The length, width, and height of a cuboidal room are respectively \(a\), \(b\), and \(c\) units, and given that \(a + b + c = 25\) and \(ab + bc + ca = 240.5\), what is the length of the longest rod that can be placed inside the room?

Q.The length, width, and height of a cuboidal room are respectively \(a\), \(b\), and \(c\) units, and given that \(a + b + c = 25\) and \(ab + bc + ca = 240.5\), what is the length of the longest rod that can be placed inside the room?

The longest rod that can be placed inside the room will be equal to the length of the room’s diagonal. \(\therefore\) The length of the rod will be \(\sqrt{a^2 + b^2 + c^2}\) units Now, \(a^2 + b^2 + c^2 = (a + b + c)^2 - 2(ab + bc + ca)\) \(= 25^2 - 2 \times 240.5\) \(= 625 - 481\) \(= 144\) \(\therefore\) The length of the rod will be \(\sqrt{144}\) units = 12 units (Answer)