Geometry & Mensuration
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Q.

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

 A.

\(1\) cm

 B.

\(\dfrac{23}{3}\) cm

 C.

\(1\) or \(\dfrac{23}{3}\) cm

 D.

None

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Solution:
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


Saikiran
 ()

Please some one can explain this doubt??