A right pyramid on a base 10m Square is 15m high.if the top 6m of the pyramid are removed.what is the volume of the remaining frustrum
The original pyramid has a volume of (1/3) * 10^2 * 15 = 500 cubic meters.
The top 6m of the pyramid is removed, leaving a new pyramid with a height of 15-6=9 meters.
Using similar triangles, we can find that the new base of the pyramid has a side length of (4/5)*10 = 8 meters.
The volume of the new pyramid is (1/3) * 8^2 * 9 = 192 cubic meters.
Therefore, the volume of the frustrum (the remaining portion of the pyramid) is 500 - 192 = 308 cubic meters.
To find the volume of the remaining frustrum (the part of the pyramid left after removing the top), we can use the formula for the volume of a frustum of a pyramid:
V = (1/3) * h * (A1 + A2 + √(A1 * A2))
where V is the volume, h is the height of the frustum, A1 is the area of the top base, and A2 is the area of the bottom base.
Given:
Height of the pyramid (original height): 15m
Height of the removed top portion: 6m
Step 1: Calculate the dimensions of the top and bottom bases.
The original base is a square with sides of 10m.
Area of the top base (A1):
A1 = side^2
A1 = 10^2
A1 = 100m^2
The dimensions of the bottom base will remain the same as the original base.
Area of the bottom base (A2):
A2 = side^2
A2 = 10^2
A2 = 100m^2
Step 2: Calculate the height of the remaining frustrum.
The height of the remaining frustrum is the original height minus the height of the removed top portion.
Height of the remaining frustrum (h):
h = original height - height of the removed top
h = 15m - 6m
h = 9m
Step 3: Calculate the volume of the remaining frustrum.
V = (1/3) * h * (A1 + A2 + √(A1 * A2))
V = (1/3) * 9m * (100m^2 + 100m^2 + √(100m^2 * 100m^2))
V = (1/3) * 9m * (200m^2 + √(10000m^4))
V = (1/3) * 9m * (200m^2 + 100m^2)
V = (1/3) * 9m * 300m^2
V = (9/3) * 9m * 100m^2
V = 3 * 9m * 100m^2
V = 27m * 100m^2
V = 2700m^3
Therefore, the volume of the remaining frustrum is 2700 cubic meters.
To find the volume of the remaining frustum, we first need to find the volume of the original pyramid, and then subtract the volume of the removed portion.
The formula for the volume of a pyramid is given by:
V = (1/3) * base area * height
Given that the base of the pyramid is a 10m square and the height is 15m, we can substitute these values into the formula to find the volume of the original pyramid.
V_original = (1/3) * 10m^2 * 15m
= (1/3) * 150m^3
= 50m^3
Since the top 6m of the pyramid is removed, the height of the remaining frustum will be 15m - 6m = 9m.
Now, we need to find the volume of the frustum (the remaining portion of the pyramid). The formula for the volume of a frustum is given by:
V_frustum = (1/3) * (A1 + A2 + √(A1 * A2)) * h
where A1 and A2 are the areas of the two bases, and h is the height of the frustum.
The area of the larger base (A1) can be calculated by squaring the side length of the base (10m):
A1 = (side length)^2
= 10m^2
= 100m^2
The area of the smaller base (A2) can be calculated similarly by subtracting the area of the removed portion from the area of the larger base:
A2 = A1 - (area of removed portion)
= A1 - (1/3) * (A1 * 6m)
= A1 - (1/3) * (100m^2 * 6m)
= A1 - (1/3) * 600m^3
= A1 - 200m^3
Now we can substitute the values into the formula to find the volume of the frustum:
V_frustum = (1/3) * (A1 + A2 + √(A1 * A2)) * h
= (1/3) * (100m^2 + (A1 - 200m^3) + √(100m^2 * (A1 - 200m^3))) * 9m
By simplifying this expression, you can find the volume of the remaining frustum.