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%opening
\newcommand{\snovdn}{Preverjanje znanja geometrije }
\newcommand{\predmdn}{mat}
\newcommand{\predmkaj}{domace\_naloge}
\newcommand{\stevilkadn}{19}
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\or j\or k\or l\or m\or n\or o\or p\or r\or s\or \v{s}%
\or t\or u\or v\or z\or \v{z}
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}
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% \renewcommand\abstractname{Povzetek}
% \date{17. november 2020}
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\title{%
\snovdn --- \stevilkadn. domača naloga
\\
\large Matematika, Gimnazija Bežigrad}
\author{\begin{tabular}{rl}
\textbf{Profesor:} & prof. Vilko Domajnko\\
\textbf{Avtor:} & Anton Luka Šijanec, 2. a
% \textbf{Avtor:} & Anton Luka Šijanec \\ & Member 2 \\ & Member 3
\end{tabular}}
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\begin{document}
\maketitle
\begin{abstract}
Ta dokument vsebuje domačo nalogo, ki zajema snov \textit{\snovdn}pri matematiki, in njene rešitve.
\end{abstract}
%\tableofcontents
\section{Učni list \textit{\snovdn}}
\begin{enumerate}[label=\textbf{\arabic*.}]
\item V enakokrakem trikotniku je zunanji kot ob osnovnici za \ang{24} manjši od zunanjega kota ob vrhu.
\begin{enumerate}[label=\textbf{\xslalph*)}]
\item Izračunaj notranje kote tega trikotnika.
$$\ang{360}=2(\gamma'-\ang{24})+\gamma' \rightarrow \gamma'=\ang{104}
\rightarrow \gamma=\ang{76}\wedge\alpha=\ang{52} \wedge \beta=\ang{52}$$
\item Izračunaj kot med simetralama obeh zunanjih kotov ob osnovnici tega trikotnika.
$$\ang{180}-\ang{128}=\ang{52}$$
\item Izračunaj kot med višino na krak in višino na osnovnico tega trikotnika.
$$180-(\ang{180}-\ang{52}-\ang{90})=\ang{142}\rightarrow\ang{180}-\ang{142}=\underline{\underline{\ang{38}}}$$
\end{enumerate}
\item
\begin{enumerate}[label=\textbf{\xslalph*)}]
\item Načrtaj trikotnik s podatki: $a=\cm{5}, t_c=\cm{4,5}, v_c=\cm{4}$.
\begin{center}
\begin{tikzpicture}
\tkzDefPoint(0,0){C} \tkzDefPoint(0,-4){V} \tkzDefPoint(0,-5){aH}
\tkzDefPoint(0,-4.5){tH}
\tkzCompass(C,V) \tkzLabelLine[below left](C,V){$v_c$}
\tkzDefMidPoint(C,V) \tkzGetPoint{M}\tkzCompass(C,M)\tkzCompass(V,M)
\tkzInterLC(C,V)(V,M) \tkzGetPoints{MAGAIN}{MX} \tkzCompass(V,MX)
\tkzDrawLine(C,MX) \tkzInterCC(M,MX)(MX,M) \tkzGetPoints{c}{d}
\tkzCompass(M,c)\tkzCompass(MX,c) \tkzCompass(M,d) \tkzCompass(MX,d)
\tkzInterLC(c,d)(C,aH) \tkzGetPoints{B}{BX} \tkzCompass(C,B)
\tkzDrawLine(C,B) \tkzLabelLine[below left](C,B){$a$}
\tkzInterLC(c,d)(C,tH) \tkzGetPoints{T}{TX} \tkzCompass(C,T)
\tkzDrawLine[red](C,T) \tkzLabelLine[below left, red](C,T){$t_c$}
\tkzDrawLine[blue](C,TX) \tkzLabelLine[left, blue](C,TX){$t_c$}
\tkzInterLC(c,d)(T,B) \tkzGetPoints{AX}{A} \tkzCompass(T,A)
\tkzDrawLine[red](A,C) \tkzLabelLine[below left, red](A,C){$b$}
\tkzInterLC(c,d)(TX,B) \tkzGetPoints{AX2}{A2} \tkzCompass(TX,A2)
\tkzDrawLine[blue](A2,C) \tkzLabelLine[below, blue](A2,C){$b'$}
\tkzLabelLine[below left, red](A,B){$c$} \tkzDrawLine(c,A2)
\tkzLabelLine[below left, blue](A2,B){$c$}
\tkzCompass[blue](C,TX) \tkzLabelPoint[blue](A2){$A$}
\tkzDrawPolygon[fill=blue!30, opacity=0.30](A2,B,C)
\tkzDrawPolygon[fill=red!30, opacity=0.30](A,B,C)
\tkzDrawPoints(C,B) \tkzLabelPoints[below left](C,B)
\tkzDrawPoints[blue](A2) \tkzDrawPoints[red](A)
\tkzLabelPoints[red, below left](A)
\end{tikzpicture}
\end{center}
\begin{center}
Obstajata \color{red}dve \color{blue}rešitvi\color{black}.
\end{center}
\item Načrtaj romb s podatki: $e+f=\cm{16}, \alpha=\ang{60}$.
\begin{center}
\begin{figure}[H]
\hspace*{-2cm}
\begin{tikzpicture}
\tkzDefPoint(0,0){O} \tkzDefPoint(16,0){C}
\tkzDrawLine(O,C) \tkzLabelLine[below](O,C){$\cm{16}$}
\tkzInterCC(O,C)(C,O) \tkzGetPoints{6}{7}
\tkzCompass(O,6) \tkzCompass(O,7) \tkzCompass(C,6) \tkzCompass(C,7)
\tkzDefMidPoint(6,C) \tkzGetPoint{3}\tkzCompass(6,3)\tkzCompass(C,3)
\tkzDrawLine(O,3) \tkzInterLC(O,C)(O,3) \tkzGetPoints{USELESS}{4}
\tkzCompass(O,4) \tkzInterCC(3,4)(4,3) \tkzGetPoints{USELESS}{1}
\tkzCompass(3,1)\tkzCompass(4,1)\tkzDefMidPoint(O,6) \tkzGetPoint{2}
\tkzCompass(O,2)\tkzCompass(6,2)\tkzDrawLine(C,2)
\tkzInterLL(O,1)(C,2) \tkzGetPoint{D} \tkzDrawLine(O,D)
\tkzInterLC(O,C)(D,C) \tkzGetPoints{USELESS}{A} \tkzDrawLine(A,D)
\tkzCompass(D,A) \tkzInterCC(A,D)(C,D) \tkzGetPoints{USELESS}{B}
\tkzCompass(A,B) \tkzCompass(C,B) \tkzDrawLine(A,B)\tkzDrawLine(C,B)
\tkzDrawPolygon[fill=red!30, opacity=.3](A,B,C,D)
\tkzLabelLine[below left](A,B){$a$}
\tkzLabelLine[below](C,B){$a$}
\tkzLabelLine[below left](C,D){$a$}
\tkzLabelLine[left](A,D){$a$}
\tkzLabelPoints[below left](O,C,D,A,B) \tkzDrawPoints(O,C,D,A,B)
\end{tikzpicture}
\end{figure}
\end{center}
\end{enumerate}
\item
\begin{enumerate}[label=\textbf{\xslalph*)}]
\item Dokaži, da je paralelogram, v katerem sta diagonali enako dolgi, pravokotnik.
$$
AC=BD\wedge BC=AD\rightarrow\bigtriangleup ABC\Tilde=\bigtriangleup BAD\rightarrow\measuredangle ABC=\measuredangle BAD\wedge\measuredangle ABC+\measuredangle BAD=\ang{180}\rightarrow
$$$$
2\cdot\measuredangle ABC=\ang{180}\rightarrow\measuredangle ABC=\ang{180}
$$
\item Enakokraki trapez z osnovnicama $a=\cm{18}$ in $c=\cm{8}$ je tangenten. Izračunaj dolžino višine na osnovnico.
$$
b=d\wedge\alpha=\beta\rightarrow\frac{a+c}{2}=b=\cm{13}\rightarrow
v_c=\sqrt{\cm{13}^2-\left(\frac{a-c}{2}\right)^2}=
$$$$
\sqrt{\cm{13}^2-\cm{5}^2}=\sqrt{\cm{144}^2}=\cm{12}
$$
\end{enumerate}
\item
\begin{enumerate}[label=\textbf{\xslalph*)}]
\item Naj bodo $A_1, A_2, ... A_9$ oglišča pravilnega devetkotnika.
\begin{enumerate}[label=\textbf{a_{\arabic*})}]
\item Izračunaj kot med daljicama $A_{1}A_{2}$ in $A_{2}A_{3}$.
$$\Sigma_\text{notranjih kotov}=
\left(n_{=8}\right)-2)\cdot\ang{180}=\ang{1080}
\rightarrow\alpha=\frac{\ang{1080}}{8}=\ang{135}$$
\item Izračunaj kot med daljicama $A_{1}A_{3}$ in $A_{3}A_{5}$.
$$\Sigma_\text{notranjih kotov}=
\left(n_{=4}\right)-2)\cdot\ang{180}=\ang{360}
\rightarrow\alpha=\frac{\ang{360}}{4}=\ang{90}$$
\end{enumerate}
\item Če enakokrakemu trikotniku $ABC$ očrtamo krožnico, pripada kraku $AC$ središčni kot \ang{75}. Izračunaj kote tega trikotnika.
$$\beta=\alpha=\frac{\ang{75}}{2}=\ang{37,5}\rightarrow\gamma=\ang{180}-2\cdot\ang{37,5}=\ang{105}$$
\end{enumerate}
\item Dokaži, da se v deltoidu diagonali sekata pod pravim kotom.
$$
\bigtriangleup ABD\Tilde=\bigtriangleup BCD\rightarrow
\overline{AP_{\left(\text{presečišče diagonal}\right)}}_{\left(v_A\right)}
=\overline{PC}_{\left(v_C\right)}\rightarrow
$$
$$
\bigtriangleup APD\Tilde=\bigtriangleup PCD\wedge P\in AC\wedge
\measuredangle CPD=\measuredangle DPA=\measuredangle APB=\measuredangle BPC
\rightarrow
$$
$$
\measuredangle CPD=\measuredangle DPA=\measuredangle APB=\measuredangle BPC
= \ang{90}
$$
\end{enumerate}
\begin{tabularx}{\textwidth}{|c||*{6}{Y|}}
\hline
\cellcolor{yellow}nadloga & \cellcolor{yellow}1 & \cellcolor{yellow}2 & \cellcolor{yellow}3 & \cellcolor{yellow}4 & \cellcolor{yellow}5 \\
\hline
\textbf{točke} & $2+2+2=\mathbf{6}$ & $3+3=\mathbf{6}$ & $3+3=\mathbf{6}$ & $(1,5+1,5)+3=\mathbf{6}$ & $\mathbf{8}$ \\
\hline
\end{tabularx}
\begin{center}
Rešiti je treba \textbf{ŠTIRI} nadloge.
$50 \% = 13$ točk, $100 \% = 26$ točk
\end{center}
\section{Zaključek}
Ta dokument je informativne narave in se lahko še spreminja. Najnovejša različica, torej PDFji in
\hologo{LaTeX}\footnote{Za izdelavo dokumenta potrebujete \texttt{f5ff} skript Antona Luke Šijanca, \texttt{TeXLive 2020} in operacijski sistem \texttt{DebIan/GNU/Linux}. Za urejanje priporočam urejevalnik besedil \texttt{vim} in pregledovalnik PDF dokumentov \texttt{evince}.}
izvorna koda, zgodovina sprememb in prejšnje različice, je na voljo v mojem šolskem Git repozitoriju na
\url{https://git.sijanec.eu/sijanec/sola-gimb-2} v mapi
\href{https://git.sijanec/sola-gimb-2/src/branch/master/\predmdn/\predmkaj/\stevilkadn/}{/\predmdn/\predmkaj/\stevilkadn/}. Povezava za ogled zadnje različice tega dokumenta v PDF obliki je \url{http://razor.arnes.si/~asija3/files/sola/gimb/2/\predmdn/\predmkaj/\stevilkadn/dokument.pdf} in/ali \url{https://git.sijanec.eu/sijanec/sola-gimb-2/raw/branch/master/\predmdn/\predmkaj/\stevilkadn/dokument.pdf}.
\if\razhroscevanje1
\section*{Razhroščevalne informacije}
Te informacije so generirane, ker je omogočeno razhroščevanje. Pred objavo dokumenta izklopite razhroščevanje. To naredite tako, da nastavite ukaz \texttt{razhroscevanje} na 0 v začetku dokumenta.
Grafi imajo natančnost \functionSamples\space točk na graf.
Konec generiranja dokumenta: \today\ ob \currenttime\footnote{To ne nakazuje dejanskega časa, ko je bil dokument napisan, temveč čas, ko je bi dokument generiran v PDF/DVI obliko. Isto velja za datum v glavi dokumenta. Če berete direktno iz LaTeX datoteke, bo to vedno današnji datum.}%\input|"date -Ins"
Dokument se je generiral R0qK1KR2 \SI{}{\second}.
\fi
% \item $$$$ aaasecgeninsaaa R0qK1KR2
\end{document}
