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OpenStudy (anonymous):

My question is attached, thanks!

OpenStudy (anonymous):

OpenStudy (anonymous):

A circle (of radius 5) with a quarter pi missing.

OpenStudy (anonymous):

The bow of the boat can travel in a circle shape until it hits the 5 foot edge of the dock or the 4 foot edge of the dock. This shape is 3/4 of a circle. You will need to calculate the area of 3/4 of a circle first. \[A = \pi r^2\] 3/4 of the area of a circle is \[\frac{3}{4}\pi r^2\] Now, something interesting happens when the boat floats toward the 4 foot side of the dock. There is 1 foot of rope to go after the bow of the boat is parallel with the 4 foot edge of the dock; this creates another fraction of a circle. This time 1/4 of a circle shape. So the total area is: \[A _{TOTAL = } A _{1} +A_{2}\] \[A _{1} = \frac{3}{4} \pi r _{5}^2\] where the radius is 5 feet, and \[A _{2} = \frac{1}{4} \pi r _{1}^2\] where the radius is 1 foot

OpenStudy (anonymous):

Alright! Do you think you could work it out to the end?

OpenStudy (anonymous):

That is up to you ;) I put all the theory there, all you have to do is understand the theory and put in the numbers.

OpenStudy (anonymous):

Amistre!!??!!

OpenStudy (amistre64):

OpenStudy (anonymous):

amistre, high-five! :D

OpenStudy (amistre64):

R=5; angle = 3pi/2 Area = 5^2 3pi ------ 2 * 2 r=1; angle = pi/2 area = 1^2 pi ------ 2 * 2 add the areas then

OpenStudy (anonymous):

The last two?

OpenStudy (amistre64):

75 1 pi(--- + ---) = 76pi/4 = abt.59.66 ft^2; which one is it rounded to? 4 4

OpenStudy (anonymous):

aha!!

OpenStudy (anonymous):

So it would be rounded to 60 ft?

OpenStudy (amistre64):

if I did it right, then id round it to 60 :)

OpenStudy (anonymous):

right?

OpenStudy (anonymous):

Thanks! I owe you one!

OpenStudy (amistre64):

:) youre welcome

OpenStudy (anonymous):

if you ever have spare time, there were two previous posts of mine that I was helped on, but the person wasn't too sure what he was doing...

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