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From Wikipedia
In geometry, a pentagon From the Greek number 5 (pente) is any fivesided polygon. A pentagon may be simple or selfintersecting. The internal angles in a simple pentagon total 540Â°. A pentagram is an example of a selfintersecting pentagon.
Regular pentagons
A regular pentagon has all sides of equal length and all interior angles are equal measure (108Â°). It has five lines of reflectional symmetry and it has rotational symmetry of order 5 (through 72Â°, 144Â°, 216Â° and 288Â°). Its SchlÃ¤fli symbol is {5}. The chords of a regular pentagon are in golden ratio to its sides.
The area of a regular convex pentagon with side length t is given by
 A = \frac{4} = \frac{5t^2 \tan(54^\circ)}{4} \approx 1.720477401 t^2.
A pentagram or pentangle is a regularstar pentagon. Its SchlÃ¤fli symbol is {5/2}. Its sides form the diagonals of a regular convex pentagon – in this arrangement the sides of the two pentagons are in the golden ratio.
When a regular pentagon is inscribed in a circle with radius R, its edge length t is given by the expression
 t = R\ {\sqrt { \frac {5\sqrt{5}}{2}} } = 2R\sin 36^\circ = 2R\sin\frac{\pi}{5} \approx 1.17557050458 R.
Derivation of the area formula
The area of any regular polygon is:
 A = \frac{1}{2}Pa
where P is the perimeter of the polygon, a is the apothem. One can then substitute the respective values for P and a, which makes the formula:
 A = \frac{1}{2} \times \frac{5t}{1} \times \frac{t\tan(54^\circ)}{2}
with t as the given side length. Then we can then rearrange the formula as:
 A = \frac{1}{2} \times \frac{5t^2\tan(54^\circ)}{2}
and then, we combine the two terms to get the final formula, which is:
 A = \frac{5t^2\tan(54^\circ)}{4}.
Derivation of the diagonal length formula
The diagonals of a regular pentagon (hereby represented by D) can be calculated based upon the golden ratio Ï† and the known side T (see discussion of the pentagon in Golden ratio):
 \frac {D}{T} = \varphi = \frac {1+ \sqrt {5} }{2} \ ,
Accordingly:
 D = T \times \varphi \ .
Chords from the circumscribing circle to the vertices
If a regular pentagon with successive vertices A, B, C, D, E is inscribed in a circle, and if P is any point on that circle between points B and C, then PA + PD = PB + PC + PE.
Construction of a regular pentagon
A variety of methods are known for constructing a regular pentagon. Some are discussed below.
Euclid's methods
A regular pentagon is constructible using a compass and straightedge, either by inscribing one in a given circle or constructing one on a given edge. This process was described by Euclid in his Elementscirca 300 BC.
Richmond's method
One method to construct a regular pentagon in a given circle is as follows:
Verification
The top panel describes the construction used in the animation above to create the side of the inscribed pentagon. The circle defining the pentagon has unit radius. Its center is located at point C and a midpoint M  is marked half way along its radius. This point is joined to the periphery vertically above the center at point D. Angle CMD  is bisected, and the bisector intersects the vertical axis at point Q. A horizontal line through Q intersects the circle at point P, and chord PD is the required side of the inscribed pentagon.
To determine the length of this side, the two right triangles DCM and QCM are depicted below the circle. Using Pythagoras' theorem and two sides, the hypotenuse of the larger triangle is found as âˆš5/2. Side h of the smaller triangle then is found using the halfangle formula:
 \tan ( \phi/2) = \frac{1\cos(\phi)}{\sin (\phi)} \ ,
where cosine and sine of Ï• are known from the larger triangle. The result is:
 h = \frac{\sqrt 5  1}{4} \ .
With this side known, attention turns to the lower diagram to find the side s of the regular pentagon. First, side a of the righthand triangle is found using Pythagoras' theorem again:
 a^2 = 1h^2 \ ; \ a = \frac{1}{2}\sqrt { \frac {5+\sqrt 5}{2}} \ .
Then s is found using Pythagoras' theorem and the lefthand triangle as:
 s^2 = (1h)^2 + a^2 = (1h)^2 + 1h^2 = 12h+h^2 + 1h^2 = 22h=22\left(\frac{\sqrt 5  1}{4}\right) \
 =\frac {5\sqrt 5}{2} \ .
The side s is therefore:
 s = \sqrt{ \frac {5\sqrt 5}{2}} \ ,
a well established result. Consequently, this construction of the pentagon is valid.
Alternative method
An alternative method is this:
 Draw a circle in which to inscribe the pentagon and mark the center point O. (This is the green circle in the diagram to the right).
 Choose a point A on the circle that will serve as one vertex of the pentagon. Draw a line through O and A.
 Construct a line perpendicular to the line OA passing through O. Mark its intersection with one side of the circle as the point B.
 Construct the point C as the midpoint of O and B.
 Draw a circle centered at C through the point A. Mark its intersection with the line OB (inside the original circle) as the point D.
 Draw a circle centered at A through the point D. Mark its intersections with the original (green) circle as the points E and F.
 Draw a circle centered at E through
Surface area is the measure of how much exposed area a solid object has, expressed in square units. Mathematical description of the surface area is considerably more involved than the definition of arc length of a curve. For polyhedra (objects with flat polygonal faces) the surface area is the sum of the areas of its faces. Smooth surfaces, such as a sphere, are assigned surface area using their representation as parametric surfaces. This definition of the surface area is based on methods of infinitesimal calculus and involves partial derivatives and double integration.
General definition of surface area was sought by Henri Lebesgue and Hermann Minkowski at the turn of the twentieth century. Their work led to the development of geometric measure theory which studies various notions of surface area for irregular objects of any dimension. An important example is the Minkowski content of a surface.
Definition of surface area
While areas of many simple surfaces have been known since antiquity, a rigorous mathematical definition of area requires a lot of care. Surface area is an assignment
 S \mapsto A(S)
of a positive real number to a certain class of surfaces that satisfies several natural requirements. The most fundamental property of the surface area is its additivity: the area of the whole is the sum of the areas of the parts. More rigorously, if a surface S is a union of finitely many pieces S_{1}, …, S_{r} which do not overlap except at their boundaries then
 A(S) = A(S_1) + \cdots + A(S_r).
Surface areas of flat polygonal shapes must agree with their geometrically defined area. Since surface area is a geometric notion, areas of congruent surfaces must be the same and area must depend only on the shape of the surface, but not on its position and orientation in space. This means that surface area is invariant under the group of Euclidean motions. These properties uniquely characterize surface area for a wide class of geometric surfaces called piecewise smooth. Such surfaces consist of finitely many pieces that can be represented in the parametric form
 S_D: \vec{r}=\vec{r}(u,v), \quad (u,v)\in D
with continuously differentiable function \vec{r}. The area of an individual piece is defined by the formula
 A(S_D) = \iint_D\left \vec{r}_u\times\vec{r}_v\right  \, du \, dv.
Thus the area of S_{D} is obtained by integrating the length of the normal vector \vec{r}_u\times\vec{r}_v to the surface over the appropriate region D in the parametric uv plane. The area of the whole surface is then obtained by adding together the areas of the pieces, using additivity of surface area. The main formula can be specialized to different classes of surfaces, giving, in particular, formulas for areas of graphs z = f(x,y) and surfaces of revolution.
One of the subtleties of surface area, as compared to arc length of curves, is that surface area cannot be defined simply as the limit of areas of polyhedral shapes approximating a given smooth surface. It was demonstrated by Hermann Schwarz that already for the cylinder, different choices of approximating flat surfaces can lead to different limiting values of the area.
Various approaches to general definition of surface area were developed in the late nineteenth and the early twentieth century by Henri Lebesgue and Hermann Minkowski. While for piecewise smooth surfaces there is a unique natural notion of surface area, if a surface is very irregular, or rough, then it may not be possible to assign any area at all to it. A typical example is given by a surface with spikes spread throughout in a dense fashion. Many surfaces of this type occur in the theory of fractals. Extensions of the notion of area which partially fulfill its function and may be defined even for very badly irregular surfaces are studied in the geometric measure theory. A specific example of such an extension is the Minkowski content of a surface.
Common formulas
In chemistry
Surface area is important in chemical kinetics. Increasing the surface area of a substance generally increases the rate of a chemical reaction. For example, iron in a fine powder will combust, while in solid blocks it is stable enough to use in structures. For different applications a minimal or maximal surface area may be desired.
In biology
The surface area of an organism is important in several considerations, such as regulation of body temperature and digestion. Animals use their teeth to grind food down into smaller particles, increasing the surface area available for digestion. The epithelial tissue lining the digestive tract contains microvilli, greatly increasing the area available for absorption. Elephants have large ears, allowing them to regulate their own body temperature. In other instances, animals will need to minimize surface area; for example, people will fold their arms over their chest when cold to minimize heat loss.
Answers:Surface Area of Cylinder: 4 r To Find the Surface Area of the Triangle Net: find the area for each triangle and then add all of them together. If the pentagonal net is regular, find the area one pentagon using A=ap (a=apothem, p=perimeter) then add those sums together.
Answers:2 pi r squared + 2 pi r h
Answers:If M is the surface enclosed by the bounding pentagon M, then by the GaussBonnet Theorem, [M] K dA + [ M] k ds = 2 (M), where (i) K = 1/r = 1 is the Gaussian curvature of the surface (which for a sphere, is constant); (ii) k = 0 along the geodesic parts of the boundary and is the multiple (10 /180) of a Dirac function at each vertex (measuring an exterior angle of 10 at the vertex); (iii) (M) = V E + F = 1 is the Euler characteristic of the surface. In this very special case, the GaussBonnet formula simplifies to A + 5 (10 /180) = 2 1, where A is the area of M. So A = (2  50/180) = (310/180) = 31 /18. Note that the total surface area of the sphere is 4 = 72 /18. So M covers somewhat less than half the entire sphere.
Answers:I won't solve the problems for you but I'll be happy to give you the formulas for surface area. SA of a Cylinder:: 2*3.14*radius*height+2*3.14*radius^2 SA of Rectangular Solid:: 2(lw)+2(hw)+2(lh)
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