Question

Hardick is painting the walls and ceiling of a cuboidal room with length, breadth and height 20m, 12m, and 10m respectively. An area of 80 sq.metre can be painted from each can. Find out how many cans of paint will be required to paint the room. (Hint: Area to be painted = 880 sq.m) ​

Answer 1

★ Given:

Length of the room = 20 m

Breadth of the room = 12 m

Height of the room = 10 m

Area that can be painted from one can = 80 m²

★ To Find:

No. of cans to be used to paint the complete room.

★ Analysis

In this question, we will first have to find the total area of the room to be painted. That is, the total area of the four walls and the ceiling. Dividing the total area with the area that can be painted from 1 can, we will get the no. of cans to be used.

★ Solution:

Length of the room = 20 m

Breadth of the room = 12 m

Height of the room = 10 m

Total area to be painted = Area of four walls + Area of ceiling

= 2 (lh + bh) + lb  

= 2 [(20 x 10) + (12 x 10)] + (20 x 12)

= 2 [200 + 120] + 240

= 2 x 320 + 240

= 640 + 240

= 880 m²

Area that can be painted from one can = 80 m²

Total area of the room to be painter = 880 m²

No. of cans required

$\sf{=\dfrac{Total\:area\:of\:room}{Area\:painted\:from\:one\:can}}$

$\sf{=\dfrac{880\:m^2}{80\:m^2}}$

$\sf{=11\:cans}$

Hence, we need 11 cans of paint to paint the cuboidal room of given dimensions.

→ Cuboid:

✳ A cuboid is a 3D figure made up of 6 rectangular faces. It has 12 edges and 8 vertices.

$\setlength{\unitlength}{0.74 cm}\begin{picture}\thicklines\put(5.6,5.4){\bf A}\put(11.1,5.4){\bf B}\put(11.2,9){\bf C}\put(5.3,8.6){\bf D}\put(3.3,10.2){\bf E}\put(3.3,7){\bf F}\put(9.25,10.35){\bf H}\put(9.35,7.35){\bf G}\put(3.5,6.1){\sf x\:cm}\put(7.7,6.3){\sf y\:cm}\put(11.3,7.45){\sf z\:cm}\put(6,6){\line(1,0){5}}\put(6,9){\line(1,0){5}}\put(11,9){\line(0,-1){3}}\put(6,6){\line(0,1){3}}\put(4,7.3){\line(1,0){5}}\put(4,10.3){\line(1,0){5}}\put(9,10.3){\line(0,-1){3}}\put(4,7.3){\line(0,1){3}}\put(6,6){\line(-3,2){2}}\put(6,9){\line(-3,2){2}}\put(11,9){\line(-3,2){2}}\put(11,6){\line(-3,2){2}}\end{picture}$

Please refer attachment to view the latex figure.

✳ LSA of cuboid = 2 (lh + bh)

✳ TSA of cuboid = 2 (lh + bh + lb)

✳ Volume of cuboid = lbh