A trapezoidal pyramid V=3168 cm cubed and height=36 cm. The lengths of the two bases of the trapezoidal base are 20 cm and 28cm. What is the height of the trapezoidal base?
138162
Volume of pyramid = 1/3 x base area x height\nArea of trapezoid = 1/2 x sum of parallel sides x height\n1/3 x [1/2 x (20 + 28) x h] x 36 = 3168\n...\nh = 11 cm.
485983
You may want to be more specific on this one... are you trying to find the largest possible area? You can inscribe lots of different trapezoids in a circle. \n\nAnd which theta are we talking about? The angle that the base makes with the circle?\n\nTry that one again,\n\nUMRmathmajor
134116
See previous answer--\n\nGives vertical height as rt27\n\nTherefore 1/2*6*rt27 is area of one side\n\nBase is 6*6 = 36
282832
If you picture your trapezoid with the shorter parallel side and the two legs as sides 4, then the only unknown is the the long parallel side.  That side is at least 4, but with two extra parts on either side (if you drop perpendiculars from the short to the long side, you'd have a rectangle and two triangles).  So the long side is 4 + 2x, where x is the length of those two parts, which is also a leg of a triangle that we created in dropping those perpendiculars.\n\nThe height of the trapezoid is the same as the other leg of the triangle.  We don't know the length of either leg, but we do know that the hypotenuse is 4.  So through the Pythagorean theorem, we have\n\n4^2 = h^2 + x^2\nh^2 = 16 - x^2\nh = sqrt(16 - x^2)\n\nBack to the trapezoid - we know the area of a trapezoid is the average of the bases times the height, so we have\n\nA = (1/2)(4 + 4 + 2x)(sqrt(16 - x^2))\n= (4 + x)(sqrt(16 - x^2))\n= (4 + x)(16 - x^2)^(1/2)\n\nTo find the maximum area, take a derivative, set it to 0, and solve for x:\n\nA' = (4 + x)[(1/2)(16 - x^2)^(-1/2)](-2x) + [(16 - x^2)^(1/2)](1)\n= (-4x - x^2)(16 - x^2)^(-1/2) + (16 - x^2)^(1/2)\n\nIf we set this to 0, then\n0 = (-4x - x^2)(16 - x^2)^(-1/2) + (16 - x^2)^(1/2)\n-(-4x - x^2)(16 - x^2)^(-1/2) = (16 - x^2)^(1/2)\n4x + x^2 = 16 - x^2\n0 = 2x^2 + 4x - 16\n0 = x^2 + 2x - 8\n0 = (x + 4)(x - 2)\nx = -4 or 2\n\nSo your long side has to be 4 + 2(2) = 8 to maximize the area.
311676
Volume is for a solid, and a pentagon is a flat shape.  Do you mean area?  And have you given us all the information.  This obviously isn't a regular pentagon because two of the sides are not equal...  What are we missing?\n\nIf you mean a pentagonal solid, I suppose you could mean the area of the base is 60 sq cm and the height is 8 cm?  In this case it is just the area of the base times the height, or 480 cubic cm.  Don't forget to express the Volume in cubic cm. (cm^3).\n\nOn the other hand, if you are given the *perimeter* of the base being 60 cm and you assume a regular pentagon, then you have a different question....\n\nHere each side is 12.  The formula for the area of a regular pentagon is:\n\nA = 1.720s^2\n\nPlugging in 12 cm, you get:\n\nA = 1.720(144) = 247.68 sq. cm.\n\nMultiply the area of the base by the height (8 cm) to get your final volume:\n\nV = 247.68 * 8 \nV = 1981.44 cubic cm \n\nAgain don't forget the cubic on the units.
3363
Find the dimensions of the base:\n36 - 4 - 4 = 28\n25 - 4 - 4 = 17\n\nSo the area of the base = 28 x 17 = 476\n\nThe height = 4\n\nSo the volume = 4 x 476 = 1904
592325
The Lourve pyramid is made of 4 triangles and a square base.\nSo the surface area of this pyramid\n= 4*(b*h)/2 + l^2\n= 2lh + l^2\nwhere the base of the triangle, b = length of a side of the square and h is the perpendicular height of the triangle.\n\nVolume = (1/3)(base area)(height)
638700
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
616157
To find the area of a trapezoid, take the sum of its bases, multiply the sum by the height of the trapezoid, and then divide the result by 2.  Then, multiply by the depth (here, .5 for 1/2 foot deep) to get the volume.
358534
Since each has eight edges it is a square pyramid\nAnd also the Lemght if the edge is given as 8 unit\nNow the altitude or axis of pyramid is sqrt(8^2+128)\nV(Volume)=1/3 * a^2 * height\nVolume of two pyramidas will be twice this that is 2V\nNow since we know the volume and base edge(a=8)\nwWe can find the axis by volume of pyr=1/3 * a^2 * axis\nconsidering a right triangle with axis and half of base making the right angle and slant height as hypotenuse you can find slant height using H^2=B^2+ALT^2\n\n\nWhile doing the problem dont calculate the values You can just fid an approximate value ad cancell terms thus making ita lot simple
259440
Base times height gives the volume of the prism.\nVolume of prramid= 1/3 x base x height =1/3 x 24=8 cubic inches.
647575
14" high, wide, and deep!  (14 is the cube root of 2744.)\n\nI just decided to add something more to my answer.  Why?  It's because just giving you the correct answer will help you to do your homework tonight, but it won't help you to learn anything!  But if I tell you how to go about solving math problems like this in the future, then you can ace your SATs, right?  Right!\n\nOK. So you know that you've got a cube.  That means that you have a geometric figure that's composed of squares, and squares--by definition--all have the same lengths on their sides.  That makes your problem way simpler, because you know you only have to calculate one side to get your answer.  You with me so far?  Good!  So you know that with a cube, you've got three dimensions--the height, the width, and the depth--and all of the dimensions will be exactly the same.\n\nWhat if you don't know how to calculate a cube root, and you also don't have a calculator?  Y'know what?  ME TOO!  I never learned how to do cube roots (or even square roots) in school, and did really well on my SATs, anyway.  What you have to remember is "factoring" and "prime numbers."  Since 2744 is an even number, you know that you have to have some 2's as your prime numbers.  Actually, you've got 3 of them.  When you take out the three 2's (which is 2X2X2, or 8), you've got 343 left.  So then you look at 343, and you guess what prime number would go into it.  3?  Nope!  3 won't go into it evenly.  And neither will 5.  So your next prime number to try is 7.  There are three 7's in 343 (7X7X7).  So there's your cube root, which is 2X7, or 14.  14 to the power of three equals 2744.  Voila!  You are a math genius, and you didn't even know that you had it in you!
663471
base times the hieght divided by 2\n\nthe base is the bottom or the triangle and the high it the top of the triangle to the middle of the base
333284
Question 20906\nSubmitted on 5/28/2002\n\nFind the area of a nonagon inscribed in a circle of radius 24cm.\n\nSolution 20906\n\nThe nonagon can be divided into nine congruent isosceles triangles. \nThe lengths of the sides are 24 cm and the vertex angle is 360/9 = 40.\nThe base of one triangle is\nB = 2 * 24 * sin (20)\nThe height of the triangle is\nh = 24 * cos (20)\nSo the area of one triangle is\nA = (1/2) * 2 * 24 * sin (20) * 24 * cos (20)\nA = 24^2 sin(20) cos(20)\nAnd the area of the nonagon is\n9 * 24^2 sin (20) cos (20)\n= 1666.1\n\n\nSubstitute in your 1 meter for the 24 centimeters and you'll get your answer.  I believe the answer should come out to somewhere around 3.353 m^2
133189
square pyramid...if square as base\n\ntriangular pyramid...if triangle as base\n\nBasically just look at the base...and add pyramid to it.\n\nEx. 5 sided regular polygon base...pentagonal pyramid.\n\nIf it isn't a regular polygon...look it up...or just leave it as pyramid I guess. LOL!\n\nTetrahedron...sound molecular...like the shape of methane molecule. I'm thinkin' this is geometry...so stick with the simplistic "pyramid".\n\nOH...BTW...if it only has one base...and is composed of multiple triangular sides...then it's a pyramid. If it has 2 bases...and any other figure for it's sides...then it's a prism.
234264
1(A) 1 L = 1000 cm^3 therefore 2L = 2000 cm^3\n1(B) 1 L = 1 Kg therefore 2L = 2Kg\n\n*******************\n\n2)\nTwo thirds less than half of what number is five sixths\n(1/2)X - (2/3) = (5/6)\n\nmultiply every term by 6\n3x - 4 = 5\n\nadd 4 to both sides\n3x = 9\nx = 3\n\n*******************\n\n3)\nvolume of a cube = side^3 = 12^3 = 1728 cubic inches\nvolume of square pyramid = (1/3)(b^2)(h)\n where b = 12 and h = 8 (because the total height = 20)\n\n v = (1/3)(144)(8) = 384 cubic inches\n\n total volume = 2112 cubic inches\n\n*******************\n\n4)\nwe do not know the relation ship between the segments, if they form a line, if they form a right triangle, or if they lie on top of each other....\n\nif they form a line segment, where "b" is between "a" and d", then the total meximum length is 12 + c. if they lie on top of each, then they have a minimum distance of |12 - c| (absolute value of 12 - "c")\n\n*******************\n\n5)\nyou would need (1 x 10^3 or 1000) to form 1 meter multiplied by 1000 to form 1 KM\n1000 x 1000 = 1,000,000 or 1 x 10^ 6\n\n*****************\n6)\nx^2 + x  : GCF = x\nx(x + 1)\n\n\n12m^2n^3 + 18mn^2 - 24m^2n^3 \n12m^2n^3 and 24m^2n^3 are like terms\ntherefore\n\n= -12m^2n^3 + 18mn^2\n\nGCF = 6mn^2\n\n6mn^2(-2mn + 3) or 6mn^2(3 - 2mn)\n\n*****************\n\n7)\n-2 2/3w - 1 1/3 = 4 \n\nadd 1 1/3 to both sides\n\n-2 2/3w = 5 1/3\n\n= -(8/3)w = (16/3)\n\nmultiply both sides by -(3/8)\n\nw = -2
239074
you can express the area as the sum of the areas of the 4 inner triangles which are each 1/2 base x height.  Each base is an edge of the quadrilateral.  The height can be determined in 2 ways and you will need both to solve the system of equations:\nfor example: to determine the height of APD, recognizwe that the height line splits APD into 2 right triangles.    Call the point where the height line intersects AD point E. The edge AD will be split between the 2 tringles proportionally:  AE = 24/69  AD and DE = 45/69 AD.  Now you can determine the height in terms of AD using pythagorean theorum.  Go from there...\n\nTry this site...