if a rhombus is inscribed in a  circle what must be true about the rhombus?
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If a rhombus is inscribed in a circle, then the rhombus is a square.  (Remember that a square is a rectangle with congruent sides, or a rhombus with congruent angles.)\n\nIf a trapezoid is inscribed in a circle, then the trapezoid is an isosceles trapezoid.
1076251
It is never-ending, continuous.  It is a symbol of rebirth, of the turning of the seasons, of life.  There are many special things about the circle.  As a sphere, it is not as subject to the elements as other 3-D shapes.  Wind and water go right over or around it.
605893
U mean Plato's CUrsed Triangle?,,\nWell, check out the gradient of the two triangle, ITS different!.\nThe difference of the gradient makes a whole in the next triangle
64159
It's debatable.  Here's an interesting column that addresses the question:\nhttp://mathforum.org/library/drmath/view/57188.html
418892
Since O is the center, this problem becomes fairly straightforward.  First, assuming <BOA = 94 degrees (not sure which of the four angles at O is 94 degrees, but say it's <BOA.  This means that the arc from B to A is also 94 degrees.  And since the chord from A to C goes through the center, it means it is a diameter.  This means that the arc from A to C is 180 degrees (half of the circle).  So that means that the arc from B to C has to be 86 degrees (180 - 94, because part of that arc from A to B is 94, the rest must add up to 180).  So now you have to use what your teacher told you about when the angle is on the circle.  Namely, that the angle on the circle (the inscribed angle) is half of the arc measure (intercepted arc).  So if the arc measure is 86 degrees, the angle must be 43 degrees.\n\nEdit: I just saw that it was <COD, not <BOA, but they are congruent because the are vertical angles, so nothing changes.
757335
x=r(t-sint) and y= r(1-cost) are the parametric equations of the cycloid where r is the radius of the rolling circle and t is the angle through which the radius of the circle has turned from its starting point. The angle t must be measured in radians.\n\nFor simplicity let r =1 so that we have the unit circle.\nwith t=60 = pi/3 we get\nx = pi/3-sin(pi/3) = pi/3 -(sqrt 3)/2\ny = 1-cos (pi/3) = 1- 1/2 = 1/2\n\ndy/dx= sint/(1-cost)\n=[(sqrt 3)/2]/(1/2) = 3^0.5 (at the point where t= 60= pi/3)\n\nFrom this , you can find the slope at any point define by the parameter t.\n\nb) The tangent is horizontal when t= pi and is perpendicular at the cusp which occurs at 2pi and multiples thereof. The maximum value  is 2r (or 2 for the unit circle). The length of the base of the arch is 2pi *r or just 2pi for the unit circle.
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It depends how you draw out the rhombus, but what ever area it covers, It will be less then the area of the page.
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Let a, b, and c be the lengths of the legs of a triangle opposite angles A, B, and C. Then the law of sines states that\na/(sinA)=b/(sinB)=c/(sinC)=2R,\n\nwhere R is the radius of the circumcircle.\n\nIn other words,\n\nIn any triangle, the ratios of the sides to the sines of the opposite angles are equal.
554629
never. in order for it to be tanget to the circle, it needs to be outside the circle. if the vertex is on the circle, it CAN have 1 tangent with an angle of 180 degrees
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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
613272
could be circle because it doesn't have any sides or angles\n\ncould be rectangle because it is not proportional (side-wise.. and only if the triangle is equilateral)\n\ncould be triangle because if all shapes are placed right-side and divided into four equal parts, like a cross-hair, the triangle is the only one without equal portions.
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Assuming that A, B, C are all on the same line.\n\nLarger circle:\nRadius = AC = 6\nArea = PI * r^2 = 36*PI\n\nSmaller circle:\nRadius = AC - AB = 6 - 2 = 4\nArea = PI * r^2 = 16*PI\n\nShaded area:\nArea = (Large A) - (Small A)\n     =  36*PI - 16*PI = 20*PI = 62.8 units^2
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A circle is the locus of a point which moves in a plane in such a way that its distance from a given fixed point is always constant.\n         The fixed point is called the centre & the constant distance is called the radius of the circle.\n           \n          A circle with centre O & radius r is denoted by C(O,r) !!!
446128
This depends on how the line is going across. If the line is horizontal and divides the circle in half, then the symbol is a 'theta', which usually stands for an angle. If it is from lower left to upper right and extends beyond the circle both ways, then it could be the symbol for the empty set or (in some books) zero. If it is close to vertical and extends beyond the circle in both ways, then it is a 'phi' and again usually represents an angle.
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An equilateral parallelogram:  Rhombus\nAn equilateral and equiangular parallelogram:  Square
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Jan 17 Note: If this is hard to understand, after sleeping on it, I may have an easier construction.\nIf you need further explanation, add another note to your question.\n\n-----------\n\nI think I have a solution.\nI'm making the assumptions that the Babylonians didn't have an acurate value for pi and used the value three. If I do that, every thing falls out.\n\nDraw a cirlce. Mark off 2 arcs. Make one the top portion of a circle say from 10 o'clock to 2 o'clock, the other the matching arc from 4 o'clock to 8 o'clock. Note: These arcs are just to get a representation of the problem and are not an acurate length.\n\nNext draw the two chords for the arcs, that is, the straight line from 10 to 2 and the other from 8 to 4.\nNow draw in the diameter of the circle from 12 o'clock to 6 o'clock.\nThe two chords cut off pieces on the ends of the diameter. Label these as length 2 (2 is the perpendicular) "Double 2 and get 4" That's the length of the two end pieces of the diameter.\n"Take 4 from 20 and get 16" This is where using 3 as a value for pi comes in. If pi 3, then a circle of circumference 60 will have a diameter of 20, so let the length of the diameter be 20. Taking 4 from 20 gives you the length of the diameter that is between the 2 chords, 16.\n\n"Square 20 get 400, square 16 get 256, take 256 from 400 get 144. Find the square root of 144. 12, the square root, is the chord."\n\nWe're obviously doing some Pythagorean theorem here so where is the triangle? \nDraw the diameter of the circle from 8 o'clock to 2 o'clock, that's the hypoteneuse as has length 20.\nThe distance between the two cords is 16. That's also the distance from 8 o'clock to 10 o'clock. Draw that line and label it 16. (Remember the drawing isn't accurate just enough to show us how things are layed out).\nThat leave the 3rd side of the triangle as the chord between 10 o'clock and 2 o'clock. \n\n20^2 - 16^2 = 12^2  so 12 is the length of the chord.\n\nHow does that sound?\n\nIn modern English, the statement of the problem would be something like this:\n\nLet AB be an arc of a circle of circumference 60. Let X be the chord connecting A and B.\nLet Y be the perpendicular bisector of the chord. If the distance from X to the circle along Y is 2, find the length of X. (You may assume that pi = 3 for the purposes of solving this problem.)
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If the figure is a square or rhombus, p = 4L\n\nIf it is an equilateral polygon with n sides, it is p = nL