a_i =\frac{180^o(n-2)}{n}\rightarrow 165^o .n= 180^on - 360^o\therefore n = 24\\
D = \frac{n(n-3)}{2}=\frac{24(24-3)}{2}\therefore \boxed{\color{red}D=252}

S_i = 180^o(n-2)\rightarrow 180^o(2n-2) = 4S_i\\360n - 360 = 4(180(n-2)\rightarrow 360n-360 = 720n-1440\\
\therefore n = 3\\
D = \frac{n(n-3)}{2} = \frac{3(3-3)}{2} \therefore\boxed{\color{red}D=0}

n = n. lados
m = n. vértices consecutivos
n_d = m.n - \frac{(m+1)(m+2)}{2}\rightarrow 21 =(4n)-\frac{(4+1)(4+2)}{2}\rightarrow \\
21 = 4n - 15\rightarrow \therefore \boxed{\color{red}n =9}

a_i = 180^o\frac{(n-2)}{2} = 180.\frac{5}{7} = \frac{900^o}{7}\\
\measuredangle FGE = \frac{1}{2}(180-\frac{900}{7}) = \frac{180^o}{7}\\
\measuredangle AGH = \frac{900}{7}-\frac{180}{7} =...

a_6 = \frac{180 (n-2)}{n}=\frac{180.4}{6}=120^o\\
H=BC\cap AF\\
\triangle AHB (equilátero) \implies HB=HA=1\\
G=PE\cap BC\rightarrow PGCD(paralelogramo)\implies GP=CD =2\\
\therefore AH = 1~ e~ HF =...

D_1-D_2 = 15\rightarrow \frac{n_1(n_1-3)}{2}-\frac{n_2(n_2-3)}{2}=15\rightarrow \boxed{n_1^2-3n_1 - n_2^2+3n_2=30}(I)\\
\boxed{n_1-n_2=3\rightarrow n_1 = 3+n_2}(II)\\
(II) em...

\frac{2(n-2)}{2} = \frac{n(n-3)}{2}\rightarrow 2n-4 = n^2-3n\\
n^2-5n+4 = 0\rightarrow n = 1 (n \geq 3)~ ou~n = 4\\
\therefore \boxed{\color{red}Quadrado}

Como \angle AOB = 120^{\circ} , a lei dos cossenos no \triangle AOB diz que:

AB^2 = 2R^2 + 2R^2 \cdot \frac 12 = 3R^2 \implies AB = R \sqrt3

Potência de M :

1 \cdot (2R-1) = 2 \cdot (R \sqrt3 -...

n+\frac{n(n-3)}{2}=28\rightarrow 2n +n^2-3n = 56\\
n^2-n-56=0\rightarrow x= -7(n\geq 3)~ ou~ \boxed{\color{red}x=8}

\frac{((n+1)((n+1)-3)}{2}=\frac{n(n-3)}{2}+6\\
n^2-2n+n-2=n^2-3n+12\rightarrow 2n=14 \therefore \boxed{\color{red}n=7}

a_i = 180^o\frac{(n-2)}{n}\\
a_e = \frac{360^o}{n}\\
\frac{a_i}{a_e}=\frac{11}{2}\rightarrow \frac{n-2}{2}=\frac{11}{2}\rightarrow 2n-4 = 22\\
\therefore n = 13\rightarrow D_m =...

n-2\rightarrow D-19\\
\frac{(n-2))(n-2-3)}{2}=\frac{n(n-3)}{2}-19\rightarrow n^2-5n-2n+10=n^2-3n-38\\
4n = 48 \therefore \boxed{\color{red}n = 12}

OO_1=d\\\frac{FF_1+CC_1}{2} = d \\
\frac{A_1A+D_1D}{2}=d\\
\frac{EE_1+B_1B}{2}=d\\
\therefore \underbrace{FF_1+CC_1+A_1A+D_1D+EE_!+B_1B}_{=18}=2(3d)\\
18 = 6d \therefore \boxed{\color{red}d = 3}...

a_i = \frac{180(n-2)}{n}=\frac{180.3}{5}=108^o\\AM = AB =ME\\\measuredangle MDE = 54^o (bissetriz~ de~ \measuredangle D)\\\begin{cases} 48^{\circ} + 2\alpha = 180^{\circ} \Leftrightarrow \alpha =...

45^{\circ}=\measuredangle EMC=\measuredangle EBC+\measuredangle ACB=\frac{1}{2}\left(\widehat{EC}+\widehat{AB}\right)\\...

S_{ai} =180^o(n-2)\\
S_{ae} =360^o\\
S_{ac} = 360^0\\180(n-2)+360+360 = 1260\rightarrow180n -360+360+360 = 1260\\
180 n = 900 \therefore \boxed{\color{red}n=5}

a_i = \frac{180^o(n-2)}{n}\\
n_1 =\frac{2n}{3}\\

\frac{180 (n-2)}{n}-10 = \frac{180( \frac{2n}{3}-2)}{\frac{2n}{3}}\\
\frac{180 (n-2)}{n}-10 = \frac{270( \frac{2n}{3}-2)}{n}\\
\cancel{180n} -360 -...

D_m = \frac{n(n-1)}{2}\\
D = \frac{n(n-2)}{3}\\
D_m=2D\rightarrow\frac{n(n-1)}{2}=2\frac{n(n-2)}{3}\rightarrow3n^2-3n=4n^2-8n\\
n^2-5n=0\therefore n(n-5) = 0 \therefore \boxed{\color{red}n=5}

r = razão\\S_i=180^o(n-2)=180^o(11-2) = 1620^o\\
\alpha = \alpha -5r (\text{menor ângulo interno}) \therefore \alpha +5r(\text{maior Ângulo interno})\\
\alpha -5r+ \alpha -4r +\alpha -3r+ \alpha...

S_{ai} = 90+90+36+4\alpha = 216+4\alpha(I)\\
n=7 \rightarrow S_{ai} = 180(7-2) = 900(II)\\
(I)em(II): 216+4\alpha = 900\therefore \alpha =171^o\\
a_i =\frac{180(n-2)}{n}\rightarrow 171n = 180n -...

m = n. lados consecutivos
n = número de lados do polígono
D_m = mn-\frac{m(m+1)}{2}=50\rightarrow 100 = 8n - 20 \rightarrow\\
\therefore n = 15\\
a_e = \frac{360^o}{n}= \frac{360^o}{15} \therefore...

a_e = \frac{360}{n} \rightarrow Menor ~ n \rightarrow Maior ~a_e\\
\frac{360}{n}-\frac{360}{n+1}=12\rightarrow360n+360 - 360n = 12n^2+12n\rightarrow\\
12n^2+12n-360 = 0\rightarrow n^2+n - 30 = 0...

\mathsf{Traçar~ DB\\
\angle DAB = 90^o - \theta=\angle AHB(\triangle AHB:isósceles)\\
B: ponto~ médio ~AM~\\
BH =BM\\
\overset{\LARGE{\frown}}{AC} =\overset{\LARGE{\frown}}{AF}\therefore \angle ABF =...

S_{ai} = {180^o(7-2)}\rightarrow 180(4) = 900\\
\measuredangle AHC = 180^o - 67^o = 113^o\\
\alpha + \beta + \theta+\omega + 90^o+90^o+ 113^o = 900^o\\
\boxed{\color{red}\alpha + \beta +...

\frac{n(n-3)}{2}=\frac{(n-1)(n-1-3)}{2} + 7\rightarrow n^2-3n=n^2-4n-n+4+14\\
2n = 4 \therefore n =9 \implies a_i = \frac{180^o(n-2)}{n} = \frac{180.7}{9}\therefore \boxed{\color{red}a_i =140^o}

∠CFI=108^∘ , portanto ∠CDF=∠DFC=180^∘−108^∘=72.

Portanto ∠FCD=180−(2⋅72^∘)=36^∘ . \rightarrow ∠GCD=108^∘+36^∘=144^∘ .

Como △GCD é isósceles \rightarrow ∠CDG=\frac{180^∘−144^∘}{2}=18^∘ .

então...

Suponha que cada equipe seja um vértice de um polígono com n vértices. Cada correspondência é representada por um lado ou diagonal do polígono, associando aleatoriamente um índice i = 1,2,…, n para...

Considere o quadrilátero PHIF, como HF||PI, temos: \angle PFH=\angle FPI=30° , além disso, é fácil ver que \angle IFH=\angle IHF=45° , então \angle IFP=45°-30°=15° e \angle HIP=\angle IHF=45°.
Seja...

\frac{n(n-3)}{2}=2(n+2)\rightarrow 4n+8 = n^2 - 3n\rightarrow \\
n^2-7n -8 = 0 \rightarrow \cancel{n = -1} ou~\boxed{\color{red}n = 8}