Area of ​​a parallelogram by side and height. How to find the area of ​​a parallelogram? Features of adjacent corners

Area of ​​a 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

1. 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
   
2. Formula for the area of ​​a triangle based on three sides and the radius of the circumcircle
   
3. 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

1. Formula for the area of ​​a square by side length
   Square area equal to the square of the length of its side.
2. 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

1. Formula for the area of ​​a parallelogram based on side length and height
   Area of ​​a parallelogram
2. 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

1. 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.
2. 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.
3. 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

1. 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,
   

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.

Example

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.

Example

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.

Example

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.)

When solving problems on this topic, except basic properties parallelogram and the corresponding formulas, you can remember and apply the following:

1. The bisector of an interior angle of a parallelogram cuts off an isosceles triangle from it
2. Bisectors of interior angles adjacent to one of the sides of a parallelogram are mutually perpendicular
3. Bisectors coming from opposite interior corners of a parallelogram are parallel to each other or lie on the same straight line
4. The sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides
5. The area of ​​a parallelogram is equal to half the product of the diagonals and the sine of the angle between them

Let us consider problems in 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 = 4, DM = 3.

Solution.

1. Triangle CMD is 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 this 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 conditions 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 relative to straight line AD. BE = CF. Therefore, straight line BC || A.D. (*)

3. Let AL be the altitude of triangle ACD, BK the altitude of triangle BCD. Since, according to the conditions 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 relative to straight line CD. AL = BK. Therefore, straight line AB || CD (**)

5. From conditions (*), (**) it follows that ABCD is a parallelogram.

Answer. Proven. ABCD is a parallelogram.

Task 3.

On sides BC and CD of the parallelogram ABCD, points M and H are marked, respectively, so that the segments BM and HD intersect at point O;<ВМD = 95 о,

Solution.

1. In 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
(Since in a right triangle the leg that lies opposite the angle of 30° is equal to half the hypotenuse).

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 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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What is a parallelogram? A parallelogram is a quadrilateral whose opposite sides are parallel in pairs.

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 – height drawn to this side.

2. If the lengths of two adjacent sides of a 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 a parallelogram are given and the angle between them is known, then the area of ​​the parallelogram is calculated by the formula:

\[ \LARGE S = \frac(1)(2) \cdot d_(1) \cdot d_(2) \cdot sin(\alpha) \]

Properties of a parallelogram

In a parallelogram, opposite sides are equal: \(AB = CD\), \(BC = AD\)

In a parallelogram, opposite angles are equal: \(\angle A = \angle C\), \(\angle B = \angle D\)

The diagonals of a parallelogram at the intersection point are divided in half \(AO = OC\) , \(BO = OD\)

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: \(\angle K B H =\angle A\) .

The bisectors of angles adjacent to one side of a parallelogram are mutually perpendicular.

The bisectors of two opposite angles of a parallelogram are parallel.

Signs of a parallelogram

A quadrilateral will be 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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