Question

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

Answer 1

Answer:

→ 7 : 9 .

Step-by-step explanation:

Let the side of a cube = a

TSA of one cube = 6a²

TSA of 3 such cubes = 3×6a² = 18a²

If 3 cubes are placed adjacent to each other, dimension of resulting cuboid is

→ length = 3a

→ breadth = a

→ height = a

Therefore, TSA = 2(lb+bh+lh)

= 2(3a×a + a×a + a×3a)

= 2(3a² + a² + 3a²)

= 2 × 7a²

= 14a².

$Ratio = \sf \frac{TSA\ of\ cuboid}{TSA\ of\ cubes} = \frac{14a^2}{18a^2} = \frac{7}{9}=\boxed{7:9}$

Hence, it is solved.

Answer 2

Answer:

→ 7 : 9 .

Step-by-step explanation:

Let the side of a cube = a

TSA of one cube = 6a²

TSA of 3 such cubes = 3×6a² = 18a²

If 3 cubes are placed adjacent to each other, dimension of resulting cuboid is

→ length = 3a

→ breadth = a

→ height = a

Therefore, TSA = 2(lb+bh+lh)

= 2(3a×a + a×a + a×3a)

= 2(3a² + a² + 3a²)

= 2 × 7a²

= 14a².

Ratio = \sf \frac{TSA\ of\ cuboid}{TSA\ of\ cubes} = \frac{14a^2}{18a^2} = \frac{7}{9}=\boxed{7:9}Ratio=TSA of cubesTSA of cuboid=18a214a2=97=7:9

Hence, it is solved.