Find the area of a regular decagon with sides "x" and an apothem of "y"?
608909
heck no. its the summer.
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
149999
A regular n-gon has the centers of its inscribed and circumscribed circles the same point. If you draw segments from this center to each of the vertex, you will get n number of radii and n number of triangles, which are all congruent. Using derivation you will arrive at the conclusion that\nA = .5aP or A = .5ans where\nA is the area, a is the apothem, s is the side and P is the perimeter\na =2A/ns\nThe radius is acquired by\nsin [180degrees(n - 2)/n] = a/r\nr = a/sin[180degrees(n - 2)/n]
342990
Take the derivative of each equation. Subtract one from the other. That'll get you the area enclosed between the two.
496924
Well, you know that each corner of the octagon lies on the circle (by definition). So, if A and B are the names of two adjacent corners, and C the center of the circle, you also know that AC=14.15, and BC=14.15. \n\nAnd you also know that a regular octagon will have its corners equally spaced around the circle. Since a circle is 360 degrees, the AC/BC angle is 360/8 = 45 degrees.\n\nFrom there you can easily calculate the length of AB using simple trigonometry, which is probably the goal of the exercise :-)
875314
Fisrt divide the regular octagon into 8 isosceles triangle. The resulting vertex angle would be 45 deg.\nDivide the isoscles into congruent right triangles with an angle of 22.5 deg, 67.5 deg and 90 deg.\nThe length of the sides opposite 22.5 deg would b equal to\nhypotenuse (radius of the circle) times sin 22.5 deg.l\nThat would be 6.046398231 cm.\nDoubling this value would result to the side of the octagon,\nand that is 12.09279646.\nMultiplying the side by 8, we get the final answer or the perimeter of the regular octagon inscribed in the circle.\nthe answer is\n96.7423717 cm.\n\nYou my use this formula for any regular octagon inscribed in a circle with radius x.\nP = 16x sine 22.5\nContact me if this is wrong.
595452
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.
117539
A trapezoid has 4 sides but you talk about 12 and 16 bases - I assume you mean 12 and 16-sided polygons. The question can be answered, as stated, if we assume that the sides of each polygon are the same length - if the sides are not the same length then the question can't be answered until you tell us the lengths of the sides. Polygons with sides that are the same length are called "regular polygons."\n\nAnswer:\n\n1) The area of a 12-sided regular polygon inscribed within a circle with a radius of 1 is 3\n\n2) The area of a 16-sided regular polygon inscribed within a circle with a radius of 1 is 3.0614674589207183
878030
How to find the area of an octagon:\n\nMeaning of words\n\nAn octagon is a polygon having eight sides.\nA polygon is a closed plain figure bounded by three or more straight sides that meet in pairs in the same number of vertices and do not intersect other than at these vertices.\nVertices is the plural of vertex.\nVertex means the point opposite the base of a figure or the point of intersection of two sides of a plain figure or angle.\n\nDifferent methods or formulas are used for calculating the area of polygons. The following method is used for calculating the area of an octagon.\n\nFormula: 2 × r2× 1.4142135 (the last number is the square root of 2)\n\nr = radius\n\nThe radius is a straight line from the centre to any point on the outside of the shape.\n\nFind the radius of the octagon, then perform the formula.\ne.g. if the radius of an octagon was 4 cm, then 2 × 42× 1.4142135 = 45.25 cm2, which is the area.
327427
Each side would have length sqrt(3), so the area would be 3 m^2.
1144014
draw a picture of the graph over the specified interval.  divide the graph into vertical rectangles.  the area under the curve will be the sum of the areas of the rectangles.  saying that youve made "n" rectangles, the only way to get the exact area is to find the sum for the limit as n approaches infinity.  work from that concept.  if you already knew that, or need more help or something, email me and i can give you more steps and a much more detailed description.  that would take a really long time though so im gonna see if this helps first.
636579
Okay, I got something really different, but it might be wrong, I used tangents and stuff to figure it out, and I assumed a dodecagon had 12 sides, but I might be wrong. Anyways, I got 38.4 for the perimeter, and 115.2 for the area.
548292
Assuming that an area may be dissected into triangles, then the area of each triangle is given by \n\nArea = (s(s-a)(s-b)(s-c))^(1/2)\n\nwhere s = (1/2)(a + b + c)\n\nand a, b, and c are the lengths of sides of each triangle -\n\nThe sought for area is (finally) the sum of all the areas of the triangles.\n\nGiven sufficient patience, any arbitrary area will be found.\n\n(Drawing a diagram while slogging through the exercise and looping around the separate triangles in the same direction can yield a shortcut whereby all of the internal perimeters eliminate each other, leaving only the perimeter of the original figure.)
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)
920664
area a to b is 1/2..\n\nhow? assume a side of the inner square as 1 so that its area would be also 1, by pythagorean theorem, half of the side of square b is equal to squareroot of 1/2, so the side of b is 2*squareroot(1/2), that makes the area to be equal to 2, getting the ratio of areas, that is 1:2 n_n
583176
A trapezoid is a 4-sided figure with one pair of parallel sides. 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.
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.