✒️CIRCLE
[tex]••••••••••••••••••••••••••••••••••••••••••••••••••[/tex]
[tex] \large\underline{\mathbb{ANSWER}:} [/tex]
[tex] \qquad \LARGE\:\:\rm{20\sqrt{21} \: meters} [/tex]
[tex]••••••••••••••••••••••••••••••••••••••••••••••••••[/tex]
[tex] \large\underline{\mathbb{SOLUTION}:} [/tex]
» Label the given points as capital letters.
- A = Main Road
- B = Gate 2
- D = Exit
- E = Gate 1
» Indicate The center of the park as C. Since BD is the diameter of the park measuring 50m, then the radius is 25m.
- [tex] CB = 25 [/tex]
- [tex] CD = 25 [/tex]
» Draw a radius from C to gate 1 (E) that is also 25m in measure.
» Since CE is perpendicular to AE, then ∆CEA is a right triangle. Solve for AE using the Pythagorean theorem.
- [tex] (A E)^2 + (C E)^2 = (C A)^2 [/tex]
- [tex] (A E)^2 + (C E)^2 = (C B + B A)^2 [/tex]
- [tex] (A E)^2 + (25)^2 = (25 + 70)^2 [/tex]
- [tex] (A E)^2 + 625 = (95)^2 [/tex]
- [tex] (A E)^2 = 9025 - 625 [/tex]
- [tex] (A E)^2 = 8400 [/tex]
- [tex] \sqrt{(A E)^2} = \sqrt{8400} [/tex]
- [tex] A E = 20\sqrt{21} [/tex]
[tex] \therefore [/tex] The distance between the Main Road tó Gate 1 is 20√21 meters
[tex]••••••••••••••••••••••••••••••••••••••••••••••••••[/tex]
(ノ^_^)ノ
