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Answer :

The longest tape required to measure the three dimensions of the room = HCF of 825, 675 and 450
Prime factorisations of 825, 675 and 450 are

$$
\begin{array}{|l}
\llap{3~~~~} 825 \\ \hline
\llap{5~~~~} 275 \\ \hline
\llap{5~~~~} 55 \\ \hline
\llap{11~~~~} 11 \\ \hline
1
\end{array}
$$
$$
\begin{array}{|l}
\llap{3~~~~} 675 \\ \hline
\llap{3~~~~} 225 \\ \hline
\llap{3~~~~} 75 \\ \hline
\llap{5~~~~} 25 \\ \hline
\llap{5~~~~} 5 \\ \hline
3
\end{array}
$$
$$
\begin{array}{|l}
\llap{2~~~~} 450 \\ \hline
\llap{3~~~~} 225 \\ \hline
\llap{3~~~~} 75 \\ \hline
\llap{5~~~~} 25 \\ \hline
\llap{5~~~~} 5 \\ \hline
3
\end{array}
$$
825 = 3 x 5 x 5 x 11

675 = 3 x 3 x 3 x 5 x 5

450 = 2 x 3 x 3 x 5 x 5

HCF = 3× 5× 5 = 75
Hence the required tape = 75cm

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