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Vilka [71]
3 years ago
15

Elaborati un desen : punctul m este mijlocul bazei triunghiuluibisoscel abc

Mathematics
1 answer:
blondinia [14]3 years ago
8 0
Uh .. Nu vorbesc limba română, dar am încercat meu cel mai bun pentru a face această întrebare.

Teorema: Fie ABC un triunghi isoscel cu AB = AC. Fie M denota punctul central al BC (adică, M este punctul pe BC pentru care MB = MC). Apoi
a) Triangle ABM este congruent cu ACM triunghi.
b) Unghi ABC = ACB Angle (unghiuri de bază sunt egale)
c) Unghi AMB = Unghi AMC = unghi drept.
d) Unghi BAM = unghiul CAM


Corolar: Prin urmare, din aceste fapte și definițiile:
Ray AM este unghiul bisectoarea unghiului BAC.Linia AM este altitudinea triunghiului ABC, prin A.Linia AM este mediatoare BSegmentul AM este mediana triunghiului ABC, prin A.Dovada # 1 din Teorema (după B & B)
Lasa bisectoarea unghiului de BAC se intersectează segmentul BC la punctul D.
Din moment ce raza AD este unghiul bisectoare, unghiul BAD = unghiul CAD.Segmentul AD = AD = sine.De asemenea, AB = AC deoarece triunghiul este isoscel.Astfel, triunghiul BAD este congruent cu CAD de SAS (side-unghi-side).
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1. The distance between the perihelion and the aphelion is 116 million miles

2. The distance from the center of Mercury’s elliptical orbit and the Sun is 12 million miles

3. The equation of the elliptical orbit of Mercury is \frac{x^{2}}{3364}}+\frac{y^{2}}{3220}=1

4. The eccentricity of the ellipse is 0.207 to the nearest thousandth

5. The value of the eccentricity tell you that the shape of the ellipse is near to the shape of the circle

Step-by-step explanation:

Let us revise the equation of the ellipse is

\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 , where the major axis is parallel to the x-axis

  • The length of the major axis is 2a
  • The coordinates of the vertices are (± a , 0)
  • The coordinates of the foci are (± c , 0) , where c² = a² - b²

∵ The Sun is located at a focus of the ellipse

∴ The sun located ate c

∵ The perihelion is the point in a planet’s orbit that is closest to the

   Sun ( it is the endpoint of the major axis that is closest to the Sun )

∴ The perihelion is located at the vertex (a , 0)

∵ The closest Mercury comes to the Sun is about 46 million miles

∴ The distance between a and c is 46 million miles

∵ The aphelion is the point in the planet’s orbit that is furthest from

   the Sun ( it is the endpoint of the major axis that is furthest from

   the Sun )

∴ The aphelion is located at the vertex (-a , 0)

∵ The farthest Mercury travels from the Sun is about 70 million miles

∴ The distance from -a to c is 70 million miles

∴ The distance between the perihelion and the aphelion =

   70 + 46 = 116 million miles

1. The distance between the perihelion and the aphelion is 116 million miles

∵ The distance between the perihelion and the aphelion is the

  length of the major axis of the ellipse

∵ The length of the major axis is 2 a

∴ 2 a = 116

- Divide both sides by 2

∴ a = 58

∴ The distance from the center of Mercury’s elliptical orbit to the

   closest end point to the sun is 58 million miles

∵ The distance between the sun and the closest endpoint is

   46 million miles

∴ The distance from the center of Mercury’s elliptical orbit and

   the Sun = 58 - 46 = 12 million miles

2. The distance from the center of Mercury’s elliptical orbit and the Sun is 12 million miles

∵ The major axis runs horizontally

∴ The equation is \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1

∵ a = 58

∵ c is the distance from the center to the focus of the ellipse

∴ c = 12

∵ c² = a² - b²

∴ (12)² = (58)² - b²

- Add b² to both sides

∴ (12)² + b² = (58)²

- Subtract (12)² from both sides

∴ b² = (58)² - (12)² = 3220

- Substitute these values in the equation

∴ \frac{x^{2}}{3364}}+\frac{y^{2}}{3220}=1

3. The equation of the elliptical orbit of Mercury is \frac{x^{2}}{3364}}+\frac{y^{2}}{3220}=1

The eccentricity (e) of an ellipse is the ratio of the distance from the

center to the foci (c) and the distance from the center to the

vertices (a) ⇒ e=\frac{c}{a}

∵ c = 12

∵ a = 58

∴ e=\frac{12}{58} = 0.207

4. The eccentricity of the ellipse is 0.207 to the nearest thousandth

If the eccentricity is zero, it is not squashed at all and so remains a circle.

If it is 1, it is completely squashed and looks like a line

∵ The eccentricity of the ellipse is 0.207

∵ This number is closed to zero than 1

∴ The shape of the ellipse is near to the shape of the circle

5. The value of the eccentricity tell you that the shape of the ellipse is near to the shape of the circle

Learn more:

You can learn more about conics section in brainly.com/question/4054269

#LearnwithBrainly

5 0
3 years ago
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