Chapter 10.5 — Exercise 10.4 — Conversion of Solids

A Sphere Recast Into a Cylinder

A sphere of radius 4.2 cm is melted and poured into a cylindrical mould of radius 6 cm. The cylinder's height is whatever height makes the two volumes match exactly, no more and no less.

Every π cancels out of this equation entirely, which is worth noticing on its own — recasting problems almost never need π's actual numeric value at all, since it appears on both sides of the volume-equality equation and divides straight out. This is worth expecting as a general pattern for the rest of this exercise too: any time two volumes are simply set equal to each other, without any surface area or painted-cost calculation mixed in, π rarely survives to the final answer at all.

Three Spheres Melted Into One

Spheres of radii 6 cm, 8 cm, and 10 cm are melted together and recast as one single larger sphere. Finding its radius uses the same volume-equality idea already used above, just with three separate volumes summed together on one side of the equation instead of just one.

The radii don't add directly — 6+8+10=24 is not the answer, and it's a tempting shortcut to reach for by mistake. Volumes add; radii only add after being cubed first and the sum cube-rooted back down, which is precisely why 1728 (not 24) is the number that actually matters here. 1728 being a recognisable perfect cube (12³) is what keeps this particular problem's final step clean.

It's worth being clear about exactly why radii can't simply be added, since the instinct to do so is a natural one. Volume grows with the cube of radius, not the radius itself — doubling a sphere's radius multiplies its volume by eight, not by two — so treating radius as something that adds linearly the way length does is a genuine category error. Cubing first, summing, then cube-rooting is the only route that respects how volume actually scales.

Two Wells, Two Different Destinations for the Dug-Out Earth

Both problems below start identically — a cylindrical well is dug out, and the earth removed is spread elsewhere — but the shape the earth is spread into is genuinely different in each case.

The second problem's embankment is really a large hollow cylinder — a ring shape in cross-section — with the well's own radius as its inner edge and the well's radius plus the embankment's width as its outer edge. Its volume isn't πR²h the way a plain cylinder's would be; it's the outer cylinder's volume minus the inner (well-shaped) hole cut straight through it, π(R²−r²)h, the same ring-shaped subtraction idea already familiar from area-of-a-ring problems.

Both wells are physically the same kind of excavation — a cylindrical hole, dug straight down — and both problems set the well's own volume equal to wherever that same earth ends up. What changes between them is only the shape of the destination: a rectangular platform in the first case, a ring-shaped embankment in the second. Once the destination shape is correctly identified, the rest of each problem is the identical volume-equals-volume substitution used everywhere else in this exercise.

Four More Conversions, Compressed

Every row in this table follows the identical two-step shape that's run through this entire exercise: find the total volume being converted, then divide by the volume of a single new piece to count how many of them fit. The lead-shots problem here is the one worth a second, closer look for what it does genuinely differently — instead of converting the vessel's entire volume, only the fraction that actually overflows (a quarter of it) gets set equal to the total volume of shots, since the rest of the water simply stays put inside the vessel.

The ice cream problem is worth a second glance too, since it's really two separate combined-solid calculations feeding into one final conversion. The cone-plus-hemisphere shape is exactly the same kind of joined solid measured back in Exercise 10.3, and only once that combined volume is found does the actual conversion step — dividing the cylinder's volume by it — even begin. Recognising that a conversion problem sometimes hides a combined-solid problem inside it is worth watching for closely, rather than assuming every "how many can be made from" problem involves nothing but a single plain shape throughout.

The silver-coin problem is worth flagging for its units alone: the coin's thickness is given in millimetres while every other measurement in the same problem is given in centimetres, so converting 2 mm to 0.2 cm before substituting anything is a necessary first step, not an optional tidy-up. Skipping that conversion would silently scale the coin's volume — and therefore the final count of coins needed — by a factor of ten in the wrong direction entirely.

The Chapter, Complete

Mensuration moved through four genuinely different skills, each building on the one before it: the introduction established the six core surface-area and volume formulas, Exercise 10.1 applied them to single solids, Exercise 10.2 and Exercise 10.3 combined multiple solids by area and by volume respectively, and this exercise closed by holding volume constant while letting shape change completely. The next chapter, Trigonometry, moves away from solids entirely, turning instead to the ratios hiding inside a right triangle's angles — a genuinely different branch of this course, built on angles rather than surfaces and volumes.