Area of ​​a parallelogram. How to find the area of ​​a parallelogram? Area of ​​a parallelogram if two sides are known

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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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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.

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