Chapter 10.4 — Exercise 10.3 — Volume of Combined Solids
Volumes of combination of solids. This is Lesson 4 of 5 in Chapter 10: Mensuration.
Adding Volumes, No Hidden Faces to Track
Surface area needed careful bookkeeping about which faces vanish where two solids meet. Volume needs none of that — every cubic centimetre inside a combined solid belongs to it regardless of where the joins happen to sit, so this exercise is really just careful addition and subtraction of whole-solid volumes, nothing more. Several problems here go a step further too, using volume to answer questions that don't look like volume questions at all on first read — a weight, a water level, a count of identical small objects.
An Iron Pillar's Weight
A cylindrical pillar (height 2.8 m, diameter 20 cm) is topped with a cone (height 42 cm) of the same radius. Finding its weight starts with volume, then converts that volume into a weight using a given material density.
radius = 10 cm, cylinder height = 280 cm, cone height = 42 cm
Volume of cylinder = πr²h = (22/7)×100×280 = 88000 cm³
Volume of cone = ⅓πr²h = ⅓×(22/7)×100×42 = 4400 cm³
Total volume = 88000 + 4400 = 92400 cm³
Weight = 92400 × 7.5 g/cm³ = 693000 g = 693 kgConverting the pillar's 2.8 m height into 280 cm before substituting is essential here — mixing metres and centimetres in the same formula would silently scale part of the answer by 100, an error that's easy to make and easy to miss without a genuine unit check first. The final weight conversion is worth noticing as its own separate step too: volume in cubic centimetres times a density given in grams per cubic centimetre gives a weight in grams, which then needs one more division by 1000 to reach the kilograms the problem actually asks for.
A Toy: Finding a Height From a Volume Ratio
A hemisphere (radius 7 cm) is topped with a cone of the same radius, and the cone's volume is given as 3/2 of the hemisphere's — an unusual starting point compared to every problem before it, since the height itself is what's actually unknown here, not a length handed over directly.
⅓πr²h = (3/2)×(⅔πr³) ⟹ h = 3r = 3×7 = 21 cm
slant height l = √(7²+21²) = √(49+441) = √490 = 7√10 ≈ 22.14 cm
Surface area = πrl + 2πr² = πr(l+2r) = (22/7)×7×(22.14+14) = 22×36.14 ≈ 794.97 cm²The volume-ratio equation collapses remarkably cleanly: every r² term cancels across both sides, leaving h expressed as a simple multiple of r alone — h=3r, regardless of what specific radius the problem had actually chosen. Only after that height is found does the problem shift entirely to surface area, needing the slant height freshly calculated from scratch — the two halves of this problem barely share any working at all beyond the radius.
It's worth being upfront about one small correction here: substituting l≈22.14 and r=7 into πr(l+2r) gives 22×36.14, which comes out to approximately 794.97 cm² when carried through carefully — a number some printings of this problem show slightly differently due to how many decimal places l was rounded to along the way. Carrying the unrounded value of l (7√10, rather than a truncated decimal) as far as possible through the calculation, and only rounding at the very final step, is the more reliable habit whenever a problem explicitly asks for an answer "correct to 2 decimal places" the way this one does.
Five More Combined-Solid Volumes
| Setup | Key step | Result |
|---|---|---|
| Largest cone cut from a cube of edge 7 cm | r=3.5, h=7 (both fixed by the cube's edge); ⅓πr²h | 89.83 cm³ |
| Cone+hemisphere (r=3.5, cone height 5) immersed in a water-filled cylindrical tub (r=5, height 9.8) | Tub volume 770 cm³ minus displaced solid's volume 154 cm³ | 616 cm³ of water left |
| Cylinder (d=7, h=10) with two conical holes (r=3, h=4) drilled out | Cylinder volume minus 2× cone volume | 309.57 cm³ |
| Spherical marbles of 1.4 cm diameter dropped into a 7 cm-diameter beaker, raising the water level by 5.6 cm | Volume of water rise ÷ volume of one marble | 150 marbles |
| Cuboid pen stand (15×10×3.5) with 3 conical depressions (r=0.5, depth 1.4) drilled in | Cuboid volume minus 3× cone volume | 523.9 cm³ of wood |
Every row here follows one of exactly two shapes: either two volumes get added (a solid built from separate pieces), or one volume gets subtracted from another (material removed, or a solid displacing liquid it's dropped into). Deciding which of the two applies is really a reading-comprehension question before it's ever a mathematics one — "immersed," "drilled," "scooped," and "hole" all signal subtraction; "mounted," "surmounted," and "stuck together" all signal addition; getting this one word wrong flips the entire calculation's sign.
The largest-cone-from-a-cube problem is worth a second look for how it fixes its own measurements: nothing about a cube's edge directly states a radius or a height, yet both get pinned down entirely by the physical constraint that the cone has to fit exactly inside. The cone's base can be no wider than the cube's face (giving the diameter, and therefore the radius, straight from the cube's edge length), and its height can be no taller than the cube itself (giving the height directly too) — a genuinely different kind of problem from the others in this table, where every measurement is simply handed over directly rather than inferred from a fitting constraint like this one.
The conical-holes and conical-depressions rows share an identical underlying shape despite their different framing — a solid drilled full circular holes reduces its own volume by exactly the volume of material removed, whether that removed material is called a "hole" in a cylinder or a "depression" in a pen stand. Multiplying a single hole's volume by however many identical holes exist, then subtracting that product once at the end, is faster and less error-prone than subtracting one hole's volume at a time, one after another.
The marbles problem is worth a second look for its underlying logic, since nothing is being added to or subtracted from a single solid here at all. Instead, the volume of water displaced (found from how far the water level physically rises) is set equal to the combined volume of however many marbles caused that rise — the number of marbles is then recovered by simple division, one marble's volume into the total displaced volume. The immersed-solid problem two rows above uses the identical displacement idea, just run in the opposite direction: there, the displaced volume is already known (it's exactly the immersed solid's own volume), and the question is how much water remains rather than how many objects caused the rise.
From Melting Nothing to Melting Everything
Every volume in this exercise stayed with its original solid — nothing melted, nothing changed shape, only combined with or removed from something else. Exercise 10.4 does exactly that instead: melting a solid down and recasting it into a completely different shape, using the one fact this exercise has already leaned on repeatedly — that volume, unlike surface area, survives being added, subtracted, or reshaped entirely without losing anything at all in between. For the six core formulas this exercise depended on throughout, revisit the chapter introduction, which every one of these problems built on directly.