Item:Q7812: Difference between revisions

@@ Property / subclass of @@
+regular convex polygon
@@ Property / subclass of: regular convex polygon / rank @@
+Normal rank
@@ Property / subclass of: regular convex polygon / qualifier @@
+follows: square
@@ Property / subclass of: regular convex polygon / qualifier @@
+followed by: regular hexagon
@@ Property / subclass of @@
+constructible polygon
@@ Property / subclass of: constructible polygon / rank @@
+Normal rank
@@ Property / subclass of @@
+dodecahedral pentagonal polytope
@@ Property / subclass of: dodecahedral pentagonal polytope / rank @@
+Normal rank
@@ Property / subclass of @@
+icosahedral pentagonal polytope
@@ Property / subclass of: icosahedral pentagonal polytope / rank @@
+Normal rank
@@ Property / subclass of @@
+equidiagonal pentagon
@@ Property / subclass of: equidiagonal pentagon / rank @@
+Normal rank
@@ Property / has quality @@
+perimeter
@@ Property / has quality: perimeter / rank @@
+Normal rank
@@ Property / has quality: perimeter / qualifier @@
+defining formula:  $P=5a$ 
P=5a
@@ Property / has quality @@
+area of plane shape
@@ Property / has quality: area of plane shape / rank @@
+Normal rank
@@ Property / has quality: area of plane shape / qualifier @@
+defining formula:  $A={\frac {5}{4}}t^{2}\mathop {\mathrm {ctg} } \,{\frac {\pi }{5}}={\frac {{\sqrt {5}}{\sqrt {5+2{\sqrt {5}}}}}{4}}t^{2}={\frac {5}{12}}Rd$ 
A=\frac{5}{4} t^2 \mathop{\mathrm{ctg}}\, \frac{\pi}{5} = \frac{\sqrt 5 \sqrt{5 + 2 \sqrt{5}}}{4} t^2 = \frac{5}{12}Rd
@@ Property / has quality @@
+inradius
@@ Property / has quality: inradius / rank @@
+Normal rank
@@ Property / has quality: inradius / qualifier @@
+defining formula:  $r={\frac {t}{2\tan(\pi /5)}}={\frac {t}{2{\sqrt {5-{\sqrt {20}}}}}}$ 
r=\frac{t}{2\tan ( \pi /5)}=\frac{t}{2\sqrt{5-\sqrt{20}}}
@@ Property / has quality @@
+circumradius
@@ Property / has quality: circumradius / rank @@
+Normal rank
@@ Property / has quality: circumradius / qualifier @@
+defining formula:  $R={\frac {a}{2}}\csc {\frac {\pi }{5}}$ 
R = \frac{a}{2}\csc\frac\pi{5}

Item:Q7812: Difference between revisions

P = 5 a {\displaystyle P=5a}

A = 5 4 t 2 c t g π 5 = 5 5 + 2 5 4 t 2 = 5 12 R d {\displaystyle A={\frac {5}{4}}t^{2}\mathop {\mathrm {ctg} } \,{\frac {\pi }{5}}={\frac {{\sqrt {5}}{\sqrt {5+2{\sqrt {5}}}}}{4}}t^{2}={\frac {5}{12}}Rd}

r = t 2 tan ⁡ ( π / 5 ) = t 2 5 − 20 {\displaystyle r={\frac {t}{2\tan(\pi /5)}}={\frac {t}{2{\sqrt {5-{\sqrt {20}}}}}}}

R = a 2 csc ⁡ π 5 {\displaystyle R={\frac {a}{2}}\csc {\frac {\pi }{5}}}

$P=5a$

$A={\frac {5}{4}}t^{2}\mathop {\mathrm {ctg} } \,{\frac {\pi }{5}}={\frac {{\sqrt {5}}{\sqrt {5+2{\sqrt {5}}}}}{4}}t^{2}={\frac {5}{12}}Rd$

$r={\frac {t}{2\tan(\pi /5)}}={\frac {t}{2{\sqrt {5-{\sqrt {20}}}}}}$

$R={\frac {a}{2}}\csc {\frac {\pi }{5}}$