In a circle, I am given the length of a chord to be 16. The portion of the radius perpendicular to the chord, which is between the chord and the perifery has the length 4. How do I determine the radius??
198826
OK, call the chord AB. The radius intersects the chord perpendicularly at the point M, which means that AM is half of AB, so AM is 8. Calling the radius R and the point on the perifery hit by this perpendicular radius D, we know that MD is 4, and therefore the perpendicular distance from the center C to M is (R-4). The triangle CMA is thusly a right triangle, with hypothenuse R and cathetes 8 and (R-4). Using Pythagoras, we get R^2 = 8^2 + (R-4)^2 or\nR^2 = 64 + R^2+16-8R. Simplifying, we get 80-8R=0 or R=10.
491825
Here's how to solve the problem:\n\nFirst find the center of the circle (call it (x1,y1).\n1. Write an expression for the distance of (x1,y1) from the first point, (1,8):  (x1-1)^2 + (y1-8)^2\n\n2. Write an expression for the distance of (x1,y1) from the second point, (-2,-1). (You should know how to do this, especially after seeing the example above.)\n\n3. Write an expression for the distance of (x1,y1) from the given line. This is a little trickier. There are probably various ways of approaching it. Here's one:\n\nRewrite the equation in slope-intercept form: y = (3/4)x +4\nThis tells us that the slope of the line is 3/4. So the slope of a line perpendicular to this line is -4/3. (The product of the two slopes has to be -1.)\n\nThe distance from (x1,y1) to the original line is measured along the direction perpendicular to that line. So it will be measured along a line with a slope of -4/3. The general equation for such a line is \ny = -(4/3)x + b, where b is the y-intercept of the line.\n\nNow we're trying to find an expression for the distance from (x1,y1) to the original line. To do this, we have to find where the perpendicular line intersects the original line. So we need to find the solution of the two equations, \ny = (3/4)x +4 and y = -(4/3)x + b.\nThis gives us two expressions for y, so we can set them equal:\n(3/4)x +4 = -(4/3)x + b\nSolving this, x = (12/25)(b-4) = (12b-48)/25\nAnd y = 9/25(b-4) + 4 = (9b +64)/25\nSo the point of intersection of the given line and the perpendicular line is ((12b-48)/25,(9b +64)/25).\n\nThe distance from (x1,y1) to this point (and to the given line) is \nsqrt((x1-(12b-48/25))^2 + (y1-(9b+64)/25)^2).\n\nHow do we find b and x1 and y1? We need to equate the three expressions for distance (this one, and the two found in items 1 and 2 above). Solving these equations will give us the center of the circle and the equation for the perpendicular to the given line.\n\nNext find the radius of the circle:\nDetermine the distance from the center to either of the two given points (using the Pythagorean theorem ... that is, using the distance formula). Call this distance D.\n\nFinally, write the equation of the circle:\n(x-x1)^2 + (y-y1)^2 = D^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.
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
477328
Here's the answer you want:  With a compass set to the radius of the circle, start marking off radii around the circle, dividing it into 6 equal parts.  Connect any 3 alternate points and you'll have your triangle.\n\nNO MATH REQUIRED!
543222
r=radius\nd=diameter\n\n1 C=2r(pi)\n2 C=d(pi)\n\n(Same thing because 2r=d)
53725
Area = Pi times the radius squared (A=pr2)\nIf you need to know how to find the radius, go here:\n\nhttp://math.about.com/library/blcircle.htm
522620
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)
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
300140
How to do this problem:\n1. Find the Radius. To do this, take the diameter and divide by two. This gives you a radius of 5.\n2. Because the Area of a circle is equal to Pi times the radius squared, and your teacher wants the answer in terms of pi, square the radius and get 25 pi.
93759
So 2*pi*radius = circumference.  And you know the circumference, so set up a little equation and solve for the radius.\n2 pi R = 4 pi\ncancel the 2 and the pi to get\nR = 2\nNow you can easily find the area using Area = pi * R^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.
330792
Assuming this means a circle...\n\ndefn/:\n\nc=circurmfernce\nA=area\nr=radius\nd=diameter\npi=pi (~3.1415926...)\nd=2*r\n\nC=2*pi*r=d*pi\nA=pi*(r^2)...(area is pi*r squared...pie are squared-no there not their round).\nA=pi*(d^2)/4.\n\njust plug and chug...\n\nC=10*pi.\nA=25*pi.\n\nanswer:\n\nit's that simple
491137
This type of problem will eventually seem EASY for you, so hang in there.  Let's begin by stressing that "a circle with radius 5" is simple to draw -- I recommend marking the 4 points that are "5 units" above, below, right and left of the center point.\n\nThat is, mark (-6,7) on your graph paper, then sketch four more points at (-1,7), (-11,7), (-6, 2) and (-6,12).  "Connect the dots" by drawing all 4 quarter-circle arcs between the points.\n\nThe equation for any circle -- in "rectangular coordinates" (which means (x,y)) -- is normally:  (x-a)^2 + (y-b)^2 = r^2, where (a,b) is the center point and r is the radius.\n\nBut what you're hoping to find is the equation of a circle in "polar coordinates" (which means (r,theta)).\n\nThe reason a sample triangle was drawn inside the given circle (I'm assuming you're looking at a textbook) is to illustrate the relationships among x, y, theta and r.  Let's inspect, using rules from trignometry:\n\nx = r cos (theta) + a\ny = r sin (theta) + b\nr = sqrt ((x-a)^2 + (y-b)^2)\ntheta = Arctan ((y-b)/(x-a))\n\nwhere (x,y) is any point on the circle, (a,b) is the circle's center, and r is the radius.\n\nThe formula for theta that I just provided may be precisely what you are looking for.  It works simply by first finding the "slope" of the line connection (x,y) and (a,b), then taking the inverse tangent of that slope, which 'converts' it to an "angle."\n\nHope this helps!
32302
A radius is half of a diameter. So the diameter is twice the radius. Multiply 15x by 2 and you get:\n\n30x
582368
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 !
369414
So R=1.6*r; R for body, r for head; and H1=K1*R is the distance from body center to the feet, K1=0.8; and h=k*r is the distance from head center to the neck, k=0.9; Now neck radius p=sqrt(r^2-h^2) and p=sqrt(R^2-H2^2), H2=K2*R is distance from body center to the neck, K2 being unknown.  Find K2;  p^2 = r^2*(1-k^2) = R^2*(1-K2^2), hence (r/R)^2*(1-k^2)=1-K2^2 and K2 = sqrt(1-(1/1.6)^2*(1-0.9^2)) = 0.9622;\nNow head volume v=v0-v1, v0=(4/3)*pi*r^3, v1=(pi/3)*r^3*(2-3h/r+(h/r)^3) is head cut-off part; v = (pi/3)*r^3*(2+3h/r-(h/r)^3) = (pi/3)*r^3*(2+2.7-0.9^3)\nThe same for body volume V=V0-V1-V2, V=(pi/3)*(1.6*r)^3 * (4-{2-3*K1+K1^3} - {2-3*K2+K2^3});\nW=v+V=r^3*C is the whole volume! To plug in numbers to find C, OK?