What is the formula for finding the area of a parallelogram. Parallelogram in problems. The formula for the area of a parallelogram by base and height
Definition of a parallelogram
Parallelogram is a quadrilateral in which opposite sides are equal and parallel.
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A parallelogram has some useful properties that make it easier to solve problems related to this figure. For example, one property is that opposite angles of a parallelogram are equal.
Consider several methods and formulas, followed by solving simple examples.
The formula for the area of a parallelogram by 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. Let's start with a generalized case without using numbers.
Let an arbitrary parallelogram with base a a a, side bb b and height h h h drawn 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 take a look at one easy problem to practice solving typical problems.
ExampleFind the area of a parallelogram in which the base equal to 10 (cm) and the height equal to 5 (cm) are known.
Solution
A=10 a=10 a =1
0
h=5 h=5 h =5
Substitute in our formula. We get:
S=10 ⋅ 5=50 S=10\cdot 5=50S=1
0
⋅
5
=
5
0
(see sq.)
Answer: 50 (see square)
The formula for the area of a parallelogram given two sides and the angle between them
In this case, the desired 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 bb b.
Now let's solve another example and use the above formula.
ExampleFind the area of a parallelogram if the side is known a a a, which is the base and with a length of 20 (see) and a perimeter pp p, numerically equal to 100 (see), the angle between adjacent sides ( a a a and bb 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 do not know only the second side of this quadrilateral. Let's find her. The perimeter of a parallelogram is given by:
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
+
2b
60=2b 60=2b 6
0
=
2b
b=30 b=30 b=3
0
The hardest part is over, it remains only 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.)
The formula for the area of a parallelogram given 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
is an acute angle between the diagonals.
The diagonals of the parallelogram are given, equal to 10 (see) and 5 (see). 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.)
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
As in Euclidean geometry, the point and the 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 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 - 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. |
Parallelogram area
Theorem 1
The area of a parallelogram is defined as the product of the length of its side times the height drawn to it.
where $a$ is the 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's draw the heights $DF$ and $AE$ (Fig. 1).
Picture 1.
It is obvious that the figure $FDAE$ is a rectangle.
\[\angle BAE=(90)^0-\angle A,\ \] \[\angle CDF=\angle D-(90)^0=(180)^0-\angle A-(90)^0 =(90)^0-\angle A=\angle BAE\]
Therefore, since $CD=AB,\ DF=AE=h$, $\triangle BAE=\triangle CDF$, by $I$ the triangle equality test. Then
So according to the rectangle area theorem:
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 those 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 $. Draw the height $DF=h$ (Fig. 2).
Figure 2.
By definition of the sine, we get
Hence
Hence, 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 height drawn to it.
Mathematically, this can be written as follows
where $a$ is the 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 times the sine of the angle between those 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 $AB=a$. Draw the height $CH=h$. Let's build it up to the parallelogram $ABCD$ (Fig. 3).
Obviously, $\triangle ACB=\triangle CDB$ by $I$. Then
So by Theorem $1$:
The theorem has been proven.
Trapezium area
Theorem 5
The area of a trapezoid is defined as half the product of the sum of the lengths of its bases times 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 the heights $BM=h$ and $KP=h$ in it, as well as the diagonal $BK$ (Fig. 4).
Figure 4
By Theorem $3$, we get
The theorem has been proven.
Task example
Example 1
Find the area of an equilateral triangle if the length of its side 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.
Parallelogram - a geometric figure, often found in the tasks of the geometry course (planimetry section). The key features of this quadrilateral are the equality of opposite angles and the presence of two pairs of parallel opposite sides. Special cases of a parallelogram are a rhombus, a rectangle, a square.
The calculation of the area of this type of polygon can be done in several ways. Let's consider each of them.
Find the area of a parallelogram if the side and height are known
To calculate the area of a parallelogram, you can use the values of its side, as well as the length of the height lowered onto it. In this case, the data obtained will be reliable both for the case of a known side - the base of the figure, and if you have the side of the figure at your disposal. In this case, the desired value will be obtained by the formula:
S = a * h(a) = b * h(b),
- S is the area to be determined,
- a, b - known (or calculated) side,
- h is the height lowered on it.
Example: the value of the base of the parallelogram is 7 cm, the length of the perpendicular dropped onto it from the opposite vertex is 3 cm.
Solution: S = a * h (a) = 7 * 3 = 21.
Find the area of a parallelogram if 2 sides and the angle between them are known
Consider the case when you know the magnitude of the two sides of the figure, as well as the degree measure of the angle that they form with each other. The data provided can also be used to find the area of the parallelogram. In this case, the formula expression will look like this:
S = a * c * sinα = a * c * sinβ,
- a - side,
- c is a known (or calculated) base,
- α, β are the angles between sides a and c.
Example: the base of a parallelogram is 10 cm, its side is 4 cm smaller. The obtuse angle of the figure is 135°.
Solution: determine the value of the second side: 10 - 4 \u003d 6 cm.
S = a * c * sinα = 10 * 6 * sin135° = 60 * sin(90° + 45°) = 60 * cos45° = 60 * √2 /2 = 30√2.
Find the area of a parallelogram if the diagonals and the angle between them are known
The presence of known values of the diagonals of a given polygon, as well as the angle that they form as a result of their intersection, allows you to determine the area of \u200b\u200bthe figure.
S = (d1*d2)/2*sinγ,
S = (d1*d2)/2*sinφ,
S is the area to be determined,
d1, d2 are known (or calculated) diagonals,
γ, φ are the angles between the diagonals d1 and d2.