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Bumek [7]
3 years ago
14

A dog is leashed to the corner of a house. How much running area does the dog have? Round your answer to the nearest square foot

, If necessary.

Mathematics
1 answer:
malfutka [58]3 years ago
4 0
The running area the dog has is 942 feet squared.
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How do you solve his with working
AlexFokin [52]
Check the picture below.

a)

so the perimeter will include "part" of the circumference of the green circle, and it will include "part" of the red encircled section, plus the endpoints where the pathway ends.

the endpoints, are just 2 meters long, as you can see 2+15+2 is 19, or the radius of the "outer radius".

let's find the circumference of the green circle, and then subtract the arc of that sector that's not part of the perimeter.

and then let's get the circumference of the red encircled section, and also subtract the arc of that sector, and then we add the endpoints and that's the perimeter.

\bf \begin{array}{cllll}&#10;\textit{circumference of a circle}\\\\ &#10;2\pi r&#10;\end{array}\qquad \qquad \qquad \qquad &#10;\begin{array}{cllll}&#10;\textit{arc's length}\\\\&#10;s=\cfrac{\theta r\pi }{180}&#10;\end{array}\\\\&#10;-------------------------------

\bf \stackrel{\stackrel{green~circle}{perimeter}}{2\pi(7.5) }~-~\stackrel{\stackrel{green~circle}{arc}}{\cfrac{(135)(7.5)\pi }{180}}~+&#10;\stackrel{\stackrel{red~section}{perimeter}}{2\pi(9.5) }~-~\stackrel{\stackrel{red~section}{arc}}{\cfrac{(135)(9.5)\pi }{180}}+\stackrel{endpoints}{2+2}&#10;\\\\\\&#10;15\pi -\cfrac{45\pi }{8}+19\pi -\cfrac{57\pi }{8}+4\implies \cfrac{85\pi }{4}+4\quad \approx \quad 70.7588438888



b)

we do about the same here as well, we get the full area of the red encircled area, and then subtract the sector with 135°, and then subtract the sector of the green circle that is 360° - 135°, or 225°, the part that wasn't included in the previous subtraction.


\bf \begin{array}{cllll}&#10;\textit{area of a circle}\\\\ &#10;\pi r^2&#10;\end{array}\qquad \qquad \qquad \qquad &#10;\begin{array}{cllll}&#10;\textit{area of a sector of a circle}\\\\&#10;s=\cfrac{\theta r^2\pi }{360}&#10;\end{array}\\\\&#10;-------------------------------

\bf \stackrel{\stackrel{red~section}{area}}{\pi(9.5^2) }~-~\stackrel{\stackrel{red~section}{sector}}{\cfrac{(135)(9.5^2)\pi }{360}}-\stackrel{\stackrel{green~circle}{sector}}{\cfrac{(225)(7.5^2)\pi }{360}}&#10;\\\\\\&#10;90.25\pi -\cfrac{1083\pi }{32}-\cfrac{1125\pi }{32}\implies \cfrac{85\pi }{4}\quad \approx\quad 66.75884

7 0
3 years ago
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