find the sum of the measures of the interior angle of a 12 sided polygon?
337862
go do your homework by yerself next time.\n\nBut becos im feeling nice today...\n\na triangle = 1 triangle , duh\na quadilateral can be divided into two triangles with vertices (points) all at the points of the quadilateral\n5 sided --> 3 triangles\n6 sided --> 4 triangles\n....\n12 sided --> 10 triangles\n1 triangle -->180 degrees\n10 triangles --> 1800 degrees
628701
Area = 1/2 ac Sin B\n\nwhere a,b,c are the sides and A B C are the opposite angles. \nFrom the 10 sided figure you get an internal angle of 360/10 - 36 degrees so the other two angles are 180-36 all divided by 2 = 144/2 = 72 degrees.\n\nand the outside edge is 240/10 = 24\n\nSo using ac = 24 and Sin 36.. plug into the formula above.. this gives you the area of one triangle, then multiply by 10
518004
You would need to disect the polygon into two shapes.  A triangle and a rectangle.  Figure the volume of those individual pieces and then add them together.
1092713
triskaidecagon\n\nor a 13-sided polygon, haha.\n\nCheck the source for polygon names from 1-20 sides  :)
548292
Assuming that an area may be dissected into triangles, then the area of each triangle is given by \n\nArea = (s(s-a)(s-b)(s-c))^(1/2)\n\nwhere s = (1/2)(a + b + c)\n\nand a, b, and c are the lengths of sides of each triangle -\n\nThe sought for area is (finally) the sum of all the areas of the triangles.\n\nGiven sufficient patience, any arbitrary area will be found.\n\n(Drawing a diagram while slogging through the exercise and looping around the separate triangles in the same direction can yield a shortcut whereby all of the internal perimeters eliminate each other, leaving only the perimeter of the original figure.)
695972
Since a rectangle has four sides, I will use the equivalent ratio of 2:6. Therefore the sum of the sides of the rectangle add up to 6. If the length of the rectangle is two, as there are two lengthwise sides, 2+2=4, 6-4=2. Therefore, the sum of the widths is two, and therefore, as 2 divided by 2=1. The width is 1, or an equivalent of that example. 1:6, 2:12, etc. Therefore, as 1 is half of the original length, 1 divided by 2 is 0.5. The length of the width in the original rectangle is 0.5.
881588
First, the isosceles triangle has sides of 10, 12, 12.  You can find the angle by law of cosines or, as I prefer, just divide the triangle into two right triangles.  The short side is 5 and the hypotnuese is 12 so the angle at G is arcsin(5/12) and the total angle FGA is twice that (since we split the original triangle in two) <FGA = 2*arcsin(5/12) = 49.249 deg.\n\nIf you make a pyramid, The radius of the hexagon (10 cm), the 12 cm side and the height form a right triangle so the height is:\n\nh = sqrt(12^2 - 10^2) = sqrt(44) = 6.633\n\nFinally, for the area, we need the height of the isosceles triangle which is the tjhird side of the right triangle in the first part. This height is:\n\nc = sqrt(12^2 - 5^2) = sqrt(119) = 10.909\n\nSo the area is: A = 3*10*sqrt(119) = 327.261
932014
The measure of the exterior angle of a triangle must be greater than the measure of either non-adjacent interior angle.\n\nThis is because the measures of the exterior angle and its adjacent interior angle must add up to 180 since they are a linear pair.  But the measures of the two non-adjacent interior angles must also add with the adjacent one to get 180 since the measures of the 3 angles of a triangle add up to 180.\n\nSo the exterior angle's measure equals the sum of the two non-adjacent interior angles, and therefore must be bigger than either one of them.\n\nGood luck, hope that helps.
884011
The sum of the measures of the EXTERIOR angles is always 360.  So subtract 162 from 180 to get an exterior angle (18) then divide 360 by 18 to see how many sides.
932841
This might help --> http://www.uwm.edu/~ericskey/TANOTES/Geometry/node15.html
18728
Again I assume you are talking about vector addition.  In this case you use trig the angle and magnitude given to find the x and y components for each vector.  Then just add all of the x componets together and all of the y components together to get the sum.  Then you can use the pathagorean theorem and arctan to find the magnitude and direction of the resultant vector.
436643
If the shape is a quadrilateral, the measures of its opposite angles WOULD add up to 180 degrees.  This is because the measure of an inscribed angle (an angle with vertex on the circle) is half the measure of its intercepted arc.  Between the two opposite angles, their combined arc is the whole circle which measures 360 degrees.  So the sum of the inscribed angles is half of that, or 180.
727910
P = NL where N is the number of sides the polygon has\n\nSo if L = 6 cm \nthen P = 6N cm
431862
There are two angles formed wherever two sides of the hexagon meet. The angle inside the hexagon is 120 degrees and the angle outside the hexagon is 360 - 120 = 240 degrees. You can see this easily by placing a point at the center of the hexagon and connecting it to the 6 points where the edges of the hexagon meet. This forms 6 equilateral triangles and every one of the angles of the 6 triangles is 60 degrees (180/3). You can see that the angle inside the hexagon where 2 sides of the hexagon meet is equal to 2 of these 60 degree angles; hence 120 degrees. Draw it and you'll see it right away.
539673
One of the most simple requirements for tiling: if a polygon tiles a plane, then in each of polygon vertexes on the plane, the sum of adjacent angles should equal 360 degrees (one of the angles may be a side of another polygon, making up 180 degrees). \n\nFor example, imagine or draw a "house-shaped" pentagon, that is a square with a right-angled triangle on top (right angle pointing up). It has two angles of 90 degrees at the bottom, two 135 degree angles where "walls" and "roof" meet, and a 90 degree angle at the top ("roof" tip). You can tile them side-by-side and then tile the same pentagons on top, flipped upside down. At the "tip of the roof", you will see that two 135 degree angles and one 90 degree angle meet. 135+135+90=360. If it would not be the case, there would be a crack in the tiling.\n\nHowever, some polygons have such angle values that it is impossible to make a sum of angles equal to 360 or 180 degrees. For example, consider regular pentagons. They have angles of 108 degrees. However you add up 108 degree angles, you cannot make up 360 or 180.
654063
The sum of all interior angles of a polygon is given by\n(2n-4)*90 degrees\nSo for pentagon n = 5 \nHence the sum of all interior angles of a pentagon would be\n540 degrees(6*90)
914112
It's usually called a quadrilateral, and the total of the internal angles of a convex quadrilateral is 360º.\n\nTo figure the sum of the internal angles of any convex polygon, the formula is (n-2)*180º, so a pentagon (five sides) has 540º, a hexagon (six sides) has 720º and so forth.