The sum of perimeters of the six faces of a cuboid is $72 \text{ cm}$ and the total surface area of the cuboid is $16 \text{ cm}^2$. Find the longest possible length that can be kept inside the cuboid

### Concept & Formula The longest possible length inside a cuboid is its spatial diagonal. We use the same algebraic identity as the previous problem, relating the sum of dimensions to the diagonal and surface area. $$ \text{Diagonal} = \sqrt{l^2 + b^2 + h^2} $$ $$ (l+b+h)^2 = l^2 + b^2 + h^2 + \text{Surface Area} $$ ### Step-by-Step Solution 1. **Analyze the sum of perimeters:** A cuboid has 6 faces: 2 of $(l \times b)$, 2 of $(b \times h)$, and 2 of $(h \times l)$. Perimeters are $2(l+b)$, $2(b+h)$, and $2(h+l)$. Sum of all 6 perimeters = $2[2(l+b) + 2(b+h) + 2(h+l)] = 4l + 4b + 4b + 4h + 4h + 4l = 8(l+b+h)$. $8(l+b+h) = 72 \implies (l+b+h) = 9 \text{ cm}$. 2. **Use the surface area given:** $2(lb + bh + hl) = 16 \text{ cm}^2$. 3. **Apply the algebraic identity to find the sum of squares:** $(l+b+h)^2 = (l^2 + b^2 + h^2) + 2(lb + bh + hl)$ $9^2 = (l^2 + b^2 + h^2) + 16$ $81 = (l^2 + b^2 + h^2) + 16 \implies (l^2 + b^2 + h^2) = 65$ 4. **Calculate the longest length (diagonal):** $\text{Diagonal} = \sqrt{65} \approx 8.062 \text{ cm}$. The closest given option is $8.05 \text{ cm}$. ### Exam Strategy & Shortcut Recognize that "sum of perimeters of 6 faces" always equals $8 \times (\text{sum of length, breadth, and height})$. Thus, $\text{Sum of sides} = \frac{72}{8} = 9$. Then, $\text{Diagonal} = \sqrt{9^2 - 16} = \sqrt{65} \approx 8.06$. ### Common Pitfall Incorrectly mapping the sum of the perimeters. Some students assume the sum of the 12 edges is $4(l+b+h)$ and equate that to 72, which leads to a completely wrong dimension sum. ### Final Answer Therefore, the correct answer is **8.05 cm**.