Three equal cubes are placed adjacently in a row. Find the ratio of to

1.	Three equal cubes are placed adjacently in a row. Find the ratio of total surface area of the new cuboid to that of the sum of the surface areas of the three cubes.
Answer» Given, Let edge length of three equal cubes = a Then, Sum of surface area of 3 cubes = 3 x 6a² = 18a² When these cubes are placed in a row adjacently they form a cuboid. Length of new cuboid formed = a + a + a = 3a Breadth of cuboid = a Height of cuboid = a Total surface area of cuboid = 2(lb x bh x hl) = 2(3a x a + a x a + a x 3a) = 2(3a² + a² + 3a²) = 2 x 7a²= 14a² Hence, = \(\frac{total\,surface\,area\,of\,new\,cuboid}{sum\,of\,surface\,area\,of\,3\,cuboids}\) = \(\frac{14}{18}\) = \(\frac{7}{9}\) = 7:9

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

Answer»

Given,

Let edge length of three equal cubes = a

Then,

Sum of surface area of 3 cubes = 3 x 6a² = 18a²

When these cubes are placed in a row adjacently they form a cuboid.

Length of new cuboid formed = a + a + a = 3a

Breadth of cuboid = a

Height of cuboid = a

Total surface area of cuboid = 2(lb x bh x hl) = 2(3a x a + a x a + a x 3a)

= 2(3a² + a² + 3a²) = 2 x 7a²= 14a²

Hence,

= \(\frac{total\,surface\,area\,of\,new\,cuboid}{sum\,of\,surface\,area\,of\,3\,cuboids}\) = \(\frac{14}{18}\) = \(\frac{7}{9}\) = 7:9