Question

find the area of four Walls of a room whose dimension are 6mx5mx3mx​

Answer 1

Given :

• Length of Wall = 6m
• Breadth of Wall = 5m
• Height of Wall = 3m

To Find :

• Area of 4 walls

Solution :

✰ As we know that, Area of 4 walls is given by 2(Length + Breadth) × Height or Perimeter of the floor × height. So let's substitute the given values in the formula to find the Area of 4 walls of the Room.

⠀

⟶⠀Area of 4 walls = 2(L + B) × H

⟶⠀Area of 4 walls = 2(6 + 5) × 3

⟶⠀Area of 4 walls = 2 × 11 × 3

⟶⠀Area of 4 walls = 2 × 33

⟶⠀Area of 4 walls = 66m²

⠀

So, Area of 4 walls is 66m²

___________________

Answer 2

❥Given :

• Length of wall 6m

• Breadth of wall 5m

• Height of wall 3m

❥To find :

• Area of four walls.

❥Solution :

Formula :

$\begin{gathered}{:} \longrightarrow \color{green}{ \boxed{ \color{aqua}\sf \: area \: of \: four\:walls = \: 2(\: l + b) \times h}}\\ \\ \end{gathered}$

So ,

Area of four wall = 2(6m + 5m) × 3m

Area of four wall = 2(11m × 3m)

Area of four wall = 2 × 33m

Area of four wall = ${66m}^{2}$

$\boxed{\huge {\therefore} \: \small\tt\blue{area \: of \: four \: wall \: = } \green{ {66m}^{2} }}$

More formulas :

$\begin{gathered}\begin{gathered} \green{\boxed { \begin{array}{ |c|c| } \hline \\ \sf \blue{Perimeter \: of \: rectangle}& \sf \red{2(length + breadth)} \\ \\ \hline \\\blue{ \sf Area \: of \: rectangle}& \sf \red{length \times breadth}\\ \\ \hline \\ \sf\blue {Length \: of \: rectangle}& \sf \red{\dfrac{Area }{ Breadth}} \\ \\ \hline \\ \sf \blue{Breadth \: of \: rectangle}& \sf \red{ \dfrac{Area}{Length} }\\ \\ \hline \\ \sf \blue{Diagonal \: of \: rectangle}& \sf \red{\sqrt{ {length}^{2} + {breadth}^{2} }} \\ \\ \end{array}}}\end{gathered}\end{gathered}$

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