geometry problem help?
489663
Reviewing the facts:\n\n1.  You're constructing a pentagon and two trapezoids by using the center of the square.  Since the sides are 1, the short parallel side of the trapezoid is 1/2.\n\n2.  Since the two trapezoids are congruent, the short parallel side is midway in the square.  So the height of each trapezoid is 1/2.\n\n3.  Since there are three areas now, and each area is equal, the area of the trapezoid is 1/3.\n\nThe area of a trapezoid is A = (1/2)h(b1 + b2).  We have everything we need now to solve for b2, the longer parallel side:\n\nA = (1/2)h(b1 + b2)\n1/3 = (1/2)(1/2)(1/2 + b2)\n1/3 = (1/4)(1/2 + b2)\n4/3 = (1/2 + b2)\nb2 = 5/6
970422
The solution is too convoluted to put it here without diagrams and a good equation editor.  If you wish, I can email a Word document detailing the solution.  I figure that the cone is (1 + 2*sqrt(2/3) + sqrt(3))*r (about 4.365 r) high and has a base radius of ((sqrt(1/2) + sqrt(4/3) + sqrt(3/2))*r or about 3.087.
932841
This might help --> http://www.uwm.edu/~ericskey/TANOTES/Geometry/node15.html
123707
Let:\n\nL = length of the rectangle\nW = width of the rectangle\nP = perimeter of the rectangle\n\nGiven:\n\nW = x\nL = 2x - 4\nP = 2L + 2W from geometry\n\nSolution:\n\nP = 2L + 2W\n64 = 2[2x - 4] + 2x\n\n2[2x - 4] + 2x = 64\n\n[4x - 8] + 2x = 64\n\n4x - 8 + 2x = 64\n\n4x + 2x = 64 + 8\n\n6x = 72\n\nx  =  72/6\n\nx  =  12 cm.\n\nTherefore:\n\nW = width = x = 12 cm. ANS.\n\nL = length = 2x - 4 = 2(12) - 4 = 24 - 4 = 20 cm. ANS.
337862
go do your homework by yerself next time.\n\nBut becos im feeling nice today...\n\na triangle = 1 triangle , duh\na quadilateral can be divided into two triangles with vertices (points) all at the points of the quadilateral\n5 sided --> 3 triangles\n6 sided --> 4 triangles\n....\n12 sided --> 10 triangles\n1 triangle -->180 degrees\n10 triangles --> 1800 degrees
802049
The answer is (D).\n\nThe perimeter is 9x (just add up the 4 sides 2x+3x+3x+x to determine this) -- which eliminates (B) and (C) as a possible answer.\n\nThe area is 6x^2 -- which eliminates (A) as a possible answer, leaving only (D).\n\nLooks like you have a trapezoid, where the base is 2x, the sides are 3x each, and the top is x.  Draw straight lines down from your top to your base, so that your trapezoid is in 3 pieces:  a right triangle with base = x/2 and hypotenuse 3x, a rectangle x by 2x, and another right triangle with base = x/2 and hypotenuse 3x.  The area of the inner rectangle is 3x^2, and you can put the two triangles together to get another x by 2x rectangle for another 3x^2 area, or a total area of 6x^2.\n\nI hope that helped.\n\n~ ♥ ~
87402
a: width\nb: length\nperimeter = 2 x (a+b)= 420 ---> a+b = 210\nwe need to find max (a x b)\n\na= 210-b ---> area = a x b = b x (210-b) = 210 - b2 = - b2 + 210b = - b2 + 2 x 105 x b - 105squared + 105squared\n= - (b2 - 2 x 105 x b + 105squared) + 105 squared\n= - (b - 105)squared + 11025\n\nSince (b - 105) squared is always >= 0,\n - (b - 105)squared is always <= 0\n  - (b - 105)squared + 11025 is always  <= 11025\n\nthe largest area will be 11025 squared feet. This happens when - (b - 105)squared = 0, which means b = 105.\n\na = 210 - b = 105\n\n\nthe shortest way is to say that a rectangle has the largest area when it's a square, so each side would be 420/4 = 105.
901368
You need to take triangles with leg length 2-sqrt(2) from each corner. That is about 0.586".\n\nThe length of each side of the octagon will be 2 * sqrt(2) - 2. This is about 0.828"\n\nThis was calculated by setting the length of each leg of each triangle to x.\n\nhypotenuse = remaining side of square\n\nx * sqrt(2) = 2 - 2x. And then solve for x.
405573
you need to use some calculus.\nget an expression for the area \ndifferentiate it\nthen find when it is a maximum.
792610
Try these;\nhttp://www.mathleague.com/help/geometry/geometry.htm\nhttp://www.coolmath.com/home.htm\nhttp://www.ics.uci.edu/~eppstein/geom.html
329126
You have to graph them and find the corner points.  One of the corner points will be the answer.\nI found the corner points to be\n(28,52) and (40,40)\nPlug each point into g = 3x+y \nFor (28,52): g = 3(28) + 52 = 136\nFor (40,40): g = 3(40) + 40 = 160\n136 is the smaller value, so (28,52) is the answer.
669681
1.  Area of rt triangle whose leg is 6ft and whose\n    hypotenuse is 10ft?\n\n    Area = 24 sq. ft.  \n\n  Explanation:\n\n    First we need to find the other leg's length...\n    Using Pythagorean Theorem,\n    \n    c**2=a**2+b**2 \n       \n    where c is the hypotenuse, a is one leg, and \n    b is the other leg.\n\n    So, 100 = 36 + b**2;  b**2 = 100-36 = 64; therefore b=8.\n    Now, Area = 1/2(base*height) since the base is one leg\n    (we'll say it is a) and the height is the other leg \n    (therefore we'll name it b) \n    the Area = 1/2(8*6) = 1/2(48)= 24 sq. ft.\n\n2.   The hypotenuse is approx. 4.9 meters.\n    \n  Explanation:\n     \n     Using Pythagorean Theorem: c**2=a**2+b**2 we find that\n     c**2 = (3.5)(3.5) + (3.5)(3.5)\n     c**2 = 12.25 + 12.25\n     c**2 = 24.25\n     c = sq. rt. (24.25)\n     c = 4.9244 meters\n     hypotenuse is approx. 4.9 meters.\n\n3. hypotenuse is approx. 20 inches long.\n\n  Explanation: Area = 1/2(x)**2 \n               where x is length of one leg.\n\n     x**2 = 2*Area;\n     x**2 = 2*98\n     x**2 = 196\n     x = sq. rt. (196)\n     x = 14 inches\n\n     Now, to find length of hypotenuse, we use the\n     Pythagorean Theorem:\n\n     c**2 = a**2 + b**2; since a=b this can be rewritten so\n     c**2 = 2(a)**2\n     c**2 = 2(14)**2 -- Remember 14 is the length of a leg.\n     c**2 = 2(196)\n     c**2 = 392\n     c    = sq. rt. (392)\n     c = 19.7989\n     c is approx. 20 inches long.\n\n\n4.   The researcher will climb approx. 10 meters \n     to reach the crack.\n\n   Explanation:\n      \n      Using the Pythagorean Theorem: c**2 = a**2 + b**2\n      If we set a = 24, then c**2 = (24)**2 + b**2,\n      but we also know that c + b = 36 \n      which means c = 36 - b.\n\n      So to solve for this we need to set\n      (36-b)**2 = (24)**2 + b**2\n      1296-72b + b**2 = (24)**2 + b**2\n      1296 - 72b = (24)**2\n      1296 - 72b = 576\n      -72b = 576 - 1296\n      -72b = -720\n         b = -720/-72\n         b = 10 \n\n      And now we have our 10 meters the researcher \n      must climb to reach the crack in the tree.\n\n\n6.    Approximately a 42 inch stick is the longest stick\n      that will fit in a box that is 24 in X 30 in X 18 in.\n\n  Explanation:\n\n      First use Pythagorean Theorem to find the diagonal\n      inside the rectangle of 24 in X 30 in.\n\n      c**2 = (24)**2 + (30)**2\n      c**2 = 576 + 900\n      c**2 = 1476\n      c = approx. 38.418 in. -- we will call it 38 in.\n\n      Now, use Pythagorean Theorem to find the diagonal\n      inside the "new and invisible" rectangle that is \n      38 in. X 18 in.\n\n      c**2 = (38)**2 + (18)**2\n      c**2 = 1476 + 324\n      c**2 = 1800\n      c = sq. rt. (1800)\n      c is approx. 42.43 in.  \n      \n      We will call "c" 42 inches long.
1017007
Ive noticed you were asking quite alot of these geometry questions. I think you should pay more attention in class and you wouldnt need to ask home work questions, and plus, just knowing the answer wont help, youll have to know how to do it, especially if this is for a final exam...
809433
1.rectangle\n\n2.trapezoid
1124064
Because you have a fixed height, h, of 30m, and the Volume  of a triangular prism is given by: V = h*A where A is the area of the triangular base, your problem reduces to maximizing A, given the perimeter of A = 210m\nAn equilateral will enclose the maximum area(see link below) So the length of each side will be 210/3 = 70 m. The area of an equilateral triangle =  S^2 x sqrt(3)/4 where S = side length. So, in your case the area = 70^2*sqrt(3)/4 = 2,121.18 m^2 and the maximum volume for your prism is 30m*2121.18 m^2 = 63,652.86 m^3\n\nFor a polygon, the general rule is: the 'closer' the shape of a polygon is to a circle, the greater the area of the interior of the polygon. This implies that a equilateral triangle, a square(see below), a regular pentagon; and in general a regular n-gon will enclose the most area. \n\nConsider a square with side length X, then the area of the square = X^2 and the perimeter is 4X. Let's keep the perimeter fixed at 4X and see what happens to the area if we let one pair of opposing sides of the rectangle be X-y and the other two be X+y (Note the perimeter will still be 4X). However, the area is now: (X+y)*(X-y) = X^2 - y^2. We see that if y>0 then area decreases, while the perimeter stays the same. We see that a square encloses the maximum area for a fixed perimeter. \n\nHowever, if you allow the height, h, to vary as well then you have 3D maximization problem; but the solution is not difficult in this case either. In general, right-angle prisms with regular polygon bases will have a maximum volume for a fixed perimeter or surface area\n\nhttp://mathforum.org/library/drmath/view/53789.html\n\nhttp://mathworld.wolfram.com/Prism.html\n\nhttp://mathworld.wolfram.com/EquilateralTriangle.html
247689
Use analytical trig (hint: sin function).  That should give you the answer you are looking for.
1137727
I like the joke!  To play along, I define:\nr = metric r  and \nd = metric d\n\nV = (pi/3)*(d^2)*(3r - d)\n\nThis will work for the tank being any fraction of a sphere, not just 1/2