Đáp án:

 \(1302 \pm 2\left( {mm} \right)\)

Giải thích các bước giải:

Ta có: 

\(\begin{array}{l}
\overline {AB}  = \dfrac{{501 + 501 + 502}}{3} \approx 501\left( {mm} \right)\\
\overline {\Delta AB}  = \dfrac{{\left| {501 - 501} \right| + \left| {501 - 501} \right| + \left| {501 - 502} \right|}}{3} \approx 0\left( {mm} \right)\\
\Delta AB = 0 + 1 = 1\left( {mm} \right)\\
 \Rightarrow AB = 501 \pm 1\left( {mm} \right)\\
\overline {BC}  = \dfrac{{801 + 801 + 802}}{3} \approx 801\left( {mm} \right)\\
\overline {\Delta BC}  = \dfrac{{\left| {801 - 801} \right| + \left| {801 - 801} \right| + \left| {801 - 802} \right|}}{3} \approx 0\left( {mm} \right)\\
\Delta BC = 0 + 1 = 1\left( {mm} \right)\\
 \Rightarrow BC = 801 \pm 1\left( {mm} \right)\\
 \Rightarrow AB + BC = 1302 \pm 2\left( {mm} \right)
\end{array}\)