The volume of a cone is S × H/3, where S is the area of the base and H is the height. According to the figure above, if AB = BC, what is the ratio of the volume of the right section of the cone (with AB as height) to the volume of the left section of the cone (with BC as height)? (Note: Figure not drawn to scale.) A. 1 : 3 B. 1 : 4 C. 1 : 6 D. 1 : 7 E. 1 : 8

(D) 1) We can create two similar triangles using AC and AB as the heights, R and x as the bases, and the edge of the cone as the hypotenuse. These triangles are similar because the angles for both are the same. The sides of similar triangles are proportional to each other.

2) Because AB = BC, and AC = AB + BC, then 2(AB) = AC.

3) Since the triangles are similar, and because 2(AB) = AC, 2x = R.

4) The volume of the left half of the cone equals the volume of the entire cone minus the volume of the right half of the cone, so the ratio we are looking for is: (volume of right half) : (volume of cone – volume of right half).

5) Volume of the entire cone = S × H/3 = (πR²) × (AC)/3 = (π(2x)²) × (2AB)/3 = 8πx²AB/3

6) Volume of the right half of the cone = S × H/3 = πx²AB/3

7) Plug these values into the ratio equation above and solve: πx²AB/3 : (8πx²AB/3 – πx²AB/3) πx²AB/3 : 7πx²AB/3 1 : 7.

The correct answer is D. ---------- Why is it that the large cone = (π(2x)²) × (2AB)/3= 8πx²AB/3. Specifically, how do you derive an 8 out of the preceding equation?

Now, let's deal with another part of your question – Why is it that the large cone = (π(2x)²) × (2AB)/3?

We use the provided formula, S × H/3, where S is the area of the base and H is the height. Knowing that AB = BC, we easily calculate the height H = AC = AB + BC = AB + AB = 2AB .

For the base we use the formula for the area of a circle. S = π × R² And now comes the last part – How do we know that R = 2x?

We know this from similarity of triangles ABD and ACE:

They are similar, because they have equal angles: they have angle A in common, and angle B = angle C as they are both right angles (AB and AC are heights of the cones and therefore are perpendicular to the bases).

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