maximum area of an isosceles triangle with each equal side 5 cm.?
921698
let the base be x\narea=1/2*x*[5^2-(x/2)^2]\n       =1/2x[25-x^2/4]\ndA/dx=1/2[x(-1/2x)+(25-x^2/4)]\nsetting it to zero\n1/4x^2=-x^2/4+25\nx^2/2=25\nx^2=50/3\nx=5rt(2/3)\nh=rt(25-50/12)\n=rt250/12\n=5/2rt(10/3)\narea=(1/2)5rt(2/3)(5/2)rt(10/3)\n=25/4rt(20/9)\n=25/12rt20\nx=10/rt3\narea=1/2*10/rt3*200/12
12258
sorry i made a mistake....this is correct...\ndivide the base by 2.  Now you have a right triangle.  the height is the same, but the base is now 3.  pygathora (spelling, whatever) theorm is A squared + B squared = C squared.  c is the side opposite the right angle, a and b are the other sides.  so 3 squared + 4 squared is 9 + 16 = 25. The square root of 25 is 5, so that is the length of the side.  the volume = area of base times height.  Area of a triangle is 1/2 base times height.  so 3 (one half of 6) times 4 = 12.  times this by height of 8 = 96 cubic units. largest rectangular side is the 6 (the longest side) times 8 = 48 units squared. hope this helped.  the lateral area is the area of the sides added together. so C times 8 = 40 (area of one side)  add the other side of equal area (the C from the second triangle we got when we split the first one) 40 + 40 = 80.  Add the third side of length 6 times 8 = 48.  80 + 48 is 128 (in units squared).  or all sides added up 5 + 5 + 6 = 16 times 8 = 128.
804994
Let the Length of the side of the rectangle be x and the length of the side of the triangle is L\n\nThen when the rectangle is inscribed it is symmetrically placed (otherwise you would not get a rectangle). So the distance from its base vertices to the vertices of the triangle are each ½(L - x)\n\nThus the width of the rectangle = ½(L - x) tan 60°\n= √(3)/2 * (L - x)\n\nThus the area of the rectangle A = length * width\n= x * √(3)/2 * (L - x)\n= √(3)/2 * (Lx - x²)\n\ndA/dx = √(3)/2 * (L - 2x)\n= 0 for stationary points\nWhence L - 2x = 0 ie x = ½L\nd²A/dx² = -√(3) < 0 for all x Thus the curve is always concave down So the stationary point is a maximum.\n\nThus the length of the rectangle is ½L and the width is √(3)L/4\n(and its area is √(3)L²/8)
599733
I believe it would be one-half times the length of each diagonal.\n\nIn other words, if the diagonals are 4 & 6 the area would be:\n\nA = 1/2 * 4 * 6 = 12\n\nIf you don't know the diagonals, but you know the length of the sides, you could construct the diagonals and calculate their dimensions using the Pythagorean theorem:  c^2 = a^2 + b^2, and then solve from there.  Keep in mind that you'll have to do some extra geometric analysis to find the length of each leg of the right triangles that are created by the intersection of the diagonals before you can use the theorem.
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
366178
Take the average between the side of an inscribed square\nand that of an inscribed hexagon. With a hexagon you\nget an isosceles triangle with base as a side and a 60\ndegree angle at the apex, leaving 120 degrees to be split\nbetween the equal pair which are then also each 60 degrees.\nSo you have an equilateral triangle whose base is equal\nto the radius of the circle and of course the base is a side\nwhich has a length of 1. \n  Now inscribe a square. There are now 4 isosceles\ntriangles with a 90 degree angle at the apex and a pair\nof 45's at the base. Draw an altitude from the apex, \nperpendicular to the base, thus bisecting the 90 at\nthe apex making another 90, 45, 45 triangle whose\nside is 1/2 the side of that of the square. The hypotenuse\nis a radius of the circle so the side of the last mentioned\ntriangle is 1/2^2 (.707). This makes the side of the\nsquare 1.414. So averaging 1 and 1.414 you get 1.207.\n Using trig we find the real answer to be 1.1755.\nSo the percentage of error is 2.67 percent from 1.207.\nI wouldn't win an engineering prize with that kind of\naccuracy, but I think I could build a pretty good\ncatapult.
776798
Every side of this triangle has a pair of ends and all the three pairs are cyclic. ie, if AB,BC & CA are the three sides of the given triangle , then the side AB has the ends (4,5),(1,1), BC has the ends  (1,1)and (-2,5) & CA has the pair of ends (4,5),(-2,5).\nBy simple inspection of these pairs we can find that the points C and A are at the same height from the abcissa hence parallel to it. By this we can conclude easily that the length of CA=6 units.\nFurther, to be an isocelous triangle in this case the point B should be on the mid line of CA, and it is normal to CA. The mid point of CA is three units away from its ends C&A. As CA is parallel to absissa we can clearly conclude that any point between (1,0)&(1,5) will be the vertex of an isosceles triangle. Here the point B is (1,1) on the said line. Thus proved that the given points form an isosceles triangle.\nWe alredy know that CA = 6\nAnd the other sides are equal to sqrt(4-1)^2 +(5-1)^2 = 5 units\nNow, we can find the area of this triangle by Heron's Formula \nie sqrt[8(8-6)(8-5)^2] =12 sq units
227708
There are two ways to do it...\n\n1.> If you know the base and the height of the triangle:\n\n(Base*Height)/2\n\nso if one of the side was 12 cm., and the perpendicular on it from the point where the other two sides meet is 6 cm., the area is = (12*6)/2 = 36 sq.cm.\n\n2.> If you know all the sides:\n\nHero's formula:\n\nFirst find s (semi-perimeter) as:\n\ns=0.5*(side1+side2+side3)\n\nnow the area= root (s * (s-side1) * (s-side2) * (s-side3))\n\nso if a triangle's sides are 3,4 and 5 cm.:\n\nsemi-perimeter = (3+4+5)/2 = 6\n\narea= root (6(6-3)*(6-4)*(6-5)) = root (6*3*2*1) = root (36) = 6 sq. cm.
322681
For the first, if one side is x, then another side would be 9-x (so that they would add up to 18).  The area is then\n\nA = x(9-x)\n= 9x - x^2\n\nTo find the maximum area, take the first derivative and set it to 0:\n\nA' = 9 - 2x\n0 = 9 - 2x\n2x = 9\nx = 9/2\n\nSince the parabola opens down, the maximum area happens when x = 9/2.\n\nSo if x = 9/2, the other side is 9 - 9/2, or 9/2 - it turns out to be a perfect square!  And the area is then (9/2)^2, or 81/4.\n\nFor the second, you already have an area and you need to find the perimeter.  So, lw = 36, and we're trying to find the smallest 2l + 2w.  Well, l = 36/w, so the perimeter can be expressed as\n\nP = 2l + 2w\n= 2(36/w) + 2w\n= 72/w + 2w\n\nWe can take a derivative and set it to 0 to find the minimum value here:\n\nP' = -72/w^2 + 2\n0 = -72/w^2 + 2\n72/w^2 = 2\nw^2 = 36\nw = 6\n\nSo if w = 6, then l = 6 as well, and the perimeter would then be 24.
1099275
Oh c'mon! Smiddy hasn't had geometry for 20 years! but this is basic. Pythagorean theorem states that on a triangle A-squared plus B-squared = C-squared where C is the hypotenuse. The hypotenuse of a triangle is the diagonal bit if it's two halves of a square and we know by definition that on a square A and B are equal. Therefore 2X squared = 8-squared or 2x^2 = 64\n\nx^2 = 32\nx = sq.rt 32\n\nthe area of the original square is x times x so you're back at 32 square centimeters
175713
Your question is too complicated to appreciate the math in it, which is supposed to be the real beauty in the first place. Anyway, I have tried figuring it out and come with equations. I do not even want to try and begin substitution and end up with more complications. I just thought once you saw the relation, you would throw those 10 points my way.\n\nAssuming a to be sides of the squares of the prism and b to be the length of the rectangles, we know that the sum of all dimensions is 14a+5b. Now I do not even want to get started why you brought it the story of the equilateral triangle. You can just say the sum  of the dimensions is 61. Anyway, since any side of the trianle, as it is equiangular hence equilateral, is twice 14a+5b and 122, the sum of the dimensions will be 61. So your first equation is 14a+5b=61\n\nYou know the volume of the prism is the length X width X heiht = a x a x b = (a^2)b This is equal to the surface area. The surface area is the sum of all the areas on the prism and that is a^2+4ab. (How about you calculate how that came up?)\n\nSo there is your second equation, (a^2)b = a^2+4ab So then from 14a+5b=61, express a in terms of b or the other way around and plug it into the second equation and experience brutal complications and figure out what the variables are. I think you know it is too easy to calculate each dimension and area after that.
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.
689719
The mode of presentation of data in the question is not sufficient to draw a single quadrilateral, there are two such quadrilaterals and hence two different answers.\n\nFIRST ANSWER: \nEqual sides i.e. sides of length 5*sqrt(2) are adjacent.\nLet the quadrilateral be ABCD. Then AB = BC = 5*sqrt(2), CD = 6, <DAB and <ABC are right angles. Drop a perpendicular to side AD from C intersecting AD at P.\n\nTotal area = area of rectangle ABCP + area of triangle CPD\n= [5*sqrt(2) x 5*sqrt(2)] + [0.5*{5*sqrt(2)-6}*5*sqrt(2)]\n= 50 + 3.7868 sq units\n= 53.7868 sq units\n\nSECOND ANSWER: \nEqual sides i.e. sides of length 5*sqrt(2) are opposite to each other.\nLet the quadrilateral be ABCD. Then AB = CD = 5*sqrt(2), BC = 6, <BCD and <ABC are right angles. \nClearly, quadrilateral ABCD is a rectangle. \nSo area = product of lengths of adjacent sides = AB x BC = 5*sqrt(2) x 6\n= 42.4264 sq units
534524
A nonagon can be broken into 9 isoceles triangles.\nThe vertex angles will be 360/9 = 40 degrees.\nThe two sides adjacent to the vertex have length 1 meter\nEach of the 9 triangles can be split into 2 right triangels with angles of 20 and 70 degrees.\nThe side opposite the 20 degree angle is sin(20) meters.\nThe side adjacent the 20 degree angle is cos(20) meters.\n\nEach of those triangles will have area (1/2)sin(20)cos(20 m^2\n\nThe total area of the 18 triagles combined is 9sin(20)cos(20).
568015
You add the areas of the six sides.\n.5 x .5 = .25 cm squared    (that's the area of one side)\n.25 cm squared x 6 sides = 1.5 cm squared
721808
If you are trying to make a triangle out of three equal lengths, all you do is devide 180 by 3.....and taddaaa you get your angles.\n\nIf does not matter how long each side is.
572270
triangle area = 1/2(height * width)\n\ncircle area = pi * radius^2