Question

16. The front wall of a room is shown in the diagram given below. If the wall is to be painted from outside at the rate of 20 per sq. m, find the cost of painting the wall. 1.2 m 1.5 m 3.6 m 2.1 m 6 m A.AL


pls answer no wrOng answer ​

Answer 1

Area to be painted = area of the walls + area of ceiling

=2(hl+hb)+lb

=[2×(7×15+7×10)+(15×10)]=500 sq m

No. of cans required =

100

500

=5

Hence, 5 cans are required to paint the walls and the ceiling of the cuboidal hall.

Answer 2

I saw how the other guy gave the wrong answer. So I'll help you. Here,

Question:

The front wall of a room is shown in the diagram given below. If the wall is to be painted from outside at the rate of 20 per sq. m, find the cost of painting the wall. I suggest you copy paste the question properly.

Step-by-step explanation:

So the question is simple.

Wall Dimensions:

Length of wall = 6m

Width of wall(Height) = 3.6m

Now, area of wall is,

Area = Length x Width

Area = (6 x 3.6)m^2

Area = 21.6m^2

Door Dimensions:

Length = 2.1m

Breadth = 1.5m

Area = Length x Width

Area = (2.1 x 1.5)m^2

Area = 3.15m^2

Window Dimensions:

Side = 1.2m

Area of square = Side^2

Area = (1.2 x 1.2)m^2

Area = 1.44m^2

Wall left to paint:

= Area of Wall - Area of Door - Area of Window

= (21.6 - 3.15 - 1.44)m^2

= 17.01m^2

Cost of painting = $20/m^2

Therefore, cost for painting the wall = 17.01m^2 x &20 = $340.2

Therefore the cost of painting the wall is $340.2

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