If a table is 40 inches long , the perimeter of the table is 120 inches. how  wide is  the table?
927623
The perimeter of a rectangle is the sum of all 4 sides, i.e. twice the width plus twice the length.\nP = 2W + 2L\n120 = 2W + 2(40), from substituting what we know\n120 = 2W + 80\n120 - 80 = 2W + 80 - 80\n40 = 2W\n40 / 2 = 2W / 2\n20 = W\nThe table is 20 inches wide.
1077978
23 on the small side.\n\nIf the small side is x, then the long side is 5x.  That makes the perimeter x+x+5x+5x or 12x.\n\n12x=276\nx=276/12\nx=23\n\nThere for the rectangle is 23cmX115cm.
185822
23*55'11.59 N\n90*43'11.59 E\nelevation 41
927866
a table is 4 x 8 figure 42 to 48 inches clearance around it or 12'x16' open area is good!
678410
Let y = the length of the base and x = the length of one of the equal legs.\nThe perimeter is the sum of all of the sides, so \nP = x + x + y or P = 2x + y\nThe leg is 8 less than 4 times the base translates into\nx = 4y - 8\nNow\nP = 2x + y\n29 = 2(4y-8) + y\n29 = 8y - 16 + y\n45 = 9y\ny = 45/9\ny = 5\nYou're looking for the length of the equal legs, so you need to find x.\nx = 4y-8\nx = 4(5) - 8\nx = 12cm
996625
Let the width be x ft and the length be y ft.\nSo perimeter is 2x + 2y = 74\nor x+y=37........(1)\n\nCalculating cost of lengths....\n2y*1 = 2y\n\nCalculating cost of widths....\n2x*4.5 = 9x\n\nTotal cost = 2y +9x = 159........(2)\n\nSolving (1) and (2) simultaneously you get x(width) = 12.14 ft and y(length) = 24.86 ft
463854
The perimeter is twice the sum of the length and the width. Given the ratio of length to width, the perimeter is necessarily eight times the width. Therefore the width is eight and the length is twenty four.
882237
2 ways:\n1. take the width in inches times the length in inches and divide by 144\n2. convert the inches to feet by dividing by 12, then take the width in feet times t he length in feet
897940
perimeter = 2 * (length + width)\n\nfor square, length = width..\n\nso, perimeter = 2 * (2 * length) = 4 * length\n\ngiven, 4 * length = 104\nso, length = 26\n\nso, each side is 26 cm long.
83837
The dimensions of the rectangle are 5.5 cm x 16.5 cm\n\nIf the rectangle is 3x longer than it is wide, then there are four units that make up two sides...1 unit width, and 3 units length.  So we have four units (4x) that make up one half of the perimeter.  That means that eight units (8x) will make up the entire perimeter.\n\n8x = 44cm, therefore x = 5.5cm\nWe already said that one side is 3x and the other is 1x, so that gives us dimensions of 5.5cm x 16.5 cm
360732
Right.  I'm not sure what the question is, but let's assume that the space we are talking about is a rectangle, and you have to figure out what the dimensions are.  Well, the perimeter, 26, is the distance around the space which has the area of 36, which for the sake of argument, we will assume is square feet.  What times what equals 36?  9 x 4, right?  So, it's possible, but not certain that we are looking at a rectangle whose dimensions are 9 feet by 4 feet.  To prove this, remember that a rectangle has four sides.  If our assumptions are correct, it has two sides of 9 feet and two of 4 feet, right? So then,  9 + 9 + 4 + 4 = 26.  It would seem that this is the right answer.  \n\nOf course 12 times 3 also equals 36.  But this time, if we substitute the 12 and the 3 as the dimensions, we would get a perimeter of 30 feet, obviously the wrong answer.  Looks like you have a rectangular space that is 9 by 4 feet.\n\nI hope this helps and that I haven't just muddied the water further.
224714
Perimeter for a square is 4*side\nSo side of square = 36/4=9 inches\n\nAnd area of square is side*side\nthus area is 9*9=81 sq inches !!!
584407
12 inches in a foot, so 16 times 12 equals 192\n\n192 inches in 16 feet.\n\n12 x 16 = 192
343312
perimeter = 2 (side1 + side2)\narea = side1 * side2
800206
The 400 feet of fencing forms the perimeter P of the rectangular area.  The perimeter of a rectangle is P = 2*W + 2*L.  The area of a rectangle is A = L*W.  Take either the W or L from the perimeter equation and solve for it in terms of the other.  For example. 2*W = P - 2*L, or W = (P - 2*L)/2.  Put this expression for W into the area formula\n\nA = L*(P - 2*L)/2\n\nA = P*L/2 - L^2\n\nTo maximize area, take the derivative of A with respect to L to get\n\nA' = P/2 - 2*L  and set it equal to zero\n\nP/2 - 2*L = 0, or  L = P/4;  This gives you the length of the area.  You can get the width from the perimiter formula W = (P-2*L)/2;\nPutting in L = P/4 gives W = (P - P/2)/2 = P*(1-.5)/2 = P/4.\n\nNotice that L = W = P/4; in other words the maximum area rectangle is a square.
740272
1/4"= 1'\n1" =4'\n12" or 1' = 4*12=48' Ans.
551690
T= L + 2B\n\nT is total surface area\nL is Perimeter x Height\nB is base area