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Solid mensuration · Lesson 1 of 3

Prisms and cylinders

Volume as cross-section × length, surface area from the net, the cylinder (curved surface 2πrh, open and closed ends), hollow pipes and boxes, and the diagonal of a cuboid.

22 minYou should already know: Plane mensuration
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A prism is a solid with the same cross-section all the way along: a cuboid, a triangular prism, a cylinder. You need plane areas for everything here.

Volume: cross-section × length

Stack identical slices: the volume is the area of one slice times how many there are.

volume=cross-section area×length\text{volume} = \text{cross-section area} \times \text{length}
Alength
PrismV=A×lengthV = A \times \text{length}
lbh
CuboidV=l×b×hV = l \times b \times h
rh
CylinderV=πr2hV = \pi r^2 h

So a cuboid has volume l×b×hl \times b \times h, and a cylinder, whose cross-section is a circle, has volume πr2h\pi r^2 h.

More: prisms and cuboids

Surface area: unfold it

The surface area is the total area of all the faces. Picture the solid opened out flat, its net, and add up the pieces. A cuboid has three pairs of equal rectangles: 2(lb+bh+lh)2(lb + bh + lh).

Cylinder: unroll the curved surfaceChange r and h
r = 3.5, h = 102πr × hwidth 2πr = 22
220curved surface, 2πrh77two ends, 2πr²297total surface area385volume, πr²h
Cut the curved surface straight down and unroll it: it's a rectangle, 22 wide (the circumference, 2πr) and 10 high, so its area is 2πrh. Add the two circular ends for the total surface area. Volume is the area of the end times the height: πr²h. (π = 22/7.)

Cut the curved surface of a cylinder straight down and unroll it: it becomes a rectangle whose width is the circumference 2πr2\pi r and whose height is hh.

CylinderFormula
curved surface2πrh2\pi rh
closed at both ends (total)2πrh+2πr2=2πr(h+r)2\pi rh + 2\pi r^2 = 2\pi r(h + r)
open at one end2πrh+πr22\pi rh + \pi r^2
volumeπr2h\pi r^2 h

More: surface area

Hollow solids: pipes and thick walls

A pipe, or a box made of thick wood, is a solid with a hole in it. The material is what’s left when you take the inside away:

material=outer volume−inner volume\begin{aligned} \text{material} &= \text{outer volume} \\ &\quad - \text{inner volume} \end{aligned}

For a pipe of length LL, outer radius RR and inner radius rr, that’s πR2L−πr2L=π(R2−r2)L\pi R^2 L - \pi r^2 L = \pi(R^2 - r^2)L. The thickness is R−rR - r, so a pipe of outer radius 5 cm and thickness 1 cm has inner radius 4 cm, and 20 cm of it uses π(25−16)×20=180π cm3\pi(25 - 16) \times 20 = 180\pi\text{ cm}^3 of metal.

Rrmetal in the pipe= outer − inner= π(R² − r²) × length
The end of a pipeThe metal is the ring between the two circles, all the way along

More: pipes, thick boxes and space left over

Working back to a missing length

When the volume or the surface area is given, write the formula with the numbers in it and solve for the unknown.

Worked example · WAEC 2022

WAEC 2022 · Paper 2 · Q4 (a)

The volume of a cylinder of radius 14 cm14\text{ cm} is 9240 cm39240\text{ cm}^3. Calculate the curved surface area of the cylinder. [Take π=227]\left[\text{Take }\pi = \frac{22}{7}\right]

  1. Find the height

    227×142×h=9240\frac{22}{7} \times 14^2 \times h = 9240, so 616h=9240616h = 9240 and h=15h = 15 cm.

    Think first. Use the volume to find hh.

  2. The curved surface

    2×227×14×15=1320 cm22 \times \frac{22}{7} \times 14 \times 15 = 1320\text{ cm}^2

More: working back from a volume

The diagonal of a cuboid

The longest rod that fits in a box runs from a bottom corner to the opposite top corner. Use Pythagoras twice: across the floor, the diagonal is l2+b2\sqrt{l^2 + b^2}; then up the height,

d=l2+b2+h2d = \sqrt{l^2 + b^2 + h^2}
lbhdd² = l² + b² + h²
Pythagoras twiceFloor diagonal first, then up to the opposite top corner

The floor diagonal, the height and dd make a right-angled triangle, so the angle θ\theta between dd and the floor has cos⁡θ=floor diagonald\cos\theta = \frac{\text{floor diagonal}}{d}.

More: diagonals and angles in a cuboid

Your turn

WAEC 2022 · Paper 2 · Q13 (a)

  1. (a)

    The diameter of a cylinder closed at both ends is 7 cm7\text{ cm}. If the total surface area is 209 cm2209\text{ cm}^2, calculate the height. [Take π=227]\left[\text{Take }\pi = \frac{22}{7}\right]

Worked solution (try it first)

(a)

  1. A closed cylinder has two circular ends and a curved side: total surface area =2πr2+2πrh= 2\pi r^2 + 2\pi rh.
  2. With r=3.5r = 3.5: 2×227×3.52+2×227×3.5×h=2092 \times \frac{22}{7} \times 3.5^2 + 2 \times \frac{22}{7} \times 3.5 \times h = 209.
  3. So 77+22h=20977 + 22h = 209, 22h=13222h = 132 and h=6h = 6 cm.

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