MathematicsGrade 7–10

Pyramid, cone and sphere

Pour three times and the prism is full: a pyramid holds exactly one third. Cones and spheres can be turned and measured in 3D.

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What is it about?

For pyramids, cones and spheres you cannot simply compute volume and surface area as “length times width times height”. The simulation shows where the formulas come from: you can rotate the solid, pour water and unfold the net.

Building blocks

  • Height hh: perpendicular distance of the apex from the base.
  • Slant height hah_a (pyramid): height of a lateral face, from the midpoint of a base edge to the apex.
  • Slant height ss (cone) or lateral edge ss (pyramid): line from the apex to the edge of the base.
  • Radius rr: for cone and sphere, the distance from the centre.

These lengths are linked by right triangles (Pythagoras’ theorem). For the square pyramid with base diagonal d=a2d = a\sqrt{2}:

ha2=h2+(a2)2,s2=h2+(d2)2=ha2+(a2)2.h_a^2 = h^2 + \left(\tfrac{a}{2}\right)^2, \qquad s^2 = h^2 + \left(\tfrac{d}{2}\right)^2 = h_a^2 + \left(\tfrac{a}{2}\right)^2 .

For the cone, the slant height satisfies s2=h2+r2s^2 = h^2 + r^2. Careful: the slant height hah_a and the lateral edge ss of a pyramid are different lines. hah_a meets the base edge at its midpoint at a right angle, ss ends at a corner.

Pyramid

With base area GG and height hh:

V=13⋅G⋅h,M=sum of the lateral faces,O=G+M.V = \tfrac{1}{3} \cdot G \cdot h, \qquad M = \text{sum of the lateral faces}, \qquad O = G + M .

For the square pyramid, G=a2G = a^2 and M=4⋅12⋅a⋅haM = 4 \cdot \tfrac{1}{2} \cdot a \cdot h_a. The water from three pyramids exactly fills the prism with the same base and height.

Cone

V=13πr2h,M=πrs,O=πr2+πrs.V = \tfrac{1}{3}\pi r^2 h, \qquad M = \pi r s, \qquad O = \pi r^2 + \pi r s .

The lateral surface unrolls into a circular sector with radius ss and arc length 2πr2\pi r. Its angle is α=360∘⋅rs\alpha = 360^\circ \cdot \tfrac{r}{s}.

Sphere

V=43πr3,O=4πr2.V = \tfrac{4}{3}\pi r^3, \qquad O = 4\pi r^2 .
  • Archimedes: cylinder (height rr, radius rr) = cone + hemisphere. The cone holds 13πr3\tfrac{1}{3}\pi r^3, the cylinder πr3\pi r^3, so the hemisphere gets 23πr3\tfrac{2}{3}\pi r^3 and the whole sphere 43πr3\tfrac{4}{3}\pi r^3.
  • From pyramids: if you cut the sphere into many pyramids of height rr, then V=13⋅O⋅rV = \tfrac{1}{3} \cdot O \cdot r. It follows that O=3V:r=4πr2O = 3V : r = 4\pi r^2.

Try it

  1. Choose the square pyramid and switch between the helper triangles. Which line is always the longest? Load pyramid
  2. Pour the pyramid three times. How much of the prism does one pyramid fill? Pouring
  3. Try the same with the cone. What changes when you change rr or hh? Pour a cone
  4. Unroll the lateral surface of the cone. Change hh and watch the angle α\alpha. Cone net
  5. Pour after Archimedes: first the cone, then the hemisphere. Archimedes
  6. Split the sphere into pyramids and rotate it. Sphere made of pyramids

Tasks

Task 1: Square pyramid

A square pyramid has base edge a=4 cma = 4\,\text{cm} and height h=5 cmh = 5\,\text{cm}. Calculate the slant height, volume, lateral surface and surface area. Load task

Show solution

ha=52+22 cm=29 cm≈5.39 cmh_a = \sqrt{5^2 + 2^2}\,\text{cm} = \sqrt{29}\,\text{cm} \approx 5.39\,\text{cm}

V=13⋅16 cm2⋅5 cm≈26.67 cm3V = \tfrac{1}{3} \cdot 16\,\text{cm}^2 \cdot 5\,\text{cm} \approx 26.67\,\text{cm}^3

M=4⋅12⋅4 cm⋅29 cm=829 cm2≈43.08 cm2M = 4 \cdot \tfrac{1}{2} \cdot 4\,\text{cm} \cdot \sqrt{29}\,\text{cm} = 8\sqrt{29}\,\text{cm}^2 \approx 43.08\,\text{cm}^2

O=16 cm2+43.08 cm2≈59.08 cm2O = 16\,\text{cm}^2 + 43.08\,\text{cm}^2 \approx 59.08\,\text{cm}^2

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Task 2: Cone

A cone has radius r=3 cmr = 3\,\text{cm} and height h=4 cmh = 4\,\text{cm}. Find ss, VV, MM and OO. Load task

Show solution

s=42+32 cm=5 cms = \sqrt{4^2 + 3^2}\,\text{cm} = 5\,\text{cm}

V=13π⋅9 cm2⋅4 cm=12π cm3≈37.70 cm3V = \tfrac{1}{3}\pi \cdot 9\,\text{cm}^2 \cdot 4\,\text{cm} = 12\pi\,\text{cm}^3 \approx 37.70\,\text{cm}^3

M=π⋅3 cm⋅5 cm=15π cm2≈47.12 cm2M = \pi \cdot 3\,\text{cm} \cdot 5\,\text{cm} = 15\pi\,\text{cm}^2 \approx 47.12\,\text{cm}^2

O=9π cm2+15π cm2=24π cm2≈75.40 cm2O = 9\pi\,\text{cm}^2 + 15\pi\,\text{cm}^2 = 24\pi\,\text{cm}^2 \approx 75.40\,\text{cm}^2

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Task 3: Sphere

A sphere has radius r=2.5 cmr = 2.5\,\text{cm}. Calculate its volume and surface area. What percentage of the enclosing cylinder (height 2r2r) does the sphere fill? Load task

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V=43π⋅(2.5 cm)3≈65.45 cm3V = \tfrac{4}{3}\pi \cdot (2.5\,\text{cm})^3 \approx 65.45\,\text{cm}^3

O=4π⋅(2.5 cm)2=25π cm2≈78.54 cm2O = 4\pi \cdot (2.5\,\text{cm})^2 = 25\pi\,\text{cm}^2 \approx 78.54\,\text{cm}^2

The cylinder has V=2πr3≈98.17 cm3V = 2\pi r^3 \approx 98.17\,\text{cm}^3. The sphere fills 23≈66.7 %\tfrac{2}{3} \approx 66.7\,\% of it.

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Ideas for the classroom

  • Use: about grades 9/10, calculating solids. Suitable as an introduction (pouring experiment), for practice (Pythagoras in solids) and for going further (Archimedes).
  • Model: the pyramid has a regular nn-gon as its base, with the apex directly above the centre. When pouring, the amount of water is calculated exactly from the volume; the drawing only approximates it.
  • Common mistakes:
    • The factor 13\tfrac{1}{3} is forgotten – correct: pyramid and cone fill only a third.
    • rr and dd are mixed up – the formulas use the radius.
    • Height hh is confused with the slant height hah_a or ss – ss is always longer than hh.