Geomerty help please?
440894
go to this website and try to understand how to get the area of the shade region \n\nhttp://www.algebralab.org/lessons/lesson.aspx?file=Geometry_AreaPerCricle.xml\n\ngood luck
544990
Well as far as a formula i can't help you there. i just happen to own a program that allows me to draw perfect shapes of any size and diminsion them accordingly (auto CADD). and at the top of your 300 ft house, or where ever it is that your measuring from the horizon at sea level is a tad over 21 miles away (21.20389 to be analy exact). first you must know the the earths circumfrance (i know i can't spell) which is 20,890,637 ft, next you have to know the distance you are away from the earth (which you do, 300 ft). now you have to know the formula for finding a line that intersects your hieght and crosses tangentally across the earths surface. know you have your triangle and apply the Pythagorean theorem. you have an answer.
27065
I've been studying this stuff recently, and I'm pretty sure this is how it goes:\n\nK is now (-2,1) and D is now (-3,-2).\n\nFor future reference on determining rotations, simply turn the graph counter-clockwise one time for 90 degrees, upside down for 180 degrees, and for 270 degrees, just turn the graph clockwise once to see what the new cooridnates are.
1017997
The equations are wrongor insufficient. Post the correct data if you want me to help/ Alternately, you could also message me.
824775
The Area of a triangle is half the area of a square. So take one verticy and take a 90degree line from that up to the other corner.\n\nOh cr@p I'm starting to forget it all now. Too long ago.\n\nIt's something like that.
861159
You need to know the base and the height of the shapes. \n\ntrapeziod: average of two bases x height\nparallelogram: base x height\ntriangle: 1/2 x base x height\n\nOr if you know how to use the sin, cos, tan, then you only need to know a few sides to calculate the base and the heights.
513176
Try http://googleearth.com\nMaster.
920579
use synthetic  divsion method \ni think this link would help\nhttp://www.purplemath.com/modules/synthdiv.htm
880255
I can just give you a link to help you solve it.\n\nhttp://www.ticalc.org/pub/83plus/basic/math/geometry/
829392
just a piont of view, but normally to divide something symmetrically you have to do it at the widest piont along the centre line. where as assymetricaly is off the centre line at a narrower piont and therefore not a symmetry,this only applies to poloygons and symmetric shapes.  hope this helps
224079
try this site... it shows step by step how to go about sloving these equations, good luck:
503725
Been a while, so I'm not sure what you're looking for exactly in "Geometric Shapes".. but these might help somewhat.\n\nSun/Plannets - Sphere\nTree Trunk - Cylindar\nPine Tree - Cone\n\n\n...and my brain stops working.  but just for humor's sake...\nGood Looking Woman - hourgalss\n\n=D good luck with the homework
459338
Go to http://mathworld.wolfram.com/Cosine.html
978664
Orthocentre = point where altitudes intersect (An altitude is the line from the vertex to the opposite side (extended if necessary))\n\nIncentre = centre of the incircle = point where perpendicular bisectors of each of the sides intersect. \nRadius of incircle = √[(s-a)(s-b)(s-c)/s] where s = semiperimeter\n\nCentroid = point where medians meet (Median is the line joining a vertex to the midpoint of the opposite sides) The centroid is the centre of gravity of the triangle and the centroid divides each median in the ratio 2:1 On a numberplane the centroid of a triangle is the point which is at the average of the sum of the three x- and y-coordinates of its vertices)\n\nOne other is:\nCircumcentre = centre of the circumcircle = point where angular bisectors meet.\nDiameter of circumcircle = a/sinA = b /sinB = c/sinC
87753
Just add up the lengths of the three sides. Works for any triangle - wether it's equilateral, isosceles or obtuse.
667922
Either use volumes of revolution or compare cross sections of the sphere and a cylinder with a couple of cones removed. The latter is how Archimedes did it.
247013
this is not so easy to answer without doing some work: usually, when you speak of tessellation, it's more likely to deal with symmetries.\n\nHere is a good site which has a couple of interesting articles and softwares on tessellations:\n\nhttp://www.scienceu.com \n(click on the "Geometry center" icon)\n\nAnyway, I'll think better about it: it's a good question.