Stay in the tricycle for the basis of the bichniy side. Zavdannya about trikutnik
VIII ... Groupy zavdan for an invitation.
Rishennya group zavdan s vicaristannyam pre-domіzhny trikutnik.
The essence of the method is to induce ancillary triumphs and vicarious powers and to reimagine elements for the residual revision of the system.
Analyze and motivate to stock up on the onset of stages:
Shukay during the analysis of an additional trikutnik.
As soon as there are new elements, for the help of which it is possible to stay with the ABC tricycle, then the meta is reached.
If it’s not possible, then, perhaps, there can be one additional tricycle, which will give all the elements.
Razberemo essence of the method on the butts.
Zavdannya 1. Build a tricycle ABC ( b= c) on a, h b .
Shukaєmo dopomіzhny trikutnik. Obviously, with such a tricycle, you can manually use a tricycle CDB.
Tse give kut S, і kut ABC. Otzhe, є a, kut B, kut C, it means you can have a trikutnik ABC. Schematically, we will write it like this:
(A, h b) → Δ CDB →< C.
(A,< B, < C) → Δ ABC.
zavdannya for independent solution:
Vikoristovuchi mіrkuvannya, similar to pointing, it is recommended to visit the trikutnik (b = c) for the offensive data:
a)< А, h b ;
b)< В, h с;
G)< В, h b ;
e)< С, h b .
Head 2. Build a tricycle according to the radius r of the inscribed stake, rose A and rose B.
Nekhai I is the center of the stake inscribed in the ABC tricycle.
(R; ½< А) → Δ AID → |AD|;
(R; ½< В) → Δ ВID → |ВD|;
(| AD | + | BD | = | AB |) → (c,< А, < В) → Δ ABC.
Head for an independent solution:
Stay in the tricycle for such items:
a) a, h c, h b; b) a, h a, h b; c) a, m a, m b;
G)< A, l A , b; д) R, h а, m a ; е) a, R, h b ;
g) b, h b, m b (de m - media, l - bisectrix, h - visoti).
independently:
Rishennya group is based on the main basis.
Mainly zavdannya:
draw the ABCD diamond along the BD and BM diagonal. (Δ BHD →< BDH → равнобедренный Δ BDA → ABCD);
awake a trapezium, from chotiroh sides.
We will stay the tricytnik behind the two sides and the kutu, we will lay them down.
Mainly zavdannya:
Have a straight-leg tricycle for two legs.
Have a rhombus on two diagonals.
Pobuduvati a rectangle on two ner_vnyh sides.
Have a parallelogram on two diagonals and cuts between them.
Stay upright on the diagonals and cut them along.
Sit down the tricycle behind the side and with the two, lay down the kutah.
Head for an independent solution:
Mainly zavdannya:
Stay tuned for the base і close kutu.
Stay in a straight-cut tricycle along the leg and close of the gostry kuta.
Stay in the rhombus behind the kut і diagonal, before passing through the top of the kut kut.
Hang the tricycle behind the top and bottom at the top.
Stay square according to the diagonal.
Stay upright trikutnik on hypotenuse and gostrom kut.
Head for an independent solution:
Mainly zavdannya:
Have a tricycle on the normal side and rose at the presentation.
Stay in the tricycle along the bichniy sides and roses at the top.
Pobuduvati tricycle for three sides.
Head for an independent solution:
Mainly zavdannya:
Stay in the tricycle for the base and the bichniy side.
Hold the diamond on the sides and diagonal.
Stay parallelogram along the two uneven sides and diagonal.
Have a parallelogram on the sides and two diagonals.
Stay upright tricycle on the leg and hypotenuse.
Head for an independent solution:
Stay tricytnyk at the top and bottom side.
Hang the tricycle behind the base and perpendicular, lowered from the end to the side.
Wake up the parallelogram according to the presentation, visibility and diagonal.
Wake up the rhombus along the height and diagonal.
Stay in the tricycle along the bichny side and the height, lowered from it.
Stay with the tricycle behind the base, hanging and bichniy sides.
literature:
B. I. Argunov, M.B. Balk "Geometric induce to the square", M, "Enlightenment" 1955r.
Glazer G. I. "History of mathematics in schools" IV - VI grades, M, "Education", 1981
І. Goldenblant "Dosvіd virіshennya geometrical problems on pobudovu" "Mathematics in schools" № 3, 1946 р
І. A. Kushnir "About one way of making the headquarters for the occasion" "Mathematics in schools" No. 2, 1984 r
A. I. Mostoviy "Zastosovuvati development ways to turn the building on to motivate" "Mathematics in schools" No. 5, 1983
A. A. Popov "Mathematics" "Chelyabinsk State Pedagogical University", 2005
E. M. Seleznyova, M. N. Serebryakova "Geometric Incentives in I - V Classes middle school"Methodical outlets. Sverdlovsk, 1974.
rіvnobedrenimє such trikotnik There are two sides to each other.
When you see the buildings according to the topic "Rivnobedreniy trikutnik" it is necessary to register with the next visit authorities:
1.
Kuti, do you lie right now? pivni sides rіvnі mіzh myself.
2.
Bisektrisi, medіany and visoti, carrying out rivnykh kutiv, Rivnі mіzh myself.
3.
The bisector, median and height, carried out to the base of the femoral tricycle, can be taken off by itself.
4.
The center is inscribed and the center of the described key lies at the height, and it means that it is on the media and bisectors, carried out to the base.
5.
Kuti, yaki є rіvnimi in a trikutnik zavzhdi gostrі.
Trikutnik є signs:
1.
Two kuta at the tricycle Rivn.
2.
Visota zbіgaєtsya s medіanoyu.
3.
The bisector is connected with the median.
4.
Visota zbіgaєtsya s bisectrix.
5.
Two legs of a tricycle rivn.
6.
Two bisectors of a tricycle Rivn.
7.
Two median tricycle rivn.
A number of tasks by topic "Rivnobedreniy trikutnik"і guided lecture їх solution.
Zavdannya 1.
At the tricycle, the height is carried out to the base, the road is 8, and the falling asleep is carried to the bichoi side yak 6: 5. To know that at the top of the tricut there is a point where the bisector is overflowing.
Decision.
Nekhai dano trikutnik ABC (Fig. 1).
1) So as AC: BC = 6: 5, then AC = 6x and BC = 5x. VN - visota, carried out to the base of the AC tricycle ABC.
So if point H is the middle of the AC (according to the power of the tricot tricot), then HC = 1/2 AC = 1/2 6x = 3x.
BC 2 = BH 2 + HC 2;
(5x) 2 = 8 2 + (3x) 2;
x = 2, todі
AC = 6x = 6 2 = 12 i
BC = 5x = 5 2 = 10.
3) So as the point of intersection of the bisectors of the tricutnik є by the center of the inscribed in the new stake, then
ВІН = r. The radius of the circumference inscribed in the trikutnik ABC is known by the formula
4) S ABC = 1/2 * (AC * BH); S ABC = 1/2 * (12 * 8) = 48;
p = 1/2 (AB + BC + AC); p = 1/2 · (10 + 10 + 12) = 16, todі ВІН = r = 48/16 = 3.
Zvidsi VO = VN - OH; VO = 8 - 3 = 5.
View: 5.
Zavdannya 2.
The bisector AD was carried out in the tricycle ABC. Areas of tricycles ABD and ADC rivni 10 and 12. Know the three times the area of the square, prompted on the hanging tricycle, carried out to the base of the AU.
Decision.
Riglenemo tricycle ABC - equestrian, AD - bisector kuta A (Fig. 2).
1) Recorded area of tricycles BAD and DAC:
S BAD = 1/2 AB AD sin α; S DAC = 1/2 AC AD sin α.
2) We know the area:
S BAD / S DAC = (1/2 AB AD sin α) / (1/2 AC AD sin α) = AB / AC.
So yak S BAD = 10, S DAC = 12, then 10/12 = AB / AC;
AB / AC = 5/6, tody nehai AB = 5x and AC = 6x.
AH = 1/2 AC = 1/2 6x = 3x.
3) Z tricycle AVN - rectangular according to Pyfagor's theorem AB 2 = AN 2 + VN 2;
25x 2 = VN 2 + 9x 2;
4) S A ВС = 1/2 AS ВН; S A B C = 1/2 6x 4x = 12x 2.
So yak S A BC = S BAD + S DAC = 10 + 12 = 22, todi 22 = 12x 2;
x 2 = 11/6; VN 2 = 16x 2 = 16 11/6 = 1/3 8 11 = 88/3.
5) The area of the square door VN 2 = 88/3; 3 88/3 = 88.
View: 88.
Zavdannya 3.
For a tricycle, the base is road 4, and the side of the road is 8. Know the square of the hanging, lowered to the side.
Decision.
For tricycle ABC - isosceles BC = 8, AC = 4 (Fig. 3).
1) VN - visota, carried out to the base of the AS of the tricycle ABC.
So, if point H is the middle of the AC (according to the power of the tricot tricot), then HC = 1/2 AC = 1/2 4 = 2.
2) Z tricytnyk VNS - rectangular according to the Pyphagorean theorem VS 2 = VN 2 + NS 2;
64 = BH 2 + 4;
3) S ABC = 1/2 (AC
1/2 AC BH = 1/2 AM BC;
AM = (AC · BH) / BC;
AM = (√60 4) / 8 = (2√15 4) / 8 = √15.
Answer: 15.
Zavdannya 4.
In the tricycle, it is lowered to a new height, equal to 16. Know the radius of the circumference of the circumference of the circumference described near the tricycle.
Decision.
For tricycle ABC - isosceles AC = 16, VN = 16 - visota, carried out to the base of the AC (Fig. 4).
1) АН = НС = 8 (according to the power of the femoral tricycle).
2) Z tricytnyk VNS - rectangular according to the theorem of Pyphagoras
BC 2 = BH 2 + HC 2;
BC 2 = 8 2 + 16 2 = (8 2) 2 + 8 2 = 8 2 4 + 8 2 = 8 2 5;
3) The ABC tricycle is clearly visible: according to the sinus theorem 2R = AB / sin C, de R is the radius of the circumference of the circumference of the ABC tricycle described near the tricycle ABC.
sin C = BH / BC
sin C = 16 / (8√5) = 2 / √5, Todi 2R = 8√5 / (2 / √5);
2R = (8√5 √5) / 2; R = 10.
Answer: 10.
Zavdannya 5.
Dovzhina visoti, carried out to the base of the ryvno-femoral tricycle, is 36, and the radius of the inscribed stake is 10. To know the area of the tricycle.
Decision.
Nekhai given a trikutnik ABC.
1) So yak the center is inscribed in the tricycle of the stake є by the point of the intersection of the bisectors, then Pro ϵ VN і АТ є bisectrix kuta A, and strum w ВІН = r = 10 (Fig. 5).
2) VO = VN - OH; BO = 36 - 10 = 26.
3) AVN tricycle is clearly visible. According to the theorem about bisektrisi kuta trikutnik
AB / AN = VO / OH;
AB / AH = 26/10 = 13/5, todі nehai AB = 13x і AN = 5x.
According to the theorem of Pyphagoras AB 2 = AN 2 + BH 2;
(13x) 2 = 36 2 + (5x) 2;
169x 2 = 25x 2 + 36 2;
144x 2 = (12 3) 2;
144x 2 = 144 9;
x = 3, todi AC = 2 AH = 10x = 10 3 = 30.
4) S ABC = 1/2 * (AC * BH); S ABC = 1/2 * (36 * 30) = 540;
View: 540.
Zavdannya 6.
A tricycle has two sides of 5 and 20. Know the bisectrix of a kut with the basis of a tricot.
Decision.
1) It is permissible that the bichni sides of the trikutnik are equal to 5, and from the subdivision - 20.
Todi 5 + 5< 20, т.е. такого треугольника не существует. Значит, АВ = ВС = 20, АС = 5 (Fig. 6).
2) Don't go LC = x, toodi BL = 20 - x. According to the theorem about bisektrisi kuta trikutnik
AB / AC = BL / LC;
20/5 = (20 - x) / x,
todі 4x = 20 - x;
Thus, LC = 4; BL = 20 - 4 = 16.
3) Skoristaєmosya by the bisectrix formula of a tricytnik kuta:
AL 2 = AB AC - BL LC,
todi AL 2 = 20 5 - 4 16 = 36;
View: 6.
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A yak will be a tricot? It’s easy to grow for additional lines, olivtsya and zoshita clits.
Pobudova tricycle tricycle will be repaired from present. Schab little ones viyshov are rivnim, the number of girls in the house is a double number.
Ділино відрізок - tricycle department - navpil.
The top of the tricycle can be vibrated on either side of the base, or evenly over the middle.
Yak to spend the night
Kuti, with the basis of a tricot tricot, can be just gostrymy. Schob trikutnik viyshov is a gostrokutny, kut at the summits is also guilty of buti gostry.
For the top of the trikutnik, he is vibrant, given from before.
Chim vishche the summit, tim less kut at the summits. Kuti at the base with a whole, apparently, grow up.
The yak will be the obtuse-angled trikutnik?
Near the top of the tricot tricycle to the base of the degree world of kut at the top, it grows.
This means that the slopes are obtuse-angled tricytes, the top is vibramoo lower.
Yak pobuduvati ravnobedreniy rectangular tricycle?
The schob will induce a rectangular tricycle, demanding the top of the vibrate on the back, half of the rectangular tricycle).
For example, if there are 6 cells before the dinner, then the top of the tricycle is at the height of 3 cells above the middle of the street. Beast to respect: at the same time, the skin at the huts, when presented, goes along the diagonal.
Pobudova rectangular tricycle can be seen from the top.
Vibrating summit, going straight uphill and going uphill and to the right. Tse - bichnі sides of the trikutnik.
Z'єdnaєmo їх і
Pobudova tricycle tricycle for an additional compass and a line without podiliv in the first place.