Do the squares have equal areas? Properties of areas of polygons Equal polygons have equal areas

VIII class: Topic 3. Areas of figures. Pythagorean theorem.

1. The concept of area. Equal-sized figures.

If length is a numerical characteristic of a line, then area is a numerical characteristic of a closed figure. Despite the fact that we are well familiar with the concept of area from everyday life, it is not easy to give a strict definition to this concept. It turns out that the area of a closed figure can be called any non-negative quantity having the following properties of measuring the areas of figures:

Equal figures have equal areas. If a given closed figure is divided into several closed figures, then the area of the figure is equal to the sum of the areas of its constituent figures (the figure in Figure 1 is divided into n figures; in this case, the area of the figure, where Si- square i-th figure).

In principle, it would be possible to come up with a set of quantities that have the formulated properties, and therefore characterize the area of the figure. But the most familiar and convenient value is the one that characterizes the area of a square as the square of its side. Let's call this “agreement” the third property of measuring the areas of figures:

The area of a square is equal to the square of its side (Figure 2).

With this definition, the area of the figures is measured in square units ( cm 2, km 2, ha=100m 2).

Figures having equal areas are called equal in size .

Comment: Equal figures have equal areas, that is, equal figures are equal in size. But equal-sized figures are not always equal (for example, Figure 3 shows a square and an isosceles triangle made up of equal right-angled triangles (by the way, such figures called equally composed ); it is clear that the square and the triangle are equal in size, but not equal, since they do not overlap).

Next, we will derive formulas for calculating the areas of all main types of polygons (including the well-known formula for finding the area of a rectangle), based on the formulated properties of measuring the areas of figures.

2. Area of a rectangle. Area of a parallelogram.

Formula for calculating the area of a rectangle: The area of a rectangle is equal to the product of its two adjacent sides (Figure 4).

Given:

ABCD- rectangle;

AD=a, AB=b.

Prove: SABCD=a× b.

Proof:

1. Extend the side AB for a segment B.P.=a, and the side AD- for a segment D.V.=b. Let's build a parallelogram APRV(Figure 4). Since Ð A=90°, APRV- rectangle. Wherein AP=a+b=AV, Þ APRV– a square with side ( a+b).

2. Let us denote B.C.Ç RV=T, CDÇ PR=Q. Then BCQP– a square with a side a, CDVT– a square with a side b, CQRT- rectangle with sides a And b.

Formula for calculating the area of a parallelogram: The area of a parallelogram is equal to the product of its height and its base (Figure 5).

Comment: The base of a parallelogram is usually called the side to which the height is drawn; It is clear that any side of a parallelogram can serve as a base.

Given:

ABCD– p/g;

B.H.^AD, HÎ AD.

Prove: SABCD=AD× B.H..

Proof:

1. Let's take it to the base AD height CF(Figure 5).

2. B.C.ïê HF, B.H.ïê CF, Þ BCFH- p/g by definition. Ð H=90°, Þ BCFH- rectangle.

3. BCFH– p/g, Þ according to the p/g property B.H.=CF, Þ D BAH=D CDF along the hypotenuse and leg ( AB=CD according to St. p/g, B.H.=CF).

4. SABCD=SABCF+S D CDF=SABCF+S D BAH=SBCFH=B.H.× B.C.=B.H.× AD. #

3. Area of a triangle.

Formula for calculating the area of a triangle: The area of a triangle is equal to half the product of its height and its base (Figure 6).

Comment: In this case, the base of the triangle is the side to which the altitude is drawn. Any of the three sides of a triangle can serve as its base.

Given:

BD^A.C., DÎ A.C..

Prove: .

Proof:

1. Let's complete D ABC to p/y ABKC by passing through the top B straight B.K.ïê A.C., and through the top C- straight CKïê AB(Figure 6).

2. D ABC=D KCB on three sides ( B.C.– general, AB=KC And A.C.=K.B. according to St. p/g), Þ https://pandia.ru/text/78/214/images/image014_34.gif" width="107" height="36">).

Corollary 2: If we consider p/u D ABC with height A.H., drawn to the hypotenuse B.C., That . Thus, in p/u D-ke height drawn to the hypotenuse is equal to the ratio of the product of its legs to the hypotenuse . This relation is quite often used when solving problems.

4. Corollaries from the formula for finding the area of a triangle: the ratio of the areas of triangles with equal heights or bases; equal triangles in figures; property of the areas of triangles formed by the diagonals of a convex quadrilateral.

From the formula for calculating the area of a triangle, two consequences follow in an elementary way:

1. Ratio of areas of triangles with equal heights equal to the ratio of their bases (in Figure 8 ).

2. Ratio of areas of triangles with equal bases equal to the ratio of their heights (in Figure 9 ).

Comment: When solving problems, triangles with a common height are very often encountered. In this case, as a rule, their bases lie on the same straight line, and the vertex opposite the bases is common (for example, in Figure 10 S 1:S 2:S 3=a:b:c). You should learn to see the total height of such triangles.

Also, the formula for calculating the area of a triangle yields useful facts that allow you to find equal triangles in figures:

1. The median of an arbitrary triangle divides it into two equal triangles (in Figure 11 at D A.B.M. and D ACM height A.H.– general, and the grounds B.M. And C.M. equal by definition of median; it follows that D A.B.M. and D ACM equal in size).

2. The diagonals of a parallelogram divide it into four equal triangles (in Figure 12 A.O.– median of the triangle ABD by the property of diagonals p/g, Þ due to the previous properties of triangles ABO And ADO equal in size; because B.O.– median of the triangle ABC, triangles ABO And BCO equal in size; because CO– median of the triangle BCD, triangles BCO And DCO equal in size; Thus, S D ADO=S D ABO=S D BCO=S D DCO).

3. The diagonals of a trapezoid divide it into four triangles; two of them, adjacent to the lateral sides, are equal in size (Figure 13).

Given:

ABCD– trapezoid;

B.C.ïê AD; A.C.Ç BD=O.

Prove: S D ABO=S D DCO.

Proof:

1. Let's draw the heights B.F. And CH(Figure 13). Then D ABD and D ACD base AD– general, and heights B.F. And CH equal; Þ S D ABD=S D ACD.

2. S D ABO=S D ABD ‑ S D AOD=S D ACD ‑ S D AOD=S D DCO. #

If you draw the diagonals of a convex quadrilateral (Figure 14), four triangles are formed, the areas of which are related by a very easy-to-remember ratio. The derivation of this relationship relies solely on the formula for calculating the area of a triangle; however, it is found quite rarely in the literature. Being useful in solving problems, the relation that will be formulated and proven below deserves close attention:

Property of the areas of triangles formed by the diagonals of a convex quadrilateral: If the diagonals of a convex quadrilateral ABCD intersect at a point O, then (Figure 14).

ABCD– convex quadrangle;

https://pandia.ru/text/78/214/images/image025_28.gif" width="149" height="20">.

Proof:

1. B.F.– total height D AOB and D BOC; Þ S D AOB:S D BOC=A.O.:CO.

2. D.H.– total height D AOD and D C.O.D.; Þ S D AOD:S D C.O.D.=A.O.:CO.

5. Ratio of areas of triangles having equal angles.

Theorem on the ratio of the areas of triangles having equal angles: The areas of triangles that have equal angles are related as the products of the sides enclosing these angles (Figure 15).

Given:

D ABC,D A 1B 1C 1;

Ð BAC=Ð B 1A 1C 1.

Prove:

Proof:

1. Lay it down on the ray AB line segment AB 2=A 1B 1, and on the beam A.C.- line segment A.C. 2=A 1C 1 (Figure 15). Then D AB 2C 2=D A 1B 1C 1 on two sides and the angle between them ( AB 2=A 1B 1 and A.C. 2=A 1C 1 by construction, and Р B 2A.C. 2=р B 1A 1C 1 by condition). Means, .

2. Connect the dots C And B 2.

3. CH– total height D AB 2C and D ABC, Þ https://pandia.ru/text/78/214/images/image033_22.gif" width="81" height="43 src=">.

6. Property of the bisector of a triangle.

Using the theorems on the ratio of the areas of triangles having equal angles, and on the ratio of the areas of triangles with equal heights, we simply prove a fact that is extremely useful in solving problems and is not directly related to the areas of figures:

Triangle bisector property: The bisector of a triangle divides the side to which it is drawn into segments proportional to the sides adjacent to them.

Given:

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Proof:

1..gif" width="72 height=40" height="40">.

3. From points 1 and 2 we get: , Þ https://pandia.ru/text/78/214/images/image041_19.gif" width="61" height="37">. #

Comment: Since the extreme members or middle members can be swapped in the correct proportion, it is more convenient to remember the property of the bisector of a triangle in the following form (Figure 16): .

7. Area of a trapezoid.

Formula for calculating the area of a trapezoid: The area of a trapezoid is equal to the product of its height and half the sum of its bases.

Given:

ABCD– trapezoid;

B.C.ïê AD;

B.H.- height.

https://pandia.ru/text/78/214/images/image044_21.gif" width="127" height="36">.

Proof:

1. Let's draw a diagonal BD and height DF(Figure 17). BHDF– rectangle, Þ B.H. = DF.

Consequence: The ratio of the areas of trapezoids with equal heights is equal to the ratio of their midlines (or the ratio of the sums of the bases).

8. Area of a quadrilateral with mutually perpendicular diagonals.

Formula for calculating the area of a quadrilateral with mutually perpendicular diagonals: The area of a quadrilateral with mutually perpendicular diagonals is equal to half the product of its diagonals.

ABCD– quadrangle;

A.C.^BD.

https://pandia.ru/text/78/214/images/image049_20.gif" width="104" height="36">.

Proof:

1. Let us denote A.C.Ç BD=O. Because the A.C.^BD, A.O.– height D ABD, A CO– height D CBD(Figures 18a and 18b for the cases of convex and non-convex quadrilaterals, respectively).

2.
(the signs “+” or “-” correspond to the cases of convex and non-convex quadrilaterals, respectively). #

The Pythagorean theorem plays an extremely important role in solving a wide variety of problems; it allows you to find the unknown side of a right triangle from its two known sides. There are many known proofs of the Pythagorean theorem. Let us present the simplest of them, based on formulas for calculating the areas of a square and a triangle:

Pythagorean theorem: In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the legs.

Given:

D ABC– p/u;

Ð A=90°.

Prove:

B.C. 2=AB 2+A.C. 2.

Proof:

1. Let us denote A.C.=a, AB=b. Let's put it on the ray AB line segment B.P.=a, and on the beam A.C.- line segment CV=b(Figure 19). Let's draw through the point P direct PRïê AV, and through the point V– straight VRïê AP. Then APRV- p/g by definition. Moreover, since Р A=90°, APRV- rectangle. And because AV=a+b=AP, APRV– a square with a side a+b, And SAPRV=(a+b)2. Next we will divide the side PR dot Q into segments PQ=b And QR=a, and the side RV– dot T into segments RT=b And TV=a.

2.D ABC=D PQB=D RTQ=D VCT on two sides, Þ Ð ACB=Ð PBQ=Ð RQT=Ð VTC, B.C.=QB=T.Q.=C.T., and https://pandia.ru/text/78/214/images/image055_17.gif" width="115" height="36">.

3. Because B.C.=QB=T.Q.=C.T., CBQT- rhombus At the same time QBC=180°-(Р ABC+Ð PBQ)=180°-(Р ABC+Ð ACB)=Ð BAC=90°; Þ CBQT- square, and SCBQT=B.C. 2.

4. . So, B.C. 2=AB 2+A.C. 2. #

The inverse Pythagorean theorem is a sign of a right triangle, i.e., it allows you to check whether the triangle is right-angled using three known sides.

Converse Pythagorean theorem: If the square of a side of a triangle is equal to the sum of the squares of its other two sides, then the triangle is right-angled and its longest side is the hypotenuse.

Given:

B.C. 2=AB 2+A.C. 2.

Prove: D ABC– p/u;

Ð A=90°.

Proof:

1. Construct a right angle A 1 and put the segments on its sides A 1B 1=AB And A 1C 1=A.C.(Figure 20). In the resulting p/u D A 1B 1C 1 by Pythagorean theorem B 1C 12=A 1B 12+A 1C 12=AB 2+A.C. 2; but according to the condition AB 2+A.C. 2=B.C. 2; Þ B 1C 12=B.C. 2, Þ B 1C 1=B.C..

2.D ABC=D A 1B 1C 1 on three sides ( A 1B 1=AB And A 1C 1=A.C. by construction, B 1C 1=B.C. from item 1), Þ Ð A=Ð A 1=90°, Þ D ABC- p/u. #

Right triangles whose side lengths are expressed in natural numbers are called Pythagorean triangles , and the triplets of the corresponding natural numbers are Pythagorean triplets . Pythagorean triplets are useful to remember (the larger of these numbers is equal to the sum of the squares of the other two). Here are some Pythagorean triples:

3, 4, 5;

5, 12, 13;

8, 15, 17;

7, 24, 25;

20, 21, 29;

12, 35, 37;

9, 40, 41.

A right triangle with sides 3, 4, 5 was used in Egypt to construct right angles, and therefore such triangle called Egyptian .

10. Heron's formula.

Heron's formula allows you to find the area of an arbitrary triangle from its three known sides and is indispensable in solving many problems.

Heron's formula: Area of a triangle with sides a, b And c is calculated using the following formula: , where is the semi-perimeter of the triangle.

Given:

B.C.=a; A.C.=b; AB=c.). Then .

4. Substitute the resulting expression for height into the formula for calculating the area of the triangle: . #

Job source: Decision 2746.-13. OGE 2017 Mathematics, I.V. Yashchenko. 36 options.

Task 11. The side of a rhombus is 12, and the distance from the point of intersection of the diagonals of the rhombus to it is 1. Find the area of this rhombus.

Solution.

The area of a rhombus can be calculated in the same way as the area of a parallelogram, that is, as the product of the height h of the rhombus by the length of the side a to which it is drawn:

In the figure, the red line together with the black line shows the height h of the rhombus, which is equal (since the lengths of the black and red lines are equal). The length of the side is a=12 also according to the conditions of the problem. We get the area of the rhombus:

Answer: 24.

Task 12. A rhombus is depicted on checkered paper with a square size of 1x1. Find the length of its longer diagonal.

Solution.

In the figure, the blue lines show the diagonals of the rhombus. It can be seen that the large diagonal is 12 cells.

Answer: 12.

Task 13. Which of the following statements are true?

1) There is a rectangle whose diagonals are mutually perpendicular.

2) All squares have equal areas.

3) One of the angles of a triangle always does not exceed 60 degrees.

In response, write down the numbers of the selected statements without spaces, commas or other additional characters.

Solution.

1) Correct. This is a rectangle that turns into a square.

“Donkey Bridge” The proof of the Pythagorean theorem was considered very difficult in the circles of students of the Middle Ages and was sometimes called Pons Asinorum “donkey bridge” or elefuga - “flight of the wretched,” since some “wretched” students who did not have serious mathematical training fled from geometry. Weak students who memorized theorems by heart, without understanding, and were therefore nicknamed “donkeys,” were unable to overcome the Pythagorean theorem, which served as an insurmountable bridge for them.

Given: ABC, C=90°, B=60°, AB=12 cm AC=10 cm Find: SABC Solve orally CA B Given: ABC, C=90°, AB=18 cm, BC=9 cm Find: B , A Answer: A=30º, B=60º Answer: 30 cm²

C² = a 2 + b 2 a b c C A B c = a 2 + b cbа In a right triangle, a and b are the legs, c is the hypotenuse. Fill the table. b =c²-a² a =c²-b² b 2 =c²-a² a 2 =c²-b²

Solution 3. ACD is rectangular, D=45° DAC=45°ACD - isosceles CD = AC = 4 SADC = 8. So the area of the entire figure S ABCB = SABC + SADC = Given: AB=2 3, BC=2, B= 90 ACD=90 BAC=3 0, D=45 Find: S ABCB. Problem 30º D C B A Area of the entire figure S ABCB = SABC + SADC 2. ABC is rectangular, SABC = 2 3; BAC=30° AC = 2BC = 4.

497 One of the diagonals of a parallelogram is its height. Find this diagonal if the perimeter of the parallelogram is 50 cm and the difference between adjacent sides is 1 cm. AD CB Given: ABCD - parallelogram, BD AD, P ABCD = 50 cm, AB-AD = 1 cm. Find: BD. Solution. Let AD=x cm, then AB=(x+1) cm. Because P ABCD =2·(AB+AD), then 50=2·(x+1+x) 25=2x+1 x=12, which means AD=12 cm, AB=13 cm. 1. AD=12 cm, AB=13 cm. 2. Find BD using the Pythagorean theorem: AB²=ВD²+AD² BD=5 (cm) 12 cm 13 cm

BC by 6 cm. Find: BC, CD, AD. " title="Problem The area of a rectangular trapezoid is 120 cm² and its height is 8 cm. Find all sides of the trapezoid if one of its bases is 6 cm larger than the other. D BC A N Given: ABCD - trapezoid, AB AD , S ABCD = 120 cm², AB = 8 cm, AD>BC by 6 cm. Find: BC, CD, AD." class="link_thumb"> 16 !} Problem The area of a rectangular trapezoid is 120 cm² and its height is 8 cm. Find all sides of the trapezoid if one of its bases is 6 cm larger than the other. D BC A N Given: ABCD - trapezoid, AB AD, S ABCD = 120 cm², AB = 8 cm, AD>BC by 6 cm. Find: BC, CD, AD. Solution. Let BC=x cm, then AD=(x+6) cm Because S ABCD = ·8·(x+6+x)=120, 4(2x+6)=120 2x+6 = 30 x = 12, which means BC 12 cm, AD=18 cm AB=8 cm, BC=12 cm, AD=18 cm Additional construction: CH AD, then ABCN is a rectangle. CH=AB=8 cm, AH=BC=12 cm, then HD=AD-AH=6 cm 12 cm 18 cm 6 cm Find CD using the Pythagorean theorem: CD2=CH2+HD2 CD=8²+6²CD=10 (cm) Answer: AB=8 cm, BC=12 cm, CD=10 cm, AD=18 cm. BC by 6 cm. Find: BC, CD, AD. "> BC by 6 cm. Find: BC, CD, AD. Solution. Let BC=x cm, then AD=(x+6) cm Because S ABCD = ·8·(x+6+x)= 120, 4(2x+6)=120 2x+6 = 30 x = 12, which means BC 12 cm, AD=18 cm 1. 2. AB=8 cm, BC=12 cm, AD=18 cm Additional formation: CH AD, then ABCN is a rectangle. CH=AB=8 cm, AH=BC=12 cm, then HD=AD-AH=6 cm 12 cm 18 cm 6 cm Find CD using the Pythagorean theorem: CD²=CH²+HD² CD=8² +6²CD=10 (cm) Answer: AB=8 cm, BC=12 cm, CD=10 cm, AD=18 cm."> BC by 6 cm. Find: BC, CD, AD. " title="Problem The area of a rectangular trapezoid is 120 cm² and its height is 8 cm. Find all sides of the trapezoid if one of its bases is 6 cm larger than the other. D BC A N Given: ABCD - trapezoid, AB AD , S ABCD = 120 cm², AB = 8 cm, AD>BC by 6 cm. Find: BC, CD, AD."> title="Problem The area of a rectangular trapezoid is 120 cm² and its height is 8 cm. Find all sides of the trapezoid if one of its bases is 6 cm larger than the other. D BC A N Given: ABCD - trapezoid, AB AD, S ABCD = 120 cm², AB = 8 cm, AD>BC by 6 cm. Find: BC, CD, AD."> !} AB C M N Given: ABC, BC=7.5 cm, AC=3.2 cm, AM BC, BN AC, AM=2.4 cm Find: BN Solution: SABC =½AM·CB=½·2.4 ·7.5=9 cm² S ABC =½BN·AC BN=2·S ABC:AC=2·9:3.2=5.625 cm Answer: 5.625 cm. Two sides of the triangle are 7.5 cm and 4 cm. Height drawn to the larger side is equal to 2.4 cm. Find the height drawn to the smaller of these sides. 470

The area of a right triangle is 168 cm². Find its legs if the ratio of their lengths is 7:12. A C B Given: ABC, C = 90º, AC: BC = 7:12, S ABC = 168 cm² Find: AC, BC. Solution: SABC =½AC·BC 168=½7x·12x 168=42x² x=2 AC=14 cm, BC=24 cm Answer: 14 cm and 24 cm. 472

Properties of Areas 10. Equal polygons have equal areas. D B A C N ABC = NFD F

Properties of areas 20. If a polygon is made up of several polygons, then its area is equal to the sum of the areas of these polygons. C B D A F

Properties of areas 30. The area of a square is equal to the square of its side. 3 cm S=9 cm 2 Using the properties of areas, find the areas of the figures

Units of area measurement 1 m 2 = 100 dm 2 1 dm 2 = 100 cm 2

Units of area measurement 1 km 2 1 ha 1 a 1 m 2 1 dm 2 1 cm 2 1 mm 2: 100: 100

Area of a rectangle b S Let us prove that S = ab a a SQUARE WITH SIDE a 2 a+b = S + a 2 + b 2 a 2 +2 ab + b 2 = 2 S + a 2 + b 2 S (a+b) 2 S 2 ab = 2 S S = ab b 2 b: 2

The floor of the room, which has the shape of a rectangle with sides of 5, 5 m and 6 m, must be covered with rectangular parquet. The length of each parquet plank is 30 cm, and the width is 5 cm. How many such planks are needed to cover the floor? 6 m 5.5 m 5 cm 30 cm

The areas of the squares built on the sides of the rectangle are 64 cm 2 and 121 cm 2. Find the area of the rectangle. 121 cm 2 S-? 64 cm 2

The sides of each of the rectangles ABCD and ARMK are equal to 6 cm and 10 cm. Find the area of the figure consisting of all points that belong to at least one of these rectangles. A 10 cm P B 6 cm 10 cm D K C 6 cm M

ABCD is a rectangle, AC is a diagonal. Find the area of triangle ABC. A a D АBC = ADC b SABC = B C

ABCD is a rectangle. Find: SABF. B CE = DE, C F E A D SABCD = Q

AB = BC = 3, AF = 5, Find: SABCDEF. B EF = 2. C 3 D E 3 A 2 5 F

S=102 C Points K, M, T and E are located 5 respectively on sides AD, AB, BC and DC of square E ABCD so that KD=7, AK=3, AM=5, BT=8, CE=5. Find the area of the quadrilateral KMTE. D T B 2 8 M 5 7 K 3 A

The area of the pentagon ABCD is 48 cm 2. Find the area and perimeter of the square ABCD. C B O A 1) 48: 3 * 4 = 64 (cm 2) SАВСD 2) AB = 8 (cm), PАВСD = 8 * 4 = 32 (cm) D

ABCD and MDKP are equal squares. AB = 8 cm. Find the area of the quadrilateral ASKM. B C 64 cm 2 8 cm 32 cm 2 D A 32 cm 2 M K 32 cm 2 R

ABCD and DСМK are squares. AB = 6 cm. Find the area of the quadrilateral OSPD. C H 6 cm A O M R D K

ABCD – rectangle; M, K, P, T are the midpoints of its sides, AB = 6 cm, AD = 12 cm. Find the area of the quadrilateral MKRT. H K 6 cm M A C R T 12 cm D

ABCD – rectangle; M, K, P, T are the midpoints of its sides, AB = 16 cm, BC = 10 cm. Find the area of the hexagon AMKSRT. C P 10 cm K B D T M 16 cm A