**Question:**

Find the length of the longest rod that can be placed in a room 12 m long, 9 m broad and 8 m high.

**Solution:**

Length of the room $=12 \mathrm{~m}$

Breadth $=9 \mathrm{~m}$

Height $=8 \mathrm{~m}$

Since the room is cuboidal in shape, the length of the longest rod that can be placed in the room will be equal to the

length of the diagonal between opposite vertices.

Length of the diagonal of the floor using the Pythagorus theorem

$=\sqrt{1^{2}+\mathrm{b}^{2}}$

$=\sqrt{(12)^{2}+(9)^{2}}$

$=\sqrt{144+81}$

$=\sqrt{225}$

$=15 \mathrm{~m}$

i.e., the length of the longest rod would be equal to the length of the diagonal of the right angle triangle of base $15 \mathrm{~m}$ and altitude $8 \mathrm{~m}$.

Similarly, using the Pythagorus theorem, length of the diagonal

$=\sqrt{15^{2}+8^{2}}$

$=\sqrt{225+64}$

$=17 \mathrm{~m}$

$\therefore$ The length of the longest rod that can be placed in the room is $17 \mathrm{~m}$.