Area of ​​a parallelogram. Solving geometry problems: solving quadrilaterals Problems finding the area of ​​a parallelogram

Theorem 1. The area of ​​a trapezoid is equal to the product of half the sum of its bases and its height:

Theorem 2. The diagonals of a trapezoid divide it into four triangles, two of which are similar and the other two have the same area:

Theorem 3. The area of ​​a parallelogram is equal to the product of the base and the height lowered by a given base, or the product of two sides and the sine of the angle between them:

Theorem 4. In a parallelogram, the sum of the squares of the diagonals is equal to the sum of the squares of its sides:

Theorem 5. The area of ​​an arbitrary convex quadrilateral is equal to half the product of its diagonals and the sine of the angle between them:

Theorem 6. The area of ​​a quadrilateral circumscribed about a circle is equal to the product of the semi-perimeter of this quadrilateral and the radius of the given circle:

Theorem 7. A quadrilateral whose vertices are the midpoints of the sides of an arbitrary convex quadrilateral is a parallelogram whose area is equal to half the area of ​​the original quadrilateral:

Theorem 8. If a convex quadrilateral has diagonals that are mutually perpendicular, then the sum of the squares of the opposite sides of this quadrilateral are equal:

AB2 + CD2 = BC2 + AD2.

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Proofs of some theorems

Proof of Theorem 2. Let ABCD be a given trapezoid, AD and BC its bases, O the intersection point of diagonals AC and BD of this trapezoid. Let us prove that triangles AOB and COD have the same area. To do this, lower the perpendiculars BP and CQ from points B and C to line AD. Then the area of ​​triangle ABD is

And the area of ​​triangle ACD is

Since BP = CQ, then S∆ABD = S∆ACD. But the area of ​​triangle AOB is the difference between the areas of triangles ABD and AOD, and the area of ​​triangle COD is the difference between the areas of triangles ACD and AOD. Therefore, the areas of triangles AOB and COD are equal, as required to be proved.

Proof of Theorem 4. Let ABCD be a parallelogram, AB = CD = a, AD = BC = b,
AC = d1, BD = d2, ∠BAD = α, ∠ADC = 180° – α. Let us apply the cosine theorem to triangle ABD:

Now applying the cosine theorem to triangle ACD, we get:

Adding the resulting equalities term by term, we obtain that Q.E.D.

Proof of Theorem 5. Let ABCD be an arbitrary convex quadrilateral, E the intersection point of its diagonals, AE = a, BE = b,
CE = c, DE = d, ∠AEB = ∠CED = ϕ, ∠BEC =
= ∠AED = 180° – ϕ. We have:

Q.E.D.

Proof of Theorem 6. Let ABCD be an arbitrary quadrilateral circumscribed about a circle, O the center of this circle, OK, OL, OM and ON the perpendiculars drawn from the point O onto the lines AB, BC, CD and AD, respectively. We have:

where r is the radius of the circle and p is the semi-perimeter of the quadrilateral ABCD.

Proof of Theorem 7. Let ABCD be an arbitrary convex quadrilateral, K, L, M and N the midpoints of the sides AB, BC, CD and AD, respectively. Since KL is the midline of triangle ABC, then line KL is parallel to line AC and Similarly, line MN is parallel to line AC and Therefore, KLMN is a parallelogram. Consider triangle KBL. Its area is equal to one quarter of the area of ​​triangle ABC. The area of ​​triangle MDN is also equal to a quarter of the area of ​​triangle ACD. Hence,

Likewise,

It means that

where does it follow that

Proof of Theorem 8. Let ABCD be an arbitrary convex quadrilateral whose diagonals are mutually perpendicular, let E be the intersection point of its diagonals,
AE = a, BE = b, CE = c, DE = d. Let us apply the Pythagorean theorem to triangles ABE and CDE:
AB2 = AE2 + BE2 = a 2 + b2,
CD2 = CE2 + DE2 = c2 + d2,
hence,
AB2 + CD2 = a 2 + b2 + c2 + d2 .
Now applying the Pythagorean theorem to triangles ADE and BCE, we obtain:
AD2 = AE2 + DE2 = a 2 + d2,
BC2 = BE2 + CE2 = b2 + c2,
where does it follow that
AD2 + BC2 = a 2 + b2 + c2 + d2 .
This means AB2 + CD2 = AD2 + BC2, which is what needed to be proven.

Problem solutions

Problem 1. A trapezoid with base angles α and β is described around the circle. Find the ratio of the area of ​​the trapezoid to the area of ​​the circle.

Solution. Let ABCD be a given trapezoid, AB and CD its bases, DK and CM the perpendiculars drawn from points C and D to line AB. The required ratio does not depend on the radius of the circle. Therefore, we will assume that the radius is 1. Then the area of ​​the circle is equal to π, let’s find the area of ​​the trapezoid. Since triangle ADK is right-angled, then

Similarly, from the right triangle BCM we find that Since a circle can be inscribed in a given trapezoid, the sums of the opposite sides are equal:
AB + CD = AD + BC,
where do we find it from?

So the area of ​​the trapezoid is

and the required ratio is equal to
Answer:

Problem 2. In a convex quadrilateral ABCD, angle A is equal to 90°, and angle C does not exceed 90°. From vertices B and D perpendiculars BE and DF are dropped onto diagonal AC. It is known that AE = CF. Prove that angle C is right.

Proof. Since angle A is 90°,
and angle C does not exceed 90°, then points E and F lie on the diagonal AC. Without loss of generality, we can assume that AE< AF (в противном случае следует повторить все нижеследующие рассуждения с заменой точек B и D). Пусть ∠ABE = α,
∠EBC = β, ∠FDA = γ, ∠FDC = δ. It is enough for us to prove that α + β + γ + δ = π. Because

from where we get that what was required to be proved.

Problem 3. The perimeter of an isosceles trapezoid circumscribed about a circle is equal to p. Find the radius of this circle if it is known that sharp corner at the base of the trapezoid is equal to α.
Solution. Let ABCD be a given isosceles trapezoid with bases AD and BC, let BH be the height of this trapezoid dropped from vertex B.
Since a circle can be inscribed in a given trapezoid, then

Hence,

From the right triangle ABH we find,

Answer:

Problem 4. Given a trapezoid ABCD with bases AD and BC. Diagonals AC and BD intersect at point O, and lines AB and CD intersect at point K. Line KO intersects sides BC and AD at points M and N, respectively, and angle BAD is 30°. It is known that a circle can be inscribed in the trapezoid ABMN and NMCD. Find the ratio of the areas of triangle BKC and trapezoid ABCD.

Solution. As is known, for an arbitrary trapezoid, a straight line connecting the point of intersection of the diagonals and the point of intersection of the extensions of the lateral sides divides each of the bases in half. So BM = MC and AN = ND. Further, since a circle can be inscribed in the trapezoid ABMN and NMCD, then
BM + AN = AB + MN,
MC + ND = CD + MN.
It follows that AB = CD, that is, the trapezoid ABCD is isosceles. The required area ratio does not depend on the scale, so we can assume that KN = x, KM = 1. From the right triangles AKN and BKM we obtain that Writing again the relation already used above
BM + AN = AB + MN ⇔

We need to calculate the ratio:

Here we used the fact that the areas of triangles AKD and BKC are related as the squares of the sides KN and KM, that is, as x2.

Answer:

Task 5. In a convex quadrilateral ABCD, the points E, F, H, G are the midpoints of the sides AB, BC, CD, DA, respectively, and O is the intersection point of the segments EH and FG. It is known that EH = a, FG = b, Find the lengths of the diagonals of the quadrilateral.

Solution. It is known that if you connect the midpoints of the sides of an arbitrary quadrilateral in series, you get a parallelogram. In our case, EFHG is a parallelogram and O is the intersection point of its diagonals. Then

Let us apply the cosine theorem to the triangle FOH:

Since FH is the midline of triangle BCD, then

Similarly, applying the cosine theorem to the triangle EFO, we obtain that

Answer:

Task 6. The lateral sides of a trapezoid are 3 and 5. It is known that a circle can be inscribed in a trapezoid. The midline of a trapezoid divides it into two parts, the ratio of their areas is equal to Find the bases of the trapezoid.

Solution. Let ABCD be a given trapezoid, AB = 3 and CD = 5 its lateral sides, points K and M the midpoints of sides AB and CD, respectively. Let, for definiteness, AD > BC, then the area of ​​the trapezoid AKMD will be more area trapezoids KBCM. Since KM is the midline of trapezoid ABCD, trapezoids AKMD and KBCM have equal heights. Since the area of ​​a trapezoid is equal to the product of half the sum of the bases and the height, the following equality is true:

Further, since a circle can be inscribed in the trapezoid ABCD, then AD + BC = AB + CD = 8. Then KM = 4 as the midline of the trapezoid ABCD. Let BC = x, then AD = 8 – x. We have:
So BC = 1 and AD = 7.

Answer: 1 and 7.

Problem 7. The base AB of the trapezoid ABCD is twice as long as the base CD and twice as long as the side AD. The length of the diagonal AC is a, and the length of the side BC is equal to b. Find the area of ​​the trapezoid.

Solution. Let E be the intersection point of the extensions of the lateral sides of the trapezoid and CD = x, then AD = x, AB = 2x. Segment CD is parallel to segment AB and is half its length, which means CD is the midline of triangle ABE. Therefore, CE = BC = b and DE = AD = x, hence AE = 2x. So, triangle ABE is isosceles (AB = AE) and AC is its median. Therefore AC is also the altitude of this triangle, which means

Since triangle DEC is similar to triangle AEB with similarity coefficient then

Answer:

Problem 8. The diagonals of the trapezoid ABCD intersect at point E. Find the area of ​​triangle BCE if the lengths of the bases of the trapezoid are AB = 30, DC = 24, sides AD = 3 and angle DAB is 60°.

Solution. Let DH be the height of the trapezoid. From triangle ADH we find that

Since the height of triangle ABC dropped from vertex C is equal to the height DH of the trapezoid, we have:

Answer:

Problem 9. In a trapezoid, the midline is 4, and the angles at one of the bases are 40° and 50°. Find the bases of the trapezoid if the segment connecting the midpoints of the bases is equal to 1.

Solution. Let ABCD be a given trapezoid, AB and CD its bases (AB< CD), M, N - середины AB и CD соответственно. Пусть также ∠ADC = 50°, ∠BCD = 40°. Средняя линия трапеции равна полусумме оснований, поэтому
AB + CD = 8. Extend the sides DA and CB to the intersection at point E. Consider the triangle ABE, in which ∠EAB = 50°. ∠EBA = 40°,
therefore, ∠AEB = 90°. The median EM of this triangle, drawn from the vertex of the right angle, is equal to half the hypotenuse: EM = AM. Let EM = x, then AM = x, DN = 4 – x. According to the condition of the problem MN = 1, therefore,
EN = x + 1. From the similarity of triangles AEM and DEN we have:

This means AB = 3 and CD = 5.

Answer: 3 and 5.

Problem 10. Convex quadrilateral ABCD is circumscribed about a circle with center at point O, with AO = OC = 1, BO = OD = 2. Find the perimeter of quadrilateral ABCD.

Solution. Let K, L, M, N be the tangent points of the circle with sides AB, BC, CD, DA, respectively, and r be the radius of the circle. Since a tangent to a circle is perpendicular to the radius drawn to the point of tangency, the triangles AKO, BKO, BLO, CLO, CMO, DMO, DNO, ANO are rectangular. Applying the Pythagorean theorem to these triangles, we obtain that

Therefore, AB = BC = CD = DA, that is, ABCD is a rhombus. The diagonals of a rhombus are perpendicular to each other, and the point of their intersection is the center of the inscribed circle. From here we easily find that the side of the rhombus is equal and therefore the perimeter of the rhombus is equal to

Answer:

Problems to solve independently

S-1. An equilateral trapezoid ABCD is circumscribed around a circle of radius r. Let E and K be the points of tangency of this circle with the sides of the trapezoid. The angle between the base AB and the side AD of the trapezoid is 60°. Prove that EK is parallel to AB and find the area of ​​the trapezoid ABEK.
S-2. In a trapezoid, the diagonals are 3 and 5, and the segment connecting the midpoints of the bases is 2. Find the area of ​​the trapezoid.
S-3. Is it possible to describe a circle around a quadrilateral ABCD if ∠ADC = 30°, AB = 3, BC = 4, AC = 6?
S-4. In the trapezoid ABCD (AB is the base), the values ​​of the angles DAB, BCD, ADC, ABD and ADB form an arithmetic progression (in the order in which they are written). Find the distance from vertex C to diagonal BD if the height of the trapezoid is h.
S-5. Given an isosceles trapezoid into which a circle is inscribed and around which the circle is circumscribed. The ratio of the height of a trapezoid to the radius of the circumscribed circle is Find the angles of the trapezoid.
S-6. The area of ​​rectangle ABCD is 48 and the length of the diagonal is 10. On the plane in which the rectangle is located, a point O is chosen so that OB = OD = 13. Find the distance from point O to the vertex of the rectangle that is farthest from it.
S-7. The perimeter of parallelogram ABCD is 26. The angle ABC is 120°. The radius of the circle inscribed in triangle BCD is Find the lengths of the sides of the parallelogram if it is known that AD > AB.
S-8. Quadrilateral ABCD is inscribed in a circle with center at point O. The radius OA is perpendicular to the radius OB, and the radius OC is perpendicular to the radius OD. The length of the perpendicular dropped from point C to line AD is equal to 9. The length of segment BC is half the length of segment AD. Find the area of ​​triangle AOB.
S-9. In a convex quadrilateral ABCD, vertices A and C are opposite, and the length of side AB is 3. Angle ABC is equal to angle BCD is equal to Find the length of side AD if you know that the area of ​​the quadrilateral is

S-10. In a convex quadrilateral ABCD the diagonals AC and BD are drawn. It is known that
AD = 2, ∠ABD = ∠ACD = 90°, and the distance between the bisector intersection point of triangle ABD and the bisector intersection point of triangle ACD is Find the length of side BC.
S-11. Let M be the intersection point of the diagonals of a convex quadrilateral ABCD, in which sides AB, AD and BC are equal. Find the angle CMD if it is known that DM = MC,
and ∠CAB ≠ ∠DBA.
S-12. In quadrilateral ABCD we know that ∠A = 74°, ∠D = 120°. Find the angle between the bisectors of angles B and C.
S-13. A circle can be inscribed in a quadrilateral ABCD. Let K be the point of intersection of its diagonals. It is known that AB > BC > KC, and the perimeter and area of ​​triangle BKC are 14 and 7, respectively. Find DC.
S-14. In a trapezoid circumscribed about a circle, it is known that BC AD, AB = CD, ∠BAD =
= 45°. Find AB if the area of ​​the trapezoid ABCD is 10.
S-15. In the trapezoid ABCD with bases AB and CD it is known that ∠CAB = 2∠DBA. Find the area of ​​the trapezoid.
S-16. In parallelogram ABCD it is known that AC = a, ∠CAB = 60°. Find the area of ​​the parallelogram.
S-17. In quadrilateral ABCD, diagonals AC and BD intersect at point K. Points L and M are the midpoints of sides BC and AD, respectively. The segment LM contains point K. The quadrilateral ABCD is such that a circle can be inscribed in it. Find the radius of this circle if AB = 3, and LK: KM = 1: 3.
S-18. In a convex quadrilateral ABCD the diagonals AC and BD are drawn. In this case ∠BAC =
= ∠BDC, and the area of ​​the circle circumscribed about the triangle BDC is equal to
a) Find the radius of the circle circumscribed about triangle ABC.
b) Knowing that BC = 3, AC = 4, ∠BAD = 90°, find the area of ​​quadrilateral ABCD.

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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Formula for the area of ​​a parallelogram

The area of ​​a parallelogram is equal to the product of its side and the height of that side.

Proof

If the parallelogram is a rectangle, then the equality is satisfied by the theorem on the area of ​​a rectangle. Next, we assume that the angles of the parallelogram are not right.

Let $\angle BAD$ be an acute angle in parallelogram $ABCD$ and $AD > AB$. Otherwise, we will rename the vertices. Then the height $BH$ from the vertex $B$ to the line $AD$ falls on the side $AD$, since the leg $AH$ is shorter than the hypotenuse $AB$, and $AB< AD$. Основание $K$ высоты $CK$ из точки $C$ на прямую $AB$ лежит на продолжении отрезка $AD$ за точку $D$, так как угол $\angle BAD$ острый, а значит $\angle CDA$ тупой. Вследствие параллельности прямых $BA$ и $CD$ $\angle BAH = \angle CDK$. В параллелограмме противоположные стороны равны, следовательно, по стороне и двум углам, треугольники $\triangle ABH = \triangle DCK$ равны.

Let's compare the area of ​​the parallelogram $ABCD$ and the area of ​​the rectangle $HBCK$. The area of ​​a parallelogram is greater by area $\triangle ABH$, but less by area $\triangle DCK$. Since these triangles are equal, their areas are equal. This means that the area of ​​a parallelogram is equal to the area of ​​a rectangle with sides length to side and the height of the parallelogram.

Formula for the area of ​​a parallelogram using sides and sine

The area of ​​a parallelogram is equal to the product of adjacent sides and the sine of the angle between them.

Proof

The height of the parallelogram $ABCD$ dropped onto side $AB$ is equal to the product of the segment $BC$ and the sine of the angle $\angle ABC$. It remains to apply the previous statement.

Formula for the area of ​​a parallelogram using the diagonals

The area of ​​a parallelogram is equal to half the product of the diagonals and the sine of the angle between them.

Proof

Let the diagonals of the parallelogram $ABCD$ intersect at the point $O$ at an angle $\alpha$. Then $AO=OC$ and $BO=OD$ by the parallelogram property. The sines of the angles that add up to $180^\circ$ are equal, $\angle AOB = \angle COD = 180^\circ - \angle BOC = 180^\circ - \angle AOD$. This means that the sines of the angles at the intersection of the diagonals are equal to $\sin \alpha$.

$S_(ABCD)=S_(\triangle AOB) + S_(\triangle BOC) + S_(\triangle COD) + S_(\triangle AOD)$

according to the axiom of area measurement. We apply the triangle area formula $S_(ABC) = \dfrac(1)(2) \cdot AB \cdot BC \sin \angle ABC$ for these triangles and angles when the diagonals intersect. The sides of each are equal to half the diagonals, and the sines are also equal. Therefore, the areas of all four triangles are equal to $S = \dfrac(1)(2) \cdot \dfrac(AC)(2) \cdot \dfrac(BD)(2) \cdot \sin \alpha = \dfrac(AC \ cdot BD)(8) \sin \alpha$. Summing up all of the above, we get

$S_(ABCD) = 4S = 4 \cdot \dfrac(AC \cdot BD)(8) \sin \alpha = \dfrac(AC \cdot BD \cdot \sin \alpha)(2)$

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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Note. This is part of a lesson with geometry problems (parallelogram section). If you need to solve a geometry problem that is not here, write about it in the forum. To indicate the action of extracting a square root in problem solutions, the symbol √ or sqrt() is used, with the radical expression indicated in parentheses.

Theoretical material

Explanations for the formulas for finding the area of ​​a parallelogram:

1. The area of ​​a parallelogram is equal to the product of the length of one of its sides and the height of that side
2. The area of ​​a parallelogram is equal to the product of its two adjacent sides and the sine of the angle between them
3. The area of ​​a parallelogram is equal to half the product of its diagonals and the sine of the angle between them

Problems on finding the area of ​​a parallelogram

Task.
In a parallelogram, the shorter height and shorter side are 9 cm and the root of 82, respectively. The larger diagonal is 15 cm. Find the area of ​​the parallelogram.

Solution.
Let us denote the smaller height of the parallelogram ABCD lowered from point B to the larger base AD as BK.
Let's find the value of the leg of a right triangle ABK formed by a smaller height, a smaller side and part of a larger base. According to the Pythagorean theorem:

AB 2 = BK 2 + AK 2
82 = 9 2 + AK 2
AK 2 = 82 - 81
AK = 1

Let us extend the upper base of the parallelogram BC and lower the height AN to it from its lower base. AN = BK as the sides of the rectangle ANBK. Let us find the leg NC of the resulting right triangle ANC.
AN 2 + NC 2 = AC 2
9 2 + NC 2 = 15 2
NC 2 = 225 - 81
NC 2 = √144
NC=12

Now let's find the larger base BC of parallelogram ABCD.
BC = NC - NB
Let's take into account that NB = AK as the sides of the rectangle, then
BC = 12 - 1 = 11

The area of ​​a parallelogram is equal to the product of the base and the height to this base.
S = ah
S = BC * BK
S = 11 * 9 = 99

Answer: 99 cm 2 .

Task

In the parallelogram ABCD, the perpendicular BO is dropped onto the diagonal AC. Find the area of ​​the parallelogram if AO=8, OC=6 and BO=4.

Solution.
Let us drop another perpendicular DK onto the diagonal AC.
Accordingly, triangles AOB and DKC, COB and AKD are pairwise equal. One of the sides is the opposite side of the parallelogram, one of the angles is a straight line, since it is perpendicular to the diagonal, and one of the remaining angles is an internal cross lying for the parallel sides of the parallelogram and the secant diagonal.

Thus, the area of ​​the parallelogram is equal to the area of ​​the indicated triangles. That is
Sparallel = 2S AOB +2S BOC

The area of ​​a right triangle is equal to half the product of the legs. Where
S = 2 (1/2 8 * 4) + 2 (1/2 6 * 4) = 56 cm 2
Answer: 56 cm 2 .