mitaliisrani wrote:A right circular cone has a height of 12 cm. If two cuts, parallel to the base, are made at a height of 2cm and 6cm from the base, find the ratio of the volumes of the three parts from top to bottom?
A) 27:98:91
B) 91:98:27
C) 98:91:23
D) 27:91:98
E) None of these

OA iA

I think we need to know the radius to solve this.

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Method 1 (not recommended)
The formula for the volume of a cone is V = (Pi* height * radius^2) * 1/3
Let take BO as "x",
we have OF = 2, thus AF = 10. Triangle AEF and ADO are in the same shape, then AF/AO = EF/BO = 10/12 =5/6, hence EF=5/6*x
Follow the same step, we get AD/AO = CD/BO = 6/12 = 1/2, hence CD = 1/2*x
For short, let's call G for grey, B for blue and O for orange
V (G+B+O) = Pi*12*x^2 *1/3 = Pi*4*x^2
V (G+B) = Pi*10*(5/6*x)^2 *1/3 = Pi* 125/54 * x^2
V (G) = Pi*6*(1/2*x)^2 *1/3 = Pi* 1/2 * x^2
They are asking us about the ratio among V (G), V (B) and V (O). Let's calculate them
V(G) = Pi*1/2*x^2 as above
V (B) = V(G+B) - V (G) = Pi*49/27*x^2
V(O) = V(G+B+O) - V(G+B) = Pi*91/54*x^2
V(G):V(B):V(O) = 1/2 : 49/27 : 91:54 = 27/54 : 98/54 : 91/54 = 27:98:91. Answer A wins.
Method 2
For those who hate equations, I have a more simple way to calculate this
Cone G, G+B, and G+B+O have the same shape and the ration among their height is 6 : 10: 12 = 1/2:5/6:1
The ratio among the volumes of those cone is the triple exponent of their height ratio, how can I get this?
The volume of a cone is calculated as V = Pi* r^2 * h * 1/3. Note that "r" is squared in this formula, so we must square the ratio among "r" multiply with ratio among heights, and the ratio among radiuses of those cones' bases is the same to that of their heights
V(G):V(G+B):V(G+B+O) = (1/2)^3 : (5/6)^3 : 1^3 = 1/8 : 125/216 : 1
V(B) = V(G+B) - V(G) = 125/216 - 1/8 = 49/108
V(O) = V(G+B+O) - V(G+B) = 1 - 125/216 = 91/216
V(G):V(B):V(O) = 1/8 : 49/108 :91/216 = 27/216 : 98/216: 91:216 = 27:98:91, Yeah, it's A.
Sorry for my bad English and expression, as well as small image ( dont know how to upload a bigger one)
Hope this could help.

Hi

this question is really difficult.

IMO A

let us look closer at it:

H1 = 12 cm
H2 = 2cm
H3 = 6cm
R1, R2, R3 are respective radiuses
@ is an angle at the base of a right circular cone( between radius and straight line segment joining the apex to the perimeter of the base)

We do not need to know exact values of the radius because we can use trigonometric function Cotangent

R1 = H1 * cot@
R2 = (H1 - H2) * cot@
R3 = (H1 - H3) * cot@

we can calculate ratio of the volumes in fallowing way:

V = pi/3 R^2 * H
ratio V3:(V2-V3):(V1-V2)

V3 = pi/3 * R3^2 * (H1 - H3) = pi/3 * ((H1 - H3) * cot@)^2 * (H1 - H3) = pi/3 * (H1 - H3)^3 * (cot@)^2 = 216 * (cot@)^2
V2 = pi/3 * R2^2 * (H1 - H2) = pi/3 * ((H1 - H2) * cot@)^2 * (H1 - H2) = pi/3 * (H1 - H2)^3 * (cot@)^2 = 1000 * (cot@)^2
V1 = pi/3 * R1^2 * H1 = pi/3 * (H1 * cot@)^2 * H1 = pi/3 * H1^3 * (cot@)^2 = 1728 * (cot@)^2

ratio is 216:784:728 which is equal to 27:98:91

hope it helps!