1
$\begingroup$

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 $d$?

$\endgroup$
  • $\begingroup$ have you tried drawing a diagram and seeing what happens? $\endgroup$ – The Integrator Jun 1 '18 at 15:56
  • $\begingroup$ What have you attempted? $\endgroup$ – Andrew Li Jun 1 '18 at 15:56
  • $\begingroup$ In the multiple choices that I had, all of them had either a sine 36 or a sine 54. So I tried law of sines, but I ended up with 2 sines, but all the options had only one sine. Sorry if my question is too trivial. $\endgroup$ – Nithya S Jun 1 '18 at 15:58
  • $\begingroup$ What do you mean by 2 sines? Also you can easily take out the value of sine 36 or sine 54 $\endgroup$ – Love Invariants Jun 1 '18 at 16:03
  • $\begingroup$ For example, one of the answer was P = 5d/sin 36. They have left it at that not taking the value of sin 36. $\endgroup$ – Nithya S Jun 1 '18 at 16:04
2
$\begingroup$

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 $$p=5s=\frac{5d}{2\cos\frac\pi 5}=\frac{5(\sqrt 5-1)d}{2}.$$

$\endgroup$
  • $\begingroup$ it should be $\sec{\frac\pi5}$. please correct it $\endgroup$ – The Integrator Jun 1 '18 at 17:07
  • 1
    $\begingroup$ @TheIntegrator: Oops! I should not compute directly on screen… Fixed. Thanks for pointing it! $\endgroup$ – Bernard Jun 1 '18 at 17:53
  • $\begingroup$ thanks for the edit , was bugging me watching that XD $\endgroup$ – The Integrator Jun 1 '18 at 17:54
  • $\begingroup$ How you evaluated $\cos\frac\pi5$ is not clear from this. $\endgroup$ – Michael Hardy Jun 1 '18 at 18:09
0
$\begingroup$

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 $$ \frac{5d(\sqrt 5 -1)} 2. $$

$\endgroup$
0
$\begingroup$

Drawing the diagram gives us the answer;

enter image description here

$\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 $p = \dfrac{5d}2\sec(36^\circ)$

$\endgroup$
  • $\begingroup$ Thank you very much. That was so clear. $\endgroup$ – Nithya S Jun 2 '18 at 0:16
  • $\begingroup$ @NithyaS if you found that helpful , consider voting and accepting the answer as the correct one. $\endgroup$ – The Integrator Jun 2 '18 at 7:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.