question_answer

Three equal cubes are placed adjacently in a row. Find the ratio of total surface area of the new cuboid (formed by adjacent placing of cubes) to that of the sum of the surface areas of the three cubes.

A)  \[7:9\]    

B)  \[49:81\]        

C)  \[9:7\]             

D)  \[27:23\]

Correct Answer: A

Solution :

(a): Let the side of the cube be = a units Total surface area of 3 cubes \[=3\times 6{{a}^{2}}\]                                           \[=18\,\,{{a}^{2}}\] Total surface area of cuboid             \[=18\,\,{{a}^{2}}-4{{a}^{2}}=14{{a}^{2}}\] \[\therefore \]      Ratio\[=\frac{14{{a}^{2}}}{18{{a}^{2}}}=7:9\]