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 About the author Rose
[tex]\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}[/tex] [tex]\begin{gathered}\begin{gathered}\bf \: To \: Find – \begin{cases} &\sf{Breadth, \: b}\end{cases}\end{gathered}\end{gathered}[/tex] [tex]\begin{gathered}\Large{\bold{{\underline{Formula \: Used – }}}} \end{gathered}[/tex] [tex] \boxed{ \red{ \sf \:Area_{(4 walls)} = 2(l + b) \times h}}[/tex] where, l = Length b = Breadth h = Height [tex]\large\underline{\sf{Solution-}}[/tex] ↝ Given that Length = 15 m Height = 6 m Area of 4 walls = 300 square meter. ↝ We know that, [tex]\rm :\longmapsto\:Area_{(4 walls)} = 2(Length + Breadth) \times Height[/tex] ↝ On substituting the values, we get [tex]\rm :\longmapsto\:300 = 2(15 + b) \times 6[/tex] [tex]\rm :\longmapsto\:25 = 15 + b[/tex] [tex]\rm :\longmapsto\:b = 10 \: m[/tex] [tex]\bf\implies \:Breadth, \: b \: = \: 10 \: m[/tex] 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) Reply
[tex]\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}[/tex]
[tex]\begin{gathered}\begin{gathered}\bf \: To \: Find – \begin{cases} &\sf{Breadth, \: b}\end{cases}\end{gathered}\end{gathered}[/tex]
[tex]\begin{gathered}\Large{\bold{{\underline{Formula \: Used – }}}} \end{gathered}[/tex]
[tex] \boxed{ \red{ \sf \:Area_{(4 walls)} = 2(l + b) \times h}}[/tex]
where,
[tex]\large\underline{\sf{Solution-}}[/tex]
↝ Given that
↝ We know that,
[tex]\rm :\longmapsto\:Area_{(4 walls)} = 2(Length + Breadth) \times Height[/tex]
↝ On substituting the values, we get
[tex]\rm :\longmapsto\:300 = 2(15 + b) \times 6[/tex]
[tex]\rm :\longmapsto\:25 = 15 + b[/tex]
[tex]\rm :\longmapsto\:b = 10 \: m[/tex]
[tex]\bf\implies \:Breadth, \: b \: = \: 10 \: m[/tex]
Additional Information :-
↝ Cube:
↝ Cuboid:
↝ Formula’s of Cube :-
↝ Formula’s of Cuboid