anonymous
  • anonymous
Find the area of the shaded portion in the square. (assuming the central point of the arc is the corresponding corner) https://media.glynlyon.com/a_matgeo_2011/7/groupi57.gif
Mathematics
  • Stacey Warren - Expert brainly.com
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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.
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
|dw:1335260783163:dw| it has to be plugged into this.
anonymous
  • anonymous
http://academicearth.org
blockcolder
  • blockcolder
A=area of square-area of quarter circle with radius 6

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anonymous
  • anonymous
is the website for knowing more
anonymous
  • anonymous
:| thanks for your time. but. that didn't quite help. at all. thank you though.
.Sam.
  • .Sam.
Area of shaded= area of square- area of sector \[Area ~of ~shaded= (6)\times(6)-\frac{1}{2}r^2 \theta\] \[Area ~of ~shaded= 36-\frac{1}{2}(6)^2 \frac{pi}{2}\] \[Area ~of ~shaded= 36-\frac{1}{2}(6)^2 \frac{pi}{2}\] \[Area ~of ~shaded= 7.722\]
jhonyy9
  • jhonyy9
will be 36-9pi because 9pi is a quarter from area of this circle with radius 6 -area of square =6*6=36 - area circle = pir^2 =36pi - a quarter of area circle = 36pi/4 =9pi - area sheped =area square - quarter of area circle =36-9pi - if we consider pi=3,14 - so than area shaped = 36-28,26=7,74
anonymous
  • anonymous
Thanks!
jhonyy9
  • jhonyy9
yw good luck was my pleasure bye
jhonyy9
  • jhonyy9
hope so much that is understandably sure

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