Solid geometry · Teacher reference · answers included

← Senior 1 · short mathematical inquiries

Teacher reference: contains answers, conditions and proofs.

10 Solid geometry · g10-10-01 · 5–10 min · TEACHER · ANSWERS

Classify by structure, not by viewpoint

Does rotating a solid change its type?

Predict: Compare a prism,pyramid,cylinder,cone,frustum andsphere; decide which have parallel bases and which have a common apex.

Compare	Computed observation
Prism Solid = Prism	Rotation changes the image, not the structural classification.
Cone Solid = Cone	Rotation changes the image, not the structural classification.
No base or apex Solid = Sphere	Rotation changes the image, not the structural classification.

Conclusion: Structure determines the class: parallel congruent bases describe prisms or cylinders; a common apex describes pyramids or cones; cutting off a smaller cone gives a frustum. A sphere has no flat base.

Conditions: Use nondegenerate solids. A curved surface is not a polygonal face, and the apparent outline changes with viewpoint.

Justification: Rotation preserves incidence, parallelism and congruence of the defining bases; it changes projected coordinates but not the solid structure.

Ask: Which defining feature can you still identify after rotating the view?

Watch for: Classifying a tilted cylinder as a cone because its image looks narrow.

Extension: Explain why a sphere cannot be classified as a many-sided prism.

Exit: A solid has two parallel congruent circular bases and a curved side. Name it.

A circular cylinder.

Before: Plane area, Pythagoras and units. · Timing: predict 1, compare 3, explain 2, exit 1 minute. Finite observations do not prove a universal claim.

10 Solid geometry · g10-10-02 · 5–10 min · TEACHER · ANSWERS

A drawing does not preserve every length

Can equal perpendicular sides look unequal in an oblique drawing?

Predict: Draw a square with its depth lengths halved. Predict the displayed side ratio and angle.

Compare	Computed observation
Half-depth square Real front length = 4; Real depth length = 4; Depth scale q = 0.5; Projected depth angle (degrees) = 45	The drawn depth is scaled; do not measure it as a real length.
Full-depth drawing Real front length = 4; Real depth length = 4; Depth scale q = 1; Projected depth angle (degrees) = 45	The drawn depth is scaled; do not measure it as a real length.
Collapsed axes Real front length = 4; Real depth length = 4; Depth scale q = 0.5; Projected depth angle (degrees) = 0	The projection has collapsed and lost depth information.

Conclusion: An oblique representation keeps the chosen front axis and scales/reorients depth. Parallel segments remain parallel, but general lengths and angles do not remain true.

Conditions: The picture uses a specified depth scale q>0 and nonparallel projected axes. A collapsed drawing cannot recover depth.

Justification: The map(x,y)↦(x+qy cosθ,qy sinθ) is linear and preserves parallel directions, but sends unit axes to lengths1,q and angleθ.

Ask: Which labels state real measurements even when the drawing is foreshortened?

Watch for: Measuring the paper slant and treating it as an actual space angle or length.

Extension: Choose a projection that collapses the square and explain what information is lost.

Exit: A depth edge of real length6 is drawn with q=1/2. What is its displayed length?

3 drawing units; the real length remains6.

Before: Plane area, Pythagoras and units. · Timing: predict 1, compare 3, explain 2, exit 1 minute. Finite observations do not prove a universal claim.

10 Solid geometry · g10-10-03 · 5–10 min · TEACHER · ANSWERS

A cone height is not its slant height

Which length belongs in the lateral-area formula?

Predict: For radius3 and vertical height4, predict the slant length and total surface area of a cone.

Compare	Computed observation
Cone net Surface to unwrap = Cone; Radius r = 3; Vertical height h = 4	Use the unwrapped length: slant height for a cone, vertical height for a cylinder.
Cylinder net Surface to unwrap = Cylinder; Radius r = 3; Vertical height h = 4	Use the unwrapped length: slant height for a cone, vertical height for a cylinder.
Zero height Surface to unwrap = Cone; Radius r = 3; Vertical height h = 0	Zero height is a degenerate boundary; the area formula has a limit, not an ordinary solid.

Conclusion: A cylinder unwraps to a rectangle of width2πr and height h. A cone unwraps to a sector with radius ℓ=√(r²+h²) and arc2πr, so its lateral area isπrℓ.

Conditions: r>0,h>0. Total surface includes the base: one for a cone,two for a cylinder. Use square units.

Justification: Cone-sector area is half its radius times arc length: (1/2)ℓ(2πr)=πrℓ. Add the circular baseπr².

Ask: Where is the vertical height after unwrapping the cone?

Watch for: Using vertical height h inπrh for the cone lateral area.

Extension: Compare the cone and cylinder total areas with the same radius and vertical height.

Exit: Find the total surface area for a cone with r=3,h=4.

ℓ=5; area=15π+9π=24π.

Before: Plane area, Pythagoras and units. · Timing: predict 1, compare 3, explain 2, exit 1 minute. Finite observations do not prove a universal claim.

10 Solid geometry · g10-10-04 · 5–10 min · TEACHER · ANSWERS

The same base and vertical height

Which height determines a prism or cone volume?

Predict: For base area12 and height5, predict the prism and cone volumes.

Compare	Computed observation
Prism Solid = Prism; Base area S = 12; Perpendicular height h = 5	Use perpendicular height, not a sloping edge.
Cone Solid = Cone; Base area S = 12; Perpendicular height h = 5	Use perpendicular height, not a sloping edge.
Zero perpendicular height Solid = Cone; Base area S = 12; Perpendicular height h = 0	Zero height collapses the solid.

Conclusion: Vprism=Sh andVcone=Sh/3 use the perpendicular height. The one-third relation is a geometric theorem, not a result proved by a fill animation.

Conditions: S>0 and h>0 for a solid; h=0 is degenerate and has zero volume. Use cubic units.

Justification: For a pyramid, a cross-section at distance z from the apex has area S(z/h)²; summing thin slices gives Sh/3. A prism has constant cross-sectional area S, giving Sh.

Ask: Would leaning the side change the perpendicular height shown?

Watch for: Using a sloping edge as height or reporting square units for volume.

Extension: Explain why equal heights alone do not fix volumes if base areas differ.

Exit: Find the cone volume when S=12,h=5.

20 cubic units.

Before: Plane area, Pythagoras and units. · Timing: predict 1, compare 3, explain 2, exit 1 minute. Finite observations do not prove a universal claim.

10 Solid geometry · g10-10-05 · 5–10 min · TEACHER · ANSWERS

A frustum is a difference of cones

Can the volume use the average of the two end areas?

Predict: Compare the correct volume with hπ(R²+r²)/2 for R=3,r=1,h=4.

Compare	Computed observation
A frustum Bottom radius R = 3; Top radius r = 1; Height h = 4	The missing cross-term matters; averaging end areas overestimates.
Cone boundary Bottom radius R = 3; Top radius r = 0; Height h = 4	Zero top radius gives a cone.
Cylinder boundary Bottom radius R = 3; Top radius r = 3; Height h = 4	Equal radii give the cylinder limit.

Conclusion: Subtract similar cones to obtain V=πh(R²+Rr+r²)/3. Simply averaging end areas omits the way intermediate cross-sections vary.

Conditions: R>0,0≤r≤R,h>0. r=0 gives a cone; r=R gives the cylinder limit.

Justification: For r<R, similarity gives total cone height H=hR/(R−r). Subtract πR²H/3−πr²(H−h)/3 and factor R³−r³. The result extends continuously to r=R.

Ask: Which missing cross-term appears after factoring the difference of cubes?

Watch for: Using the trapezium area rule directly for volume.

Extension: Show that the average-end-area estimate is too large when R≠r.

Exit: Find the volume for R=3,r=1,h=4.

52π/3 cubic units.

Before: Plane area, Pythagoras and units. · Timing: predict 1, compare 3, explain 2, exit 1 minute. Finite observations do not prove a universal claim.

10 Solid geometry · g10-10-06 · 5–10 min · TEACHER · ANSWERS

Double a sphere: four or eight times?

Do surface area and volume scale at the same rate?

Predict: Double a sphere radius. Predict its area ratio and volume ratio.

Compare	Computed observation
Same radius Original radius r = 1; Radius scale k = 1	Area uses the square factor; volume uses the cube factor.
Double radius Original radius r = 1; Radius scale k = 2	Area uses the square factor; volume uses the cube factor.
Collapse to a point Original radius r = 1; Radius scale k = 0	The scaled object is a point, not a sphere.

Conclusion: A sphere has surface area4πr² and volume4πr³/3. Scaling radius by k changes area by k² and volume by k³.

Conditions: r>0,k>0 for spheres; k=0 collapses to a point, not a sphere. Area and volume use different units.

Justification: Substitute kr into the formulas and divide by the original positive area or volume; constants cancel and leave k²,k³.

Ask: Which power of radius belongs to each physical dimension?

Watch for: Doubling area or volume merely because radius doubles.

Extension: If surface area grows by9, find the volume factor.

Exit: A sphere radius triples. State the area and volume factors.

Area9,volume27.

Before: Plane area, Pythagoras and units. · Timing: predict 1, compare 3, explain 2, exit 1 minute. Finite observations do not prove a universal claim.