since the area of a circular segment is
a = 1/2 r^2 (θ-sinθ)
you need θ such that cos(θ/2) = x/8
so, you need to solve
1/2 * 64 (2*arccos(x/8)-sin(2*arccos(x/8))) = 64π/3
x = 2.12 inches
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3.5 mi/hr = 0.058 mi/min
If the swim subtends an angle θ, then the distance swum is 2x, then
x/8 = sin θ/2
x = 8 sinθ/2
the distance jogged is then 8π - 8θ
total time is then
(16 sin(θ/2))/.0583 + (8π - 8θ)/.0875
If you poke around at the graph, you can prolly figure the max/min for the domain 0 <= θ <= π
http://www.wolframalpha.com/input/?i=%2816+sin%28%CE%B8%2F2%29%29%2F.0583+%2B+%288%CF%80+-+8%CE%B8%29%2F.0875
Problem 1.) Three friends want to share a circular 16-inch pizza equally by exactly only two parallel cuts. How far from the center must the cuts be? Hint: The pizza is a circle of radius 8 inches.
Problem 2.) You are at the southernmost point of a circular lake of radius 8 miles. Your plan is to swim a straight course to another point on the shore of the lake, then jog to the northernmost point. You can jog 150% as fast as you can swim. With a detailed explanation of your steps, find, in minutes, the least and the most time this trip will take if you swim at 3.5 miles per hour.
2 answers
LOL ITS 33,3
2 answers