Rizags
  • Rizags
Help with area of circles question! Drawing Below:
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Rizags
  • Rizags
|dw:1449448867137:dw|
CMorris2004
  • CMorris2004
Where is the circle. nvm
CMorris2004
  • CMorris2004
?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

Rizags
  • Rizags
Assume that the quadrilateral is a square, and its sides are the diameters of the circle
tkhunny
  • tkhunny
Please provide an actual problem statement and your best work.
CMorris2004
  • CMorris2004
yep
Rizags
  • Rizags
how can I find the area of the shaded regions WITHOUT calculus? also the side is 1 unit
Rizags
  • Rizags
first thing i tried to do was find the total area of all the semis and subtract from the total square area: \[1-2\pi(0.5)^2\]
Rizags
  • Rizags
But then i'm left with that middle area
Rizags
  • Rizags
If i can get the area of the middle portion, i have the answer
tkhunny
  • tkhunny
Ummm... Prove to me that there IS a "middle portion".
Rizags
  • Rizags
I think I messed up the diagram and there shouldn't be a middle portion...|dw:1449449371478:dw|
tkhunny
  • tkhunny
Indeed. And, since that is zero, what is your answer?
Rizags
  • Rizags
area of the square minus the area of the four semicircles without overlap
tkhunny
  • tkhunny
Which is what?
Rizags
  • Rizags
actually its the opposite, so \[2\pi0.5^2-1\]
Rizags
  • Rizags
is that correct?
tkhunny
  • tkhunny
I don't think so. How did you arrive at that?
Rizags
  • Rizags
the area of one semicircle is \[0.5\pi r^2=0.5(\pi)(0.5)^2\] since there are four of them, I multiply this by 4 to get \[2(\pi)(0.5)^2\]this would be the area of the four semicircles without overlap. The area of the square (1) is simply the area of the four semicircles with overlap. So subtracting these two should give me the overlap i thought
tkhunny
  • tkhunny
Area of Square = 1 Area of the four semi-circles is \(4\cdot\dfrac{1}{2}\pi r^{2}\) with \(r = 1/2\). This counts each leaf twice. In fact, counting each leaf is what is wanted. Each half-leaf is \(\dfrac{1}{2}r^{2}\left(\alpha - sin(\alpha)\right)\) with \(r = 1/2\) and \(\alpha = \pi/2\). You must sum 8 of these.
Rizags
  • Rizags
the area of one semicircle is \[0.5\pi r^2=0.5(\pi)(0.5)^2\] since there are four of them, I multiply this by 4 to get \[2(\pi)(0.5)^2\]this would be the area of the four semicircles without overlap. The area of the square (1) is simply the area of the four semicircles with overlap. So subtracting these two should give me the overlap i thought
tkhunny
  • tkhunny
Okay, here's what is confusing me... "Is that correct?" You tell me if it is correct. Subtracting the square from the non-overlapped circles produces the correct calculation. Why do you doubt. You also did not simplify. That would have helped. Note that your answer is verified by knowing about circular chords and calculating directly. \(8\cdot \dfrac{1}{2}\cdot\left(\dfrac{1}{2}\right)^{2}\cdot\left(\dfrac{\pi}{2} - 1\right) = \dfrac{\pi}{2} - 1\) as well. Good work. Have more faith in your accurate and correct thinking. (and simplify)
mathstudent55
  • mathstudent55
|dw:1449454780848:dw|