Question

Three cubes each of edge $10cm$ are joined to form a cuboid. Find the following:
(i) The surface area of resulting cuboid.
(ii) The ratio of surface are of resulting cuboid to the sum of the surface areas of three cube

Answer 1

Three cubes of edge $10cm$ are joined to form a cuboid-

Therefore,

Length of cuboid $\left(l\right)=10+10+10=30cm$

Breadt of cuboid $\left(b\right)=10cm$

Height of cuboid $\left(h\right)=10cm$

$\therefore$ Total surface area of cuboid $\left(A\right)=2(lb+bh+hl)$

$\Rightarrow A=2[(30\times 10)+(10\times 10)+(10\times 30)]$

$\Rightarrow A=2(300+100+300)$

$\Rightarrow A=1400{cm}^2$

Hence the surface area of resulting cuboid is $1400{cm}^2$.

As the side of each small cube is $10cm$.

Therefore, the area of each small cube $\left(a\right)$ will be-

$a=6{\left(\text{side}\right)}^2$

$\Rightarrow a=6\times {10}^2=600{cm}^2$

Sum of surface area of all three cubes $=600+600+600=1800{cm}^2$

Now, ratio of surface are of resulting cuboid to the sum of the surface areas of three cube $=\frac{1400}{1800}=\frac{7}{9}$

Hence the ratio of surface are of resulting cuboid to the sum of the surface areas of three cube is $7:9$.