An architect is designing a public storage building with 2 different sized storage units the ceilings are 8 feet high in each storage unit
a. All storage units have square floors the volume will be 1800 what are the dimensions of each small unit
b. The ratio of dimensions of the architect drawing to the dimensions of the actual small storage is 1 inch to 72 inches. How many inches are the architects drawing would represent the actual width of each small unit in part A?
c. large units have rectangular floors .the volume of each of the small units is one third the volume of the larger unit ,what is the volume of the larger unit?
d. the dimensions of the rectangular floor of a large storage unit are whole numbers greater than or equal to the floor dimension of a small unit in part A, what could the floor dimensions of a large unit be?
a. V = Af*h = 1800Ft^3,
Af * 8 = 1800,
Af = 225Ft^2 = 15x15Ft = Area of floor,
Dimensions = 15x15x8Ft.
b. 15Ft = 180 in.
Width = 180in / 72 = 2.5 in.
c. Vs = Vl/3 = 1800Ft^3,
Vl/3 = 1800,
Vl = 5400 Ft^3 = Volume of large unit.
d. Vl = Af*h = 5400Ft^3,
Af*8 = 5400,
Af = 675Ft^2 = L*W,
Let W = 15Ft,
15L = 675Ft^2,
L = 45Ft.
Floor = 45 x 15 Ft.
hi my name is gabby and i need help with a problem for my math homework that my math teacher gave me for my threed period theacher and the problem is the rectangular floor and the area i 1800 but i don't now what the width is
a. To find the dimensions of each small unit, we can start by finding the area of the square floor since the height is given as 8 feet.
Let's assume the length and width of the square floor are both equal and denoted by "x". The formula for the volume of a rectangular solid is V = length × width × height. Since the height is 8 feet, we have:
V = x × x × 8 = 1800
To solve for "x", we can divide both sides of the equation by 8:
x × x = 1800 / 8
x² = 225
To find the square root of both sides, we get:
x = √225
x = 15
Therefore, the dimensions of each small unit are 15 feet by 15 feet.
b. To find the number of inches the architect's drawing would represent for the actual width of each small unit in part a, we know that the ratio of dimensions on the drawing to the actual dimensions is 1 inch to 72 inches.
Since the width of the small unit is 15 feet, we need to convert it to inches by multiplying by 12 (as there are 12 inches in 1 foot):
Width in inches = 15 feet × 12 inches/foot = 180 inches.
Now, applying the ratio of 1 inch to 72 inches:
Inches on architect's drawing = Width in inches ÷ 72 = 180 inches ÷ 72 = 2.5 inches.
Therefore, the architect's drawing would represent a width of 2.5 inches for each small unit.
c. Since the volume of each small unit is one-third the volume of the larger unit, we can set up an equation using the given volume of the small units:
Volume of small unit = (length × width × height) = 1800
Volume of large unit = 3 × (length × width × height) [because it is three times the volume of a small unit]
We are told that the height is the same for all units, which is 8 feet.
Therefore, the volume of the large unit can be represented as:
Volume of large unit = 3 × (length × width × 8)
Since the volume of each small unit is 1800, we have:
3 × (length × width × 8) = 1800
Now we can solve for the volume of the larger unit:
24 × (length × width) = 1800
length × width = 1800 / 24
length × width = 75
So, the volume of the larger unit is 75.
d. Since the floor dimensions of the large storage unit are whole numbers greater than or equal to the floor dimensions of a small unit in part a (which are 15 feet by 15 feet), we need to find pairs of whole numbers that have a product of 75 (since the volume of the larger unit is 75).
Possible floor dimensions include:
1. Length = 1, Width = 75, Or
2. Length = 3, Width = 25, Or
3. Length = 5, Width = 15, Or
4. Length = 15, Width = 5, Or
5. Length = 25, Width = 3, Or
6. Length = 75, Width = 1.
These are a few examples of the floor dimensions for a large storage unit that satisfy the given conditions. There may be other possible solutions as well.