Find the dimensions of a rectangle whose perimeter is 26 meters and whose area is 40 square meters.?
456567
5 meters by 8 meters\n\n5+5+8+8=26\n\n5x8=40\n\n\nThat was easy
789618
a) 24 in.\n\nSince a square has four sides...  You must add six for each side (i.e. 6 + 6 + 6 + 6) to get perimeter.
915466
Let a and b the lengths of the sides of the rectangle (with a the side with the semi-circle).\nThe perimeter is 2b+a(pi/2+1)=16, and the area ab+pi/8*a^2. Substituting b in the area (from the perimeter equation), the area is 8a-a^2*(1/2+pi/8). \nSetting the derivative to 0, we have a=32/(4+pi). Then b comes from the perimeter constraint as b=8*(1-(2+pi)/(4+pi)).\nNumerically this gives a=4.48, b=2.24 and an area of 17.92
191768
r u sure its not 40 m less, if it is here:\nLet w=width\n    w+40=length\nw(w+40)=21^2\nw^2+40w=441\nw^2+40w-441=0\n(w+49)(w-9)=0\nw=-49 or w=9\nthe width is 9 m and the length is 13 m
955861
area: l*w = 256\n-> 4w*w = 256\n->w(sqaure) = 64\n->w=8 (width)\n->l = 4w = 4*8 = 32(length). \n\nThere you go.
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.
783277
let the height be x\nthe cross section will be (40-2x)*x\ndA/dx=40-4x\nsetting this to 0\nx=10\nso the length of the sides will be 10 cm each for max area of x section
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.
936604
let the equal sides be x\n1/2x^2=area=20 given\nx^2=40\nx=rt40\nusing the pythagoras theorem\nc^2=a^2+b^2\nc=rt(a^2+b^2)\nthird side=rt(40+40)\n=rt80\n=4rt5 cm
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.
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.
979599
L = length\nW = width\n\nLW = Area\n1) LW = 35\n\n2L + 2W = Perimeter\n2) 2L + 2W = 27\n\nSolve equation 1 for L and substitute into equation 2\n\nLW = 35\nL = 35/W\n\n2L + 2W = 27\n2(35/W) + 2W = 27\n70/W + 2W = 27\n70 + 2W^2 = 27W\n2W^2 - 27W + 70 = 0\n(2W - 7)(W - 10) = 0\nW = 7/2, 10\n\nL = 35/W\n\nIf W = 3.5, L = 10 and if W = 10, L = 3.5\n\nAssume L is longer than W: L = 10, W = 3.5
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
958726
width = x\nlength = 2x +2\nperimeter = 52\n\nperimeter = 2(length) + 2(width)\n\n52 = 2(2x + 2) + 2(x)\n52 = 4x +4 +2x\n52 = 6x +4\n48 = 6x\n8 = x\n\nbecause x = 8 then:\n\nwidth = 8\nlength = 2(8) + 2 \nlength = 16 + 2\nlength = 18\n\nlength = 18, width = 8\n\nto check it add up the numbers to see if they equal the perimeter of 52\n\n18 + 18 + 8 + 8 = 52 True.  answer is correct
136459
Find the factors of 56 (or all the possible measures of sides that would equal an area of 56). \n2 and 28\n4 and 14\n7 and 8\n\nThen add 2l + 2w to get the perimeters.\n2(2) + 2(28) = 60\n2(4) + 2(14) = 38\n2(7) + 2(8) = 30\n\nSo the least perimeter would be 30 cm.
300586
Adding. example if the lenght of rectangle is 4 and width is 2 then to work out the perimeter you do this: 4+2+4+2= 12 = perimeter.
975485
H*W= area\nheight times width\n\none side has a height of 8 and its width is 9 thus the area would be 72