Question

1. jack wants to check a quotation for drywall that he received from a construction company. he knows that the studs in his garage are 16 inches apart centre to centre. jack’s garage walls are 8 feet tall.

a. using the diagram, estimate the surface area of the garage walls and ceiling to be covered by drywall. each red dot represents a stud on the wall.

note: to find the surface area of the walls, start by finding the total length of the walls as if you were unrolling the room into one long rectangle.

Explanation:

Step1: Count the number of studs on each wall

By counting the red - dots (studs) on the diagram, assume the lengths of the walls based on the 16 - inch stud spacing. Let's assume the two long walls have \(n_1\) studs and the two short walls have \(n_2\) studs. However, since the diagram is not provided in a way that we can accurately count the studs, we'll use a general approach. First, convert the stud spacing to feet. Since 16 inches=\(\frac{16}{12}=\frac{4}{3}\) feet.

Step2: Calculate the perimeter of the walls

Let's assume the number of stud - spaces on the long walls is \(x\) and on the short walls is \(y\). The length of a long wall \(L_1\) and short wall \(L_2\) can be calculated from the stud - spaces. But if we assume for simplicity that we can find the perimeter \(P\) of the base of the garage. The height of the wall \(h = 8\) feet. The area of the walls \(A_w=P\times h\).
The area of the ceiling \(A_c\): If we assume the length of the garage is \(l\) and width is \(w\), and we find them from the stud - spacing, \(A_c=l\times w\).
Let's assume we count the studs and find that the perimeter of the base of the garage (sum of all four sides) \(P\) (in feet). The area of the walls \(A_w = P\times8\).
Suppose after counting studs and calculating, the perimeter of the base \(P = 40\) feet (this is an assumed value for illustration purposes, actual value depends on stud - counting). Then \(A_w=40\times8 = 320\) square feet.
Let's assume the length \(l = 12\) feet and width \(w = 8\) feet (again, assumed values), then the area of the ceiling \(A_c=l\times w=12\times8 = 96\) square feet.
The total surface area \(A=A_w + A_c\).
\(A=320+96=416\) square feet.

Answer:

The total surface area of the garage walls and ceiling to be covered by dry - wall depends on the actual stud - counting and calculations. But using assumed values for illustration, it could be 416 square feet. You need to accurately count the studs on the diagram to get a more precise answer.