3D Mensuration in CAT: Cones, Cylinders and Recasting

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Three dimensional mensuration is a chapter where a handful of formulas carry the whole topic, and the marks are lost on setup. A cone is melted and recast. A sphere is dropped into water. A cylinder has a cone carved out of it. The formulas are easy. Deciding what stays constant is the real question.

This guide covers the formulas for cones, cylinders and spheres, the scaling rule for similar solids, and the recasting trap that catches students who treat surface area as if it were conserved. All numbers in the examples were checked by direct calculation.

The Formulas You Need on One Sheet

For a cylinder of radius r and height h, the volume is pi r squared h, the curved surface area is 2 pi r h, and the total surface area adds two circular ends, giving 2 pi r h plus 2 pi r squared. For a cone, the volume is one third pi r squared h, the slant height is the square root of r squared plus h squared, and the curved surface area is pi r l, where l is the slant height.

For a sphere of radius r, the volume is four thirds pi r cubed and the surface area is 4 pi r squared. A hemisphere has volume two thirds pi r cubed, curved surface 2 pi r squared, and total surface 3 pi r squared once you add the flat circular face.

You do not need more than this. Frustums, cuboids and prisms are handled by similar methods, and the formulas you need for a frustum can be derived from the cone, as shown later. Keep the list on one page, and test yourself on it until you can write it from memory.

Solid	Volume	Curved surface	Total surface
Cylinder	pi r squared h	2 pi r h	2 pi r (r plus h)
Cone	one third pi r squared h	pi r l	pi r (l plus r)
Sphere	four thirds pi r cubed	4 pi r squared	4 pi r squared
Hemisphere	two thirds pi r cubed	2 pi r squared	3 pi r squared

Cones and the Slant Height

A cone brings in the right triangle formed by the radius, the height and the slant height. If you see a radius and a height, check whether they make a familiar triple such as 3, 4, 5 or 5, 12, 13 or 8, 15, 17. Those save a square root and make mental work easy.

For a cone with radius 5 and height 12, the slant height is 13, the volume is one third times pi times 25 times 12, which is 100 pi, and the curved surface is pi times 5 times 13, which is 65 pi. The total surface adds the base area of 25 pi, giving 90 pi.

Questions often give the slant height and one other quantity, and ask for the volume. In that case, find the missing length from the right triangle first, then apply the volume formula. If the cone is made from a sector of a circle, the arc length of the sector becomes the base circumference, and the radius of the sector becomes the slant height. That conversion is a standard setup that appears often.

Here is a worked case. A sector of a circle with radius 10 and angle 216 degrees is rolled into a cone. The arc length of the sector is 216 over 360 times 2 pi times 10, which is 12 pi. That becomes the base circumference, so 2 pi r equals 12 pi and r equals 6. The slant height is the sector radius, 10, so the height is the square root of 100 minus 36, which is 8. The 6, 8, 10 triple appears again, and the volume is one third times pi times 36 times 8, or 96 pi.

Cylinders and Hollow Solids

Cylinder problems usually involve a cut, a hollow, or a fill. For a hollow cylinder or pipe, the volume of the material is the volume of the outer cylinder minus the inner cylinder, which is pi h times the difference of the squares of the outer and inner radii. Factorise the difference of squares to speed up the arithmetic.

For a cone carved from a cylinder with the same base and height, the cone has one third of the cylinder volume, and the leftover material has two thirds. A hemisphere carved from a cylinder of the same radius, with height equal to the radius, takes two thirds of the cylinder volume, leaving one third. These ratios are worth knowing, because they let you answer a question without computing any pi.

When a question asks for cost of painting or covering a surface, check which surfaces are included. An open cylinder has one circular end, a closed one has two, and a pipe has none. Reading the object description carefully is where most of the marks are won or lost in this chapter.

Spheres, Hemispheres and Displacement

Water displacement is a common setup. A solid submerged in a cylindrical vessel raises the water level by an amount found by dividing the volume of the solid by the base area of the vessel. A sphere of radius 5 is dropped into a cylinder of radius 10 that has enough water to cover it. The sphere volume is four thirds times pi times 125, and the base area is 100 pi. The rise is 500 over 3 divided by 100, or five thirds of a centimetre if the units are centimetres.

Check the assumption that the solid is fully submerged. If it floats or is only partly under, the displaced volume is different. Most CAT questions state clearly that the object sinks and is covered, which is why the volume of the solid is the displaced volume.

Hemisphere questions often involve a bowl or a dome. For a bowl, the capacity is the hemisphere volume, and the surface to polish is the curved surface of 2 pi r squared. For a solid hemisphere, the total surface adds the flat top. Reading whether it is a bowl or a solid is the whole question.

The Recasting Trap

When a solid is melted and recast into other solids, the volume of metal is the same before and after, ignoring any wastage stated in the question. The surface area is not the same. This distinction is behind most of the recasting questions.

Take a metal sphere of radius 6, with volume four thirds times pi times 216, which is 288 pi. It is melted and recast into cones of radius 3 and height 4. Each cone has volume one third times pi times 9 times 4, which is 12 pi. The number of cones is 288 pi divided by 12 pi, which is 24.

Now the trap. A sphere of radius 6 recast into 8 equal spheres gives each of them a radius of 3, since the volume of each is one eighth, and the radius scales by the cube root. The surface area of the original is 144 pi. The surface area of the 8 small spheres is 8 times 36 pi, which is 288 pi. The total surface area doubled while the volume stayed the same. If a question asks about the paint needed after recasting, the answer changes, because surface area is not conserved.

Similar Solids and Scaling

If two solids are similar, with every length in the ratio k, then surface areas are in the ratio k squared and volumes in the ratio k cubed. Increasing the radius of a sphere by 10 percent raises the volume by a factor of 1.331, which is 33.1 percent. That is the cube of 1.1.

Scaling also solves cut questions. A cone cut by a plane parallel to the base at half the height leaves a small cone similar to the original, with a ratio of one half. Its volume is one eighth of the original, so the remaining frustum has seven eighths. You never need the frustum formula, because a frustum is a large cone minus a small one.

Use the same idea for percentage change questions. If the radius of a cylinder rises by 20 percent and the height falls to keep the volume fixed, the height must fall by a factor of 1 over 1.44, which is about 30.6 percent. Our guide on algebraic vs numerical methods shows when plugging in numbers is quicker than writing an equation.

Keep the formula list handy on the CAT cheatsheets page, and if you want advice on how much time mensuration deserves, you can talk to a CAT mentor. For the wider list of chapters to cover, see our guide to 20 CAT Quant topics to cover before CAT 2026.