Perimeter of a regular pentagon in terms of its diagonal If the length of one of the diagonals of a regular pentagon is $d,$ how can we represent the perimeter of the pentagon in terms of $d$?
 A: As the perpendicular bisectors of the sides of  the regular pentagon are its  axes of symmetry, a fast angles  computation shows the angle made by a side and the adjacent diagonal is equal to π/5, so we deduce that, denoting $s$ the length of a side, $d$ the length of a diagonal
$$\frac d2=s\cos\frac\pi 5$$
 and the perimeter $p$ is
$$p=5s=\frac{5d}{2\cos\frac\pi 5}=\frac{5(\sqrt 5-1)d}{2}.$$
A: Drawing the diagram gives us the answer;

$\beta $ and $\gamma$ can be found as shown below;
let $\beta  =x$
in $\triangle$ABE , ${AB} = AE\implies  \angle{ ABE} = \angle AEB = 108 - x$
$2(108-x)+108 = 180$
$x = 72^\circ$
In $\triangle BED$, $EB =BD = d$ and $\angle BED =\angle BDE = 72^\circ$
$\implies \gamma= 180 - 2(72) = 36^\circ$
Now applying the sine rule gives us;
$\dfrac{\sin(36)}{s}=\dfrac{\sin(72)}{d}=\dfrac{\sin(72)}{d}$
where $s= ED$
$s= d\cdot \dfrac{\sin(36)}{\sin(72)}$
$s= d\cdot \dfrac{\sin(36)}{2\sin(36)\cos(36)}$
$s= \dfrac{d}2\cdot \sec(36)$
So the total perimeter is $p = \dfrac{5d}2\sec(36^\circ)$
A: In Euclid you have a theorem that considers two angles subtended by an arc of a circle: one with its vertex at the center of the circle, and the other with its vertex and any point on the circle – any point at all other circle; it doesn't matter which one. The theorem says the measure of the latter angle is exactly half that of the former.
Circumscribe a circle around the regular pentagon. Look at the point on the circle that is an endpoint of the diagonal you are considering. That will be the vertex of several angles: (1) the angle between the tangent line to the circle and one of the sides; (2) the angle between that side and one of the diagonals of the pentagon; (3) the angle between that diagonal and the next diagonal; (4) the angle between that last diagonal and the next side of the pentagon; (5) the angle between that last side and the tangent line. The five arcs of the circle are equally long; therefore the measures of these five angles are equal; therefore each is $180^\circ/5 = 36^\circ.$
Those two diagonals and one side have three angles adding up to $180^\circ$ and two of them have equal measures and the third is $36^\circ;$ therefore the other two are $72^\circ.$
So how long is the short side of that isosceles triangle as a function of the length of one of the diagonals? Call the length of the diagonal $d$ and the length of the side we seek $s.$
If you bisect one of those $72^\circ$ angles you cut off two other triangles, the smaller of which has the same three angles and hence the same shape as the whole triangle, with two sides of length $s.$ The third side of that small triangle must have length $\dfrac s d\cdot s,$ by similarity. The larger of those two triangles cut off by bisection has a side of length $s.$ That side of length $s,$ plus the aforementioned side of length $\dfrac s d \cdot s,$ make up a diagonal of length $d.$ Therefore
$$
s + \frac s d \cdot s = d
$$
$$
\frac s d + \left( \frac s d\right)^2 = 1
$$
Solving this quadratic equation for $s/d$ yields
$$
\frac s d = \frac{\sqrt 5 -1} 2.
$$
Hence
$$
s = \frac {d(\sqrt 5 -1 )} 2
$$
and the circumference of the pentagon is therefore
$$
\frac{5d(\sqrt 5 -1)} 2.
$$
