一个表格设置产生不同的背景宽度

Question

我必须使用相同的表格设置来装箱。两个问题：

（1）灰度背景的宽度不同。

（2）无法通过 添加多行方程\begin{align*}。（错误： \begin{align}仅允许在段落中使用。）

我需要两个不同的\SetTblrInner[mytblr]and吗\SetTblrInner[mytblr2]？下面是上述错误的示例：

\documentclass{scrartcl}
\usepackage[margin=1.0cm]{geometry}

\renewcommand{\familydefault}{\sfdefault}
\usepackage{tikz}
\usetikzlibrary{angles,
                positioning,
                quotes,
                shapes.geometric,
                }
\tikzset{baseline=(current bounding box.center),
    my angle/.style = {draw, font=\small,
                       angle eccentricity=1.55, angle radius=3mm},
    base/.style = {draw, thick},
    dashline/.style = {densely dashed, thick},
    every edge quotes/.style = {auto=right, inner sep=1pt, font=\small},
    Rect/.style args = {#1/#2}{base, minimum height=#1, minimum width=#2},
    Trap/.style args = {#1/#2}{base, trapezium,
                        trapezium left angle=#1, trapezium right angle=#2,
                        minimum width=30mm}
        }
\usepackage{tcolorbox}
\tcbset{fonttitle=\bfseries, coltitle=white, colback=white,
        colframe=black,nobeforeafter,sharp corners,
        sidebyside, sidebyside gap=3mm,
        lower separated=false
        }

\usepackage{sfmath}
\usepackage{tabularray}
 \usepackage{amsmath}
\NewTblrEnviron{mytblr} % <--- added by @Jasper Habicht
\SetTblrInner[mytblr]{             
    hline{1,2,Z} = 1pt, vline{1,Z} = 1pt,
    hline{even[3-X]} = {dashed, fg=gray, 0.5pt},
    colspec = {Q[l, bg=gray!10, font=\small\bfseries]
               X[c, bg=gray!10]
               Q[c, wd=33mm]},
    colsep  = 3pt,
    cell{1}{1} = {c=3}{l},
    cell{odd[3]}{1} = {c=2}{c, fg=black, mode=math},
    row{1}  = {bg=gray!30, mode=text},
    row{2-Z}= {belowsep=4pt},
    row{odd[3-Z]} = {ht = 19mm}
                      }

\begin{document}
\noindent
\begin{tcolorbox}[title = Title]
    \begin{mytblr}{}
    Square            &   &   \\
    Area                &   &   \\
    \begin{align*}
    A&=a\cdot a\\
     & = a^{2}
    \end{align*}&=a\cdot a  a^2    &   &   
            \tikz{\node (a) [Rect=14mm/14mm] {};
                  \path (a.south west)  to["a"] (a.south east)
                        (a.south east)  to["a"] (a.north east);
                }   \\
    Circumference       &   &   \\
   u=4\cdot a           &   &   
            \tikz{\node (a) [Rect=14mm/14mm] {};
                  \path (a.south west)  to["a"] (a.south east)
                        (a.south east)  to["a"] (a.north east);
                }   \\
    \end{mytblr}
    
\medskip
            \begin{mytblr}{}
    Pythagorean Theorem                    &   &   \\
    \textnormal{In any right-angled\\
     triangle it holds that}                       &   &   \\
    a^{2}+b^{2}=c^{2}  &   & 
            \tikz{[very thick]
\draw (-1.5,0)coordinate[](A)
    --(1.2,0)coordinate[](B)
    --(-1.5,1.2)coordinate[](C)
    --cycle;
% rechte Winkel
    \pic[draw,fill=gray!10,fill opacity=0.5,angle radius=4mm,"$\pmb{\cdot}$" opacity=1]
      {angle=B--A--C};
% Streckenbezeichnungen
\path (A) -- node [midway,below] {\scriptsize b} (B);
\path (A) -- node [midway,left] {\scriptsize a} (C);
\path (B) -- node [midway,above] {\scriptsize c} (C);}   \\
    \end{mytblr}
\end{tcolorbox}
\end{document}

Answer 1

像这样？

您的代码中有三个错误：

• 而不是align*你应该使用aligned
• 对于“勾股定理”中的两行文本，您应该删除\textnormal花括号前的命令（参见下面的 MWE）
• 之后的虚假数学代码align*（删除它！）

无关：对于右上角的标记，我将使用：

• 对于形状边框的标记，你应该使用我在 ma 答案中绘制图像时使用的方法，然后你在“正方形”部分复制
• 我要添加的角度样式tikzset（您可以根据自己的意愿更改样式名称）

\documentclass{scrartcl}
\usepackage[margin=1.0cm]{geometry}

\renewcommand{\familydefault}{\sfdefault}
\usepackage{tikz}
\usetikzlibrary{angles,
                positioning,
                quotes,
                shapes.geometric,
                }
\tikzset{baseline=(current bounding box.center),
    my angle/.style = {draw, font=\small,
                       angle eccentricity=1.55, angle radius=3mm},
    base/.style = {draw, thick},
    dashline/.style = {densely dashed, thick},
    every edge quotes/.style = {auto=right, inner sep=1pt, font=\small},
    Rect/.style args = {#1/#2}{base, minimum height=#1, minimum width=#2},
    Trap/.style args = {#1/#2}{base, trapezium,
                        trapezium left angle=#1, trapezium right angle=#2,
                        minimum width=30mm},
    ANG/.style = {draw, fill=gray!10, angle radius=4mm, 
                  angle eccentricity=#1},
        }
\usepackage{tcolorbox}
\tcbset{fonttitle=\bfseries, coltitle=white, colback=white,
        colframe=black,nobeforeafter,sharp corners,
        sidebyside, sidebyside gap=3mm,
        lower separated=false
        }

\usepackage{sfmath}
\usepackage{tabularray}
 \usepackage{amsmath}
\NewTblrEnviron{mytblr} % <--- added by @Jasper Habicht
\SetTblrInner[mytblr]{
    hline{1,2,Z} = 1pt, vline{1,Z} = 1pt,
    hline{even[3-X]} = {dashed, fg=gray, 0.5pt},
    colspec = {Q[l, bg=gray!10, font=\small\bfseries]
               X[c, bg=gray!10]
               Q[c, wd=33mm]},
    colsep  = 3pt,
    cell{1}{1} = {c=3}{l},
    cell{odd[3]}{1} = {c=2}{c, fg=black, mode=math},
    row{1}  = {bg=gray!30, mode=text},
    row{2-Z}= {belowsep=4pt},
    row{odd[3-Z]} = {ht = 19mm}
                      }

\begin{document}
\noindent
\begin{tcolorbox}[title = Title]
    \begin{mytblr}{}
    Square              &   &   \\
    Area                &   &   \\
    \begin{aligned}
    A&=a\cdot a\\
     & = a^{2}
    \end{aligned}       &   &
            \tikz{\node (a) [Rect=14mm/14mm] {};
                  \path (a.south west)  to["a"] (a.south east)
                        (a.south east)  to["a"] (a.north east);
                }   \\
    Circumference       &   &   \\
   u=4\cdot a           &   &
            \tikz{\node (a) [Rect=14mm/14mm] {};
                  \path (a.south west)  to["a"] (a.south east)
                        (a.south east)  to["a"] (a.north east);
                }   \\
    \end{mytblr}

\medskip
            \begin{mytblr}{}
    Pythagorean Theorem                     &   &   \\
    {In any right-angled\\
     triangle it holds that}                &   &   \\
    a^{2}+b^{2}=c^{2}  &   &
            \tikz{[very thick]
    \draw (0,0)coordinate[](A)
                -- (3,0)coordinate[](B)
                -- (0,1.5)coordinate[](C)
                -- cycle;
% rechte Winkel
    \pic[ANG=0.5, "$\pmb{\cdot}$"] {right angle=B--A--C};
% Streckenbezeichnungen
    \path   (A) edge ["$b$"] (B)
            (B) edge ["$c$"] (C)
            (C) edge ["$a$"] (A);
            }   \\
    \end{mytblr}
\end{tcolorbox}
\end{document}