how do you find the area and the perimeter of a parallegram?
103731
The area and perimeter would both be found using the same formula as finding it for a rectangle. Because the angles and lengths must be equal to be a parallegram if you cut off the angled part (trianle) from one end and put it on the other you would have a regular rectangle.\n\n2l+2h = perimeter\nl*w = area\n\nDo it the easy way!!!
172067
Perimeter is the distance around a figure. The area is the amount of space inside the shape. Good luck!
190089
use heron's formula: \nArea=SQRT(s(s-a)(s-b)(s-c)), \nwhere s=(a+b+c)/2 or perimeter/2. \n\nyou only need to know the sides of the triangles. \nfrom (0,9) to (4,6), the distnace is a=5\nfrom (0,9) to (-1,12) the distance is b=sqrt(10)~3.16\nfrom (4,6) to (-1,12) the distance is c=sqrt(61)~7.81\n\nso, s=(5+3.16+7.81)/2=7.985\n\nplugging these 4 numbers into the formula gives \nSQRT(7.985(7.985-5) (7.985-3.16) (7.985-7.81))=4.486\n\nso the area is approximately 4.486
552905
The AREA of an isosceles triangle is found by taking 1/2 the sum of the long and short parallel sides, times the height. One way to compare this with the parallelogram is to take two copies, turn one 180 degrees and glue them along the equal (nonparallel) edges. \nYou now have a larger parallelogram - with one side composed of the sum of the original short and long parallel edges. Use the usual formula for the area of this parallelogram, then divide by 2 (since it is composed of two equal copies).
906875
The trick is turning the word problem into equations.\n(Key: length = l; width = w)\n\nThey tell you the length is 3 feet more than twice the width, so:\nl = 2w + 3\n\nThen they tell you the perimeter (which is twice the width and twice the length) is 36, so:\n2l + 2w = 36\n\nYou should be able to take it from there.
549797
enclose the ireegular shape with a rectangle and find the empty spaces. then subtract that with the area of the enclosed rectangle to get ur answer.
598751
L=length w=width\n\nSo L=3w\n\nThe formula for perimeter is 2L+2W=P\nNow, you just plug in the variables and solve\n\n2(3W)+2W=64 --multiply 2 time 3W\n6W+2W=64    --add 6W and 2W\n8W=64    --Divide 8 on each side\nW=8\n\nWidth is 8 units\n\n--hope that helped
238936
The apothem of a regular polygon is the distance from the center of the polygon to the midpoint of one of the sides.  It is used to calculate the area of the regular polygon:\n\nA = 1/2 a p  \n\nA = area\na = length of apothem\np = perimeter
850259
here:http://www.cut-the-knot.org/Curriculum/Geometry/AreaOfParallelogram.shtml
698034
a ) you effectively have a perimeter of 180m because 160 plus 20. \n\n so you take 180/4 because it;s retangle.  so your area would be 45 x 45 = 2025 m^2
259191
Perimeter\ntwo sides are 8 and the other two sides are 16. \n\n8 times 2 is 16 and 16 times 2 is 32. \n\n16+32= 48(unit) \n\n                Area\njust multiply them:) \n\n8 times 16= 128(unit) squared\n\nDoes that help?
530421
decagon: S x 10\nrhombus: S x 4\nparallelogram: (L x2)+(W x2)\n\nragular fugures are figures that have the same lengths in all the sides ( which is the case with the decagon) so you just add up all the sides or multiply the side by the number of sides the figure has.\n\na rhmbus has all the sides of equal length so again you just add all the sides or multiply the side by 4.\n\nthe opposite sides of the parallelograms are of the same length so you multiply the lenght and width of the parallelogram by 2 and add the results and thats it.\n\ngood luck.
573267
It's the formula for the area of a regular polygon\n\nArea = 1/2 times apothem times perimeter\n\n(apothem = perpendicular distance from center to midpoint of a side)
48612
Let r be the radius of the circle and s the length of the square's sides.\ncircle perimeter=2*pi*r, square perimeter=4*s\n2*pi*r=4*s => s=pi*r/2\ncircle area=pi*r^2, square area=s^2\ns^2=(pi*r/2)^2=pi^2*r^2/4\nBut pi<4 => pi^2*r^2/4<pi*r^2 => square area < circle area
551690
T= L + 2B\n\nT is total surface area\nL is Perimeter x Height\nB is base area
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
912405
Let x be the radius of the semicircle, then the rectangle has base 2r and height h.  The total perimeter of the window is\n\npi*r + 2r + 2h = 20. \n\nSo h = (20 - pi*r - 2r) / 2\n\nThe area is (pi*r^2)/2 + 2rh.  Replace h with the above so that the area is a function of r.  Differentiate to find the maximum.