВАD is acute. Then cos VAD = 3 / 5.

3. From the triangle ABD, using the cosine theorem, we find the square of the diagonal BD.

ВD 2 = АВ 2 + АD 2 – 2 · АВ · ВD · cos ВAD.

ВD 2 = 8 2 + 15 2 – 2 8 15 3 / 5 = 145.

Answer: 145.

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Area of ​​a parallelogram

Theorem 1

The area of ​​a parallelogram is defined as the product of the length of its side and the height drawn to it.

where $a$ is a side of the parallelogram, $h$ is the height drawn to this side.

Proof.

Let us be given a parallelogram $ABCD$ with $AD=BC=a$. Let us draw the heights $DF$ and $AE$ (Fig. 1).

Picture 1.

Obviously, the $FDAE$ figure is a rectangle.

\[\angle BAE=(90)^0-\angle A,\ \] \[\angle CDF=\angle D-(90)^0=(180)^0-\angle A-(90)^0 =(90)^0-\angle A=\angle BAE\]

Consequently, since $CD=AB,\ DF=AE=h$, by the $I$ criterion for the equality of triangles $\triangle BAE=\triangle CDF$. Then

So, according to the theorem on the area of ​​a rectangle:

The theorem has been proven.

Theorem 2

The area of ​​a parallelogram is defined as the product of the length of its adjacent sides times the sine of the angle between these sides.

Mathematically this can be written as follows

where $a,\ b$ are the sides of the parallelogram, $\alpha $ is the angle between them.

Proof.

Let us be given a parallelogram $ABCD$ with $BC=a,\ CD=b,\ \angle C=\alpha $. Let us draw the height $DF=h$ (Fig. 2).

Figure 2.

By definition of sine, we get

Hence

So, by Theorem $1$:

The theorem has been proven.

Area of ​​a triangle

Theorem 3

The area of ​​a triangle is defined as half the product of the length of its side and the altitude drawn to it.

Mathematically this can be written as follows

where $a$ is a side of the triangle, $h$ is the height drawn to this side.

Proof.

Figure 3.

So, by Theorem $1$:

The theorem has been proven.

Theorem 4

The area of ​​a triangle is defined as half the product of the length of its adjacent sides and the sine of the angle between these sides.

Mathematically this can be written as follows

where $a,\b$ are the sides of the triangle, $\alpha$ is the angle between them.

Proof.

Let us be given a triangle $ABC$ with $AB=a$. Let's find the height $CH=h$. Let's build it up to a parallelogram $ABCD$ (Fig. 3).

Obviously, by the $I$ criterion for the equality of triangles, $\triangle ACB=\triangle CDB$. Then

So, by Theorem $1$:

The theorem has been proven.

Area of ​​trapezoid

Theorem 5

The area of ​​a trapezoid is defined as half the product of the sum of the lengths of its bases and its height.

Mathematically this can be written as follows

Proof.

Let us be given a trapezoid $ABCK$, where $AK=a,\ BC=b$. Let us draw in it the heights $BM=h$ and $KP=h$, as well as the diagonal $BK$ (Fig. 4).

Figure 4.

By Theorem $3$, we get

The theorem has been proven.

Sample task

Example 1

Find the area of ​​an equilateral triangle if its side length is $a.$

Solution.

Since the triangle is equilateral, all its angles are equal to $(60)^0$.

Then, by Theorem $4$, we have

Answer:$\frac(a^2\sqrt(3))(4)$.

Note that the result of this problem can be used to find the area of ​​any equilateral triangle with a given side.

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). A perpendicular drawn from an arbitrary point on the opposite side to a line containing this side is called the height of a 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 =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, which is formed by vectors a and b, is equal to the modulus of the product of the given vectors, namely:

Let's consider example No. 1: Given a parallelogram, the side of which is 7 cm and the height is 3 cm. How to find the area of ​​a parallelogram, we need a formula for the solution.

Thus S= 7x3. S=21. Answer: 21 cm 2.

Consider example No. 2: Given bases are 6 and 7 cm, and also given an angle between the bases of 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 = 0.5, respectively S = 6*7*0.5=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

Definition of parallelogram

Parallelogram is a quadrilateral in which opposite sides are equal and parallel.

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The parallelogram has some useful properties that make it easier to solve problems involving this figure. For example, one of the properties is that opposite angles of a parallelogram are equal.

Let's consider several methods and formulas followed by solving simple examples.

Formula for the area of ​​a parallelogram based on its base and height

This method of finding the area is probably one of the most basic and simple, since it is almost identical to the formula for finding the area of ​​a triangle with a few exceptions. First, let's look at the generalized case without using numbers.

Let an arbitrary parallelogram with a base be given a a a, side b b b and height h h h, carried to our base. Then the formula for the area of ​​this parallelogram is:

S = a ⋅ h S=a\cdot h S=a ⋅h

A a a- base;
h h h- height.

Let's look at one easy problem to practice solving typical problems.

Find the area of ​​a parallelogram in which the base is known to be 10 (cm) and the height is 5 (cm).

Solution

A = 10 a=10 a =1 0
h = 5 h=5 h =5

We substitute it into our formula. We get:
S = 10 ⋅ 5 = 50 S=10\cdot 5=50S=1 0 ⋅ 5 = 5 0 (see sq.)

Answer: 50 (see sq.)

Formula for the area of ​​a parallelogram based on two sides and the angle between them

In this case, the required value is found as follows:

S = a ⋅ b ⋅ sin ⁡ (α) S=a\cdot b\cdot\sin(\alpha)S=a ⋅b ⋅sin(α)

A, b a, b a, b- sides of a parallelogram;
α\alpha α - angle between sides a a a And b b b.

Now let's solve another example and use the formula described above.

Find the area of ​​a parallelogram if the side is known a a a, which is the base and with a length of 20 (cm) and a perimeter p p p, numerically equal to 100 (cm), the angle between adjacent sides ( a a a And b b b) is equal to 30 degrees.

Solution

A = 20 a=20 a =2 0
p = 100 p=100 p =1 0 0
α = 3 0 ∘ \alpha=30^(\circ)α = 3 0 ∘

To find the answer, we only know the second side of this quadrilateral. Let's find her. The perimeter of a parallelogram is given by the formula:
p = a + a + b + b p=a+a+b+b p =a+a+b+b
100 = 20 + 20 + b + b 100=20+20+b+b1 0 0 = 2 0 + 2 0 + b+b
100 = 40 + 2b 100=40+2b 1 0 0 = 4 0 + 2 b
60 = 2b 60=2b 6 0 = 2 b
b = 30 b=30 b =3 0

The hardest part is over, all that remains is to substitute our values ​​for the sides and the angle between them:
S = 20 ⋅ 30 ⋅ sin ⁡ (3 0 ∘) = 300 S=20\cdot 30\cdot\sin(30^(\circ))=300S=2 0 ⋅ 3 0 ⋅ sin(3 0 ∘ ) = 3 0 0 (see sq.)

Answer: 300 (see sq.)

Formula for the area of ​​a parallelogram based on the diagonals and the angle between them

S = 1 2 ⋅ D ⋅ d ⋅ sin ⁡ (α) S=\frac(1)(2)\cdot D\cdot d\cdot\sin(\alpha)S=2 1 ​ ⋅ D⋅d⋅sin(α)

D D D- large diagonal;
d d d- small diagonal;
α\alpha α - acute angle between diagonals.

Given are the diagonals of a parallelogram equal to 10 (cm) and 5 (cm). The angle between them is 30 degrees. Calculate its area.

Solution

D=10 D=10 D=1 0
d = 5 d=5 d =5
α = 3 0 ∘ \alpha=30^(\circ)α = 3 0 ∘

S = 1 2 ⋅ 10 ⋅ 5 ⋅ sin ⁡ (3 0 ∘) = 12.5 S=\frac(1)(2)\cdot 10 \cdot 5 \cdot\sin(30^(\circ))=12.5S=2 1 ​ ⋅ 1 0 ⋅ 5 ⋅ sin(3 0 ∘ ) = 1 2 . 5 (see sq.)

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:

1. The sides that are opposite are identical in pairs.
2. 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:

1. , lowered to one side, are perpendicular;
2. opposite vertices have parallel bisectors;
3. 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 has been 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:

1. a and b, α - sides and the angle between them;
2. d 1 and d 2, γ - diagonals and at the point of their intersection;
3. 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
along the sides and one of the diagonals	 Conclusion The parallelogram, as one of the key figures of geometry, is used in life, for example, in construction when calculating the area of ​​a site or other measurements. Therefore, knowledge about the distinctive features and methods of calculating its various parameters can be useful at any time in life.