How to find the length of the side of a pentagon inscribed into unit circle?
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.
1031322
Perimeter of a circle = 2 *pi* radius or = pi *diameter\n\nFrom your picture the diameter of that semicircle = 8 \n\nso the perimeter of the circle = 8 * pi\n\nperimeter of semicircle (half circle) = 8*pi / 2 = 4* pi\n\nAlso, you can find the radius of the circle = diameter/2 = 8/2= 4\n\nThe Width of that rectangular = 1+ radius = 1+4 = 5\n\nThe two bottom short edges = the length of the rectangular - diameter of the semicircle = 10-8=2\n \nThe perimeter of the shaded area = Length of the rectangular + 2 * width  +  two bottom short edges + perimeter of the  semicircle = 10+2*5+2+4*pi=22 + 4* pi\n\nthe answer is a.\n\nHave fun !
1062497
What everyone has said is correct, but look at it this way: Consider a square. We know from the theorem of pythagoras that the ratio of the hypotenuese to the length of one side = sqrt(2) or, if you like, the ratio of one side to the hypotenuese = 1/sqrt(2). These constant ratios are independent of the size of the square. We have the ratio of the perimeter of the square to the hypotenuese = 4/sqrt(2) = 2*sqrt(2), which is independent of the size of the square.  Similarly the ratio of the circumference of a given circle to the longest chord(the diameter) has an exact value, and similarly this value is independent of the size of the circle. As has been mentioned, for any regular n-gon(square, pentagon, hexagon etc) the ratio of the perimeter to the length of the longest line segment that can be drawn within the n-gon will have a specific value depending on n, and this value can be determined. As we let n approach infinity, the n-gon approaches a circle and the ratio of the perimeter of the n-gon to the longest "chord" approaches a specific limit and this limit is pi, the ratio of the circumference to the diameter.  And, as mentioned above, analytic techniques can (and have) determined the value of the constant pi to a trillion or so digits.\n\nFootnote: SK, determining the limit of a series or even that the limit exists does not necessarily require calculus.\nFor example, the sum of a geometric series such as Sum[1/(2^n)] = 1/2 + 1/4 + 1/8+....+1/2^n is proved without calculus to be 1- 1/2^n and the limit of it (without calculus) as n---> inf. = 1. Check out the below link regarding pi.\n\n http://mathworld.wolfram.com/ArchimedesAlgorithm.html
237156
Pi * r^2\n\n(Pi R Squared)\n\nwhere R is the radius (which is also equivalent to half the diameter)
950225
Let a side of the square be 'x' and a side of the pentagon be 'y'.\nFirst, is the pentagon a regular pentagon or a pentagon drawn at random? If it isn't regular, then the ratio is impossible to determine as each side of the pentagon can be different. So I'll assume that it's regular. In a regular pentagon, all the sides are equal.\nPerimeter of square = 4x\nPerimeter of regular pentagon = 5y\nFrom question,\n4x = 5y\n4x/y = 5\n4(x/y) = 5\nx/y = 5/4\nThe ratio of a side of a square to that of a side of the pentagon is 5:4.\nA side of the square must be larger than a side of the pentagon if the perimeters are equal.
234891
I'm assuming that you are going to wrap the sheet to form a cone. \n\nTo simplify things, let x be the radius of the semicircle.\nThe circumfrence of the full circle is 2(pi)x, so the length of the semicircle arc is (pi)x. When you wrap the semicircle to form a cone, the base of the cone will have circumfrence of (pi)x, so its radius is x/2. The cone you get will have a sloping side of length x.\n\nThe volume of the cone is (pi)(r^2)h/3.\n\nThe height of the cone is third side of the triangle formed by the sloping side of the code (length x) and a radius of the base (x/2), so its length is sqrt(x^2 - (x/2)^2) = (x)sqrt(3)/2.\n\nSo the volume is ((pi)((x/2)^2)(x)sqrt(3)/2)/3 = \n\n((pi)((x^3)sqrt(3))/24 \n\nSubstituting x = 140mm  = 14 cm you have:\n\n((pi)((14^3)sqrt(3))/24 = 622.1 cc\n\n(Check arthmetic)
566727
I'm afraid that with the given information, this would be impossible to determine. The only thing one could conclude from this information is that this polygon has an even number of sides. \n\nIt could be a square with sides of 3cm each\nIt could be a hexagon with sides of 2cm each\nIt could be an octagon with sides of 1.5cm each\nIt could be a decagon with sides of 1.2cm each\nIt could be a 12-gon with sides of 1cm each\n\nThe list is infinite unless you are given more specific information, such as the length of each side or the angles within the shape.
903414
Because the diagonal creates two right triangles you can use the Pythagorean theorem to find the length of the other side:\n\n5^2 +x^2 = 9^2\n25 + x^2 = 81\nx^2 = 56\nsqrt of 56 = 7.48\n\nNow that you have the length of the other side you can use the formula    area = length x width:\n\n5 x 7.48 = 37.4 feet
831242
This is just the inside of the unit circle including its circumference.\nThe unit circle (circle at the origin with radius=1) is\nx^2 + y^2 = 1\nAnd you want x^2 + y^2 ≤ 1\nso its the circle plus its inside.\n( if it had been x^2 + y^2 ≥ 1 it would be the circle plus everything outside )\nIf you wanted x in terms of y, you just get |x| ≤ √(1-y^2) .
686717
base area depends on what pentagonal figure you have as base. The link in source will help.
890883
1)One leg of a right triangle is 12 cm long. The diference between the other two sides is 2 cm.\nFind the sides of the triangle. (Pythagorean Theorem)\n\nWhat we can take from this problem is that one LEG is 12 cm long, while the other LEG is unknown (we will call it X) and the HYPOTENUSE -- the longest side of a right triangle -- is also unknown (we will designate it X+2, since it differs 2 cm from the unknown leg).\n\nSetting up Pythagorean Theorem as always:\n\n "Leg squared" + "Leg squared" = "Hypotenuse squared"\n\n  12^2 + x^2 = (x+2)^2\n\nExpanding:\n\n  144 + x^2 = x^2 + 4x + 4\n\nCollecting like terms:\n\n  4x = 140\n\nDivide by 4:\n\n   x = 35\n\nThe unknown leg is 35 cm, while the hypotenuse is 37.  Thus, we can also call the triplet (12,35,37) a "Pythagorean Triple" (of which there are infinitely many!).\n\n\n2) The hypotenuse of a right triangle is 58 cm long. The diference between the other two sides\nis 2 cm. Find the sides of the triangle. (Pythagorean Theorem\n\nIn this case, we know the hypotenuse, but do not know the lengths of the legs.  For convenience, we will again use the expressions X and (X+2) for each side's length.\n\nBy P.T.:\n\n x^2 + (x+2)^2 = 58^2\n\nExpanding:\n\n x^2 + x^2 + 4x + 4 = 3364\n\nCollect like terms:\n\n 2 x^2 + 4x - 3360 = 0\n\nDivide by the GCF (2):\n\n  x^2 + 2x - 1680 = 0\n\nCan we factor this trinomial easily?  We could, if we knew integer factors of -1680 (one positive, one negative) that combine to make 2.\n\nLuckily, -1680 can be factored into -2*2*2*2*3*5*7, and these can be grouped into two sets:\n      2*3*7 = 42\n -2*2*2*5 = -40\n\nFactor:\n\n (x+42) (x-40) = 0\n\nAlthough two solutions come from these factors (-42 and 40), we may only consider the positive answer, since lengths of triangle's legs are always positive!\n\n Therefore, x=40 (Leg 1), and Leg 2 is 42 cm.
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
634060
The length of one side of the rectangle can be related to the Circumference of the circle.\nTo get the perimeter of a circle, called the circumference of the circle, one simply has to multiply 2 times pi (about 3.14) times the radius of the particular circle. \nThe circumference of a circle can be calculated from its diameter using the formula:\nc = πd\nOr, substituting the radius for the diameter:\nc = 2πr\nWhere r is the radius and d is the diameter of the circle, and π (the Greek letter pi) is the constant 3.141 592 653 589 793...
437123
Looks like you're missing a few things here.  To find an arc length, you need a radius along with the measurements you gave.  If you had a length of one of the sides, say AP, then you could find the arc length by taking a ratio of that length compared to the circumference of a circle:\n\narc length / circumference = angle / 360\n\nIf you solve for the arc length and plug in your known values, you'd have\n\narc length = 2*pi*AB * 25 / 360\n\nFrom this, you can see that angle CPB is irrelevant.\n\nAlso, it's not clear that this is a circle.  You list two angles, and the only thing we can figure is that they both share a vertex at P.  It's not clear if A is between B and C or outside of it.  But again, it doesn't really matter how angle CPB plays into this, since the arc length is really a function of the angle it contains and the radius of the circle.
336458
the sides are 2 units (3-1=2) \nand the ends are 10 units. (6-(-4)=10)\n\narea = side times end = 2*10 = 20\nperimeter = 2 * side + 2 * end = 2*2 + 2*10=4+20=24
942591
Chord length = R (see dimensions of rectangle ie flat part of slice)\nSo the size of the angle (θ) subtended at centre of circle is 60° as all sides of the triangle so formed = R.\nSo θ = π/3\n\nSo length of side of rectangle forming curved surface = length of Arc\n= Rθ\n= πR/3\n\nAlso area of segment at either end = ½R²(θ - sinθ)\n= ½R²(π/3 - sinπ/3)\n= ½R²(π/3 - √3/2)\n\nSo total surface area = Area of curved part + area of flat rectangle + 2 * area of end segment\n= πR/3 * 4R + R*4R + 2 * ½R²(π/3 - √3/2)\n= R²(5π/3 + 4 - √3/2)\n\nVolume = Area of segment * height\n= ½R²(π/3 - √3/2) * 4R\n= 2R³(π/3 - √3/2)
650879
Use string, paper, and a ruler.  The string can represent the perimeter, and you can draw a grid on the circle to represent the units of area.  Just don't get too in depth on the origins of pi.