Anonymous
Anonymous asked in Science & MathematicsMathematics · 12 months ago

Circle Geometry Math Problem help????

Alex plans to walk around a circular lake that has diameter 8 km. Points E and H are on opposite sides of the lake and lie on a straight line through the center of the lake. Each point is 7 km from the center. In the route EFGH, EF and GH are tangents to the lake and FG is an arc along the shore of the lake. If Alex starts at point E, walks to point F, walks to point G, and ends at point H, what is the total distance he will walk?

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  • 12 months ago
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    You can imagine the coordinates of the points. The point E is the origin. → E (0 ; 0) C (7 ; 0) H (14 ; 0)

    The typical equation of a circle is: (x - xo)² + (y - yo)² = R² → where:

    xo: abscissa of center → (7) in your case

    yo: ordinate of center → (0) in your case

    R: radius of circle → 4 in your case

    The equation of your circle is: (x - 7)² + y² = 16

    The typical equation of a line is: y = mx + y₀ → where m: slope and where y₀: y-intercept

    The equation of the line (EF) is: y = mx + y₀

    As the line (EF) passes through the origin, then: y₀ = 0

    The equation of the line (EF) become: y = mx

    Restart from the circle (to find the intersection with the line)

    y² = - x² + 14x - 33 → we've just seen that: y = mx

    - x² + 14x - 33 = m²x²

    x² + m²x² - 14x + 33 = 0

    x².(m² + 1) - 14x + 33 = 0

    Δ = (- 14)² - 4.[(m² + 1) * 33]

    Δ = 196 - 132m² - 132

    Δ = - 132m² + 64

    The line (EF) is tangent to the circle, so there is only one common point between the circle and the line, so there is only one solution, so: Δ = 0

    - 132m² + 64 = 0

    m² = 16/33

    m = ± 4/(√33)

    In your case, the line (EF) has a negative slope, you keep only the negative solution: m = - 4/(√33)

    The equation of the line (EF) become: y = - [4/(√33)].x

    Recall:

    x₁ = (- b - √Δ) / 2a

    x₂ = (- b + √Δ) / 2a

    As Δ = 0 → x₁ = x₂ = x₀ = - b/2a

    x₀ = 14/[2.(m² + 1))

    x₀ = 7/(m² + 1) → recall: m² = 16/33

    x₀ = 7/[(16/33) + (33/33)]

    x₀ = 33/7 ← this is the abcissa of the point F

    Recall: y = - [4/(√33)].x

    y = - [4/(√33)] * (33/7)

    y = - (4/7).√33 ← this is the ordinate of the point F

    → Point F [33/7 ; - (4/7).√33]

    Then, let's calculate the distance [EF]

    EF² = xF² + yF²

    EF² = (33/7)² + [- (4/7).√33]²

    EF² = 1617/49

    EF = (√1617)/7

    Whatever the triangle, the sum of the 3 angles is always 180 °. In the triangle AEF, you can write:

    90 + α + β = 180

    α + β = 90

    β = 90 - α

    …and you can see that:

    β + γ + β = 180

    2β + γ = 180 → recall: β = 90 - α

    2.(90 - α) + γ = 180

    180 - 2α + γ = 180

    γ = 2α

    tan(α) = yF/xF

    tan(α) = [- (4/7).√33] / (33/7)

    tan(α) = (- 4√33) / 33

    α ≈ 34.85 ° → recall: γ = 2α

    γ = 69.7°

    Then, let's calculate the distance [FG]

    FG = (π.d) * (γ/360) → where γ in degrees and where d is the diameter of the circle

    FG ≈ 8 * π * (69.7/360)

    FG ≈ 4.866 km

    EH = EF + FG + GH → but you can see that: GH = EF

    EH = 2.EF + FG

    EH = (2/7).√1617 + 4.866

    EH ≈ 16.355 km

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  • EF = GH = sqrt(7^2 - 4^2) = sqrt(49 - 16) = sqrt(33)

    arcFG = 2 * pi * 4 * (2t / (2pi)) = 16 * pi * t / (2 * pi) = 8t

    sin(t) = 4/7

    sin(2t) = 2sin(t)cos(t) = 2 * (4/7) * sqrt(1 - (4/7)^2) = (8/7) * (1/7) * sqrt(49 - 16) = 8 * sqrt(33) / 49

    2t = arcsin(8 * sqrt(33) / 49)

    2 * sqrt(33) + 8 * arcsin(8 * sqrt(33) / 49) =>

    21.22105455563941120897910329879

    • Pope
      Lv 7
      12 months agoReport

      Re: 2 * sqrt(33) + 8 * arcsin(8 * sqrt(33) / 49)

      The second term is 8(2t) = 16t.
      It should be 8t, as you showed above.
      Am I wrong?

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  • 12 months ago

    EF = 7 - 8/2 = 3

    GH = 7 - 8/2 = 3

    FG = πr = 4π

    Total distance: 6 +4π

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  • 12 months ago

    This should help:

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