find the possible dimensions of each corral?
958161
Being adjacent they share a common side. Assuming the corrals are identical, then if we let x be the length of the common side and y be the length of the other side, the amount of fencing required is 3x+4y = 240ft. The area of each corral is xy =1000sq ft. So y = 1000/x. Substitute this value in the first equation and you have:\n\n3x+4(1000/x)=240\n\n3x^2 +4000 =240x\n\n3x^2 - 240x+4000=0\n\nFrom the quadratic formula, x= 56.33ft. or x= 23.67ft.\n\nIn the first case the corral would have dimensions:\n\n56.33ft and 1000/56.33 = 17.75ft (approximately)\n\nIn the 2nd case, the dimensions are:\n\n23.67ft. and 1000/23.67 = 42.25ft (apprx)\n\nIn both cases 240ft of fencing would be used. Check:\n\n3x+4y=240ft\n\nCase 1: 3(56.33)+4(17.75)= 239.99 = 240\n\nCase 2: 3(23.67)+4(42.25) = 240.01 = 240\n\nIf you wanted the corrals to be closer to squares you would choose the 2nd dimensions.
47718
So you've got an eight-sided figure and each side is 72-feet.  If you draw horizontal and vertical lines from each vertex (the angles where sides meet) to the other side of the octagon, you'll see that the octagon can be split up into one large square (in the center), four rectangles (up, down, left and right), and four triangles (in the corners).  You following me?  Get out a pencil and get to it!  Okay.  So what are lengths of your square's sides?  You'll note that the square has 72-foot sides, just like the original octagon.  Since you know how to find the area of the square, you're on your way.\n\nNext.  you have four rectangles, each with a length of 72-feet, but the short end - the width, how do we figure out the width of the rectangle?  Okay.  Look at your triangles.  Each triangle has one side of 72 feet - the side which is the outside of the octagon - and then two equal sides - it's an isosceles triangle with a 90-degree angle.\n\nOkay, time to call on pythagoras.  Pythagorean theorem states that a^2 + b^2 = c^2.  What that means is this: If you take the length of sides a and b (the ones we don't know) and square them, and add them together, the answer will be 72-squared (side C).  Too bad I'm not there with you to make this part easier to understand.  We know that a and b are equal in this case because our triangle is isosceles.  Therefore, 2x^2 is equal to 72^2.  This gives us a messy equation wherein our answer is, x is ROUGHLY equal to 51 feet.\n\nSo.  You have a square with sides 72 feet long.  You have 4 rectangles each with length 72 and width 51.  You have 4 triangles each with base 51 feet and height 72 feet.  You know that squares and rectangles have an area equal to length times width, and triangles have an area equal to one-half base times height.\n\nCalculate the area for your 9 figures and add them together.  THAT will be the square footage of your octagon.\n\nWhat... did you think I was going to do it ALL for you?
847266
you might try addition and subtraction this is not complicated just add table-space-table till you reach the room dimensions ( in each direction )\n\nno matter what shape the tables are in the formula they are effectively square or rectangular ( actual dimension plus the 3' of clearance )
241144
to calculate the area of a 2-dimensional space you always multiply length x width, or in this case height\n\nC
900541
Land area: 7,663,942,028,116.63 m^2\nWater area: 416,522,509,501.38 m^2\nTotal area: 8,080,464,511,718.13 m^2
941173
area = 1/2 base times height\nBase times height = 88  (44*2)\nLenght of the legs are 11 and 8
481063
What is the shape of the lot?
904752
here's a link, page down and see if you can follow the steps. :\ngood luck.\n\nhttp://cstl.syr.edu/fipse/Algebra/Unit2/expo.htm
671808
4 sides and corners in a square
438309
You are taking away x feet on each side.\nThus, the length is 6-2x and the width is 8-2x and the height is x.\nV = Lwh\nV = (6-2x)(8-2x)x
1004359
You might as well have asked what are the dimensions of the universe.\n\nYou need to narrow down to state, prison or jail etc.  Every state does prison time differently and every county does jail time slightly different as well.
239395
Go to   \nhttp://www.sportsknowhow.com/basketball/dimensions/basketball-court-dimensions.html
460116
It depends what it looks like.\nWithout seeing the actual picture I cannot find the area.
200496
does this help?\nhttp://www.bballgoals.com/bbgcourtdimensions.html
1150077
Any length greater or equal the circumference of a 1-acre circle. \n\nCalculate the acre as square feet, square yards or square meters.\n\nDivide by pi and calculate the square root to get the radius, then multiply by 2 pi for the circle circumference.
269106
Lot is not a standard size like acre or hectare.\n\nA lot can be 20' by 20' or 10' by 350' or any other number of combinations.
1009332
Convert everything to inches, then multiply.  LxHxW.  They need to be the same units, though.