Finding the area of a track

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pretzelboi

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I'm super confused and have a mat test tomorrow please help! Thank you!!


Wake County Middle School needs to mow the grass in the middle of the track. The interior straight segments are approximately 84 meters and the curved semicircles are about 115 meters. How much area will need to be mowed? Round the solution to the nearest hundredths place.
 
pretzelboi said:
I'm super confused and have a mat test tomorrow please help! Thank you!!


Wake County Middle School needs to mow the grass in the middle of the track. The interior straight segments are approximately 84 meters and the curved semicircles are about 115 meters. How much area will need to be mowed? Round the solution to the nearest hundredths place.
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This is a poorly worded problem! Did you post the EXACT problem as it was given to you? Do you have a sketch to go with this problem?

The interior straight segments are approximately 84 meters - are those the lengths of each straight section?

If we assume that the "115 meters" is the length of each semi-circular parts - then the diameter of the circular section is (115/pi =) 73.21 meters.

With these assumed dimensions - can you draw an approximate sketch of the track?
 
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Now that is much clearer. So the diameter of the circular arc is ~73.21 meters.

Divide the track into one big rectangle and two semicircular patches?

What are the lengths of the sides of the rectangle? What is the area of the rectangle?

What are the areas of the semicircular patches?
 
how did you get 73.21 and for the rectangle i did 73.21*84 to get the area of the rectangle and i got 6149.64
 
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pretzelboi said:
how did [Subhotosh] get 73.21 [for the diameter of the circle] …
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Hello pretzelboi. The two semicircles may be joined together, to make a complete circle. We're given half the circumference of that circle (115), so the full circumference is 230.

The formula for a circle's circumference (C) in terms of diameter (d) is:
\[C = \pi \cdot d\]
\[230 = \pi \cdot d\]

Solve for d. You've been asked to report the total area rounded to the nearest hundredth of a square meter, so I would carry four decimal places in all intermediate calculations. Round only the final answer to two places. If we were to round everything to two places, then we'd be using values containing a bit of error in the calculations, and those results might not be valid to two places. By carrying four decimal places until the end, we can be confident that the round-off error won't affect digits in the first two decimal places.

In other words, use d=73.2113 for calculating the area of the rectangular part, and use 36.6056 for the radius, when calculating the area of the circular part.

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amathproblemthatneedsolve said:
The rectangular part should be C= pi (3.14x73.2113) what do you get??
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Hi amptns. I'm not following your post above (must be typos, C is not pi).

The area of a rectangle is length × width, and we're given the length (84). The width of the rectangle is the same as the diameter of the semicircle (73.2113).

I get 6149.7492 m2, for the rectangular area (another intermediate result, rounded to four places).

Also, when working with the circle, I would not round pi to two places. Using the pi key on a scientific calculator is better, or use pi=3.1416, if you must round. Intermediate results need to carry extra digits, to avoid round-off error (see post #8).

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A = pi · r2

A = 3.1416(36.6056)2 ≈ 4209.6496 m[su]2[/sup]

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