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Question

A right circular cone is divided into three parts by trisecting its height by two planes drawn parallel to the base. Show that the volume of the three portions starting from the top are in the ratio 1 : 7 : 19. [CBSE 2017]

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Solution

Let ABC be a right circular cone of height 3h and base radius r. This cone is cut by two planes such that AQ = QP = PO = h.



Since ABO~AEP (AA Similarity)
AOAP=BOEP3h2h=rr1r1=2r3 .....1
Also, ABO~AGQ (AA Similarity)
AOAQ=BOGQ3hh=rr2r2=r3 .....2
Now,
Volume of cone AGF,
V1=13πr22h =13πr32h From 2 =127πr2h
Voulme of the frustum GFDE,
V2=13πr12+r22+r1r2h =13π4r29+r29+2r29h From 1 and 2 =727πr2h
Voulme of the frustum EDCB,
V3=13πr2+r12+r1rh =13πr2+4r29+2r23h From 1 and 2 =1927πr2h
∴ Required ratio = V1:V2:V3=127πr2h:727πr2h:1927πr2h=1:7:19

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