IMO A=1:3
V=SH/3=piR^2H/3
now, x(smaller radius)=R/2
V2(right one)=pi(R/2)^2H/3......(1)
V1(left side of cone)=V-V2 =piR^2H/4.....(2)
this gives, V2:V1=1:3
Tough Geometry Problem
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earth@work
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earth@work
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Sorry, there was a miss calculation in solution above....Took height for V2 as H instead of H/2...have corrected below
V=SH/3=piR^2H/3
now, x(smaller radius)=R/2
V2(right one)=pi(R/2)^2(H/2)/3......(1)
V1(left side of cone)=V-V2 =1/3*piR^2H/4*7/8.....(2)
this gives, V2:V1=1:7
So answer is D=1:7
V=SH/3=piR^2H/3
now, x(smaller radius)=R/2
V2(right one)=pi(R/2)^2(H/2)/3......(1)
V1(left side of cone)=V-V2 =1/3*piR^2H/4*7/8.....(2)
this gives, V2:V1=1:7
So answer is D=1:7
Given:
Let the radius of the smaller base be x
Let the radius of the larger base be y
Given AB/AC = 1/2, or AC = 2AB
Using the property of similar triangles,
x/y = AB/AC = 1/2
or y = 2x
Ratio of the volumes of the small (to the right and the large cone):
Vs/Vl = (pi * x^2 * AB/3)/(pi * y^2 * AC/3)
Vs/Vl = 1/8 (after simplification)
Vs/(Vl-Vs) = 1/7
Therefore, D is correct
Let the radius of the smaller base be x
Let the radius of the larger base be y
Given AB/AC = 1/2, or AC = 2AB
Using the property of similar triangles,
x/y = AB/AC = 1/2
or y = 2x
Ratio of the volumes of the small (to the right and the large cone):
Vs/Vl = (pi * x^2 * AB/3)/(pi * y^2 * AC/3)
Vs/Vl = 1/8 (after simplification)
Vs/(Vl-Vs) = 1/7
Therefore, D is correct
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earth@work
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Hi smohsin, attaching a figure with clarificationsmohsin wrote:Thanks! However, could you please further elaborate on how you got x=R/2 and height of right cone= 1/2 of left cone.
Thanks for your help.
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