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(A) Let the edge of each cube be $a$.When three such cubes are placed adjacently in a row,they form a cuboid.The dimensions of this new cuboid are: Le

Vedclass · Accepted Answer

$7:9$

Vedclass · Answer

(A) Let the edge of each cube be $a$.When three such cubes are placed adjacently in a row,they form a cuboid.The dimensions of this new cuboid are: Length $(l) = 3a$,Breadth $(b) = a$,and Height $(h) = a$.The total surface area of the new cuboid is given by the formula $2(lb + bh + lh)$.Substituting the values: $2(3a 	imes a + a 	imes a + 3a 	imes a) = 2(3a^2 + a^2 + 3a^2) = 2(7a^2) = 14a^2$.The surface area of one cube is $6a^2$. Therefore,the sum of the surface areas of the three cubes is $3 	imes 6a^2 = 18a^2$.The required ratio is $\frac{14a^2}{18a^2} = \frac{7}{9}$,which is $7:9$.

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

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