Parallelogram in problems. Area of a parallelogram How to calculate the area of a parallelogram knowing its sides
A parallelogram is a geometric figure that is often found in problems in a geometry course (section planimetry). 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 rhombus, rectangle, square.
Calculating the area of this type of polygon can be done in several ways. Let's look at 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 at your disposal the side side of the figure. In this case, the required value will be obtained using the formula:
S = a * h (a) = b * h (b),
- S is the area that should have been determined,
- a, b – known (or calculated) side,
- h is the height lowered onto it.
Example: the value of the base of a 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
Let's consider the case when you know the sizes of two sides of a figure, as well as the degree measure of the angle that they form between themselves. The data provided can also be used to find the area of a parallelogram. In this case, the formula expression will look like this:
S = a * c * sinα = a * c * sinβ,
- a – side,
- c – known (or calculated) base,
- α, β – angles between sides a and c.
Example: the base of a parallelogram is 10 cm, its side is 4 cm less. The obtuse angle of the figure is 135°.
Solution: determine the value of the second side: 10 – 4 = 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 us to determine the area of the figure.
S = (d1*d2)/2*sinγ,
S = (d1*d2)/2*sinφ,
S is the area to be determined,
d1, d2 – known (or calculated by calculations) diagonals,
γ, φ – angles between diagonals d1 and d2.
Parallelogram is a quadrilateral whose sides are parallel in pairs.
In this figure, opposite sides and angles are equal to each other. The diagonals of a parallelogram intersect at one point and bisect it. Formulas for the area of a parallelogram allow you to find the value using the sides, height and diagonals. A parallelogram can also be presented in special cases. They are considered a rectangle, square and rhombus.
First, let's look at an example of calculating the area of a parallelogram by height and the side to which it is lowered.
This case is considered a classic and does not require additional investigation. It’s better to consider the formula for calculating the area through two sides and the angle between them. The same method is used in calculations. If the sides and the angle between them are given, then the area is calculated as follows: 
Suppose we are given a parallelogram with sides a = 4 cm, b = 6 cm. The angle between them is α = 30°. Let's find the area: 
Area of a parallelogram through diagonals
The formula for the area of a parallelogram using the diagonals allows you to quickly find the value.
For calculations, you will need the size of the angle located between the diagonals.
Let's consider an example of calculating the area of a parallelogram using diagonals. Let a parallelogram be given with diagonals D = 7 cm, d = 5 cm. The angle between them is α = 30°. Let's substitute the data into the formula: 
An example of calculating the area of a parallelogram through the diagonal gave us an excellent result - 8.75.
Knowing the formula for the area of a parallelogram through the diagonal, you can solve many interesting problems. Let's look at one of them.
Task: Given a parallelogram with an area of 92 square meters. see Point F is located in the middle of its side BC. Let's let's find the area trapezoid ADFB, which will lie in our parallelogram. First, let's draw everything we received according to the conditions.
Let's get to the solution: 
According to our conditions, ah =92, and accordingly, the area of our trapezoid will be equal to 
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.
Definition of parallelogram
Parallelogram is a quadrilateral in which opposite sides are equal and parallel.
Online calculator
The parallelogram has some beneficial properties, which simplify solving problems associated with 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.
ExampleFind 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.
ExampleFind 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.)
Square geometric figure - a numerical characteristic of a geometric figure showing the size of this figure (part of the surface limited by the closed contour of this figure). The size of the area is expressed by the number of square units contained in it.
Triangle area formulas
- Formula for the area of a triangle by 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 - Formula for the area of a triangle based on three sides and the radius of the circumcircle
- Formula for the area of a triangle based on three sides and the radius of the inscribed circle
Area of a triangle is equal to the product of the semi-perimeter of the triangle and the radius of the inscribed circle. where S is the area of the triangle,
- lengths of the sides of the triangle,
- height of the triangle,
- the angle between the sides and,
- radius of the inscribed circle,
R - radius of the circumscribed circle,
Square area formulas
- Formula for the area of a square by side length
Square area equal to the square of the length of its side. - Formula for the area of a square along the diagonal length
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,
- length of the side of the square,
- length of the diagonal of the square.
Rectangle area formula
- Area of a rectangle equal to the product of the lengths of its two adjacent sides
where S is the area of the rectangle,
- lengths of the sides of the rectangle.
Parallelogram area formulas
- Formula for the area of a parallelogram based on side length and height
Area of a parallelogram - Formula for the area of a parallelogram based on two sides and the angle between them
Area of a parallelogram is equal to the product of the lengths of its sides multiplied by the sine of the angle between them.a b sin α
where S is the area of the parallelogram,
- lengths of the sides of the parallelogram,
- length of parallelogram height,
- the angle between the sides of the parallelogram.
Formulas for the area of a rhombus
- Formula for the area of a rhombus based on side length and height
Area of a rhombus equal to the product of the length of its side and the length of the height lowered to this side. - Formula for the area of a rhombus based on side length and angle
Area of a rhombus 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. - Formula for the area of a rhombus based on the lengths of its diagonals
Area of a rhombus 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,
- length of the height of the rhombus,
- the angle between the sides of the rhombus,
1, 2 - lengths of diagonals.
Trapezoid area formulas
- Heron's formula for trapezoid
Where S is the area of the trapezoid,
- lengths of the bases of the trapezoid,
- lengths of the sides of the trapezoid,
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