what would b e3  the radius of a circle with a diameter of 8 inch?
363400
diameter is two times the radius so the radius is half of the diameter. \n\n4 inches\n\ntosmith
522790
Circumference = 2*radius*pi\n\nTherefore, the radius is 30\n\nand the diameter is 60..\n\nHope that helps...
543954
1/3 x pie x radius (power of 2) x height
554627
the circumference of a circle can be calculated by using the formula:\n circumference of circle=2*pie*r\n where pie=22/7;it is equal to 3.1415;\n r=radius of the circle.\nThanks.
195018
Draw all this out so you can see it - it won't be to scale, but don't let that fool you.....\n\n\ndiameter = 50cm (radius = 25cm)\nchord = 20cm from diameter (thus it's 20cm from the center)\n\nOK, you'll need to assume that the line segment forming the diameter touches the chord at point B on the circle - i.e. we'll say the diameter is line segment AB and the chord is line segment DB. \n\nC is the center of the circle, so we have radius CB = 25cm. \n\nAlso, we'll have line segment CE drawn perpendicular from the center (C) to point E on the chord, forming a right angle and thus *bisecting* chord DB.\n\nYou have the hypotenuse of triangle ECB (radius CB) as 25cm, and side EC as 20cm (distance from the center to the chord - always perpendicular).\n\nLine segment EB will be the second side of a right triangle, so your formula:\n\nEC^2 + EB^2 = CB^2\n20^2 + EB^2 = 25^2\nEB^2 = 25^2 - 20^2\nEB^2 = 625 - 400 = 225\nEB = 15cm\n\nSince EB = half the length of the chord DB, DB = 2*15cm = 30cm.
71470
3.1416 X radius squared\n12" circle= 6" radius\n6" X 6" = ? X 3.1416(pi)= ?\nGot the picture?\nNot going to do your homework for you.\n\n(sorry Lucky Lady but pi X diameter is the circumference)
840317
if it is a true arch the length would equal the diameter,\nso radius would be half the diameter (length)
692007
its quite simple the radius is half way threw the circle so its the radius 3 squared which equals 9 and 9 times pi or 3.14 equals 88.7364 but u mite have to round it so if u round to a whole number it would be 89 but to the nearest tenth it would be 88.7
303304
Use the well known formulas that relate the radius and diameter of a circle to its circumfrence.\n\nC = pi * D\nC = 2 * pi * R\n\nFor pi, you can use 3.14 or 22/7, or whatever value your teacher wants you to use.
579127
what is the diameter....? Just multiply the diameter times pi. (3.14)
900589
The largest radius would be 5 in., and here's why:\n\nYou have a square baking dish, with each side of the baking dish having a length of 10 in.\n\nAnd that means that the distance from the middle of the pan to the edge is 5 in.\n\nSo the largest circle that could be inside the baking pan would be a circle whose center is at the center of the pan and whose radius goes out to the edge of the pan.\n\n~ ♥ ~
799820
the radius is 1/2 the distance across a circle, while the diameter is twice the radius and is a straight line passing through the center of the circle to another side of the circle.
143664
radius= diameter/2\narea=(pi)(radius)^2\n\narea of an 18 inch pizza= (pi)(9)^2= 81(pi)\narea of a 14 inch pizza= (pi)(7)^2= 49(pi)\n\ndifference= 81pi-49pi= 32pi= 100.5 square inches
811157
Your question is not well posed, but I'm going to restate it in the way that I'm guessing you're thinking.  \n\nSuppose you have a 20 inch diameter circle.  Take two points on the circle and draw a straight line segment between them that is precisely 8 inches in length (this is termed a "secant", or a "chord").  What is now the arc-length of the segment of circle between those two points?\n\nIf that is your question, then here is how I work the problem:\n\nYou have defined an equilateral triangle, two edges of which are the radii of the circle, 10 inches each, and one which is the chord of 8 inches.  We can now determine the ANGLE between the two 10 inch edges using basic trigonometry.  The easiest way for me to think about this, offhand is to divide the equilateral triangle in half, making two right triangles.  The hypotenuse is 10 inches, the opposite leg is 4 inches, and therefore the sin function of the angle is equal to 4/10.  the angle is therefore (approximately) 23.6 degrees.  Since we "trigged" a right triangle that was half of the original equilateral triangle, we need to double the result, yielding 47.2 degrees.\n\nThe angle between the two radii that subtend the arc who's chord is 8 inches (on a 20 inch circle) is 47.2 degrees.\n\n47.2 degrees represents 47.2/360 of the circle (or 0.13 of the circle).  Therefore the arc length of the segment in question is given by 0.13 x the circumfrence.\n\nThe circumfrence is pi times the diameter or 3.14 x 20 = 62.8 inches, therefore the arc length is given by 0.13 x 62.8 =\n8.2 inches\n\nThe answer here isn't terribly surprising.  The 20 inch diameter circle is quite large compared to the short segment of an 8 inch straight line.  The "gentle curvature" of the circle compared to the line means that the arc length of the circular segment isn't going to be much longer than the straight line.  If  you took it to the extreme and compared the circular segment from a circle who's DIAMETER was about 8 inches, then the arc length would become nearly half the circle, 3.14 x 8 / 2 =12.5 inches.  On the other hand, if the diameter of the circle got really really big (like the diameter of the earth for example) then you'd not be able to tell the difference between the straight 8 inch rod and the curve of the circle at all.  The answer would approach exactly 8 inches.\n\nI hope that answered your question!
39475
Diameter of a circle, I'm assuming?\n\nDiameter = 2 * Radius\nDiameter = Circumference / (2 * pi)\nArea = pi * (diameter / 2)^2
946857
it is simply base area*height\n\nour base area is a circle for a perfect cylinder, which would be pi*(r^2), where r is your radius.\n\nYou need to convert your diameter into meters. There are 1000 millimeters to a meter. So your diameter is .150 meters, and because your radius is half that, your radius is .075 meters. \n\nplug in your values into pi*(r^2)*h, and there you go!
789377
Pi by diameter\n( 22 / 7 ) X diameter