Question

3 equal cubes are placed adjacently in a row . find the ratio of the total surface area of the resulting cuboid to that of the sum of the surface area of 3 cubes .​

Answer 1

Let each side of cube = a cm

surface area of cube = 6a² cm²

sum of the 3 cubes = 6a² ⁼ 18a² cm²

length of resulting cuboid = 3a cm

breadth of resulting cuboid = a cm

height of resulting cuboid = a cm

total surface area = 2 ( lb + bh + hl )

• 2 ( 3a*a + a*a + a *3a )
• 2 ( 3a² + a + 3a² )
• 2 ( 7a² )
• 14a² cm²

ratio = t.s.a of resulting cuboid / area of cube

ratio = 18a²/ 14a²

ratio = 7/9

ratio = 7:9

Answer 2

Answer:

Let the side of cube be x .

Then TSA of one cube = 6x

2

TSA of three cubes = 18x

2

Also Dimensions of cuboid = (3x,x,x)

So TSA of cuboid = 2(3x×x+3x×x+x×x)=14x

2

So

Ratio of the total surface area of the resulting cuboid to that of the sum of the total surface areas of the three cubes=

18x 2

14×2= 7.9 ratio

So, the answer is 7.9 ratio