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.
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Obra drove 160Ï€ meters along a circular track....
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DavidG@VeritasPrep
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We're given: r^2 * π = 57,600 * πVincen 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.
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
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Scott@TargetTestPrep
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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.
Since the area enclosed by the circular track is 57,600Ï€ square meters, the radius of the track, r, can be determined by the equation:
Ï€r^2 = 57,600Ï€
r^2 = 57,600
r = √57,600 = 240
Therefore, the circumference (or length) of the track is 2Ï€r = 2Ï€(240) = 480Ï€ meters. So Obra drove
160Ï€/480Ï€ = 1/3 or 33.3% of the circular track.
Answer: E
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