Question:
Three equal cubes are placed adjacently in a row. Find the ratio of a total surface area of the new cuboid to that of the sum of the surface areas of the three cubes.
Solution:
Length of the new cuboid = 3a
Breadth of the cuboid = a
Height of the new cuboid = a
The Total surface area of the new cuboid (TSA) = 2(lb + bh + hl)
(TSA)1 = 2(3a ∗ a + a ∗ a + a ∗ 3a)
$(\mathrm{TSA})_{1}=14 \mathrm{a}^{2}$
The Total Surface area of three cubes
$(\mathrm{TSA})_{2}=3 * 6 \mathrm{a}^{2}$
$(\text { TSA })_{2}=18 a^{2}$
Therefore, $(\text { TSA })_{1} /(\text { TSA })_{2}=14 a^{2} / 18 a^{2}$
Therefore, Ratio is 7: 9
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