school water tank is in the shape of a frustum of a cone. The height of the tank is 7.2 m and the top and bottom radii are 6m and 12 m respectively.
(a) Calculate the area of the curved surface of the tank, correct to 2 decimal places.
(b) Find the capacity of the tank, in liters, correct to the nearest liter.
(c) On a certain day, the tank was filled with water. If the school has 500 students and each student uses an average of 40 liters of water per day, determine the number of days that the students would use the water.
(a) Why did the math book go to therapy? Because it had too many problems! Let's calculate the area of the curved surface of the tank.
To find the curved surface area of a frustum of a cone, we need to calculate the slant height first. The slant height can be found using the Pythagorean theorem in the triangle formed by the height, the slant height, and the radius difference.
The height of the tank is h = 7.2 m.
The top radius is r1 = 6 m.
The bottom radius is r2 = 12 m.
The slant height (l) can be found by taking the square root of the height squared plus the difference in radii squared.
l = sqrt(h^2 + (r2 - r1)^2)
Plugging in the values, we have:
l = sqrt(7.2^2 + (12 - 6)^2)
= sqrt(51.84 + 36)
= sqrt(87.84)
≈ 9.37 m (rounded to 2 decimal places)
Now, we can calculate the curved surface area (A) using the formula:
A = π(r1 + r2)l
A = π(6 + 12) * 9.37
= π(18) * 9.37
= 56.55π
So, the area of the curved surface of the tank is approximately 56.55π square meters.
(b) Now, let's find the capacity of the tank in liters. We know the formula for the volume of a frustum of a cone is given by:
V = (1/3)πh (r1^2 + r2^2 + r1r2)
V = (1/3) * π * 7.2 * (6^2 + 12^2 + 6 * 12)
= (1/3) * π * 7.2 * (36 + 144 + 72)
= (1/3) * π * 7.2 * 252
≈ 6336.4π cubic meters
To convert the volume to liters, we multiply by 1000:
V_liters = 6336.4π * 1000
≈ 19913968.98 liters
So, the capacity of the tank is approximately 19913968.98 liters (rounded to the nearest liter).
(c) Let's calculate how many days the students can use the water. We know that each student uses an average of 40 liters of water per day, and there are 500 students.
Total water used per day = 40 liters * 500 students = 20000 liters
Number of days = Total capacity of the tank / Total water used per day
= 19913969 liters / 20000 liters
≈ 995.70 days
So, the students would be able to use the water for approximately 995.70 days (rounded to the nearest whole day).
(a) To calculate the area of the curved surface of the tank, we need to find the slant height of the frustum of the cone.
The slant height can be found using the Pythagorean theorem:
slant height^2 = (height^2) + (difference in radii)^2
= (7.2^2) + (12 - 6)^2
= 51.84 + 36
= 87.84
Taking the square root of 87.84, we find:
slant height = √87.84 ≈ 9.37 m
Now, we can calculate the curved surface area using the formula for the lateral surface area of a frustum of a cone.
Curved surface area = π(r1 + r2) × slant height
= π(6 + 12) × 9.37
= 3.14 × 18 × 9.37
≈ 529.81 m²
Therefore, the area of the curved surface of the tank is approximately 529.81 m².
(b) To find the capacity of the tank, we need to calculate the volume and convert it to liters.
The volume of a frustum of a cone can be calculated using the formula:
Volume = 1/3 × π × (r1^2 + r2^2 + r1 × r2) × height
= 1/3 × 3.14 × (6^2 + 12^2 + 6 × 12) × 7.2
= 0.333 × 3.14 × (36 + 144 + 72) × 7.2
= 0.333 × 3.14 × 252 × 7.2
≈ 1668.28 m³
To convert cubic meters to liters, we multiply the volume by 1000:
Capacity = 1668.28 × 1000
≈ 1,668,280 liters
Therefore, the capacity of the tank is approximately 1,668,280 liters.
(c) Given that each student uses an average of 40 liters of water per day, we can find the number of days the students would use the water by dividing the total capacity of the tank by the total amount used per day.
Number of days = Capacity / (number of students × amount used per student per day)
= 1,668,280 / (500 × 40)
= 1,668,280 / 20,000
= 83.41 days
Therefore, the students would use the water for approximately 83.41 days.
To find the answers to these questions, we need to calculate the curved surface area of the tank, the capacity of the tank, and the number of days the water would last based on the average water usage of the students.
(a) Calculating the area of the curved surface of the tank:
The curved surface of a frustum of a cone can be obtained by subtracting the area of the smaller cone (top) from the area of the larger cone (bottom).
First, let's calculate the slant height of the frustum. We can use the Pythagorean theorem to find the slant height (l) of the frustum:
l^2 = (12 - 6)^2 + 7.2^2
l^2 = 36 + 51.84
l^2 = 87.84
l = sqrt(87.84) = 9.37
Now, calculating the curved surface area (A) of the frustum:
A = π(R + r)l
A = π(12 + 6) * 9.37
A ≈ 452.39
Therefore, the area of the curved surface of the tank is approximately 452.39 square meters.
(b) Finding the capacity of the tank:
The capacity of a frustum of a cone can be calculated using the formula:
V = (1/3)πh(R^2 + r^2 + Rr)
V = (1/3)π * 7.2(12^2 + 6^2 + 12 * 6)
V ≈ 2418.48 cubic meters
Since we need to find the capacity in liters, we convert cubic meters to liters:
1 cubic meter = 1000 liters
So, the capacity in liters = V * 1000
Capacity ≈ 2418.48 * 1000 ≈ 2,418,480 liters
Therefore, the capacity of the tank is approximately 2,418,480 liters.
(c) Determining the number of days the students would use the water:
The number of days can be calculated by dividing the capacity of the tank by the average daily water usage of the students.
Number of days = Capacity / (Number of students * Average daily water usage)
Number of days ≈ 2,418,480 / (500 * 40)
Number of days ≈ 2,418,480 / 20,000
Number of days ≈ 120.92
Therefore, the number of days that the water would last for 500 students, with each student using an average of 40 liters per day, is approximately 121 days.
I'm sure you know how to find the volume and area of a cone
Think of the frustrum as a large cone with its top cut off.
So, find the area or volume of both the entire cone and the cut-off part.
To get the numbers for the frustrum, just subtract the cut-off part from the whole cone.
If you get stuck, come on back and show what work you have.