Question

Three equal cubes are placed adjacently in a row. Find the ratio of the total surface area of the new cubiod to that of the sum of the surface areas of the three cubes.

Answer 1

Let edge of each equal cubes = x
Then, surface area of one cube = $6x^2$
and surface area of three cubes = $3\times 6x^2=18x^2$
By placing the cubes in a row,
The length of newly formed cubiod (l) = 3x
Breadth (b) = x
and height (h) = x
$\therefore$ Surface area of the cubiod so formed
= $2(l\times b+b\times h+h\times l)$
= $2(3x\times x+x\times x+x\times 3x)$
= $2(3x^2+x^2+3x^2)=2\times 7x^2=14x^2$
$\therefore$ Ratio = $14x^2:18x^2$ = 7:9