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Question

The sum of the lengths of the twelve edges of a rectangular box is 140, and the distance from one corner of the box to the farthest corner is 21. What is the total surface area of the box?

Accepted Answer

Let a, b, and c be the dimensions of the box. It is given that 140=4a+4b+4c and 21=a^2+b^2+c^2 hence 35=a+b+c (1) and441=a^2+b^2+c^2 (2). Square both sides of (1) and combine with (2) to obtain align* 1225 & = (a+b+c)^2 &= a^2+b^2+c^2+2ab+2bc+2ca &= 441+2ab+2bc+2ca. align* Thus the surface area of the box is 2ab+2bc+2ca=1225-441=784.