The process of training students in the eighth grade elective classes in mathematics

2. It is known that AOB = 35 °, 50 ° = BOC. Find the angle AOC. For each of the possible cases make a drawing with a ruler and a protractor.

A negative number is not a solution, as a measure of degree angle - a positive number. Reply: 15o or 85o.

Homework

1) Points A, B and C lie on a straight line. It is known that AB = BC = 10cm 25cm. What can be the length of the segment AC? (Answer: 35cm or 15cm).

2) It is known that AOB = 45 °, 25 ° = BOC. Find the angle AOC. For each of the possible cases make a drawing with a ruler and a protractor. (Answer 700 or 20 °).

3) On the line are the points A, B, C and D. Find the length of the segment with endpoints at the midpoints of AB and CD, if AC = 5, BD = 7.

2 class. (Problems in the subject triangles)

(Generalization tasks homework)

1. On the line containing the segment AB, we take a point C so that AC = s, AB = a. Find the length of the segment Sun

Decision.

1) Let the point A lies between C and B. Then by the axiom of measurement segments BC = AB + AC, where BC = a + c.

2) Let the point B lies between A and C. Then AC = AB + BC and BC = c - a.

3) If the point C lies between points A and B, then AB = AC + BC and BC = a - c.

Obviously, the cases 2) and 3) are incompatible because the lengths of the segment BC are opposite, and the length of the segment - the number is positive. Thus, one of these cases is not the answer.

Answer: BC = a + or VS = \ a - \.

2. Beam from the output of the vertex (ab). Find (as), if (ab) = a (bc) =.

The solution:

Main

1. Find the angles of an isosceles triangle if one of its angles equal to 80 °.

solution: 1) Let B = 800, then C = 800,

Since the triangle ABC - isosceles.

Since the sum of the angles in 1800, the

A + B + C = 1800

A 1600 = 1800

A = 200

2) Let A = 800, then

A + B + C = 1800

B + C = 1000

since B = C, as ABC - an isosceles triangle, then B = C = 500

Answer: 200, 800, 800 or 500, 500, and 800.

3. Find the side of an isosceles triangle if one of its sides is 10, and the perimeter 26.

1) let AB = 10, then AC = 10, since the triangle ABC - isosceles. Then

AB + AC + BC = 26

BC = 26/10/10 = 6.

2) Let BC = 10, Then

AB + AC + BC = 26

AB + AC = 16

as AB = AC, as ABC - an isosceles triangle, then AB = AC = 8.

Answer: 10, 10, and 6 or 8, 8, and 10.

3.Cherez vertices A and C of the triangle ABC draw a line perpendicular to the bisector of the angle ABC intersects the line CB and BA at the points K and M. Find AB if BM = 8, RC = 1.

Given:

ABC, ABL = LBC, AK BL, CM BL, BM = 8, KC = 1.

Find AB.

The solution:

Consider the triangle ABK:

Isosceles triangle, as in - height

And while bisector therefore AB = VC.

Consider triangle Navy:

Isosceles as VQ - height and bisector simultaneously, thus BM = Sun

1) Suppose that AB <BC, then

BK = BC = KC = 8-1 = 7

AB = BK = 7

2) Assume that AB> BC, then

BK = BC = KC = 8 +1 = 9

AB = BK = 9

Answer: 7 or 9.

4. Bisector of the angle at the base of an isosceles triangle intersects the side at an acute angle a. Find the angles of the triangle. For what values ??of a problem has two solutions?

Decision. Denote the angle you through 2x.

Depending on which of the corners - weight or AES - taken as a get different solutions.

1) If the AEC = a, x + 2x + a = 180 °, where x = 60 ° - and B = 120 ° -. But as in the <90 °, and> 45 °. Moreover, C = -60 °.

2) If the AEC = a, then this angle as external to the triangle AEC is the sum of the two interior angles are not adjacent to it. That is, a = Sx and x =. Then B = C = 180 ° -. But as in the <90 °, and <135 °, which is stated in the problem (the angle a - acute).

Answer: for a> 45 ° the problem has two solutions:

A = B = 120 ° -, C = -60 °,

or A = B = C = 180 ° -.

Homework:

1. Find the angles of an isosceles triangle if one of its exterior angles is 100°.

The solution:

Let the external angle at A = 1000, C + B = 1000, by the theorem on the outer corner of the triangle. Since the triangle ABC - an isosceles triangle, then B = C = 500 = 800 = 1800-1000

2) let the exterior angle at B = 1000, B = 800, so C = 800, since the triangle ABC - isosceles.

Since the sum of the angles in 1800, the

A + B + C = 1800

A 1600 = 1800

A = 200

Answer: 200, 800, 800 or 500, 500, and 800.

2. Find the angles of an isosceles triangle, if one of them is a.

If the angle is equal to a, then A = C =, if A = C = A, then B = 1800-2a. answer, and a or 1800-2a, 1800-2a and a.

3. The perimeter of an isosceles triangle is equal to P, one of the parties is a. Find the second side of the triangle.

The solution:

1) let AB = a, then AC = a, since the triangle ABC - isosceles. Then AB + AC + BC = p, BC = p and a = p-2a. 2) Let BC = a, then AB + AC + BC = p AB + AC = p as well as AB = AC, as ABC - an isosceles triangle, then AB-AC =.

A: P-2a, p-2a, and, or, and and.

3 class. (Tasks in the subject circle)

1. Given two circles with a common center and a radius of 3 and 7. Find the radius of the circle tangent to each of these circles.

2. In the plane, there are two circles. What is the radius of the circle tangent to the given circles and having a center on a line passing through the centers, if the radii of these circles and the distance between their centers, respectively: a) 1,3,5 b) 5,2,1 c) 3 , 4.5? How many solutions is the problem?

The solution:

Each item can be 4 possible arrangements third circle on the two data because data centers of the circles do not match.

Let B be the center of a circle with a radius of 1, N-center of a circle with a radius of 3, E is the radius of the desired circle. Then

AB = BC = 1, ND = MN = 3, BN = 5.

1) BN = BC + CN,

CN = 5-1 = 4

CM = CN + NM = 4 +3 = 7

EM = CM = 3,5.

2) BN = BC + CD + DN,

CD = BN-BC-DN = 5-1-3 = 1,

AD = AC + CD = 2 +1 = 3,

ED = AD = 1,5.

3) BM = BC + CD + DN,

CD = BM-BC-DN = 5-1-3 = 1

ED = CD = 0,5

4) AM = AB + BN + NM = 1 +5 +3 = 9

EM = AM = 4,5.

Let A-center of a circle with a radius of 2, B-center of a circle with a radius of 5, C-desired circle.

Then AN = AM = 2, BD = BK = 5, AB = 1.

1) CK =?

DN = DB-BA-AN = 5-1-2 = 2

KM = KD-DN-NM = 10-2-4 = 4,

CK = KM = 2.

2) CD =?

DN = DB-BA-AN = 5-1-2 = 2

CD = DN = 1.

3) CK =?

DN = DB-BA-AN = 5-1-2 = 2

KN = DK-DN = 10-2 = 8

CK = KN = 4.

4) CM =?

DM = DN + NM = 2 +4 = 6,

CM = DM = 3.

Let A-center of a circle with a radius of 3, B-center of a circle with a radius of 4, C-desired circle. Then

AD = AM = 3, BN = BK = 4, AB = 5.

1) CN =?

BM = AB-AM = 5-3 = 2

AN = AB-BN = 5-4 = 1

MN = AB-BM-AN = 5-2-1 = 2

CN = MN = 1.

2) CK =?

KM = KN-NM = 8-2 = 6,

CK = KM = 3.

3) CD =?

CD = DM-NM = 6-2 = 4, CD = DM = 2.

4) CK =?

DK = DM + NK-NM = 6 +8-2 = 12

CD = DK = 6.

Answer: a) 3.5 or 0.5 or 1.5 or 4.5;

B) 1 or 2 or 3 or 4;

B) 1 or 2 or 3 or 6.

3. At the vertices of the triangle centers are located three mutually tangent circles. The radius of the circles, if the triangle are 5,6,7. How many solutions is the problem?

SOLUTION:

Touch points are located on the line joining points, that is, AB, BC and AC.

1) Let AC = 5, BC = 6, AB = 7, AD = x, then OS = CP = 5 s, BM = PB = 7's, but CP + PB = 6 x 5 x +7- = 6, AD = x = 3, CP = 5 x = 2, PB = 7 x = 4.

2) Let AC = 5, BC = 7, AB = 6, BM = x, then AM = AD = x-6, SD = CP = x-7, but with AO + = 5 x 7 + x-6 = 5

BM = x = 9, AO = x 6 = 3, PB = X-7 = 2.

3) If AC = 7, BC = 6, AB = 5

BM = x, then

AM = AD = x-5, SD = CP = x-6, but

SB + AB = 7 x 7 + x 6 = 7

BM = x = 9, AO = x 5 = 4, PB = X 6 = 3.

4) Let AC = 6, BC = 5, AB = 7

BM = x, then

AM = AD = x-7, SD = CP = x-5, but

SB + AB = 6, x-7 + x 5 = 6

BM = x = 9, AO = x 7 = 2, PB = X 5 = 4.

A: 2,3,4 or 2,3,9 or 3,4,9 or 2,4,9.

Homework.

1. Given two circles with a common center and the radius R and K (k). Find the radius of the circle tangent to each of these circles. (The solution is analogous problem in the class.)

Answer: or.

2. On the line are the points A, B, C and D, at what AB = 2,

CD = 3. Segments AC and BD are the diameters of the two circles. Find the distance between the centers of the circles

The solution:

A: 2.5 or.

4 class (quads)

1. Given three points A, B, C, do not lie on a straight line. Construct a point M such that the points A, B, C, M are the vertices of a parallelogram.

parameter in this problem is the uncertainty of which of the three segments: AB, AC or BC to take a diagonal of a parallelogram.

Building:

3) AC

4) The mid-AU

5) In

6) M: M In, = BM

7) AB, AM, Sun, CM.

AVSM quadrilateral parallelogram by definition. Construction in the other two cases is similar to the first.

2. Bisector A parallelogram ABCD intersects the side BC at point C. Find the perimeter of the parallelogram, if VC = 15cm COP = 9cm.

The solution:

Isosceles triangle ABK from the equality of the angles at the base. Therefore, AB = EC = 15cm.

1) To let the point is between B and C, then

BC = VC + CC = 15 +9 = 24 (cm)

P = 2 (AB + BC) = 2 (15 +24) = 78 (cm)

2) to let the point is outside of the segment BC, then

BC = EC-CS = 9.15 = 6 (cm)

P = 2 (AB + BC) = 2 (15 +6) = 42 (cm)

A: 78cm or 42cm.

3. On the parallelogram ABCD is known that the angle ABD is 40 ° and that the centers of the circles circumscribed about triangles ABC and CDA, lie on the diagonal BD. Find the angle DBC.

The solution:

Since the centers of the circles circumscribed about triangles ABC and CDA, lie on the diagonal BD, then the points of intersection of the perpendicular are on BD, we find that BD AU, therefore, ABCD - or diamond centers of the circles lie at the intersection of the diagonals, and then ABCD - rectangle.

If the diamond, then the angle DBC = 400, as the diagonal of the diamond are the bisectors of the angles.

If the rectangle, the angle DBC = 500, because the angle ABC = 900.

A: 400 or 500.

Homework.

1. Find the perimeter of the parallelogram if the bisector of one of the angles of a parallelogram divides the side into segments 7cm and 14cm.

(Answer: 70cm or 42cm).

2. Of the parallelogram with a line intersecting the two opposite sides, cut a diamond. The remaining parallelogram in the same way again cut diamond. And again from the remaining parallelogram again cut diamond. The result is a parallelogram with sides 1 and 2. Find the source side of the parallelogram.

Let AO = 1, AM = 2.

MF = MN = DN = DP = AO = 1

AD = AM + MN + DN = 2 +1 +1 = 4 = OP = OB

AB = AO + OB = 1 +4 = 5

2) Let AO = 2, AM = 1.

MF = MN = DN = DP = AO = 2

AD = AM + MN + DN = 1 +2 +2 = 5 = OP = OB

AB = AO + OB = 2 +5 = 7

3) Let AN = 2, AD = 1.

MN = MK = KB = CB = AD = 1

AB = AN + MN + MK + KB = 2 +1 +1 +1 = 5

4) Let AN = 1, AD = 2.

MN = MK = KB = CB = AD = 2

AB = AN + MN + MK + KB = 1 +2 +2 +2 = 7

5) Let AN = 1, AM = 2.

MF = MD = DS = AN = 1

AD = AM + MD = 2 +1 = 3 = NS = NK = KB

AB = AN + NK + KB = 1 +3 +3 = 7

6) Let AN = 2, AM = 1.

MF = MD = DS = AN = 2

AD = AM + MD = 1 +2 = 3 = NS = NK = KB

AB = AN + NK + KB = 2 +3 +3 = 8

Answer: 4 and 5 or 5 and 7, or 1 and 5, or 2 and 7 or 3 and 7, or 3, and 8.

5 class (quads)

1. In an isosceles right triangle inscribed in a rectangle so that the two vertices are on the hypotenuse and the other two - the other two sides. What are the sides of the rectangle, if you know that they are both 5-2, and the hypotenuse of a triangle is 45 cm?

The solution:

Triangles AKM and DOM-isosceles, the equality of the angles at the base.

Consequently the CM and AK = RR RH.

1) Suppose, then heading 5x, and CR = 2.

AB = AC + CS + RH = MC + CS + RH = 2 Ч 5 Ч 5 Ч

12 x 45 =

x = 3.75

MK = 5x = 18.75 and CR = 2 = 7.5

2) Let, then

MK = 2, and CO = 5x.

AB = AC + CS + RH = MC + CS + RH = 2 x 2 x 5

45 = 9?

x = 5

MK = 5x = 25 and CR = 2x = 10

Answer: 18.75 and 05.07, or 25 and 10.

2. Diagonal trapezoid divides it into two isosceles triangles. Angle at the base of one of these triangles is equal to 40 °. find angles trapeze.

The solution:

Obviously, in the triangle FAC VC-base, as the angle from the stupid, in the triangle ABK all angles acute, so we can consider three cases:

AB = VC

then CF = SCR = 40 ° C = 1000,

PCA = FAC as internal cross lying,

PCA = WAC = 400, AVC = 1000.

A = 400, B = 1400, C = 1000, K = 800.

2) BK = AK, then the SLE = CF = 40 ° C = 1000,

PCA = FAC as internal cross lying,

AVC = WAC = 700.

A = 700, B = 1100, C = 1000, K = 800.

3. AB = AC, FAC = SCR = 40 ° C = 1000,

PCA = FAC as internal cross lying,

then ABK = WCA = 400, WAC = 1000.

A = 1000, B = 800, C = 1000, K = 800.

The result was a parallelogram, a trapezoid is given by the condition, then such an event can not be.

Another solution is obtained by considering the trapezoid, which had the more reason one corner of the island, and the other - stupid.

BC = VC, AB = VC. A = 400, then PCA = 400

PCA = FAC as internal cross lying, VKS = C = 700.

A = 400, B = 1400, C = 700, A = 1100.

A: 400, 1400, 1000, 800, or 700, 1100, 1000, 800, or 400, 1400, 700, 1100.

3. Bisectors of the angles A and D of the parallelogram ABCD intersect line BC at points E and F, respectively. Find the sides of a parallelogram, if its perimeter is equal to p, and we know that.

Solution: 1.rassmotrim first case where the point of intersection of the bisectors lies inside the parallelogram.

AB = BE = CF = CD, because the triangles ABE and FDC isosceles, AD = CD as part of a parallelogram, BAD = BEA Both internal Crosswise lying at BC | | AB and AC bisector.

Let EF = x

= AB + BE + FC-EF

= 3BE-x =

Hence x = *.

AB = BE =

BC = BE + CF-x =.

2. Now consider the case where the point of intersection of the bisectors is outside the parallelogram.

Here, as in the previous solution: AB = BE = CF = CD similarly.

Let

EF = x

= AB + BE + FC + EF

= 3BE + x =

Hence x =

AB = BE =

BC = BE + CF + x =.

Answer: if the point of intersection of the bisectors lies outside the parallelogram, then

AB =, BC =;

if the point of intersection of the bisectors - inside the parallelogram, then

AB =, BC =

Homework:

1. attitude angles A and B, adjacent to the side of the trapezoid ABCK well 2:3. diagonal AC divides the trapezoid into two isosceles triangles. Find the angles of a trapezoid.

SOLUTION: Then A = 2, B = 3. as the sum of the interior angles is equal to one-way in 1800, then 2 +3 x = 1800, x = 360, then A = 720, B = 1080.

BAC = BCA = 360, and BAC = 360.

1) AC = SC SCA = BAC = 360, ACK = 1080

A = 720, B = 1080, C = 1440, K = 360.

2) AC = AC = ASC ASC = 720,

A = 720, B = 1080, C = 1080, K = 720.

3) when the SC = AK get a parallelogram.

A: 720, 1080, 1440, 360, 720, 1080, 1080, 720.

2. The sides of the trapezoid are 17cm and 10cm, find a trapeze, if you know that its height is 8 cm, and the average line of 30cm.

The solution:

Consider the AVE and the NRC.

Rectangular. By the Pythagorean theorem in ABE:

AE2 = 100-64 = 36

AE = 6.

By the Pythagorean theorem to the NRC:

PH2 = 289-64 = 225,

AE = 15.

BC = EN as VSNE a parallelogram (opposite sides are parallel.)

1) MC = (BC + AR) =

(BC + AE + EN + NA) = (BC + AE + BC + NO) = (2VS 6 +15) = 30

BC = 19.5 cm

AP = 40.5 cm

2) MC = (BC + AR) = (BC + EN + HP-AE) = (BC + BC + HP-AE) = (2VS-6 +15) = 30

BC = 25.5 cm

AP = 34.5 cm

A: 25.5 cm and 34.5 cm or 19.5 cm and 40.5 cm.

6 lessons (the circle and t Pythagoras)

1. The two sides of a right triangle are equal to 3m. and 4m. Find a third party.

1) Let AC = 4 m, BC = 3m. by the Pythagorean theorem in the triangle ABC:

AB2 = 16 +9 = 25

AB = 5

2) Let AB = 4 m, BC = 3m.

by the Pythagorean theorem in the triangle ABC:

AC2 = 9.16 = 7

AC =

Answer: 5 or

2. The distance between the centers of two circles is equal to 10k. One of the circles has a radius of 5k, the second - 6k. Intersects the smaller circle at points A and B, and for the most at C. Find the length of the chord AB, knowing that AB = 2 * Sun

The solution:

O1O = 10k, 5k = O1M, o2c = 6k

Therefore, CM = 1k and 4k = O1S by the Pythagorean theorem in triangle O1AS: AC2 = 25k-16k = 9k, AC = 3, AB = 6k.

2. Let AB = 2x by the Pythagorean theorem in triangle O2EV

O2E2 = 25k2-x2

O2E =

by the Pythagorean theorem in triangle O1MO2

O1M2 100k2 =-4x2

O1M = 2

O1M = 2O2E

CO1 = 6k = CM + MO1 = EO2 + MO1

= 6k + EO2 MO1 = 3O2E

O2E 2k

x = k

AB = 2 K.

Answer: 2 to or 6k.

3. The perimeter of the trapezoid is equal to 112. The point of tangency of the circle inscribed in a trapezoid divides one of the sides into segments of length 8 and 18. Find a trapeze.

The solution: NC = CE = 8, the equation for the hypotenuse and a leg triangle NOC and EOC. ED = MD = 18, the equation for the hypotenuse and a leg triangle DOE and DOM. KD = MD-MC = 8.18 = 10

By the Pythagorean theorem for triangles SKD:

SC = r = 12

AB = BN + AM (as well as CD = NC + MD)

BN + AM + AB = 60 (since 112-8-8-18-18 = 60),

Then AB = 30.

By the Pythagorean theorem for triangles AVL:

AL =

1.P = AB + BN + NC + CD + DM + ML + AL =

= 30 + BN +8 +26 +18 + ML +18 = 112

ML = BN = 6, BC = 8 +6 = 14, AD = 18 +6 +18 = 42.

2 P = AB + BN + NC + CD + DM + ML-AL = 30 + BN +8 +26 +18 + ML-18 = 112

ML = BN = 24, BC = 24 +8 = 32, AD = 24-18 +18 = 24.

Answer: 14 and 42 or 24 and 32.

Homework:

1. The two sides of the triangle are 25cm and 30cm. Find a third party, if the altitude drawn to it is 24cm.

SOLUTION:

AB = 30cm, BC = 25 cm, BH = 24cm.

Triangles APD and VSN - rectangular.

By the Pythagorean theorem to the AVN:

AH2 = 900-576 = 324

AH = 18 (cm)

By the Pythagorean theorem in VSN:

CH2 = 625-576 = 49

CH = 7 (cm).

Since it is not mentioned, or obtuse-angled triangle, then we can consider two cases:

1) an acute:

AC = AH + CH = 18 +7 = 25 (cm).

2) obtuse:

AC = AN-RR = 7.18 = 11 (cm).

A: 25cm or 11cm.

7 Employment (circle etc Pythagoras)

1. Length of adjacent sides of a quadrilateral inscribed in a circle differ by 1. Length of the smallest of them as equals I. Find the radius of the circle. solution:

1) BC = 1, then AB = BC = 2, AD = 1

AC =

OS =

2) BC = 1, then AB = BC = 2, AD = 3, with the radius

BT = Let OM = a

By the Pythagorean theorem of trans-ka AOM

K =

Of trans-ka OCR:

K =

2.25 = 0.25 +3-2 and

a = k =

answer:

2. Given segment of length 20. Three circles with radii of 4 are centered at the ends or in the middle. Find the radius of the fourth circle tangent to three given.

1. Solution: k-desired radius.

OO1O2-isosceles triangle, with sides equal to (a-4), then the height of OA is also the median.

By the Pythagorean theorem:

From AOO2

OA2 = (K-4) 2-25

From AOO3

OA2 = (K +4) 2-225

-8K-8K-25 = 225, 16k = 200, K = 12.5

2. Let K be the desired radius, oo2 = a, then a = 4 + and, by the Pythagorean theorem for triangles OO2O3

a2 = ((a +4) +4) 2-100

16a = 36,

a = 2.25,

k = 6.25.

Answer: 6.25 or 12.5.

Homework:

1. Find the height of an isosceles triangle with base a and the radius of the circle R.

Decision.

Since the vertex lying opposite the base, may lie in one of the two arcs of the circumscribed circle (ie in different half-planes in the line containing the base of the triangle), then the task will be to have two different solutions:

1) If the angle opposite to the base, acute (B), the distance from the center of the circle to the bottom of OM = N-R, where H - height VM held to the ground. Then by the Pythagorean theorem for triangles AOM: AO2 = R2 = () 2 + (HR) 2, from which we obtain a quadratic equation in H: H2-2HR + = 0. The roots of this equation

H1, 2 = R ±.

If the angle opposite to the base, obtuse (R), the distance from the center of the circle is equal to the base of the OM = R-H, and consequently, R2 = () 2 + (RH) 2 that will lead to the same quadratic equation. Thus, the quadratic equation itself provided two different solutions of this problem.

Reply: R ±.

8.1 Dido problem. (Example rectangles). (For students in the textbook Pogorelov AV)

In the last lesson, let's see what all the same land acquired Dido. The legend was this:

In the IX century. BC Phoenician princess Dido, fleeing the persecution of his brother, went to the west along the coast of the Mediterranean Sea to look for a refuge, it has attracted a place on the coast of the Gulf of Tunis. Dido led negotiations with the local leader Yarbom the sale of land. She asked a very small area - "as much as you can surround bovine skin." Dido persuaded Yarba, and the transaction took place. Dido then cut up by the bull's skin into small ribbons, tied them together and surrounded by a large area on which the fortress and the city of Carthage.

The task of finding among all closed curves with a given perimeter that which covers the maximum area, called the problem of Dido.

Dido problem is formulated in this way: "at any of F, for a given perimeter, the area will be the greatest?"

Suppose that Dido was a rope length p m

She was offered a choice of several plots of rectangular shape.

This all sections of rectangular shape with a perimeter p. Which of them will have the largest area?

To start, let's get the length of the rope is 100 m, then

If one of the parties - x, then the other-50.

Figuring the square, we get

X (50-x) = 50x + x2 = 625 - (x2-50x 625) = 625 - (25-x) 2

The difference would be greatest if the (25-x) 2 = 0, ie, x = 25, ie, if the quadrilateral square.

Now consider the general case where the perimeter p.

If one of the parties - x, then the other-s.

Figuring the square, we get

X (x) = x * 2 = () 2 - (x2 + x + () 2) = () 2 - (-x) 2

The difference would be greatest if (-x) 2 = 0, ie, x =, ie if a quadrilateral - square.

Thus it turns out that of all the rectangles with the same perimeter of the square has the largest area.

8.2 Dido problem. (Example parallelograms). (For students in the textbook Atanasyan LS, etc.)

In the last lesson, let's see what all the same land acquired Dido. The legend was this:

In the IX century. BC Phoenician princess Dido, fleeing the persecution of his brother, went to the west along the coast of the Mediterranean Sea to look for a refuge, it has attracted a place on the coast of the Gulf of Tunis. Dido led negotiations with the local leader Yarbom the sale of land. She asked a very small area - "as much as you can surround bovine skin." Dido persuaded Yarba, and the transaction took place. Dido then cut up by the bull's skin into small ribbons, tied them together and surrounded by a large area on which the fortress and the city of Carthage.

The task of finding among all closed curves with a given perimeter that which covers the maximum area, called the problem of Dido.

Dido problem is formulated in this way: "at any of F, for a given perimeter, the area will be the greatest?"

Let us consider this problem on the example of the parallelograms.

Consider the different types of parallelograms with equal side lengths.

Since the area of a parallelogram is equal to a * b * sin a ^ b, then the largest area is obtained by sin a ^ b = 0, that is, the angle of the line. That is the largest area of a rectangle.

Consider the different types of rectangles:

This all sections of rectangular shape with a perimeter p. Which of them will have the largest area?

To start, let's get the length of the rope is 100 m, then

If one of the parties - x, then the other-50.

Figuring the square, we get

X (50-x) = 50x + x2 = 625 - (x2-50x 625) = 625 - (25-x) 2

The difference would be greatest if the (25-x) 2 = 0, ie, x = 25, ie, if the quadrilateral square.

Now consider the general case where the perimeter p.

If one of the parties - x, then the other-s.

Figuring the square, we get

X (x) = x * 2 = () 2 - (x2 + x + () 2) = () 2 - (-x) 2

The difference would be greatest if (-x) 2 = 0, ie, x =, ie if a quadrilateral - square.

Thus it turns out that of all parallelograms with the same perimeter of the square has the largest area.



Conclusion

* Begin to use the task with the geometrical parameters can be from the very early period of the study of geometry.

* The use of such tasks does not allow students to "zakostenet" in their abilities and skills of geometric knowledge.

* Problems with the geometrical parameters are creative and can not be included in the mandatory minimum, they should be attributed to the problems of the "advanced" level.

Most often the student to really think about the lesson just once. Lessons are on the scheme: "warm up" students, checking homework, repetition of the previous lesson, an explanation of the new material, the initial consolidation, the application of acquired knowledge to solve problems involving the previously learned material. The limited time frame lesson teacher (you have time to do everything planned) and the time of the study subjects (it must be remembered that the delay in this lesson will cause further delay), the focus of teacher and student to achieve short-term goals (successfully write their own or control work, pass a test) - All this does not contribute to the problems in class creative or technically challenging nature. However, these are the problems give the student to a deeper understanding of the material, see the "highlight" in solving geometric problems.

In optional classes teacher can not stick to subjects under the sections and to be creative, making its program of extracurricular activities.

Therefore, to date research related to the development of the content and methodology of the electives, are relevant.

The presented thesis provides a practical course to solve problems with the geometrical parameters. This elective course is designed for students of secondary schools in which the geometry is taught from textbooks and AV Pogorelov Atanasyan LS

The second part of the thesis contains all the necessary theoretical material, plan elective "parameters in the geometry" and summaries of specific studies. Elective course is designed for eight sessions of 2 hours. Tasks selected topics (learning parameters in the geometry simple tasks, solving construction problems in the subject triangles, solving problems on the circle, the solution of problems in the subject line quadrilaterals (2 classes), solution of problems in the theme of the circle and r Pythagoras (2 classes ) solution Dido), and related to the geometry of the binding rate of eighth grade, that is, the teacher does not have to tell the students in advance elective information from the geometry of the eighth grade. Tasks in each subject range from simple to complex. At the final session of the course to solve the problem of Dido, which was mentioned in the first lesson.

The study of elective course "parameters in the geometry" promotes spatial, logical and creative thinking and mathematical skills.

This course is also designed to raise sustained interest in geometry, as it solved problems differ from problems solved in the course of studying the geometry of the textbooks and AV Pogorelov Atanasyan LS

During such tasks more need not only be specific knowledge of the geometry, but also the ability to build the correct drawings to see how different pieces can be placed relative to each other, and do not stop at one embodiment, as there may be others.

Thus, the solution of problems with the geometric parameters of the problem confronts students consider various consequences when considering the various options, which is the actual problem in our daily life.



Bibliography

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