Obra drove 160Ï€ meters along a circular track. If the area enclosed by the circular track on which she drove is 57,600Ï€ square meters, what percentage of the circular track did Obra drive? A. 6.67% B. 12.5% C. 18.75% D. 25% E. 33.3% The OA is the option E. What is the equation I should set to get an answer here? Experts, I ask for your help. Thanks in advanced.

Vincen wrote:Obra drove 160Ï€ meters along a circular track. If the area enclosed by the circular track on which she drove is 57,600Ï€ square meters, what percentage of the circular track did Obra drive? A. 6.67% B. 12.5% C. 18.75% D. 25% E. 33.3% The OA is the option E. What is the equation I should set to get an answer here? Experts, I ask for your help. Thanks in advanced. We're given: r^2 * Ï€ = 57,600 * Ï€ r^2 = 57,600 r = 240. (Hopefully, you recognize that 24^2 = 576.) If r = 240, then the circumference = 2Ï€r = 2*240*Ï€ = 480Ï€. She's gone 160Ï€, or 160Ï€/480Ï€ = 1/3 = 33.33...%. The answer is E

Vincen wrote:Obra drove 160π meters along a circular track. If the area enclosed by the circular track on which she drove is 57,600π square meters, what percentage of the circular track did Obra drive? A. 6.67% B. 12.5% C. 18.75% D. 25% E. 33.3% The OA is the option E. What is the equation I should set to get an answer here? Experts, I ask for your help. Thanks in advanced. Since the area enclosed by the circular track is 57,600π square meters, the radius of the track, r, can be determined by the equation: πr^2 = 57,600π r^2 = 57,600 r = √57,600 = 240 Therefore, the circumference (or length) of the track is 2πr = 2π(240) = 480π meters. So Obra drove 160π/480π = 1/3 or 33.3% of the circular track. Answer: E

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