Question

Reema is painting the walls and ceiling of a cuboidal room with length, breadth and
height of 12m, 8m and 5m respectively. If each can of paint covers 8 sq m of area. How
many cans of paint will be required to paint the room?

Answer 1

${\underline{\underline{\sf{Understanding \: The \: Question:}}}}$

Here Quetsion is from Surface areas and Volume. It is given that Reema wants to paint a cuboidal room in which she wants to paint 4 alternate side walls and ceiling only. [ Remember that ground area is not included.] It has length 12 m, Breadth 8m and height 5 m. It is also given that one can of paint can paint 8 m². We have to find Number of can required to paint.

${\underline{\underline{\sf{Given :}}}}$

• Length of room = 12 m
• Breadth of the room = 8 m
• Height of the room = 5 m
• Area paint of one can = 8 m²

${\underline{\underline{\sf{To \: find :}}}}$

• Number of cans required to paint.

$\rule{200}{2}$

${\underline{\underline{\sf{Solution :}}}}$

●First, here we have to find Total Surface Area of room. So, T.S.A will be :-

${\sf{\ratio{\longmapsto{Total \: Surface \: Area = 2(lb + bh + lh)}}}}$

Here

• l = Lenght
• b = Breadth
• h = Height

DIAGRAM

$\setlength{\unitlength}{0.74 cm}\begin{picture}\thicklines\put(5.6,5.4){\bf }\put(11.1,5.4){\bf }\put(11.2,9){\bf }\put(5.3,8.6){\bf }\put(3.3,10.2){\bf }\put(3.3,7){\bf }\put(9.25,10.35){\bf }\put(9.35,7.35){\bf }\put(3.5,6.1){\sf 8\:cm}\put(7.7,6.3){\sf 12\:cm}\put(11.3,7.45){\sf 5\:cm}\put(6,6){\line(1,0){5}}\put(6,9){\line(1,0){5}}\put(11,9){\line(0,-1){3}}\put(6,6){\line(0,1){3}}\put(4,7.3){\line(1,0){5}}\put(4,10.3){\line(1,0){5}}\put(9,10.3){\line(0,-1){3}}\put(4,7.3){\line(0,1){3}}\put(6,6){\line(-3,2){2}}\put(6,9){\line(-3,2){2}}\put(11,9){\line(-3,2){2}}\put(11,6){\line(-3,2){2}}\end{picture}$

${\sf{\ratio{\longmapsto{Total \: Surface \: Area = 2(12 \times 8 + 8 \times 5 + 12 \times 5)}}}} \\ \\ {\sf{\ratio{\longmapsto{Total \: Surface \: Area = 2(96 + 40 + 60)}}}} \\ \\ {\sf{\ratio{\longmapsto{Total \: Surface \: Area = 2(196)}}}} \\ \\ { \underline{ \boxed{\sf{\ratio{\longmapsto{Total \: Surface \: Area = 392 \: {m}^{2} }}}}}}$

$\rule{200}{2}$

● Secondly we have to find Area of the base of room. Base of room is in the form of Rectangle. So, area is

DIAGRAM

$\setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\multiput(0,0)(5,0){2}{\line(0,1){3}}\multiput(0,0)(0,3){2}{\line(1,0){5}}\put(0.03,0.02){\framebox(0.25,0.25)}\put(0.03,2.75){\framebox(0.25,0.25)}\put(4.74,2.75){\framebox(0.25,0.25)}\put(4.74,0.02){\framebox(0.25,0.25)}\multiput(2.1,-0.7)(0,4.2){2}{\sf\large 12 cm}\multiput(-1.4,1.4)(6.8,0){2}{\sf\large 8 \: cm}\put(-0.5,-0.4){\bf }\put(-0.5,3.2){\bf }\put(5.3,-0.4){\bf }\put(5.3,3.2){\bf }\end{picture}$

${\sf{\ratio{\longmapsto{Area \: of \: the \: Base = Length \times Breadth}}}} \\ \\ {\sf{\ratio{\longmapsto{Area \: of \: the \: Base = 12 \times 8}}}} \\ \\ { \underline{ \boxed{\sf{\ratio{\longmapsto{Area \: of \: the \: Base = 96 \: {m}^{2} }}}}}}$

$\rule{200}{2}$

● Next, we have to find Area of four walls and ceiling of the room. The area is

${\sf{\ratio{\longmapsto{Area \: required = T.S.A - Area \: of \: Base }}}} \\ \\ {\sf{\ratio{\longmapsto{Area \: required =392 \: {m}^{2} - 96 \: {m}^{2} }}}} \\ \\ { \underline{ \boxed{\sf{\ratio{\longmapsto{Area \: required =296 \: {m}^{2} }}}}}}$

So, area she have to paint is 296 m².

$\rule{200}{2}$

● Now, No. of Can required is :-

${\sf{\ratio{\longmapsto{No. of \: Cans = \cfrac{ Area \: of \: the \: Hall}{ Area \: Painted \: by \: one Can} }}}}$

${\sf{\ratio{\longmapsto{ No. of \: Cans = \cfrac{296}{8} }}}} \\ \\ { \large{ \underline{ \boxed{ \red{\sf{\ratio{\longmapsto{ No. of \: Cans = 37 \: Cans }}}}}}}}$

$\rule{350}{2}$

● More to know :-

• Total Surface area = 2 (Length x Breadth + breadth x height + Length x height)
• Lateral Surface area = 2 height(length + breadth)
• Volume of the cuboid = (length × breadth × height)

Answer 2

Answer:

37 cans

Step-by-step explanation:

pls mark brainliest