Geometry & Mensuration

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The external length, breadth and height of a closed box are 10 cm, 9 cm and 7 cm respectively. The total inner surface area of the box is 262 sq. cm. If the walls of the box are of uniform thickness \(t\) cm, then \(t\) equals


\(1\) cm


\(\dfrac{23}{3}\) cm


\(1\) or \(\dfrac{23}{3}\) cm



 Hide Ans

Option(A) is correct

The inner dimensions will be \((10-2t), (9-2t)\) and \((7-2t).\) 

Thus the inner surface area will be

\(=2 \times [(10 - 2t)(9 - 2t) + (10 - 2t)(7 - 2t) + (9 - 2t)(7 - 2t)]\)

\(= 262\)
This will boil down to a quadratic which can be solved and we will get two values of \(t\). One should not solve the quadratic but make use of the options. Substituting \(t = 1\),  we  see  that  it  satisfies  the  above  equation.
Obviously \(\dfrac{23}{3}\)is not worth considering as it is greater than 7 and the thickness has to be less than the least
outer dimension, i.e. 7.

(2) Comment(s)

Shreyash Okhade

why we have taken 2t in the solution y we cannot simply take t as it is


Please some one can explain this doubt??