Solid geometry and spatial reasoning

Covering a box and filling it are different jobs

Imagine a closed rectangular box measuring 3 cm by 2 cm by 4 cm. Covering its six faces takes 52 cm² of material in an ideal thin-surface model; its interior geometric volume is 24 cm³. The square and cubic units tell you which question each answer solves. We will explain both numbers by identifying the parts of the box.

Before calculating, predict what changes if only the height doubles. The capacity doubles, but the top and bottom keep the same areas. Later, compare that change with doubling every dimension, which gives different area and volume factors.

Choose the preparation you need, then return to the spatial arguments

You need rectangle and triangle areas, circle circumference and disk area, consistent length units, squares and roots, and the right-triangle distance relationship. Coordinate triples extend the ordered pairs used in the coordinate lesson. The linked preparation lessons let you revisit each idea before it enters a formula.

For a first pass, compare the box faces, cylinder surface parts and pyramid heights. Use the exact labels and accompanying tables. Then return to lines and planes, the spatial-distance derivation and cross-sections when the shape roles feel familiar. The later proof uses two perpendicular steps; a perspective picture alone does not establish them.

Lines that miss one another need not be parallel

Work in Euclidean three-dimensional space. A point specifies a position; a line extends indefinitely in two directions; a plane extends in two independent directions. A finite segment or the parallelogram used to sketch a plane is only part of that object. Three noncollinear points determine a unique plane, while three collinear points lie in many planes. Four points need not be coplanar: (0, 0, 0), (1, 0, 0) and (0, 1, 0) determine the plane z = 0, which does not contain (0, 0, 1).

Two distinct lines in one plane either intersect at one point or are parallel and never meet. In space there is another possibility: skew lines do not meet and are not coplanar. For example, the x-axis consists of (t, 0, 0), while a line through (0, 0, 1) in the y-direction consists of (0, u, 1), with t and u real. They cannot meet because their z-coordinates differ, and their directions are not parallel. A projection may make them cross on paper without giving them a shared spatial point.

In this lesson, perpendicular lines meet at a right angle. Skew lines can have orthogonal directions, as in the example above, but they do not meet and are not called perpendicular lines under this convention. A line perpendicular to a plane meets it at a foot F and is perpendicular to every line in that plane through F. Being perpendicular to just one line of the plane is insufficient: the y-axis is perpendicular to the x-axis, yet both lie in z = 0. The z-axis, by contrast, is perpendicular to the whole plane z = 0 at the origin.

Two distinct planes are either parallel, with no common point, or meet in an entire line. The planes z = 0 and z = 2 are parallel. The planes x = 0 and z = 0 meet along the y-axis, where both coordinates are zero and y can vary. The fact that a sketch shows one crossing point does not reduce their intersection to one point. Space therefore requires both position and direction information, rather than applying every planar rule unchanged.

Lengths, boundary areas and enclosed volumes answer different questions

A closed solid includes its interior and boundary. Its surface area measures that boundary, while its volume measures the three-dimensional region inside. A polyhedron has finitely many polygonal faces; its total surface area is their area sum. A cuboid and a pyramid are polyhedra. A circular cylinder and a cone have curved boundary parts, and a sphere is a curved surface. We use sphere for the boundary at fixed distance r > 0 from a centre, and closed ball for that boundary together with its interior, just as circle and disk were distinguished in the preceding lesson.

Use a common length unit before combining dimensions. Areas have square units and volumes have cubic units. Since 1 m = 100 cm, a square metre is 100² = 10000 cm² and a cubic metre is 100³ = 1000000 cm³. A litre is 1000 cm³ = 1 dm³; it is a volume unit, not a surface-area unit. Choosing a larger length unit changes the numerical area and volume by the square and cube of the conversion factor.

The solid formulas below require positive physical dimensions. A cuboid needs a > 0, b > 0 and h > 0; a right circular cylinder needs r > 0 and h > 0; the specified rectangular pyramid needs positive base sides and perpendicular height. A zero dimension collapses a requested three-dimensional solid to a lower-dimensional boundary case. A negative proposed radius or side is invalid before multiplication or squaring. Signed coordinates and coordinate differences can be negative, but they are positions and directed changes rather than proposed physical lengths.

State which boundary is included. A closed cylinder has two end disks as well as its curved side. An ideal open-top container requires a different material-area sum, and its capacity uses interior dimensions rather than an unspecified outer volume. The formulas describe ideal geometric regions; a problem about a wall, coating or hollow object must specify which surfaces or dimensions it means.

A cuboid's six faces and uniform cross-section explain its formulas

For the 3-by-2-by-4 cm box, one pair of faces has area 6 cm² each, another has 12 cm² each, and the third has 8 cm² each. The six-face sum is 2(6 + 12 + 8) = 52 cm². Four layers of a 6 cm² base give the established volume 24 cm³. Which pair stays unchanged if only the height changes?

A cuboid is a rectangular prism whose three edge directions are mutually perpendicular. Let its positive dimensions be a, b and h. Opposite faces are congruent rectangles: two have area ab, two ah and two bh. Adding the six face areas gives S = 2(ab + ah + bh). Each face is counted once, including the hidden ones. Its established volume formula is V = abh: the rectangular base has area ab and the perpendicular height is h. Counting unit cubes illustrates integer dimensions; the formula also applies to positive fractional or real dimensions.

A net records selected cuts and unfolds the faces into the plane. For a cuboid, place a strip of four side rectangles with widths a, b, a and b and common height h. Attach one a-by-b end rectangle above the first a-by-h rectangle and the other below it, along that side rectangle's opposite length-a edges. The strip folds around the four sides and the attached ends close the top and bottom. The six planar rectangles have disjoint interiors, so their area sum equals the boundary area. A list of six rectangles or an overlapping layout is not automatically a valid net: the joins, matching lengths and fold need to be checked.

More generally, a right prism has two congruent polygonal bases in parallel planes and rectangular side faces. With base area B, base perimeter p and positive perpendicular height h, its volume is V = Bh. If the base is a simple polygon without holes, each base side of length e gives a rectangular side face of area eh; summing gives lateral area ph. Including both bases gives total S = 2B + ph. The height is the distance between the base planes, equal to the joining edge lengths for a right prism. We do not substitute a slanted joining edge into this right-prism surface formula.

For example, a right triangular prism whose base is a right triangle with legs 3 and 4 has B = 6 and perimeter p = 3 + 4 + 5 = 12 in consistent units. With prism height 5, V = 30 cubic units and S = 2 × 6 + 12 × 5 = 72 square units. The triangular base's perpendicular height is one of its legs, whereas the prism's height measures separation of its two base planes. Their shared name does not make them the same length.

A cylinder's curved side unrolls to a circumference-by-height rectangle

Take a closed right circular cylinder with radius 2 cm and height 3 cm. If you cut its curved side vertically and flatten it without stretching, its width is the circumference 4π cm and its height is 3 cm. Predict whether the two end disks belong in the curved-side area or must be counted separately.

A right circular cylinder has two congruent disk bases in parallel planes, with their centres joined by an axis perpendicular to those planes. Require radius r > 0 and height h > 0. Its established volume formula is V = πr²h: the base disk has area πr² and each cross-section parallel to the bases has the same disk shape and area. This is the circular counterpart of the prism formula, using the circle-area result already established; a rigorous derivation for the curved solid goes beyond counting finitely many polygonal pieces.

Cut the curved surface along one straight generator parallel to the axis and unroll it. The resulting rectangle has height h and width equal to the base circumference, 2πr. Thus curved-surface area is L = 2πrh. Adding the two end-disk areas gives total closed-cylinder area S = 2πrh + 2πr². The generator length equals the perpendicular height because this cylinder is right. The top and bottom disks are included once each, and their boundary circles add no extra area.

With r = 2 and h = 3 in one length unit, V = 12π cubic units, L = 12π square units and S = 20π square units. The equal numerical coefficients for V and L in this example do not make the quantities interchangeable: one has cubic units and the other square units. If a specified model excludes the top disk but keeps the bottom, its material area is L + πr² = 16π square units, not the closed total.

A diameter must first be halved to obtain r. Keep π exact until an approximation is requested. Radius zero or height zero fails the positive solid specification. In particular, substituting h = 0 into the closed-surface expression would count two coincident end disks as if they were distinct boundary pieces; that substitution is not a surface-area verdict for a valid three-dimensional cylinder.

A rectangular pyramid has two face-slant lengths, not one

Take a rectangular base 6 cm by 4 cm and an apex 3 cm directly above its centre. From the centre to the midpoint of a 6 cm edge, the horizontal offset is 2 cm; to the midpoint of a 4 cm edge, it is 3 cm. Those two right triangles have the same vertical height but different horizontal legs. Predict whether their sloping face altitudes will be equal.

A pyramid has a polygonal base and triangular faces joining its base edges to an apex outside the base plane. Here use a right rectangular pyramid with positive base sides a and b and positive perpendicular height h; its apex is directly above the base centre. In coordinates, the base corners are (0, 0, 0), (a, 0, 0), (a, b, 0) and (0, b, 0), while the apex is T = (a/2, b/2, h). Its perpendicular-height foot is the centre C = (a/2, b/2, 0). This explicit apex position is part of the model, rather than something inferred from a perspective picture.

Let ℓₐ be the altitude within a triangular face whose base edge has length a. That altitude joins T to the midpoint M of the edge; the horizontal distance CM is b/2. The perpendicular height TC = h and CM form a right triangle, so ℓₐ² = h² + (b/2)². The segment TM is perpendicular to that edge: its horizontal part is perpendicular to the edge and its vertical part is perpendicular to the base plane. Similarly, for a face on an edge of length b, ℓᵦ² = h² + (a/2)². Both face slants use positive square roots after a, b and h pass their guards.

The two faces on length-a edges have combined area 2 × (aℓₐ/2) = aℓₐ; those on length-b edges have combined area bℓᵦ. Including the rectangular base gives S = ab + aℓₐ + bℓᵦ. If a ≠ b, the face slants generally differ, so replacing both by a single unspecified slant loses the correct face areas. A lateral edge from T to a base corner has another length e, with e² = h² + (a/2)² + (b/2)². It is not the perpendicular height of either triangular face.

Use the established pyramid-volume result V = Bh/3, where B is the base area and h is perpendicular distance to its plane. Here V = abh/3. It is one third of a prism's volume for the same base area and perpendicular height, not because a pyramid looks one third as large in a drawing. We use the general volume result without claiming a full Cavalieri or integration proof in this lesson. Zero height makes the apex coplanar with the base; a zero base side or negative dimension also fails this nondegenerate solid specification.

Cone and sphere formulas need their own shape conditions

A right circular cone has a disk base of positive radius r and an apex directly above its centre at positive perpendicular height h. Its slant height ℓ runs from the apex to a base-circle point, with ℓ = √(r² + h²). Use the standard cone formulas V = πr²h/3 for enclosed volume, L = πrℓ for curved-surface area and S = πrℓ + πr² for the total closed boundary. Slant height belongs in curved area; perpendicular height belongs in volume. These formulas do not describe an unspecified oblique cone by merely relabelling its slanted length h.

For a sphere of positive radius r, use surface area S = 4πr² and the closed ball's volume V = 4πr³/3. Points on the sphere have distance r from its centre; points strictly inside the ball have smaller distance. The sphere has no flat base or apex. Its surface-area formula is not obtained by adding the area of a finite collection of flat faces. We accept the cone and sphere results here; their general curved-area and volume derivations require arguments beyond the finite planar proofs used earlier.

For a right cone with r = 3 and h = 4, ℓ = 5, so L = 15π, S = 24π and V = 12π in the appropriate square and cubic units. For a ball with r = 3, the sphere area is 36π square units and the ball volume is 36π cubic units. Equality of those numerical coefficients is accidental and unit-dependent. A zero radius or zero cone height is a collapsed boundary case, not another accepted positive-dimension solid.

Spatial distance follows from two perpendicular-component calculations

Use P = (1, −2, 3) and Q = (4, 2, 15) in one common length unit. Their horizontal changes are 3 and 4, so the horizontal distance to H = (4, 2, 3) is 5. The vertical step from H to Q is 12, and the direct distance is 13. The two diagrams separate the 3–4–5 horizontal step from the 5–12–13 vertical step before we write the general formula.

Choose mutually perpendicular Cartesian axes with a common physical length unit on all three. For P = (x₁, y₁, z₁) and Q = (x₂, y₂, z₂), define signed changes Δx = x₂ − x₁, Δy = y₂ − y₁ and Δz = z₂ − z₁. The component lengths are their absolute values. Let H = (x₂, y₂, z₁), the point with Q's first two coordinates in P's horizontal plane. The preceding planar distance formula gives PH² = (Δx)² + (Δy)², including its aligned and coincident cases.

If PH > 0 and Δz ≠ 0, segment HQ is perpendicular to that horizontal plane and therefore to PH at H. Triangle PHQ is nondegenerate, so the Pythagorean theorem gives PQ² = PH² + HQ² = (Δx)² + (Δy)² + (Δz)². Distance is nonnegative, hence PQ = √((Δx)² + (Δy)² + (Δz)²). This derivation uses the right-triangle result twice, through its planar distance formula and then a vertical component, without assuming a spatial distance formula as an earlier proof step.

Handle the boundary components directly. If Δz = 0, then Q = H and the spatial distance is the planar distance PH. If PH = 0, then P = H and the distance is the vertical line length |Δz|. If both are zero, P and Q coincide and PQ = 0. The single formula agrees with all these cases, but no nondegenerate right triangle is asserted at a collapsed step. Reversing P and Q negates every change while preserving its square, so the distance is symmetric.

A cuboid's opposite-corner distance follows as √(a² + b² + h²), after its physical dimensions are accepted as positive. A route along three perpendicular edges has length a + b + h and is longer than that diagonal: its square includes the additional positive terms 2ab + 2ah + 2bh. More generally, for point components, the route |Δx| + |Δy| + |Δz| equals the direct distance only when at most one component is nonzero. Explain whether a problem asks for a joining segment, a specified route or an enclosed volume.

A cross-section cuts a solid; a projection changes the view

A plane cross-section is the part of a solid lying in a chosen plane. An interior cut parallel to a cuboid's a-by-b base gives an a-by-b rectangle. An interior cut parallel to a right cylinder's bases gives a disk of the same radius r. A plane through the cylinder's axis cuts its closed solid in a rectangle with dimensions 2r and h. These sections answer different questions; a cylinder's outline in a perspective drawing is not one of its plane sections by definition.

For the centred rectangular pyramid, a cut parallel to the base at height t with 0 ≤ t < h gives a rectangle whose sides are a(1 − t/h) and b(1 − t/h). Similar triangles in vertical planes through the apex give the common length factor (h − t)/h. The section area is ab(1 − t/h)². At t = h the section is just the apex, not a positive-area rectangle; outside 0 ≤ t ≤ h it is empty. At half-height the side lengths are halved and the area is a quarter of the base area, illustrating why pyramid sections are not uniform like prism sections.

For a closed ball of radius r > 0, let d ≥ 0 be the perpendicular distance from its centre to a cutting plane. For 0 ≤ d < r, the section is a disk of radius √(r² − d²), by the centre-to-plane perpendicular and the planar right-triangle relationship; the central case d = 0 uses radius r directly. The sphere's boundary section is the circle around that disk. At d = r only one point remains, and at d > r there is no section. Check the distance interval before using a real square root or announcing a positive-area disk.

A projection maps spatial points onto a viewing plane and can hide depth. For the orthogonal projection onto the xy-plane, retain x and y and discard z. The segment from (0, 0, 0) to (3, 4, 12) has projected length 5 but spatial length 13. Points (0, 0, 0) and (0, 0, 5) project to the same point while remaining 5 units apart. Other viewing directions can similarly hide separation. Apparent angles, crossings and equal lengths in a picture do not prove their spatial counterparts. Use the stated coordinates, perpendicularity or incidence conditions as the certificate.

A uniform length factor squares for area and cubes for volume

Double every dimension of the 3-by-2-by-4 cm box. Each rectangular face gets two factors of 2, so total area becomes 4 × 52 = 208 cm². Volume gets three factors of 2, becoming 8 × 24 = 192 cm³. Doubling only its height would give volume 48 cm³ and surface area 92 cm², because the two 6 cm² end faces stay unchanged.

A positive uniform enlargement k > 0 multiplies every length by k while preserving angle and shape relations. For a cuboid, S′ = 2((ka)(kb) + (ka)(kh) + (kb)(kh)) = k²S, while V′ = (ka)(kb)(kh) = k³V. A right prism's base area receives k², its perimeter k and its height k, producing the same area and volume factors. The formulas show why a length factor cannot be applied unchanged to every quantity.

The curved-solid formulas agree. Replacing r and h by kr and kh gives cylinder area k²(2πrh + 2πr²) and volume k³πr²h. For the pyramid, each face slant also receives k, since √((kh)² + (kb/2)²) = k√(h² + (b/2)²) when k > 0. Its base and triangular face areas therefore receive k², and abh/3 receives k³. Cone slants and the sphere formulas give the same factors.

Doubling all dimensions gives 4 times the area and 8 times the volume. An enlargement by 3/2 gives area factor 9/4 and volume factor 27/8. Conversely, similar solids with volume ratio 8 have length ratio 2 and area ratio 4, selecting the positive real cube root for the length factor. This inference needs known similarity; a volume ratio alone between arbitrary solids does not determine their surface-area ratio.

Changing just one dimension is not a uniform enlargement. For a fixed-radius cylinder, doubling h doubles its volume and curved-side area, but the two end-disk areas are unchanged, so its total surface area does not generally double. Zero scaling collapses a solid. A negative multiplier is not a positive physical length scale in this lesson; if coordinates are transformed by a reflection or a negative dilation, lengths would use its magnitude and that would require a separately stated transformation.

Return from the formula to the solid being described

Begin with the shape and its hypotheses, then identify which lengths are given. A cylinder diameter is not its radius; a pyramid edge or face slant is not its perpendicular height. For a rectangular base, preserve which face is built on which base side. Convert lengths to one unit and reject nonpositive physical dimensions before products, squares, roots or a solid verdict. A formal numerical expression for a collapsed object is not automatically the boundary area of its lower-dimensional limit.

Surface-area addition needs the exposed boundary. If two cuboids with disjoint interiors are joined along an entire common rectangular face of area F, their combined volume is the sum of their volumes, while their exterior area is S₁ + S₂ − 2F: the shared face was counted on each cuboid and is now internal. If their interiors overlap, volume addition also double-counts that overlap and needs correction. Stating the joins or intersections makes the chosen calculation reviewable.

Keep exact fractions, roots and π expressions until rounding is requested. A display can approximate √29 or draw a circle as an ellipse to suggest perspective, but its pixels do not establish exact spatial distances or angles. A decimal may instead be an exact mathematical input, in which case it belongs in the exact arithmetic. Rounded measurements support results at their stated precision; they do not independently establish an ideal right angle, a centred apex or uniform similarity.

A useful final check combines dimensions and meaning. Surface areas should have square units and positive values for accepted solids; volumes should have cubic units and positive values. A face slant or lateral edge in a positive-height centred pyramid is longer than its perpendicular height. A cuboid diagonal exceeds each edge. These checks can catch a mismatched role or unit, while the actual formula and its hypotheses supply the proof.

Explore four scenes while preserving their physical and coordinate meanings

The cuboid scene has eight (a, b, h) choices: (3, 2, 4), (1, 1, 1), (1, 2, 2), (1/2, 3/2, 2), (2, 2, 5/2), (3, 2, 0), (0, 2, 3) and (−1, 2, 3). Accepted positive dimensions give base area, total face area, volume and exact opposite-corner distance. A zero height or base side is a collapsed boundary case, and a negative proposed side is invalid. These rejected requests do not receive an invented nondegenerate cuboid.

The right-cylinder scene has eight (r, h) choices: (2, 3), (1, 2), (3/2, 5/2), (1/2, 4), (3, 1), (2, 0), (0, 3) and (−1, 3). Both dimensions must be positive before base area, curved area, closed total area and volume are calculated. Exact results retain π. The drawing's apparent elliptical base is a spatial view of a specified circular disk, rather than a change to its radius or area formula.

The centred rectangular-pyramid scene has eight (a, b, h) choices: (6, 4, 3), (4, 4, 3/2), (3, 2, 4), (1/2, 3/2, 2), (2, 2, 1), (3, 2, 0), (0, 2, 3) and (−1, 2, 3). Its accepted apex is (a/2, b/2, h) above a base in z = 0. Separate face-slant squares h² + (b/2)² and h² + (a/2)² preserve the a-edge and b-edge roles. Surface area adds the four triangular faces and base, while volume uses the perpendicular height. Nonpositive dimensions are rejected before these solid calculations.

The spatial-distance scene has eight point pairs: (0, 0, 0) to (3, 4, 12); (1, −2, 3) to (4, 2, 15) and its reverse; (0, 0, 0) to (3, 4, 0); (0, 0, 0) to (0, 0, 5); (2, −1, 3) to itself; (0, 0, 0) to (1/2, 1/2, 1/2); and (0, 0, 0) to (16, 12, −15). Negative coordinates and zero component changes remain valid point data. The last pair has exact squared distance 625 and distance 25, even though this scene's chosen projection places the endpoints at the same displayed position. Exact coordinates determine the result; projected overlap alone does not establish coincident spatial points.

Worked examples

Try it yourself

1. Classify the x-axis relative to each line (0, u, 1), (t, 1, 0) and the y-axis, with t and u real. Are the y-axis and the plane z = 0 perpendicular just because the y-axis is perpendicular to the x-axis? Describe the intersections of planes z = 0 with z = 2 and with x = 0. Do the points (0, 0, 0), (1, 0, 0), (0, 1, 0) and (0, 0, 1) lie in one plane?

   Show the answer and reasoning

   The first line is skew to the x-axis; the second is parallel to it; the y-axis meets it perpendicularly. The y-axis lies in z = 0 and is not perpendicular to that plane. The planes z = 0 and z = 2 are parallel, while x = 0 and z = 0 meet along the y-axis. The four listed points are not coplanar.

   The first pair has different coordinate directions and cannot meet because one line has z = 0 and the other z = 1, so the lines are nonparallel and skew. The second pair shares the x-direction but has constant y-values 0 and 1, so the lines are distinct, coplanar and parallel. The coordinate x- and y-axes meet at the origin at 90°. Plane perpendicularity requires the perpendicular direction to every line in the plane through the foot, rather than just one; the y-axis is itself a line of the plane. The plane equations give either incompatible z-values or x = z = 0 with arbitrary y. Finally, the first three noncollinear points determine z = 0, and the fourth has z = 1.

2. A cuboid has positive dimensions a = 1/2, b = 3/2 and h = 2 in one common unit. Find its total surface area, volume and opposite-corner distance exactly. Would the proposals (3, 2, 0), (0, 2, 3) or (−1, 2, 3) give accepted cuboids by the same procedure?

   Show the answer and reasoning

   S = 19/2 square units, V = 3/2 cubic units and the diagonal is √26/2 units. None of the later proposals is an accepted nondegenerate cuboid: the first two contain a zero dimension and the last a negative side.

   The face products are ab = 3/4, ah = 1 and bh = 3, so S = 2(3/4 + 1 + 3) = 19/2. The base-area times height is (3/4) × 2 = 3/2. The diagonal square is (1/2)² + (3/2)² + 2² = 1/4 + 9/4 + 4 = 13/2 = 26/4, whose positive root is √26/2. The square and cubic units differ from the diagonal's length unit. A zero input collapses the requested solid, and a negative proposed length is invalid; neither is repaired by obtaining a formal product or squared value.

3. A right circular cylinder has diameter 6 cm and perpendicular height 5 cm. Find its curved-surface area, total closed surface area and volume exactly. If an ideal material model keeps its curved side and bottom but omits the top disk, what area remains? Why do r = −3 or h = 0 fail the positive-dimension solid specification?

   Show the answer and reasoning

   The radius is 3 cm. Curved area is 30π cm², closed total area 48π cm² and volume 45π cm³. Omitting just the top disk leaves 39π cm². A negative radius is invalid and zero height collapses the cylinder, so neither receives the closed-solid verdict.

   Halve the diameter before substitution. Then L = 2π × 3 × 5 = 30π, and two end disks add 2π × 3² = 18π, giving S = 48π. Volume is π × 3² × 5 = 45π. The open-top material model retains only one disk, so its area is 30π + 9π = 39π. Area has square units while enclosed volume has cubic units. Squaring −3 would erase its prohibited sign rather than establish a radius. At h = 0 the two end disks coincide; mechanically adding their areas is not the boundary area of an accepted three-dimensional cylinder.

4. A centred right square pyramid has a = b = 4 cm and perpendicular height h = 3/2 cm. Find its common face slant, total surface area and volume. Is that common face slant a reason to assume equal face slants for every centred rectangular pyramid? Why is an apex-to-corner edge a different length?

   Show the answer and reasoning

   The common face slant is 5/2 cm, total surface area is 36 cm² and volume is 8 cm³. Equal slants follow here from a = b; unequal rectangular base sides need their separate slant expressions. An apex-to-corner edge has length √41/2 cm and is not a face altitude.

   Both midpoint offsets are 2, so ℓ² = (3/2)² + 2² = 9/4 + 4 = 25/4 and ℓ = 5/2. The two pairs of triangular faces contribute 4(5/2) + 4(5/2) = 20 cm², and the base adds 16, giving S = 36. Volume uses h: V = 4 × 4 × (3/2)/3 = 8. For a general rectangle, the offsets are b/2 and a/2 and need not agree. The corner edge includes both offsets: e² = 9/4 + 4 + 4 = 41/4, giving e = √41/2, longer than the face altitude because it runs to a corner rather than a midpoint.

5. A right circular cone has radius 3 cm and perpendicular height 4 cm. Find its slant height, curved area, total closed area and volume. Separately find the area of a radius-3 cm sphere and the volume of its closed ball. Explain which results are accepted standard formulas here and why equal numerical area and volume coefficients are not an equality of quantities.

   Show the answer and reasoning

   For the cone, ℓ = 5 cm, curved area is 15π cm², total area is 24π cm² and volume is 12π cm³. The sphere area is 36π cm² and ball volume is 36π cm³. The curved-solid area and volume formulas are established results used here; matching numerical coefficients with different units do not make area equal to volume.

   The right-cone slant is the positive root of 3² + 4² = 25. Use πrℓ = 15π for curved area, add base area 9π for a closed total 24π, and use perpendicular height in πr²h/3 = 12π for volume. Sphere area is 4π × 3² = 36π and ball volume is (4/3)π × 3³ = 36π. The two 36π values have square and cubic centimetres respectively, so they measure different things and would receive different conversion factors in metres. This calculation uses the accepted formulas without pretending to prove them by a finite flat-face dissection.

6. Find the exact distances from (0, 0, 0) to (3, 4, 0), to (0, 0, 5), and to (1/2, 1/2, 1/2); also find the distance from (2, −1, 3) to itself. Which cases require a nondegenerate vertical-step triangle? Separately, does projected length 5 in the xy-plane establish that the spatial segment from (0, 0, 0) to (3, 4, 12) has length 5?

   Show the answer and reasoning

   The distances are 5, 5, √3/2 and 0 units, respectively. Only the fractional-coordinate pair has both a positive horizontal component distance and a nonzero vertical component, giving the stated nondegenerate vertical-step triangle. The separate segment has actual length 13; its xy-projected length 5 omits its vertical component 12.

   The squared distances are 9 + 16 + 0 = 25; 0 + 0 + 25 = 25; 1/4 + 1/4 + 1/4 = 3/4; and 0. The first pair is planar, so no second nondegenerate triangle is needed; the second is purely vertical, so its horizontal step collapses; the coincident pair has no positive step. In the fractional case, the horizontal square is 1/2 > 0 and Δz = 1/2 ≠ 0, allowing both steps. For (3, 4, 12), the true square is 9 + 16 + 144 = 169, while xy projection retains only 9 + 16. Distances come from spatial coordinates and the established perpendicular axes, rather than apparent separation on a projected drawing.

7. A centred rectangular pyramid has a = 6, b = 4 and h = 3. Find the parallel-to-base section at half-height. A closed ball has radius 5: classify its section by a plane at centre distance 3, 5 or 6, distinguishing the sphere boundary from the ball region. Finally, what area and volume factors follow when all dimensions of a solid are enlarged by 3/2, and do those factors follow from increasing only its height?

   Show the answer and reasoning

   The pyramid section is a 3-by-2 rectangle of area 6 square units. At distance 3 the ball section is a disk of radius 4 and area 16π, while the sphere section is its boundary circle; at distance 5 only a point remains, and at distance 6 the section is empty. A positive uniform length factor 3/2 gives area factor 9/4 and volume factor 27/8. Changing only height does not establish that similarity.

   At t = h/2, the section's length factor is 1 − t/h = 1/2, so its sides are 3 and 2 and its area is one quarter of base area 24. For the ball, 0 ≤ 3 < 5 gives section-radius square 25 − 9 = 16 and positive radius 4. The ball includes the disk's interior, while the sphere includes only points at exact centre distance 5 and cuts in the bounding circle. Equality d = r gives one point; d > r gives none and no positive disk-root verdict. Under uniform k = 3/2, area receives k² = 9/4 and volume k³ = 27/8. Increasing only height leaves some dimensions and face areas unchanged, so the uniform-scaling conclusion cannot be assumed.

What to carry forward

• Spatial lines need both direction and position information: distinct coplanar lines intersect or are parallel, while noncoplanar nonintersecting lines are skew. A perpendicular to a plane is perpendicular to every line of that plane through its foot.
• Surface area measures a boundary in square units; volume measures an enclosed region in cubic units. Convert lengths first, include only the specified surfaces, and require positive physical dimensions for nondegenerate solids.
• A cuboid has S = 2(ab + ah + bh) and V = abh. For a right prism with base area B, perimeter p and perpendicular height h, total S = 2B + ph and V = Bh; matching faces and joins are part of a valid net.
• A right circular cylinder has curved area 2πrh, closed total area 2πrh + 2πr² and volume πr²h. The radius is half the diameter, and a zero height does not justify counting coincident end disks as a closed solid.
• A centred rectangular pyramid uses separate face slants √(h² + (b/2)²) and √(h² + (a/2)²); its total area is ab + aℓₐ + bℓᵦ and its established volume is abh/3. Cone and sphere formulas are used with their own shape conditions and proof scope.
• Spatial distance is √((Δx)² + (Δy)² + (Δz)²) for perpendicular Cartesian axes with common units. Planar, aligned and coincident cases remain valid without false nondegenerate-triangle claims; a projected picture can hide separation and angles.
• A positive uniform enlargement k gives area factor k² and volume factor k³. Cross-sections depend on the cutting plane, and changing one dimension or merely observing a similar picture does not establish uniform similarity.