A solid right circular cone of height $27$ cm is cut into pieces along a plane parallel to its base at a height of $18$ cm from the base. If the difference in volume of the two pieces is $225$ cc, the volume, in cc, of the original cone is

$243$

$232$

$256$

$264$

✅ Correct Option: 1

We have a cone that's been sliced horizontally, creating two pieces:

Original cone: Height = 27 cm, base radius = R

Cut location: 18 cm from the base (so 9 cm from the top)

Result: Two pieces with volume difference = 225 cc

When we cut the cone parallel to its base, the smaller piece on top is similar to the original cone.

Using similar triangles:

Original cone: Height = 27 cm, Radius = R

Top piece: Height = 9 cm, Radius = r

Since the triangles are similar: $\frac{r}{R} = \frac{9}{27} = \frac{1}{3}$

Therefore: $r = \frac{R}{3}$

Volume of original cone: $V_{original} = \frac{1}{3}\pi R^2 \times 27 = 9\pi R^2$

Volume of top piece (small cone): $V_{top} = \frac{1}{3}\pi r^2 \times 9 = 3\pi r^2$

Since $r = \frac{R}{3}$ : $V_{top} = 3\pi \left(\frac{R}{3}\right)^2 = 3\pi \times \frac{R^2}{9} = \frac{\pi R^2}{3}$

Volume of bottom piece (truncated cone): $V_{bottom} = V_{original} - V_{top}$

$V_{bottom} = 9\pi R^2 - \frac{\pi R^2}{3}$

$= \frac{27\pi R^2 - \pi R^2}{3} = \frac{26\pi R^2}{3}$

The difference in volume of the two pieces is 225 cc.

Since the bottom piece is larger: $V_{bottom} - V_{top} = 225$

$\frac{26\pi R^2}{3} - \frac{\pi R^2}{3} = 225$

$\frac{25\pi R^2}{3} = 225$

$\pi R^2 = 225 \times \frac{3}{25} = \frac{675}{25} = 27$

$V_{original} = 9\pi R^2 = 9 \times 27 = 243$ cc

We use similar triangles to find radius relationships in cone problems. The volume of a frustum equals the volume of the large cone minus the volume of the small cone.

Answer: 243 cc

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A solid right circular cone of height $27$ cm is cut into pieces along a plane parallel to its base at a height of $18$ cm from the base. If the difference in volume of the two pieces is $225$ cc, the volume, in cc, of the original cone is