Question

Daniel is painting the walls and ceiling of a cuboid hall with l, b and h of 15m, 10m and 7m respectively. From each can of paint \[100{m^2}\] of area is painted. How many cans of paint will she need to paint the room?

Answer 1

Hint: We are given with the dimensions of a cuboidal room. In order to paint the walls and ceiling of the room only the base of the room is to be left. So we will paint 4 walls and the ceiling. Now we will find the areas of the walls and ceiling. Then on adding them we will get the total area that is to be painted. And after finding the area we will divide the area by the area that is getting painted in one can in order to find the number of cans required to paint the room.

Complete Step by Step Solution:
Given is a cuboid room that Daniel is painting. So in order to paint we should have the idea of what area of room we have to paint. So there are 4 walls and 1 ceiling. We have drawn the diagram of this room.

Now
\[are{a_{front{\text{ }}wall{\text{ }} + back{\text{ }}wall}} = 2\left( {l \times h} \right) = 2\left( {15 \times 7} \right) = 210{m^2}\]
\[are{a_{Two{\text{ }}side{\text{ }}walls}} = 2\left( {h \times b} \right) = 2\left( {7 \times 10} \right) = 140{m^2}\]
\[are{a_{ceiling}} = l \times b = 15 \times 10 = 150{m^2}\]
Total area to be painted is given by,
\[{\text{Total area to be painted = }}are{a_{front{\text{ }}wall{\text{ }} + back{\text{ }}wall}} + are{a_{Two{\text{ }}side{\text{ }}walls}} + are{a_{ceiling}}\]
Putting the areas we get,
\[{\text{Total area to be painted = 210 + 140 + 150 = 500}}{{\text{m}}^2}\]
Now number of cans required to paint the room = \[\dfrac{{Total{\text{ }}area{\text{ }}to{\text{ }}be{\text{ }}pa\operatorname{int} ed}}{{Area{\text{ }}painted{\text{ }}in{\text{ }}one{\text{ }}can}}\]
\[ \Rightarrow \dfrac{{500}}{{100}} = 5\]

So, Daniel will require 5 cans to paint her room.

Note:
Note that we are finding the number of cans required so don’t multiply the area with the area painted in one can. That is not the way to solve the question. Also we have considered the ceiling but don’t consider the base of the room. Also if in problem any window or door is given we will remove the area of that part also.