Question

Three equal cubes are placed together in a row to form a cuboid. Find the
ratio of the total surface area of the new cuboid to that of the surface area ofthe three cubes.

Answer 1

if 3 cubes of edge a units are joined
then
length of cuboid = a
breadth of cuboid=3a
height of cuboid=a
total surface area of cuboid= 2(lb+BH+LH)
$2(a \times 3a + 3a \times a + a \times a) \\ 2(3 {a}^{2} + 3 {a}^{2} + {a}^{2} ) \\ 2 \times 7 {a}^{2} \\ 14 {a}^{2} \\ tsa \: of \: cubes = 6 {a}^{2} + 6 {a}^{2} + 6 {a}^{2} \\ \: \: \: \: \: \: \: \: = 18 {a}^{2} \\ ratio = \frac{14 {a}^{2} }{18 {a}^{2} } \\ = \frac{7}{9}$

Answer 2

Total surface area of three cube is3•6l^2=18l^2
Total surface area of new cuboid is
2(3l•l)+(l•l)+(3l•l)=2(3l^2)+(l^2)+(3l^2)
=2(7l^2)
=14l^2
The ratio is 14l^2:18l^2
=7:9