What is the area of the smallest circle sector in the circle x2 + y2 = 4 that has corner points in \n(0,0) , (√3,1), (√2,√2)\nThe area is a rational number times pi
212308
Use A = 1/2(r^2)theta (radians).  The sector formed by the radius from the origin to (sqrt2,sqrt2) and the radius from the origin to (2,0) has an angle of 45 deg.  The sector formed by the angle of the radius from the origin to (sqrt3,1) and the x-axis has an angle of 30 deg.\n45 deg. - 30 deg, = 15 deg or pi/12 radians.  The area is (1/2)(r^2)pi/12 = pi(r^2)/24 for any sector with a central angle of pi/12.  Since r^2 = 4, A= 4pi/24 = pi/6.
243099
Imagine the square of string:\n\nWhat will you do to make it a circle? You would have to bend the sides to form an arc like the way you would bend an archer's bow. For it to become a circle, you have to bend the four sides equally so that the corners will no longer be present.\n\nSo your new circle now will have an area bigger than the square. If you overlap the 2 figures together, you would have a big circle and inside it is a square whose 4 corners touch 4 equidistant points on the circle.\n\nArea of the circle: 64/pi OR 20.3718
418063
Circumference\n\nC = 2 π r\n\nC = 2(3.141592654)(1)\n\nC = 6.283185307\n\nC = 6.28 Feet rounded to two decimal places\n\nThe answer is C - 6.28\n\n- - - - - - - - - - - - - - - -\n\nArea of a circle\n\nA = π r²\n\nA = 3.141592654 (1)²\n\nA = 3.14 Squared feet. Rounded to two decimal places\n\nThe answer is A - 3.14\n\n- - - - - - - s-
114598
Ok, If A is a point on the circle and C is the middle of the circle then AC is the RADIUS of the circle.  Which means any point on the circle to C WILL ALWAYS be 12.  So if the excess of the circle to B is 8, then Add the RADIUS to the distance to B, and we arrive at the result of 20. Which is AC + (Distance of circle to B)\n\n20
339128
determine the square of 6 - sqrt[(x-2)^2 + y^2]\n\n-->\n\n36 + [(x-2)^2 + y^2] - 12*sqrt[(x-2)^2 + y^2]\n= 40 + x^2 + 2x + y^2 - 12*sqrt[(x-2)^2 + y^2]
291604
A circle has 360 degrees. Now imagine that from the icosagon you can make 20 equally-sized triangles. The angle made by each triangle is 360/20=18 degrees. The other two angles of each triangle are both 81 degrees.\n\nThe basic formula for the area of a triangle is 1/2 base x height. How do we find these? Well, imagine that from each of the 20 triangles we make two additional right triangles. In doing this we know several things:\n\n-The hypotenuse is equal to 6.5 cm (the radius).\n-The angles of the smaller triangle are now 9 degrees, 81 degrees, and 90 degrees.\n-The sine of 9 degrees is equal to the opposite side divided by the hypotenuse. If we multiply this value by 6.6 we get the length of the opposite side ("1/2 base").\n-The sine of 81 degrees multiplied by 6.5 gives us the "height".\n-Once you compute each value (1/2 base and height) you multiply them by each other and you get the area of the triangle.\n-Multiply the area of that triangle by 20, since you have 20 of them. Voila.
167921
well that question cant be answered b/c only a circle has a radius nothing else
327684
the function will form an area under the curve (between the curve and the x-axis). if you rotate that area about the x-axis, it forms something like a cylinder with varying radii. if you slice that "cylinder" perpendicular to the x-axis, you see a series of circles with the radius equal to the y-value of the function at that point (x-value).\n\nthe area of that circle is π*r^2 with the r having a value as stated above. \n\nfor #1, r = cos(3x)  or\n\narea = A(x) = π*r^2 = π*[cos(3x)]^2\n\nthe volume = ∫A(x)dx \n\nwhere ∫ = integral\n\ntherefore, for #1\n\nV = ∫π*[cos(3x)]^2 = πx/2 + πsin(6x)/12 \n\nevaluating this integral for the limits (0 to π/12)\n\nV = [π^2/24 + π/12] - [0 - 0]\nV = [π^2/24 + π/12] = 0.673032 or 0.67\n\n--------------------------------\n\nradius = 1/x\n\nA = = πr^2 = π/(x^2)\n\nV = ∫π/(x^2) = -π/x\n\nevaluating at the limits (3.5 to 5)\n\nV = -π/5 - (-π/3.5) = -0.6283 + 0.8976 = 0.2693 or 0.27
230018
Area = Pi x R squared\nArea = Pi x (x - 6) squared\nArea = Pi x (x^2 - 12x + 36)\nArea = Pix^2 -Pi12x + Pi36\nArea = 3.14x^2 - 37.68x + 113.04\n\n(Using the approximation 3.14 for Pi)\n\nOf course, it has been a while and I'm a little rusty...
663945
One way to reason is this.  Let n be any even integer greater than or equal to 4.  Divide the circle into n equal wedges (by drawing n/2 diameter lines with angles between successive lines equal to 360/n degrees).  Now, cut the wedges out and make a "rectangle" with them: n/2 wedges with the points downward and the arcs on top of adjacent pairs of wedges touching each other, then the other n/2 wedges with the points upward and adjacent arcs touching, filling the spaces of the first n/2 wedges.  What you end up with is almost a rectangle with top and bottom each made of n/2 360/n-degree arcs, and the lateral sides slanting slightly, made from the long edges of the wedges on each end.\n\nYou can estimate the area of this figure (which is the same as the area of the original circle of radius r) by pretending it IS a rectangle.  The length is the circumference of the circle, or 2*pi*r/2 (since half the circumference is the top, half is the bottom) (by definition of pi).  The width is the length of a wedge, or r.  So the area is about pi*r^2.  \n\nTo turn the estimate into an exact value, consider what happens as n goes to infinity, that is, it increases without bound.\n\nAs n goes to infinity, the slanted edges aproach being verticle and the number of arcs making up the top and bottom approaches infinity, and the arcs themselves approach straight horizontal lines.  Thus, the figure approaches a rectangle of exactly area pi*r^2.  However, for each n, you have the same circle of the same area, so the area of that circle is pi*r^2.
673075
alright, ill tell you, but not without explaining it.  im giving you the benefit of the doubt that you are not just cheating on your homework :)\n\nthe standard equation of a circle, using only variables is\n(x-h)^2 + (y-k)^2 = r^2\nwhere the center is at (h,k) and the radius is r\n\nthat means that when you look at the equation youre given, the x coordinate of the center is whatever number is subtracted from the x\n\n(x+1) ---> (x- -1)  so  h = -1\n\nyou use the same process to determine "k", which is the y coordinate of the center\n\n(y+3) ---> (y- -3)  so  k = -3\n\nthat makes the center of the circle at (-1, -3)\n\n\nfinding the radius is also simple using the standard equation. here it is again, just for reference:\n(x-h)^2 + (y-k)^2 = r^2\n\nsee how you can use that to get this?\n9/4 = r^2\n\nso just take the square root of both sides . . .\nr = +/- 3/2\n\nbecause you took the square root, the 3/2 could technically be either positive or negative, but since youre talking about the radius of a circle and you can't have a negative distance, you only take the positive answer.\n\nso r=3/2
635885
well using the y = function, you may need to create a positive and negative graph\n\nfor a circle with a center of 0, 0:\n\nsquare root ( x^2 - r^2)  (replace r with your desired radius)\nand\n- square root( x^2 - r^2)
468191
It's a circle of radius 4, centered at the origin.\n\nThe equation of a circle centered at (a, b) is:\n\n(x - a)² + (y - b)² = R²\n\nWhen the circle is centered at the origin, (a, b) = (0, 0), or:\n\nx² + y² = R²\n\nTherefore, y is simply given as:\n\ny = ±sqrt( R² - x² )\n\nIn this case, R² = 16, so R = 4. The plot you want to make is one such that every point on the plot is 4 units away from the origin.
43364
Circle equations take the form of (x-h)^2+(y-k)^2=r^2, where the center of the circle is at (h,k) and r is the radius.  \nBecause the information tells you where the center of the circle is, you can plug that back into the equation: \n(x-(-1)^2+(y-5)^2=r^2,\nwhich simplifies to (x+1)^2+(y-5)^2=r^2.  \nNow you can plug the into the equation the other point that is given.  \n(-4+1)^2+(-6-5)^2=r^2\nYou can evaluate this to get 130=r^2.  Plug this back into your equation (x-h)^2+(y-k)^2=r^2 to get\n(x+1)^2+(y-5)^2=130.
454188
Area of figure = Area of equilateral triangle + Area of arc AB + Area of arc BC.\n------------------------------------------------------------------------------------\nNow, to find the area of an equilateral triangle, use the formula\n\nA = ((s^2)/4)*(sqrt3) where s is the length of the side. Here, S= 7, so now solve the equation.\n\n((s^2)/4)*(sqrt3)\n=((7^2)/4)*(sqrt3)\n=(49/4)*(sqrt3)\n=12.25*sqrt3\n------------------------------------------------------------------------------------\nTo find the area of the arc, you need to see angle ABC as a central angle for a circle. Basically, see ABC as a sector of a circle where AB is the radius. Each angle in an equilateral triangle is 60 degrees.\n\nIn order to find the area of a sector, use the formula\n\na = (area of circle)*(d/360) where d is the number of degrees. Here, d equals 60. The area of a circle is (pi)(r^2) where r is the radius of a circle. Here, r equals 7. Now solve.\n\na = ((pi)(r^2))*(d/360)\na = ((pi)(7^2))*(60/260)\na = ((pi)(49))*(1/6)\na = 49pi * (1/6)\na = (49(pi))/6\n\nThe area of the sector is (49pi)/6. This area, however, includes the equilateral triangle. So, in order to find the area of the arc, we need to subtract the area of the triangle from the area of the sector.\n\na = ((49pi)/6) - 12.25(sqrt3)\nSimplify this on paper to get\na = (98pi - 147sqrt3)/12\n\nThe area of arc AB equals the area of arc BC. So, to get the total area of the two arcs, multiply by two to get\n\narea of two arcs = (98pi - 147sqrt3)/6\n------------------------------------------------------------------------------------\nNow add the area of the triangle and the two sectors.\n\nA = (12.25pi * sqrt3) + (98pi - 147sqrt3)/6\nSimplify this to get a total area of the figure...\n\nA = (196pi - 147sqrt3) / 12\n------------------------------------------------------------------------------------
359507
1) Let sides of each square be x\nTherefore dimensions of box are (4 - 2x) by (3 - 2x) by x\nV =  x(4 - 2x)(3 - 2x)\n= 12x  - 14x² + 4x³\ndV/dx = 12 - 28x + 12x²\n= 4(3 - 7x + 3x²)\n= 0 for stationary points\nSo x = (7 ± √(49 - 36))/6\n= (7 ± √(13))/6\n= (7 + √(13))/6, (7 - √(13))/6\n\nd²V/dx² = -28 + 12x > 0 when x = (7 + √(13))/6\nand < 0 when x = (7 - √(13))/6\n\nTherefore maximum value when x = (7 - √(13))/6 ≈0.566 units.\nSo Vmax = (7 - √(13))/6 *(4 - 2*(7 - √(13))/6) * (3 - 2*(7 - √(13))/6)\n≈3.032 units³\n\n2) y = √(x + 1)\nLet P (t, √(t + 1)) be on the curve and let Q = (3, 0)\n\nThus |PQ|² = (3 - t)² + (0 - √(t + 1))²\n= 10 - 5t + t²\n\nIf |PQ| is minimum so is |PQ|²\n\nd|PQ|²/dt = -5 + 2t = 0 for stationary points and as this is a quadratic with a> 0 then it is a minimum\n\nSo t = 2.5\nWhence the required point is (2.5, √(3.5))
400396
.............6\n........______\n....../.|...........|&#92;\n...../..|...........|..&#92; 2\n..../_|_____|__&#92;\n\n\nThe triangle on the right is a 30-60-90, so the shorter side must be 2/1 = 1.  So the length of the bottom part is 6+1+1 or 8.  The height must be 1√3, so the area is 1/2(6+8)√3 = 7√3.\n\nGopal screwed up when he did 2*2cos60.  That's 2*2*1/2 = 2, not 4.