A toy is in the form of a cone mounted on a hemisphere of diameter 7 cm. The…

Concept: The volume of a composite solid is the sum of the volumes of the pieces it is built from. A cone of base radius r and height h has volume \(\frac{1}{3}\pi r^{2}h\), and a hemisphere of radius r has volume \(\frac{2}{3}\pi r^{3}\). A hemisphere's own height equals its radius, so when a cone stands on a hemisphere of the same radius the solid's total height is the cone's height plus that radius — the cone's height is never the total height.

Application: Fix the shared radius first, then split the total height between the two pieces before any volume is computed.

1. The hemisphere's diameter is 7 cm, so the shared radius is \(r=\frac{7}{2}=3.5\) cm.

2. The hemisphere accounts for 3.5 cm of the total height, so the cone's height is \(h=14.5-3.5=11\) cm.

3. Cone: \(\frac{1}{3}\pi r^{2}h=\frac{1}{3}\times\frac{22}{7}\times(3.5)^{2}\times 11=\frac{423.5}{3}\) cm3.

4. Hemisphere: \(\frac{2}{3}\pi r^{3}=\frac{2}{3}\times\frac{22}{7}\times(3.5)^{3}=\frac{269.5}{3}\) cm3.

5. Adding the two pieces: \(\frac{423.5}{3}+\frac{269.5}{3}=\frac{693}{3}=231\) cm3.

Cross-check: Factor before substituting: \(\frac{1}{3}\pi r^{2}h+\frac{2}{3}\pi r^{3}=\frac{1}{3}\pi r^{2}(h+2r)\). Here \(h+2r=11+7=18\) cm, so the volume is \(\frac{1}{3}\times\frac{22}{7}\times 12.25\times 18=38.5\times 6=231\) cm3 — the same figure reached by an independent route.

Common slips:

• Reading 14.5 cm as the cone's height instead of 11 cm counts the hemisphere's 3.5 cm of height twice.

• Using \(\frac{4}{3}\pi r^{3}\) for a full sphere in place of \(\frac{2}{3}\pi r^{3}\) for a hemisphere doubles the curved part.

• Treating the given 7 cm as the radius rather than the diameter changes both the squared-radius factor and the cone height inferred from the fixed total height, so the error cannot be corrected by one fixed multiplier.

The toy's volume is therefore 231 cm3.