In the circle of 20 cms I draw a chord of 5 cms. What will be the arc length over the chord?
132935
Formula for Arc length: (2pir) x (arc length/360)\n\n(2)(pi)(10) x (5/360)=\n62.8 x 0.01388=\n\n.87 cm
36754
This seems like a homework problem, so I'll give a few hints.\n\nThe arclength is radius times the angle subtended.\n\nTo find the angle subtended, consider the isosceles triangle created inside the circle where the chord is the hypotenuse and the radius of the circle is the two other legs. \n\nSince the isosceles triangle created is not a right triangle you need to use a specific formula to find the angle "C" subtended by the chord.
214635
100 cm, its a 3:4:5 right triangle.\n\n[lay the cylinder out flat, if you will.  4 times around is 80 cm, and the width is 60 cm.  So the string has to form the hypotenuse of the 60 cm by 80 cm sided triangle.  60 divided by 20 is 3, 80 divided by 20 is 4, and a right triangle is famously 3:4:5, so 5 times 20 is 100 cm]
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.
432460
a) Assume the following:\n\nAngles a= 10 deg, b= 80 Deg, c=90 deg \nHypotenuse length = 2.5 m\nheight = 2.5 * cos 80 Deg= .434 m\nlength = 2.5 * sin 80 Deg= 2.462 m\nacceleration = "g" * Sin 10 Degree
56003
It is the length divided by the radius
315306
150 cm.\n\nExplanation: each turn has a length equal to the diagonal of a rectangle of 24 and 18 cm, where 18 = height / 5 turns.\nThe length of the diagonal is therefore 30 cm.\n\nSo the total length is 5x30 = 150 cm
360683
Easy. Draw triangles.\n\nFirst, draw a line 35m downfield. Now draw a line at an angle of 25 degrees to the right for 15m.\n\nNow extend the first (35m) line indefinitely in its original direction. Now draw a line from the far end of the 25-degree line perpendicular to the first line. You now have one little right triangle with a hypotenuse of 15m and an angle of 25 degrees.\n\nFigure out how long the extension is from the point of turning to the perpendicular: 15 cos 25.\n\nThe extended line in the original direction thus has length\n\n35 + 15 cos 25\n\nThe prependicular piece has length\n\n15 sin 25\n\nNow draw one long line from the original starting point to the ending point. You have a large right triangle with sides\n\n15 sin 25\nand\n35 + 15 cos 25\n\nYou can figure out the hypotenuse and angles of that big triangle.
524837
A short element of the curve is dL=sqrt(dr^2+(r*dt)^2), r*dt elementary length of the arc with radius r and polar angle t (your theta), dr elementary increase of r for dt.\nThus dL=sqrt(r’^2+r^2)*dt.\nIf  r=a*(1-cost), then dL=a*sqrt((sint)^2+(1-cost)^2)=a*sqrt(2)*sqrt(1-cost)*dt. \nFortunately 1-cost=2*(sin(t/2))^2 or dL=2a*sin(t/2)*dt, L=-4a*cos(t/2) for t=0 until 2pi, thus L=-4*a*(-1-1)=8*a – Oh! check it poor kid!\nIf r=(sin(t/2))^2=0.5*(1-cost), then a=0.5 for t=0 until pi. It must be done likewise. Thus L=-4*0.5*(0-1)=2
203052
The angle is 36 degrees (10%). The chord length is 30.901699 cm = 2*50*cos(90-36/2) cm.
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
154430
It is a problem in long division with a remainder. You need to simply divide the whole integers (everything before the decimal point) and keep the remainder as the inches.  Thus your problem becomes 20 feet and 6 inches, you can check your answer by reversing the problem.  In 20 feet there are 20 times 12 inches, or 240 inches with 6.138 inches left over.  Thus the answer is 20 feet and 6 inches (rounding to whole inches as in your example). \n\n-LOVE YOUR NIEGHBOR AS YOURSELF
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You can't. It depends on which way the diameter is turned in the circle. There are an infinite amount of solutions. Tell your teacher to be more specific.\n\ntosmith
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180°\n\nYesterday it was asked about a 5-pointed star.\n\nWell let us look at an n-pointed star\n\n1. Draw the circumscribing circle\n\n2. Join the vertices and create a regular n sides polygon inside the circumscribing circle\n\n3. Join each vertex to the centre\n\n4. Each point of the star is half the ange at the centree subtended on the same side.\n\n5. Each angle in the centre is equal as they are all subtended by equal chords as the sides of a regular polygon are equal.\n\n6. The sum of the angles at the centre of the circle = 360°\n\n7 So the sum of the angles at the circumference \n= sum of the angles of the star \n= ½ of 360°\n= 180°\n\nBONUS\n8 So each angle = 180°/n\n\nIf n = 5 each angle = 36°\nIf n = 7 each angle = (25 5/7)°
59323
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]
159029
draw a big circle...\ncut a piece of string that stretches across the circle, this is the diameter...\ncut another, and one more...got 3?\nglue them to the outside of the circle...\ndoes it enclose the circle?...do ya need more string???\nthats why pi (not pie) is 3.14......
650821
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.