Mensuration 3D

Before any particular solid, learn the most basic rule: volume of a box = length × breadth × height. Why? Consider a box 5 cm long, 4 cm wide and 3 cm high. One layer on the floor holds 5 × 4 = 20 small cubes of 1 cm³. There are 3 such layers, so the box holds 20 × 3 = 60 cubes, that is 60 cm³. Volume is base area × height, and this rule applies to every prism and cylinder later.

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In the exam, questions of this type are rarely about the formula. They are about units. The dimensions may be in metres and the answer wanted in litres, or the sides in inches and the area wanted in square feet. So build one fixed habit:

• Step 1: Bring all three lengths to one unit.
• Step 2: Multiply to get the volume (or area).
• Step 3: Convert the final answer using 1 L = 1000 cm³ and 1 m³ = 1000 L = 1 kL.

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Radhika's water tank is 2 m × 1 m × 0.5 m. Volume = 1 m³ = 1000 litres. If water costs ₹4 per 100 litres, filling it costs 1000100 × 4 = ₹40.

How many objects fit, or are needed. "How many buckets of 5 L fill a 1 m³ tank?" Number = total volumevolume of one = 10005 = 200. Bricks in a wall work the same way: volume of wallvolume of one brick, with both volumes in the same unit.

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Area in different units. If a carton is measured in inches and the area is asked in square feet, either convert each side to feet first (60 in = 5 ft), or find the area in square inches and divide by 144, not by 12.

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A cuboid is a box with six rectangular faces: a brick, a matchbox, a room. It has length l, breadth b and height h. Opposite faces are equal, so the six faces come in three pairs: l × b (top and bottom), b × h (the two sides) and l × h (front and back).

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Surface areas.

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Diagonal. The longest rod that fits in a room runs from one floor corner to the opposite ceiling corner. First the floor diagonal = √(l² + b²); then, with the height, d = √(l² + b² + h²). Radhika's cupboard measures 6 ft × 2 ft × 3 ft, so the longest stick it can hold is √(36 + 4 + 9) = √49 = 7 ft.

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Edges. A cuboid has 12 edges, four each of l, b and h. Sum of all edges = 4(l + b + h).

A very useful identity. (l + b + h)² = l² + b² + h² + 2(lb + bh + hl). In words:

So when the sum of the dimensions and the diagonal are given, TSA comes in one line.

Three adjacent faces. If the areas of the three faces meeting at a corner are x = lb, y = bh and z = hl, then xyz = l²b²h², so V = √(xyz). And TSA = 2(x + y + z).

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Open box from a sheet. Cut squares of side x from the four corners of an L × B sheet and fold up the flaps. The box is (L − 2x) × (B − 2x) × x.

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A cube is a cuboid in which l = b = h = a. So no new formulas are needed; put a in place of l, b and h:

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The usual question gives one measurement and asks for another. Always go through the side a: find the side first, and everything else follows. Radhika's dice-shaped gift box has a volume of 125 cm³. Side = ∛125 = 5 cm, so TSA = 6 × 25 = 150 cm² and the longest diagonal = 5√3 cm.

Cubes joined end to end. n cubes of side a joined in a row make a cuboid na × a × a. Its TSA = 2(na·a + a·a + na·a) = (4n + 2)a². Two cubes give 10a², three cubes give 14a². You can also count it the other way: each joint hides two faces, so the total is 6na² − 2(n − 1)a², which is the same expression.

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Cutting a cube into small cubes. A cube cut into n³ equal small cubes needs (n − 1) cuts along each of the three directions, so 3(n − 1) cuts in all. For 27 = 3³ pieces, that is 3 × 2 = 6 cuts.

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Ratios and scaling. For two cubes, sides a₁ : a₂ give areas a₁² : a₂² and volumes a₁³ : a₂³. To go from a volume ratio to an area ratio, take the cube root first and then square.

A piece cut out of a cube. When a smaller cube of side a is removed from a corner, the volume falls by a³, but the surface area does not change. The removed corner took away three squares of area a² from the outside, and the cut exposes three new squares of area a² inside. Removing a corner leaves the total surface area the same.

If the small cube is scooped out of the middle of one face instead, one square a² is lost and five squares a² are exposed, so the surface area rises by 4a². For a cube cut away along an edge, it rises by 2a².

Radhika has a cube of side 12 cm. A cube of side 3 cm is cut from one corner. The volume left is 1,728 − 27 = 1,701 cm³, and the surface area is still 6 × 144 = 864 cm².

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Look at a cuboid again. It is a rectangle (the base) pushed straight up through a height h. A right prism is the same idea with any base: triangle, square, trapezium or hexagon. It has two identical bases, one at the bottom and one at the top, joined by rectangular side faces. A cuboid is just a rectangular prism, and a cube is a square prism of height a.

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Because the solid is a base pushed up, the formulas are simple:

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So every prism question is really a 2D question about the base. You need these base areas from Mensuration 2D:

• Right triangle: 12 × leg × leg
• Any triangle with sides a, b, c: Heron's formula √(s(s − a)(s − b)(s − c)), s = a + b + c2
• Equilateral triangle of side a: √34a²
• Regular hexagon of side a: 6 × √34a² = 3√32a²
• Trapezium: 12 × (sum of parallel sides) × distance between them

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Radhika's tent is a triangular prism. The front triangle has base 4 m and height 3 m, and the tent is 5 m long. Volume of air = 12 × 4 × 3 × 5 = 30 m³. Each sloping side is √(2² + 3²) = √13 m, so the canvas for the two slopes is 2 × √13 × 5 = 10√13 m².

Be careful with the word "height". In a prism it means the length of the prism, the distance between the two bases, and not the height of the triangle in the base. Many questions use both, so label them separately.

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Euler's formula for prisms: a prism with an n-sided base has n + 2 faces, 2n vertices and 3n edges. A triangular prism has 5 faces, 6 vertices and 9 edges.

A pyramid has a base (square, triangle, rectangle…), and all its side faces are triangles meeting at one point at the top, called the apex. In a right (regular) pyramid the apex is exactly above the centre of a regular base, so all the side triangles are identical.

There are three lengths here, and you must not mix them up:

• Height (h): from the apex straight down to the centre O of the base.
• Slant height (l): from the apex to the midpoint of a base edge. It is the height of each side triangle.
• Lateral edge (e): from the apex to a corner of the base.

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For a square base of side a, look at the right triangle formed by the apex, O and the midpoint of an edge. Its legs are h and a2, so l² = h² + (a2)². For the lateral edge use half the diagonal: e² = h² + (a√22)² = h² + a²2.

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Formulas.

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Radhika makes a paper pyramid on a square of side 6 cm, with height 4 cm. Slant height = √(4² + 3²) = 5 cm. LSA = 12 × 24 × 5 = 60 cm², TSA = 60 + 36 = 96 cm², and volume = 13 × 36 × 4 = 48 cm³.

Triangular base. For an equilateral base of side a, the centre is the centroid. Its distance to the midpoint of an edge (the inradius) is a2√3, and to a corner (the circumradius) it is a√3. So l² = h² + (a2√3)².

Regular tetrahedron. A tetrahedron is a triangular pyramid. In a regular tetrahedron all four faces are equilateral triangles of side a (4 faces, 4 vertices, 6 edges). Its results:

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Now the round solids begin. A cylinder is a prism whose base is a circle: a can, a pillar, a roller. So the prism rules carry straight over:

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Radhika's water bottle has radius 3.5 cm and height 20 cm. Capacity = 227 × 3.5 × 3.5 × 20 = 770 cm³ = 0.77 litre. Area of the label = 2 × 227 × 3.5 × 20 = 440 cm².

Going backwards. Many questions give two of CSA, TSA, V, r and h and ask for a third. These links help:

• VCSA = πr²h2πrh = r2, so r = 2VCSA.
• CSATSA = hr + h. If CSA : TSA = 5 : 8, then h : r = 5 : 3.
• If r + h is given, TSA = 2πr(r + h) gives r directly.

Rolling a sheet into a cylinder. The edge that goes round becomes the circumference (2πr = that edge); the other edge becomes the height. Rolling along the longer side gives the bigger volume, because volume grows with the square of the radius.

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Roller questions. In one revolution a roller presses its curved surface on the ground once. Area covered = CSA × number of revolutions.

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Ratios and changes. V ∝ r²h and CSA ∝ rh. If r is increased by 10% with h fixed, V becomes 1.21 times (21% up). If r falls by 20%, h must become 10.8² = 1.5625 times (56.25% more) to keep V the same.

A cylinder cut into two halves. Cut a solid cylinder along a plane through its axis. Each half has: half of the curved surface (πrh), two half-discs (together πr²) and the new flat rectangular face of size 2r × h. So

The volume of each half is 12πr²h. The new rectangle appears on both halves, so the total surface area of the two pieces rises by 2 × 2rh = 4rh compared with the original cylinder. A half-pipe, a gutter or a semicircular trough is the same half-cylinder.

Radhika's cylinder has r = 3.5 cm and h = 10 cm. Each half has volume 12 × 227 × 12.25 × 10 = 192.5 cm³, and the total surface area goes up by 4 × 3.5 × 10 = 140 cm² when it is cut.

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A pipe is a cylinder with a smaller cylinder taken out of its middle. If the outer radius is R and the inner radius is r, the metal forms a ring of area π(R² − r²) at each end, and that ring runs along the whole length h.

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For surface area, count every surface that is actually there:

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Radhika's steel glass has an inner radius of 3 cm and walls 0.5 cm thick, so R = 3.5 cm. If it is 10 cm tall (leave out the bottom), steel used = 227 × (3.5² − 3²) × 10 = 227 × 3.25 × 10 ≈ 102 cm³.

Well. Digging a well removes a solid cylinder of earth: volume = πr²h. If the earth is spread around the well as an embankment, it becomes a hollow cylinder; we will meet this in the water and flow type.

Hollow box. The same idea works for a cuboid: wood or metal used = outer volume − inner volume. For a closed box of thickness t, every inner dimension is 2t less: inner box = (l − 2t)(b − 2t)(h − 2t). For an open box (no lid), the length and breadth still lose 2t, but the height loses only t.

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Weight. Mass = volume of material × density. Watch the units: to change g to kg, divide by 1000.

A cone is to a cylinder what a pyramid is to a prism. It has the same circular base and the same height, but it narrows to a point. So:

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Why πrl? Cut the curved surface along a slant and open it out. You get a sector of a circle of radius l whose arc is the base circumference 2πr. Area of that sector = 12 × arc × radius = 12 × 2πr × l = πrl.

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The same picture tells you how to make a cone from a sector. If a sector of radius R and angle θ is folded into a cone, then slant height = R and 2πr = arc = θ360 × 2πR, so r = θ360R. A semicircle (θ = 180°) gives r = R2; a quarter circle gives r = R4.

Rotating a right triangle. Turn a right triangle about one of its legs and it sweeps out a cone. That leg becomes the height, the other leg the radius, and the hypotenuse the slant height. Radhika turns a 6-8-10 triangle about the 8 cm side: h = 8, r = 6, V = 13π × 36 × 8 = 96π cm³. Turned about the 6 cm side it gives h = 6, r = 8 and V = 128π cm³, which is a different cone.

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Cone and cylinder together. Same r and h ⇒ V_cone : V_cyl = 1 : 3. For a cone of radius 3r and a cylinder of radius r with the same height, the ratio is 13 × 9 : 1 = 3 : 1. Write V ∝ r²h for both and keep the 13 on the cone side only.

Funnels, caps and cones made from a sector. A funnel, a party cap or a paper cup is made by rolling a sector of a circle. Three things carry over:

• the radius R of the sector becomes the slant height l of the cone;
• the arc of the sector becomes the circumference of the base, so 2πr = θ360 × 2πR, which gives r = θ360 × R;
• the area of the sector becomes the curved surface area πrl.

Then h = √l² − r² follows from the right triangle. A semicircle (θ = 180°) always gives r = R2, and therefore h = √32R.

Radhika rolls a sector of radius 10 cm and angle 144° into a cap. Base radius r = 144360 × 10 = 4 cm, slant height 10 cm, and height h = √100 − 16 = √84 cm. The curved surface of the cap is π × 4 × 10 = 40π cm², the same as the area of the sector.

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Cut a cone with a plane parallel to its base. The top piece is a small cone; the bottom piece, shaped like a bucket or a lampshade, is a frustum.

The small cone is similar to the whole cone. Every length of it (height, radius, slant height) is the same fraction k of the matching length of the whole cone. So by the scaling rule:

• heights k : 1, radii k : 1, slant heights k : 1
• curved surface areas k² : 1
• volumes k³ : 1

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Radhika cuts a cone at the middle of its height. Then k = 12, so the small cone has 18 of the volume and the frustum has 78: small cone : frustum = 1 : 7. If she cuts at one-quarter of the height from the top, k = 14, so small cone : whole = 1 : 64 and small cone : frustum = 1 : 63.

Read the position of the cut carefully. In a 40 cm cone, a cut "10 cm from the vertex" gives k = 14. A cut "10 cm from the base" leaves a small cone of height 30 cm, so k = 34 and its volume is 2764 of the whole.

Three or more parts. If three cuts divide the height into four equal parts, the cones measured from the top have heights 1, 2, 3, 4 (in parts), so their volumes are 1, 8, 27, 64. The four pieces are 1, 8 − 1, 27 − 8, 64 − 27, that is 1 : 7 : 19 : 37. Each piece is the difference of two consecutive cubes.

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Frustum formulas (R = bottom radius, r = top radius, h = height):

An open bucket has the CSA and the bottom circle only.

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The same logic works for a pyramid cut parallel to its base. Volume of a pyramid frustum = h3(A₁ + A₂ + √(A₁A₂)), where A₁ and A₂ are the areas of the two ends, and the small pyramid on top is again k³ of the whole.

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A sphere is the set of all points at the same distance r from a centre: a ball, a marble, a globe. It has no faces, no edges and no vertices, only one curved surface.

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An easy way to hold the surface area in memory: it is exactly four times the area of the sphere's greatest circle (πr²). Archimedes also showed that a sphere fits inside a cylinder of radius r and height 2r, and the sphere's volume is 23 of that cylinder's: 23 × πr² × 2r = 43πr³.

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Radhika's cricket ball has a diameter of 7 cm, so r = 3.5 cm. Leather needed = 4 × 227 × 3.5 × 3.5 = 154 cm². Volume = 43 × 227 × 3.5³ = 17923 cm³.

Spherical shell (a hollow ball with outer radius R and inner radius r):

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Scaling. Here the rule is at its cleanest: area ∝ r² and volume ∝ r³.

• Radius +30% (× 1.3) ⇒ area × 1.69 (+69%), volume × 2.197 (+119.7%)
• Radius −20% (× 0.8) ⇒ volume × 0.512 ⇒ a decrease of 48.8%
• If the surface area increases by 96%, the radius factor is √1.96 = 1.4, so the volume factor is 1.4³ = 2.744 (+174.4%)

Radius increase given as a number. If r becomes r + a, the surface area increases by 4π[(r + a)² − r²] = 4πa(2r + a). Use this to find r.

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Numerically equal. If the volume equals the surface area in number, 43πr³ = 4πr² ⇒ r = 3. If V = k × SA, then r3 = k ⇒ r = 3k.

Cut a sphere through its centre and you get two hemispheres. Each one has half the volume and half the curved surface of the sphere, and it also has a new flat circular face.

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So the TSA of a hemisphere is not half the TSA of a sphere. Radhika cuts a ball of radius 3.5 cm into two halves. The ball had 4πr² = 154 cm² of surface. The two halves together have 2 × 3πr² = 6πr² = 231 cm², an increase of 2πr² = 77 cm², which is 50% of the original.

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Hemispherical bowl (a hollow hemisphere with outer radius R, inner radius r and thickness t = R − r):

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Dome and painting questions. A dome is a hemisphere without the floor, so only the curved surface 2πr² is painted. If the circumference of the base is given, use 2πr = circumference to get r first.

Comparisons at the same radius: V_sphere : V_hemisphere = 2 : 1, and TSA_hemisphere : TSA_sphere = 3πr² : 4πr² = 3 : 4.

Real objects are often two solids joined together. A tent is a cylinder with a cone on top, an ice-cream is a cone with a hemisphere on it, a capsule is a cylinder with a hemisphere at each end, and a water tank is a cylinder with a hemispherical top. Two rules handle all of them:

• Volume: add (or subtract) the volumes of the parts.
• Surface area: add only the surfaces you can actually see or touch. The face where two parts are joined is hidden inside, so do not count it. A tent has no floor, and a toy has no hidden base to count.

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Radhika's tent has a cylindrical part of radius 7 m and height 4 m, with a conical top of slant height 10 m. Canvas = CSA of cylinder + CSA of cone = 2 × 227 × 7 × 4 + 227 × 7 × 10 = 176 + 220 = 396 m². The circle where the cone sits on the cylinder is not canvas, and neither is the floor.

Capsule (a cylinder of length h with hemispheres of radius r at both ends): total length = h + 2r. Surface = 2πrh + 4πr²; volume = πr²h + 43πr³.

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Cloth for a tent. If the cloth is w metres wide, the length of cloth needed = areaw.

People in a tent. If each person needs a m² of floor and b m³ of air, then floor area = n × a and volume = n × b. For a conical tent, h = 3Vbase area = 3 × nbna = 3ba.

Comparing two solids with the same radius r. Write each volume in r and h and compare:

• Cone = hemisphere in volume ⇒ 13πr²h = 23πr³ ⇒ h = 2r
• Cylinder = hemisphere ⇒ πr²h = 23πr³ ⇒ h = 23r
• Cone and hemisphere with equal CSA ⇒ πrl = 2πr² ⇒ l = 2r, h = √3 r
• Same r and the same height (h = r for the hemisphere): cylinder : hemisphere : cone = 3 : 2 : 1

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Melting and recasting. When metal (or wax, clay or ice) is melted and reshaped, the shape changes but the amount of material does not. So this type has just one equation:

• One solid into another: write both volumes and solve for the unknown length.
• One big into many small: number of small solids = volume of bigvolume of one small.
• Many into one: add all the volumes, then find the new side or radius.
• If x% is lost, use (100 − x)% of the original volume.

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Radhika melts a wax cube of side 6 cm and makes small cubes of side 2 cm. Number = 2168 = 27. There is a quicker way to see it: the side ratio is 3 : 1, so the volume ratio is 27 : 1.

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Drawing a wire. A wire is a long, thin cylinder: πr²L = volume of the metal. Keep the units the same throughout. A radius in mm with a length in metres is where most students go wrong.

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Same shape, scaled. For similar solids, number of small ones = (big lengthsmall length)³. A sphere of radius 6 made into spheres of radius 2 gives 3³ = 27 balls, and their total surface area is 3 times that of the big one (27 balls × 19 of the area each).

Inscribed shapes (the largest solid inside another). Here nothing is melted; one solid just fits inside another and touches it. Draw the cross-section and match the lengths:

• Largest sphere in a cube of side a: diameter = a ⇒ r = a2
• Largest cube in a sphere of radius R: body diagonal = 2R ⇒ a√3 = 2R
• Cuboid in a sphere: √(l² + b² + h²) = 2R
• Largest cylinder or cone in a cube: r = a2, h = a
• Sphere in a cylinder that just holds it: cylinder r = R, h = 2R, and the sphere is 23 of the cylinder
• Sphere touching the base and the slant of a cone (radius r, height h, slant height l): radius = rhr + l
• Regular tetrahedron of edge a in a sphere: R = a√64
• Cube in a hemisphere of radius r (base on the flat face): a = r√(23)

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Displacement. When a solid is dropped into water and goes fully under (with no overflow, in a vessel whose horizontal cross-section is the same all the way up), it pushes the water up by exactly its own volume. If it is only partly under or floats, use just its submerged volume. The water does not care about the shape of the solid, only its volume. So:

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• Cylindrical container of radius R: πR² × h = volume of the object ⇒ h = volumeπR²
• Rectangular container l × b: l × b × h = volume of the object
• If the container is already full, the water that overflows = volume of the object.
• Counting: number of balls dropped = base area × risevolume of one ball.

Radhika drops a stone of volume 770 cm³ into a cylindrical jar of radius 7 cm. Rise = 770154 = 5 cm.

Flow through a pipe or canal. Water moving at speed v through a pipe of cross-section area A delivers A × v of water in every unit of time. So:

[[fig:1]]

Keep the units together. If the speed is in km/h, change it to metres per minute or metres per hour to match the time. A canal of radius 2 m with water flowing at 3 km/h delivers π × 4 × 3000 m³ every hour.

Filling and emptying. Time = volumerate. If a tank of 6 m³ is filled at 40 litres per minute, time = 600040 = 150 minutes. A half-full tank needs only half the volume.

Digging and spreading earth. Earth dug out of a pit (or a well) has a fixed volume. If it is spread evenly on the land around, rise in level = volume of eartharea on which it is spread. Do not include the area of the pit itself in the spreading area.

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A room, a hall, a classroom, a godown or a boundary wall is a cuboid, so every question in this type is really a cuboid question set in a practical context. The answer depends on two things: which faces are counted, and whether all the lengths are in the same unit.

Painting, whitewashing and plastering. The four walls have area 2(l + b) × h. The ceiling adds l × b, and the floor is counted only when the job is on the floor (cement, tiles). Doors and windows are cut out of the wall area. Then cost = area × rate per m².

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Air in a room. The space inside is the volume l × b × h. If every person needs v m³ of air, the number of persons that can sit or sleep in it is l × b × hv.

Bricks in a wall. A wall is a long, thin cuboid. Number of bricks = volume of the wall ÷ volume of one brick, once both are written in the same unit (centimetres are easiest). If the question says that a part of the wall is mortar, the bricks fill only the rest of the volume.

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Radhika's garden wall is 5 m long, 2 m high and 20 cm thick. In centimetres the wall is 500 × 200 × 20 = 20,00,000 cm³. A brick of 20 × 10 × 8 cm has volume 1,600 cm³, so the wall needs 20,00,0001,600 = 1,250 bricks.

Sheet metal and tanks. An open tank has no top, so the sheet needed is lb + 2(l + b)h. A closed tank needs the full 2(lb + bh + hl). If a circular opening of radius r is left in the top, subtract πr² from the area.

[[fig:2]]

Her classroom is 9 m × 6 m × 4 m, so its volume is 216 m³. At 6 m³ of air per student it holds 216 ÷ 6 = 36 students.

A cube or cuboid is painted on all its faces and then cut into equal small cubes. The question asks how many of the small cubes have 3, 2, 1 or no painted face. You do not need to count cube by cube. The position of a small cube decides the answer: a cube at a corner touches three outer faces, a cube on an edge touches two, a cube in the middle of a face touches one, and a cube inside touches none.

Take a cube of side n cut into n³ unit cubes (n ≥ 2). Then:

Check with n = 3: 8 + 12 + 6 + 1 = 27 = 3³. The factor (n − 2) appears because the two outer layers are used up by the paint, and what is left in each direction is n − 2 cubes.

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The unpainted cubes form a smaller cube of side (n − 2) in the middle. This one idea gives the other counts too: layer by layer, the top layer has 4 corner cubes (3 faces), 4(n − 2) edge cubes (2 faces) and (n − 2)² cubes with one face. Every middle layer has 4 edge cubes (2 faces), 4(n − 2) cubes with one face and (n − 2)² unpainted cubes.

[[fig:1]]

A cuboid of a × b × c, painted and cut into unit cubes, works the same way, with the inner block being (a − 2)(b − 2)(c − 2):

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Only some faces painted. Do not use the formulas. Think of the painted faces as layers. If only the top and bottom faces are painted, every cube in the top and bottom layers has exactly one painted face, which is 2n² cubes.

Cuts and new surface. To cut a cube into n³ equal cubes you need (n − 1) cuts in each of the three directions. Every cut makes two new faces, so the total surface area of all the small cubes is n times the surface area of the big cube. For n = 4, it is 4 times.

Radhika has a cube of side 6 cm, painted on all faces and cut into 1 cm cubes. The cubes with no paint are (6 − 2)³ = 64, those with one painted face are 6 × 4² = 96, and those with two painted faces are 12 × 4 = 48.