Question

determinati nr laturilor unui poligon regulat care are nr diagonalelor egal cu 9, 20 ,54 si 135​

Answer 1

 

Explicatii:

Intr-un poligon regulat dintr-un varf nu putem duce diagonale:

- la la el insusi  (este un nonsen)

- la varfurile vecine  (s-ar confunda cu laturile)

Rezulta ca intr-un poligon regulat cu n laturi si n varfuri

de la oricare varf putem duce diagonale doar la (n - 3) varfuri.

Daca vom calcula numarul de diagonale = n(n - 3) gresim deoarece fiecare diagonala a fost prinsa in calcul de 2 ori.

Rezulta ca formula corecta pentru calculul numarului de diagonale este:

nd = n(n-1)/2\\

.

$\displaystyle\bf\\a)\\\\\frac{n(n-3)}{2}=9\\\\n(n-3)=18\\\\n^2-3n-18=0\\\\(-3n=-6n+3n)\\\\(n^2-6n)+(3n-18)=0\\\\n(n-6)+3(n-6)=0\\\\(n-6)(n+3)=0\\\\n_1=6\\\\n2=-3\notin N\\\\\implies~n=6~laturi\\\\b)\\\\\frac{n(n-3)}{2}=20\\\\n(n-3)=40\\\\n^2-3n-40=0\\\\(-3n=-8n+5n)\\\\(n^2-8n)+(5n-40)=0\\\\n(n-8)+5(n-8)=0\\\\(n-8)(n+5)=0\\\\n_1=8\\\\n2=-5\notin N\\\\\implies~n=8~laturi$

.

$\displaystyle\bf\\c)\\\\\frac{n(n-3)}{2}=54\\\\n(n-3)=108\\\\n^2-3n-108=0\\\\(-3n=-12n+9n)\\\\(n^2-12n)+(9n-108)=0\\\\n(n-12)+9(n-12)=0\\\\(n-12)(n+9)=0\\\\n_1=12\\\\n2=-9\notin N\\\\\implies~n=12~laturi$

.

$\displaystyle\bf\\d)\\\\\frac{n(n-3)}{2}=135\\\\n(n-3)=270\\\\n^2-3n-270=0\\\\(-3n=-18n+15n)\\\\(n^2-18n)+(15n-270)=0\\\\n(n-18)+15(n-18)=0\\\\(n-18)(n+15)=0\\\\n_1=18\\\\n2=-15\notin N\\\\\implies~n=18~laturi$