Q.3 The length, breadth and height of a room are 825 cm, 675 cm and 450 cm respectively. Find the longest tape which can measure the three dimensions of the room exactly.

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