Q29. - Geometry

Question

A room28ft long and  wide has walls 8ft high. 

1. What is the total wall area?
2.  How many gallon cans of paint should be bought to paint the walls if 1 gal of paint covers. $300f{t}^2$

Step-by-Step Solution

Answer

1. Total area of the wall is  $768f{t}^2$
2. Total gallon is required $2.56$ gallon.

Step 1. Given information.

 Length of the room$\left(l\right)=28ft$

Width of the room  $\left(b\right)=20ft$

Height of the wall $\left(h\right)=8ft$

Step 2. Concept Used.

If b is the base of the room h, is the height of the room and i is the length of the room.

Then, the   area of the room is $A=b\times h$

Step 3. Let’s calculate the area of the room.

The dimension of room has four rectangles.

Thus,

The area of the room is

 $\begin{array}{l}A=(28\times 8)+(20\times 8)+(28\times 8)+(20\times 8)\\ A=16(28+20)\\ A=768f{t}^2\end{array}$

So,   the total area of the wall is$768f{t}^2$ .

 

 

 

 If  cover =$\frac{768}{300}=2.56$1 gallon 

  cover =   $\frac{768}{300}=2.56$ gallon.

 

 Therefore, total area of the wall i$768f{t}^2$s And total $2.56$ gallon is required.