Question

The area of the four walls of a room is 300m^2. Its length and height are 15 m and 6 m respectively. find its breadth​

Answer 1

$\begin{gathered}\begin{gathered}\bf\: Given-\begin{cases} &\sf{Area_{(4 walls)} = 300 \: {m}^{2} } \\ &\sf{Length, \: l \: = 15 \: m}\\ &\sf{Height, \: h \: = 6 \: m} \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf \: To \: Find - \begin{cases} &\sf{Breadth, \: b}\end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\Large{\bold{{\underline{Formula \: Used - }}}}  \end{gathered}$

$\boxed{ \red{ \sf \:Area_{(4 walls)} = 2(l + b) \times h}}$

where,

• l = Length

• b = Breadth

• h = Height

$\large\underline{\sf{Solution-}}$

↝ Given that

• Length = 15 m

• Height = 6 m

• Area of 4 walls = 300 square meter.

↝ We know that,

$\rm :\longmapsto\:Area_{(4 walls)} = 2(Length + Breadth) \times Height$

↝ On substituting the values, we get

$\rm :\longmapsto\:300 = 2(15 + b) \times 6$

$\rm :\longmapsto\:25 = 15 + b$

$\rm :\longmapsto\:b = 10 \: m$

$\bf\implies \:Breadth, \: b \: = \: 10 \: m$

Additional Information :-

↝ Cube: 

• A cube has six faces, eight vertices and twelve edges. All the faces of the cube are in square shape and are of equal length.

↝ Cuboid: 

• A cuboid has six faces, eight vertices and twelve edges. The faces of the cuboid are parallel. But not all the faces of a cuboid are equal in length.

↝ Formula's of Cube :-

• Total Surface Area = 6(side)²

• Curved Surface Area = 4(side)²

• Volume of Cube = (side)³

• Diagonal of a cube = √3(side)

• Perimeter of cube = 12 x side

↝ Formula's of Cuboid

• Total Surface area = 2 (Length x Breadth + breadth x height + Length x height)

• Curved Surface area = 2 height(length + breadth)

• Volume of the cuboid = (length × breadth × height)

• Diagonal of the cuboid =√(l² + b² + h²)

• Perimeter of cuboid = 4 (length + breadth + height)