Question

3
Two cubes each of volume 64 cm are joined end to end. Find the surface area of the resulting cuboid​

Answer 1

Answer:

volume of a cube =side³

volume =64 cm³ (given)

side of the cube = cube root of 64=4 cm ( as 4×4×4=64)

when we join two cubes then the two joining faces are not included in tsa

so we have to find the area of only 5 faces of each cube

now ,each side of cube is a square

so area of the 5 square sides of a cube =5×(side²)

=5×(4²l=5×16=80cm²

now the two cubes are same so the area of the cubes = 2×80=160cm²

so the area of the cuboid by joining the two cubes excluding the two joined sides =160 cm²

Answer 2

$\bf \underline{Given} :$ ↴

$\sf\implies Two \: cubes \: each \: of \: volume \: 64 \: cm ^{3} \: are \: joined \: end \: to \: end.$

$\bf \underline{To \: find} :$ ↴

$\sf\implies The \: surface \: area \: of \: resulting \: cuboid.$

$\bf \underline{\underline{Solution}}$

$\sf First \: of \: all, we \: have \: to \: find \: each \: side \: of \: the \: cube.$

$\sf\implies Here, volume \: of \: 1 \: cube = 64 \: cm ^{3}$

$\red{ \sf\implies Volume \: of \: cube = (Side)^{3}}$

$\sf\implies \sqrt[3]{64} = Side$

$\sf\implies 4 = Side$

$\purple{ \sf\implies Each \: side \: of \: the \: cube = 4 \: cm }$

$\sf Now, the \: dimensions \: of \: the \: cuboid \: are :$

$\sf\implies Length \: of \: cuboid \: (Side + Side) = 4 + 4 = 8 \: cm$

$\sf\implies Breadth \: of \: cuboid =4 \: cm$

$\sf\implies Height \: of \: cuboid =4 \: cm$

$\sf Finding \: the \: total \: Surface \: area \: of \: resulting \: cuboid.$

$\red{ \sf\implies TSA \: of \: cuboid = 2(LB+BH+HL)}$

$\sf \underline{Where},$

$\sf\implies \underline{L = Length},\: \underline{B = Breadth} \: and \: \underline{ H = Height}$

$\sf\implies 2(LB+BH+HL)$

$\sf\implies 2(8 \times 4+4 \times 4+4 \times 8)$

$\sf\implies 2(32+16+32)$

$\sf\implies 2(80)$

$\purple{\sf\implies 160 \: cm ^{2}}$

$\underline{ \boxed{ \pink{\sf \: Hence, surface \: area \: of \: the \: resulting \: cuboid \: is \: 160 \: cm^{2}}}}$