From the top of a cone of base radius $12$ cm and height $20$ cm, a small cone of base radius $3$ cm is to be cut off. How far down the vertex is the cut to be made? Find the volume of the frustum so obtained.

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Solution

Given, $r_{1} = 12 c m, h_{1} = 20 c m, r_{2} = 3 c m, h_{2} = ?$
We know, $\frac{r_{1}}{r_{2}} = \frac{h_{1}}{h_{2}}$
$∴ \frac{12}{3} = \frac{20}{h_{2}}$
$∴ h_{2} = \frac{3 \times 20}{12} = 5 c m$
$∴$ The cut to be made at $5$ cm from the vertex.
Volume of the frustum $= \frac{1}{3} π h (r_{1}^{2} + r_{2}^{2} + r_{1} r_{2})$
$= \frac{1}{3} \times \frac{22}{7} \times 15 (12^{2} + 3^{2} + 12 \times 3) = \frac{110}{7} \times (144 + 9 + 36)$
$= \frac{110}{7} \times 189 = 2, 970$
$∴$ Volume of the frustum $= 2, 970 c m^{3}$ .

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From the top of a cone of base radius 12cm and height 20cm, a small cone of base radius 3cm is to be cut off. How far down the vertex is the cut to be made? Find the volume of the frustum so obtained.

Given, r1=12cm,h1=20cm,r2=3cm,h2=?We know, r1r2=h1h2∴123=20h2∴h2=3×2012=5cm∴ The cut to be made at 5cm from the vertex.Volume of the frustum=13πh(r21+r22+r1r2)=13×227×15(122+32+12×3)=1107×(144+9+36)=1107×189=2,970∴Volume of the frustum=2,970cm3.

From the top of a cone of base radius $12$ cm and height $20$ cm, a small cone of base radius $3$ cm is to be cut off. How far down the vertex is the cut to be made? Find the volume of the frustum so obtained.