suggest a lab activity to find the length of an arc?
927947
build a big arc out of paper mache.\ngive the lengths of straight lines to the end of the arc. \n\nhave students gague the length by "eyeballing it"\n\nthen use a piece of rope to demonstrate the actual length..\n\n.
323670
measure each diagonal, come up with the area of one triangle X it by 4, then work out the area of the base simple!
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.
1092151
You'll have to give us a better idea of what you're trying to do.  An angular error of 1 second of arc corresponds to a deflection of 5 microns over a length of one metre.  It's difficult just making things smooth to 5 microns in a workshop, so any angle you measure will probably have an error in excess of 1 second anyway.\n\nI doubt any solution we can provide would fall under most definitions of "cheap".
749836
The 14 foot line is called a secant.  Bisect it, and draw a radius through the point of bisection.  You now have two identical right triangles, with hypotenuses of 12 feet, and known sides of 7 feet.  Use trigonometry to calculate the angle at the center of the circle of either triangle, and then a simple proportion (involving the circle's total circumference) will give you the arc length you want.
1144014
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.
715307
Heres one good example. You are working as a lanscpaer. A person wants a 20 foot tree in their yard. You go to pick out a tree but you can't find the height of the trees except for one. You can use trig to find the hight of the other trees
508595
You have two ways:\nOld one: Thomson divide the arc into strips at equal intervals and solve each.\nNew one: Integrate from y=0, first root to y=0 second root
358092
Check the Omega online catalog. They have an extensive supply of measuring devices for lab, process, and manufacturing uses. Customer service is top notch.\n\nwww.omega.com
978661
Have fun:\nhttp://www.math.nyu.edu/~griffith/teaching/06numanalysis/labs/bisection/
290531
a trip to the ruler factory. hahahahaha.\nsorry but they just wont be interested and will cry if they cant measure by themselves. just make it as fun as possible. videos w/ little animals, etc are good.
939597
measure the distance between the beginning of the positive transition to the end of the negative transition.\nThis will be the wave length.
823299
How about drawing a bunch of straight lines of differing length, from a fixed point, so that the tips create the image of an arc?
163366
Yes with a gaussmeter.
990570
You could use the law of cosines to find the lengths of the diagonals.\n\nThe measure of interior angle B = 128 4/7 degrees.\nAC^2 = AB^2 + BC^2 - 2(AB)(BC)(cos B).\nSince the heptagon is regular, BC = AB, so you can substitute:\nAC^2 = 2AB^2 + - 2(AB)^2(cos B)\nAC^2 = 2AB^2(1 - cos B)\nAC^2 = 2AB^2(1 - cos (128 4/7))\nAC = sqrt [2AB^2(1 - cos (128 4/7))]\n\nThe measure of angle BCD = 128 4/7 degrees.\nThe measure of angle BCA = 25 5/7 degrees because that triangle is isosceles.\nTherefore, the measure of angle DCA is 102 6/7 degrees.\n\nAD^2 = AC^2 + CD^2 - 2(AC)(CD)(cos DCA)\n\nAnd so on...\n\nCheating with a calculator, it's this:\n\nAC = sqrt [2AB^2(1 - (-0.6234898))]\nAC = sqrt [2AB^2(1.6234898)]\nAC = sqrt [AB^2(3.2469796)]\nAC = sqrt [3.2469796](AB)\nAC = 1.8019377358(AB)\n\nThe measure of angle BCD = 128 4/7 degrees.\nThe measure of angle BCA = 25 5/7 degrees because that triangle is isosceles.\nTherefore, the measure of angle DCA is 102 6/7 degrees.\n\nAD^2 = AC^2 + CD^2 - 2(AC)(CD)(cos DCA)\nAgain, CD = AB and you've already found AC, so substitute:\nAD^2 = [1.8019377358(AB)]^2 + AB^2 - 2[1.8019377358(AB)](AB)(cos DCA)\nAD^2 = 3.2469796(AB^2) + AB^2 - 3.60838754716(AB)^2(cos DCA)\nAD^2 = 4.2469796(AB^2) - 3.60838754716(AB)^2(cos DCA)\nAD^2 = 4.2469796(AB^2) - 3.60838754716(AB)^2(cos 102 6/7)\nAD^2 = 4.2469796(AB^2) - 3.60838754716(AB)^2(-0.22252)\nAD^2 = 4.2469796(AB^2) + 0.8019377358(AB)^2\nAD^2 = 4.2469796(AB^2) + 0.8019377358(AB)^2\nAD^2 = 5.04891734(AB)^2\nAD = sqrt [5.04891734(AB)^2]\nAD = 2.2469796(AB)\n\n1/AC + 1/AD = 1/1.8019377358(AB) + 1/2.2469796(AB)\n1/AC + 1/AD = 2.2469796/4.04891734(AB) + 1.8019377358/4.04891734(AB)\n1/AC + 1/AD = 4.04891734/4.04891734(AB)\n1/AC + 1/AD = 1/AB\n\nI like this problem, and I'll give it a shot without a calculator later this afternoon.
534712
Actually sometimes. A secant passes through a circle at two points, a tangent at one. You COULD have a secant hit the tangent at a point other than the point of tangency, and there is indeed a theorem around for determining the relative lengths of lines and so forth, but many questions dealing with area and arc length put the vertex at the point of tangency for simplicity's sake.
408251
Measure the diameter and multiply by pi (3.1415)