
The length, breadth and height of a room are \[5\text{ }m,\text{ }4\text{ }m\text{ }and\text{ }3\text{ }m\] respectively. Find the cost of white washing the walls of the room and ceiling at the rate of Rs. 7.50 per ${{m}^{2}}$.
Answer
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Hint: Assume length, breadth and height of the room be $l,b\text{ and }h$ respectively. Find out the surface area that is going to be white wash the walls of room and ceiling by applying the formula of curved surface area of a cuboid. Then multiply the total area obtained with the rate of white washing.
“Complete step-by-step answer:”
A cuboid is a $3D$ shape. Cuboids have 6 faces, 12 vertices and 8 edges. Cuboids are made from 6 rectangles which are placed at right angles. A cuboid that uses all square faces is a cube. Total surface area of a cuboid is the area of six rectangles and curved surface area is the area of 4 rectangles containing height.
T.S.A of cuboid $=2\left( lb+bh+hl \right)$
and, C.S.A of cuboid $=2\times \left( l+b \right)\times h$
Now, we come to the question. As it is given that we are white washing the room, so it is obvious that we will not white wash the floor. So, the total surface area we have to white wash is the sum of the curved surface area and area of the ceiling.
Therefore, area to be white washed
$\begin{align}
& =2\times \left( l+b \right)\times h+l\times b \\
& =2\times \left( 5+4 \right)\times 3+5\times 4 \\
& =54+20 \\
& =74\text{ }{{m}^{2}} \\
\end{align}$
Now, cost of white washing
$\begin{align}
& =\text{total area to be white washed }\times \text{ rate per unit area} \\
& \text{=74}\times \text{7}\text{.5} \\
& \text{=555} \\
\end{align}$
Hence, the total cost of white washing is Rs. 555.
Note: We have not considered the total surface area of the room because the floor is not white washed. So, curved surface area and area of ceiling is considered, as we have been asked in the question.Students should remember the formulas of curved surface and total surface of cuboid.
“Complete step-by-step answer:”
A cuboid is a $3D$ shape. Cuboids have 6 faces, 12 vertices and 8 edges. Cuboids are made from 6 rectangles which are placed at right angles. A cuboid that uses all square faces is a cube. Total surface area of a cuboid is the area of six rectangles and curved surface area is the area of 4 rectangles containing height.
T.S.A of cuboid $=2\left( lb+bh+hl \right)$
and, C.S.A of cuboid $=2\times \left( l+b \right)\times h$
Now, we come to the question. As it is given that we are white washing the room, so it is obvious that we will not white wash the floor. So, the total surface area we have to white wash is the sum of the curved surface area and area of the ceiling.
Therefore, area to be white washed
$\begin{align}
& =2\times \left( l+b \right)\times h+l\times b \\
& =2\times \left( 5+4 \right)\times 3+5\times 4 \\
& =54+20 \\
& =74\text{ }{{m}^{2}} \\
\end{align}$
Now, cost of white washing
$\begin{align}
& =\text{total area to be white washed }\times \text{ rate per unit area} \\
& \text{=74}\times \text{7}\text{.5} \\
& \text{=555} \\
\end{align}$
Hence, the total cost of white washing is Rs. 555.
Note: We have not considered the total surface area of the room because the floor is not white washed. So, curved surface area and area of ceiling is considered, as we have been asked in the question.Students should remember the formulas of curved surface and total surface of cuboid.
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