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A cone of radius 8 cm and height 12 cm is divided into two parts by a plane through the mid-point of its axis parallel to its base. Find the ratio of the volumes of two parts.

Answers (1)

Answer 1:7

Solution

            When a cone is divided into two parts by a plane through the mid-point the image formed is

           

            In figure \Delta AGE:\Delta EFC        Q\angle E is common angle \angle F=\angle G=90^{\circ}

            So the corresponding sides are in equal ratio.

                        \frac{EF}{FG}=\frac{FC}{GA}

                      \frac{ 6}{12} =\frac{ FC}{}8

                        FC = 4 cm

            Volume of cone EDC= \frac{1}{3}\pi r_{2}^2h

                        = \frac{1}{3}\times 3.14 \times (4)^2 \times 6=100.48cm^3

            Volume of frustum of cone

ABCD =\frac{1}{3}\pi h[r_{1}^2+r_{2}^2 +r_{1}r_{2}]

                        =\frac{1}{3}\times 3.14 \times 6[(8)^2+(4)^2 +8 \times 4]

                        =6.28 [64+16+32]

                       = 6.28[112] = 703.36

            Volume of cone EDC : volume of ABCD

                        100.48 : 703.36

                        1 : 7

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