Question

The length of cuboid is two times of its breadth and four times of its height. If the volume of cuboid is 64m³, find its total surface of area.

Answer 1

Answer

The total surface area of the cuboid is 112m²

$\bf\large\underline{Given}$

• The length of the cuboid is 2 times of its breadth and 4 times of its height

$\bf\large\underline{To \ Find}$

• The total surface area of the cuboid

$\bf\large\underline{Formula \ to \ be \ used}$

$\bullet \: \: \sf Volume \: of \:cuboid = l \times h \times b \\ \bullet \: \: \sf Surface \: area = 2(lh + lb + hb) \\ \sf where : \\ \sf l \longrightarrow \: length \: of \: cuboid \\ \sf h\longrightarrow \: \: height of \: cuboid \\ \sf b \longrightarrow breadth \: of \: cuboid$

$\bf\large\underline{Solution}$

We are given ,

The length is 2 times the breadth

$\sf\implies l = 2b \\\\ \sf\implies b = \dfrac{l}{2}$

And again ,

The length is 4 times the height

$\sf \implies l = 4h \\\\ \sf\implies h = \dfrac{l}{4}$

Also we are given ,

$\sf Volume \: of \: cuboid = 64m {}^{3} \\ \\ \implies \sf l \times h \times b = 64 {m}^{3} \\ \\ \implies \sf l \times \frac{l}{2} \times \frac{l}{4} = 64 {m}^{3} \\ \\ \sf \implies l {}^{3} = 64x8 {m}^{3} \\ \\ \sf \implies l {}^{3} = 4^{3} \times 2^{3} m ^{3} \\ \\ \sf\implies l ^{3} = ( {8m)}^{3} \\ \\ \sf \implies l = 8m$

Therefore , the length of the cuboid is 8m

Thus the breadth and height is

$\sf h = \dfrac{8m}{2} \: \: and \: \: \: b = \dfrac{8m}{4} \\ \\ \sf \implies h = 4m \: \: and \: \: \implies b = 2m$

Now , calculating the surface area :

$\sf S.A \: of \: cuboid = 2(8 \times 4+ 8 \times 2 + 4 \times 2) \\ \\ \sf \implies S.A \: of \: cuboid = 2(32 + 16 + 8) \\ \\ \sf \implies S.A \: of \: cuboid =2 \times 56 \\ \\ \sf \implies S.A \: of \: cuboid = 112m {}^{2}$

Ther , required surface area of the cuboid is 112m²

Answer 2

$\large\bf Given:-$

• The length of cuboid is two times it's breadth.

• The length of cuboid is four times it's height.

• Volume of cuboid = 64m³

$\large\bf To\:Find$

• The total surface area of the cuboid

$\large \bf Solution$

Let the length of the cuboid be x

Given length is 2 times its breadth

$\rm\implies l = 2b$

$\rm\implies b = \dfrac{l}{2}$

$\rm\implies b = \dfrac{x}{2}$

Given, length is 4 times its height

$\rm\implies l = 4h$

$\rm\implies h = \dfrac{l}{4}$

$\rm\implies h = \dfrac{x}{4}$

Volume of cuboid = l × b × h = 64cm³

$\rm\implies x \times \dfrac{x}{2} \times \dfrac{x}{4} = 64$

$\rm\implies \dfrac{ { \: \: x}^{3} }{8} = 64$

$\rm\implies { { \: \: x}^{3} } = 64 \times 8$

$\rm\implies { { \: \: x}^{3} } = 512$

$\rm\implies x = \sqrt[3]{512}$

$\rm\implies x = 8$

$\rm Length = x = 8m$

$\rm Breadth = \dfrac{x}{2} = \dfrac{8}{2} = 4m$

$\rm height= \dfrac{x}{4} = \dfrac{8}{4} = 2m$

Now;

Surface area of cuboid = 2(lb + bh + hl)

$\rm \longrightarrow2 \{(8 \times 4 )+ (4 \times 2) + (2 \times 8) \}$

$\rm \longrightarrow2 (32+ 8+ 16)$

$\rm \longrightarrow2 (56)$

$\rm \longrightarrow112$

Therefore:-

$\boxed{ \bf \therefore Surface \: area \: of \: cuboid \: = 112 {m}^{2} }$