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Get access to the detailed solutions to the previous years questions asked in IIM IPMAT exam
Given that, The surface area of a closed rectangular box, which is inscribed in a sphere, is 846 sq cm
So, 2(lb + bh + hl) = 846
And 4(l + b + h) = 144
(l + b + h) = 36
(l + b + h)2 = l2 + b2 + h2 + 2(lb + bh + hl)
1296 = (l2 + b2 + h2) + 846
450 = l2 + b2 + h2
We are told that this cuboid is inscribed in a sphere, the body diagonal of the cuboid equals the diameter of the sphere, this can be visualised as:
