laying a piece of pipe in a lawn. the lawn is rectangle, 15feet long and 8feet wide.?
483088
this is very easy, just use the pythagorean therom (i might spell it wrong) the formula is: a^2 + b^2 = c^2 (c is the length of the pipe)\n\n15^2 + 8^2 = c^2\n\n225 + 64 = c^2\n\n289 = c^2\n\n17 = c\n\nyou need 17 feet of pipe
331831
The dimesnions of the rectangle are 14 feet by 34 feet.\n\nYou know that the perimeter is 96 feet and that each long side is 20 feet more than the short sides.\n\nAs there are 2 long sides the total length of both these sides must be 40 feet more than the total length of the 2 short sides (20 x 2 = 40).\n\nIf all the sides were the same length as the short sides then the rectangle would have a perimeter of 56 feet (we know it's 96 feet to start with and that the additional length on the two long sides is 40 feet and 96 - 40 = 56).\n\nEach side would therefore be 14 feet long since 56 divided by 4 = 14.\n\nNow that we know the length of the short sides all we need to do is add 20 to get the length of the long sides, 14 + 20 = 34.\n\nTo check it you can add 14 + 14 (the two short sides) + 34 + 34 (the two long sides) = 96.\n\nI hope that makes sense, it's 1:25AM here and I'm getting tired.
726535
Length of wire = 9in+9in+11in+11in\n                  =40in\n\nLength of Square = 40in divide by 4\n                          = 10in
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.
358814
width of pool = 32\nwidth of fence therefore = 32 + 7 + 7 = 46\nlength of pool = 18\nlength of fence therefore = 18 + 7 + 7 = 32\n\nperimeter = 2(width) + 2(length)\np = 2(46) + 2(32)\np = 92 + 64\np = 156\n\n156 feet of fencing is needed.
278523
you need to figure out the inside and outside diameter of each pipe example 1 7/8 inside dia, will fit over 1 7/8 outside dia, hope this helps good luck
942923
Without using calculus, draw the rectangle with a pretty river for one side.  Label the two congruent ends of fence x.  Since they add up to 2x, the remaining piece of fence is 5000 - 2x long.\n\nThe area is now x(5000-2x) or 5000x - 2x^2\n\nIf you graphed this it would be an upside down parabola,and its highest point would be the vertex.  The formula for the x coordinate for the vertex is x = -b/2a\n\nFor this, it's x = -5000 over 2(-2) or -5000/-4 or 1250\n\nSo x = 1250 and the two ends are 1250 ft each leaving 2500 ft for the third side.  \n\nHope that helped.
1073502
Once again you will see that this is a rectangle.\nA = length times width\nA = 7*2\nA = 14
794849
I am not sure if I understand your question correctly, or else your question is not complete.\nFor example, a usual question of this sort is to find the maximum area of a rectangle for a given perimeter, the classical fence problem, as follows:\n\nYou are given a specific length of fence L to make en enclosure of maximum area.  What would be the dimensions of the rectangle.\n\nHere we take the length of one side as x, then the other side will be (L-2x)/2.  The area is thus A(x)=x(L-2x)/2.\n\nTo obtain the value of x for a maximum area, we differentiate A with respect to x, \ndA(x)/dx = d(Lx/2-x^2)=L/2-2x\nfor maximum area, dA(x)/dx=0, so we equate\nL/2-2x=0\nand solving for x gives x=L/4.\n\nThat means that the maximum area rectangle for a given perimeter is a square!\n\nHope that helps with your project.\nPost if you need other information, but please post details, especially how far you are with your project, so we can help you better.
844322
Let width = x, length = 2x.\nPerimeter = x + 2x + x + 2x = 240\nSo 6x = 240\nSo x = 40\nSo width is 40, length is 80, so area = 3200 square feet.
644833
if x is width, then 30-x is length \n(so that 2* (width_length)=60),\nand area is indeed 30x-x^2, which is maximized at x=15.\n\nGenerally, a rectangle with a given perimeter will have max area if it is square.
647438
1 sq yd = 9 sq ft.  So, divide by 9.
608232
Rec first  8 X 4 = 32\n32 X 16 = 512  this is Area of rectangle.\n\nCircles pi X r^2 = Area\n3.14159 X 8 X 8 = 201\n\n512 - (2 X 201) = shaded area\n109.87 = shaded area\n\nOK ?
939304
A meter is about 3 inches longer than a yard
894264
You have a 5ft wide path around the garden.\nLength of garden + path = 34 + 2(5) = 44ft\nBreadth of garden + path = 63 + 2(5) = 73 ft\nArea of garden + path = 44 x 73 = 3212 sq ft\nArea of garden = 34 x 63 = 2142 sq ft\nArea of path = 3212 - 2142 = 1070 sq ft\n\nImagine it as a big rectangle, 44 ft by 73 ft, with a smaller rectangle 34 ft by 63 ft in it. You add 10 ft respectively to the length and breadth of the garden as there is 5ft of path to the left and right, top and bottom of the garden (hence the word 'surround').\nIt would be easier if I could draw the diagram for you. Thus I am trying to describe it here.\n\nHope this helps :)
1136755
The wire is 7/8 m long and if it has to be cut into 14 pieces then \n(7/8) / 14 = 1/16\n\nTherefore, the length of each is 1/16m
1032617
1 yard = 3 feet.\n1 feet = 2 handspans approx (act. less.)\n\n1 yard = 6 handspans approx.\n1 yard < 6 handspans. (actually)