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Solid mensuration · Lesson 2 of 3

Cones and pyramids

Volume as one third of base × height, the slant height from Pythagoras, the curved surface πrl, and a cone made by bending a sector.

18 minYou should already know: Plane mensuration
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A pyramid rises from a flat base to a point (the vertex). A cone is a pyramid with a circular base. Both hold exactly one third as much as the prism with the same base and height:

volume=13×base area×height\text{volume} = \frac13 \times \text{base area} \times \text{height}
hbase
PyramidV=13×base×hV = \frac13 \times \text{base} \times h
hrl
ConeV=13πr2hV = \frac13\pi r^2 h

So a cone has volume 13πr2h\frac13\pi r^2 h. In both drawings the dashed height hh goes straight down from the top to the centre of the base, at right angles to it.

More: volumes of cones and pyramids

Heights and slant heights

Three lengths in a cone make a right-angled triangle:

  • the height hh, from the vertex straight down to the centre of the base;
  • the base radius rr;
  • the slant height ll, from the vertex down the side to the edge of the base.
l2=r2+h2l^2 = r^2 + h^2
hrll² = r² + h²
Height, radius, slant heightA right-angled triangle inside every cone

In a right pyramid, the vertex is directly above the centre of the base, where the diagonals cross. The right-angled triangle is made by the slant edge, half the base’s diagonal and the height.

Worked example · WAEC 2014

WAEC 2014 · Paper 2 · Q5 (a, c)

The diagram shows a right pyramid with a rectangular base WXYZWXYZ and vertex OO. If ∣WX∣=8 cm|WX| = 8\text{ cm}, ∣ZW∣=6 cm|ZW| = 6\text{ cm} and ∣OX∣=13 cm|OX| = 13\text{ cm}, calculate the:

8 cm6 cm13 cmWXYZO
Not to scale.

height of the pyramid;

volume of the pyramid.

  1. Half the diagonal

    • The diagonal: ∣ZX∣=82+62=10{|ZX| = \sqrt{8^2 + 6^2} = 10} cm.
    • So from the centre MM to the corner XX is 5 cm.

    Think first. The base is 8 cm by 6 cm. How long is the diagonal?

  2. (a) The height

    ∣OM∣=132−52=144=12|OM| = \sqrt{13^2 - 5^2} = \sqrt{144} = 12 cm.

    Think first. Triangle OMXOMX has a right angle at MM and hypotenuse OX=13OX = 13.

  3. (c) The volume

    13×(8×6)×12=192 cm3\frac13 \times (8 \times 6) \times 12 = 192\text{ cm}^3

More: the height of a pyramid

The surface of a cone

curved surface=πrltotal, solid cone=πrl+πr2\begin{aligned} \text{curved surface} &= \pi r l \\ \text{total, solid cone} &= \pi r l + \pi r^2 \end{aligned}

More: the surface of a cone

A cone from a sector

Cut a sector out of a sheet and bend it until its straight edges meet. You get a cone.

Bend a sector into a coneChange the angle of the sector
135°sector: radius 28hrl = 28the cone
10.5base radius r = 135/360 × 2825.96height h = √(l² − r²)924curved surface πrl = sector area2998volume ⅓πr²h
Join the two straight edges. The sector's radius becomes the slant height l = 28. Its arc (66) goes round the base, so 2πr = 135/360 × 2πl, which gives r = 135/360 × 28 = 10.5. Then h = √(28² − 10.5²) = 25.96. The vertical angle at the top is 2 × sin⁻¹(r/l) ≈ 44°.

As you bend it:

  • the sector’s radius becomes the cone’s slant height ll;
  • the sector’s arc becomes the circumference of the base: 2πr=θ360×2πl2\pi r = \frac{\theta}{360} \times 2\pi l, so r=θ360×lr = \frac{\theta}{360} \times l;
  • the sector’s area becomes the cone’s curved surface.
θllarc → base circler = θ ÷ 360 × l
Bending a sectorThe radius becomes the slant height; the arc becomes the base circle

More: a cone from a sector

Worked example · WAEC 2022

WAEC 2022 · Paper 2 · Q8 (b)

A sector of angle 220∘220^\circ is removed from a thin circular metal sheet of radius 63 cm63\text{ cm}. It is then folded with the straight edges meeting to form a right circular cone. Calculate, correct to one decimal place, the: (i) base radius; (ii) volume, of the cone. [Take π=227]\left[\text{Take }\pi = \frac{22}{7}\right]

  1. (i) The base radius

    The arc becomes the base’s circumference, so r=220360×63=38.5{r = \frac{220}{360} \times 63 = 38.5} cm.

    Think first. What does the sector's arc become?

  2. The height

    • The slant height is the sheet’s radius, 63 cm.
    • h=632−38.52=2486.75{h = \sqrt{63^2 - 38.5^2} = \sqrt{2486.75}}.
    • h≈49.867{h \approx 49.867} cm.

    Think first. What is the slant height? Then use Pythagoras.

  3. (ii) The volume

    13×227×38.52×49.867≈77 435.6 cm3\begin{aligned} &\tfrac13 \times \tfrac{22}{7} \times 38.5^2 \times 49.867 \\ &\approx 77\,435.6\text{ cm}^3 \end{aligned}

Your turn

WAEC 2021 · Paper 2 · Q3

The curved surface area of a cone is 242 cm2242\text{ cm}^2. If the slant height is 4 cm4\text{ cm} more than the radius, calculate, correct to one decimal place, the: [Take π=227]\left[\text{Take }\pi = \frac{22}{7}\right]

  1. (a)

    radius;

  2. (b)

    height;

  3. (c)

    volume of the cone.

Worked solution (try it first)

(a)

  1. Let the radius be rr.
  2. Then l=r+4l = r + 4.
  3. Curved surface: 227×r(r+4)=242\frac{22}{7} \times r(r + 4) = 242, so r2+4r=77r^2 + 4r = 77 and r2+4r−77=0r^2 + 4r - 77 = 0.
  4. Factorise: (r+11)(r−7)=0(r + 11)(r - 7) = 0.
  5. The radius is positive, so r=7.0r = 7.0 cm.

(b)

  1. l=11l = 11 cm, so h=112−72h = \sqrt{11^2 - 7^2}
    =72= \sqrt{72}
    ≈8.5\approx 8.5 cm.

(c)

  1. Volume =13×227×72×72= \frac13 \times \frac{22}{7} \times 7^2 \times \sqrt{72}
    ≈435.6 cm3\approx 435.6\text{ cm}^3.

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