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? Relevance

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

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 • Login to reply the answers
• 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

• 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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• EF = 7 - 8/2 = 3

GH = 7 - 8/2 = 3

FG = πr = 4π

Total distance: 6 +4π

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• This should help: • Login to reply the answers