A hexagon need not have any lines of symmetry. Or, it can have just one line of symmetry. A regular hexagon has six lines of symmetry, including three along the lines bisecting the angles and three along the lines formed by bisecting the sides. A regular hexagon has a rotational order of 6.
A regular pentagon doesn't tessellate because of the way the sides are joined together and hes shape of the regular pentagon.Something along those lines anyway.....
There are several options: A hexagon based pyramid A pentagonal based bipyramid A triangular based prism + quadrilateral pyramid (attached along a quadrilateral face) A triangular pyramid on a triangular prism (attached along a triangular face)
If it is lying along its length, and he cross section is, for example, a regular pentagon, then it will have no vertical lines.
Think of your regular trapezoid (half a hexagon). Now cut it in half with a line perpendicular to the top and bottom. You can see the right angles along one side. The top and bottom lines are parallel. The seond angle on the bottom is acute, and the second top is obtuse. Can't be bothered to provide diagrams, though. Draw them yourself.
If the forces are all normal (at right angles to) the sides the resultant is 0 (they all cancel each other out ).
It could be an irregular hexagon: for example, a regular hexagon that has been stretched along one diagonal.It could be an irregular hexagon.
A hexagon need not have any lines of symmetry. Or, it can have just one line of symmetry. A regular hexagon has six lines of symmetry, including three along the lines bisecting the angles and three along the lines formed by bisecting the sides. A regular hexagon has a rotational order of 6.
In that case, the resultant will either be the sum or the difference of the two forces (if they act in the same direction) or their difference (if they act in opposite directions).
A hexagon need not have any lines of symmetry. Or, it can have just one line of symmetry. A regular hexagon has six lines of symmetry, including three along the lines bisecting the angles and three along the lines formed by bisecting the sides. A regular hexagon has a rotational order of 6.
It is approx 1087.6 N along the bisector of the two lines of action.
Only an equilateral triangle, square and a regular hexagon can be used to make regular tessellations but there are innumerable polygonal and non-polygonal shapes which will tessellate by themselves, and others which will tessellate along with other shapes.
ABCD is a squre. forces of magnitudes 1,2,3,P, and Q units act along AB, BC, CD, DA and AC respectively. find the value of P and Q so that the resultant of five forces is a couple
A resultant force is the equivalent force, of two or more individual forces that act on the same object. For example, if somebody pulls with a force of 100 N due north, and somebody else pulls with a force of 100 N due east, the resultant force would be about 141 N, due north-east. Calculations are done with vectors; specifically, any vector has to be separated into components along the x-axis and the y-axis (and the z-axis, in a 3-dimensional situation), and the components are added individually.
Resolving a force into components along mutually perpendicular directions requires the calculation of the cosine and sine of the angle made by the force with one of them. The resultant of two two forces acting at right angles to one another is in the direction whose tangent is proportional to the forces.
Two unequal forces The resultant force of two unequal forces clearly depends upon the magnitude of these forces and also depends on the angle of interception. One can clearly apply trigonometrical theorem for the same. such as if force x and force y of x1 and y1 magnitude is acting on an object N and if the angle between these forces is (180) degrees this means they are acting on opposite direction , and resultant force will be in direction of force x1 ir x1 > y1 or y1 if y1 < x1 So these unequal forces must be studied along with the angle of interception Hope this will give you a fair idea for calculation ( force is a vector and not scalier)
A regular hexagon with one vertex at the origin, and a side along the x-axis and of length s has vertices at: (0, 0) (s, 0) (1.5*s, 0.5*s*√3) (s, s*√3) (0, s*√3) (-0.5*s, s*√3) Since you now have both endpoints of each line segment, their equations are easy to find.