3.2 Solid Mensuration, Measurement, and Unit Conversion

3.2 Solid Mensuration, Measurement, and Unit Conversion

Mensuration items are pure formula recognition plus disciplined bookkeeping of units. Nothing here needs a calculator if you keep $\pi$ symbolic or use the friendly approximation $\pi \approx \frac{22}{7}$ when the radius is a multiple of $7$. This section is deliberately about solids and units; the perimeter and area of flat figures belong to the plane-geometry section.

1. The core solids

Here $B$ is the base area, $P$ the base perimeter, $h$ the vertical height and $\ell$ the slant height.

Slant height, derived not given

The slant height is the hypotenuse of a right triangle standing inside the solid:

• Cone: the legs are the radius and the vertical height, so $\ell = \sqrt{r^2 + h^2}$. A cone with $r = 6$ and $h = 8$ has $\ell = \sqrt{36+64} = 10$.
• Square pyramid: the legs are the vertical height and half the base edge, so $\ell = \sqrt{h^2 + \left(\frac{s}{2}\right)^2}$. A pyramid with $s = 10$ and $h = 12$ has $\ell = \sqrt{144 + 25} = 13$.

Curved-surface formulas ($\pi r\ell$, $\frac{1}{2}P\ell$) always want $\ell$; volume formulas always want $h$. Swapping them is the single most common mensuration error.

Space diagonals

        +-----------+          D = sqrt(l^2 + w^2 + h^2)
       /|          /|
      / |         / |          Cube:  face diagonal = s*sqrt(2)
     +-----------+  |                 space diagonal = s*sqrt(3)
     |  +--------|--+
     | /   D     | /
     |/          |/
     +-----------+

A crate measuring $3 \times 4 \times 12$ m accepts a rod of length $\sqrt{9+16+144} = \sqrt{169} = 13$ m — the longest straight object that fits.

Composite solids

Split, compute, then add or subtract. A grain silo formed by a cylinder of radius $3$ m and height $10$ m capped by a hemisphere of the same radius holds

$$V = \pi(3)^2(10) + \tfrac{2}{3}\pi(3)^3 = 90\pi + 18\pi = 108\pi \approx 339.3\ \text{m}^3.$$

Note the hemisphere's flat face disappears inside the joint, so it is not counted in surface area.

2. The scaling law — the classic trap

Multiply every linear dimension of a figure by $k$ and lengths scale by $k$, areas by $k^2$, volumes by $k^3$.

"Doubling the tank doubles the water" is false: doubling all dimensions gives eight times the capacity. Read the wording precisely — if only the radius of a cylinder doubles while the height is unchanged, volume rises by $2^2 = 4$, because $r$ appears squared and $h$ does not change at all.

3. Metric prefixes and conversion

Moving down the ladder multiplies by $10$ each step; moving up divides by $10$. The order kilo–hecto–deka–base–deci–centi–milli is worth reciting once before the exam.

Common metric–English equivalents (memorize the exact ones):

• $1$ inch $= 2.54$ cm exactly; $1$ foot $= 30.48$ cm; $1$ mile $\approx 1.609$ km
• $1$ kg $\approx 2.205$ lb; $1$ lb $\approx 453.6$ g
• $1$ US gallon $\approx 3.785$ L; $1$ L $\approx 0.264$ US gallon
• $1$ hectare $= 10{,}000$ m$^2$; $1$ km$^2 = 100$ hectares

Why $1$ m$^2 = 10{,}000$ cm$^2$

Because the conversion factor is raised to the same power as the unit. $1\ \text{m} = 100\ \text{cm}$, so

$$1\ \text{m}^2 = (100\ \text{cm})^2 = 10{,}000\ \text{cm}^2,\qquad 1\ \text{m}^3 = (100\ \text{cm})^3 = 1{,}000{,}000\ \text{cm}^3.$$

Also fix in memory: $1\ \text{cm}^3 = 1\ \text{mL}$, $1\ \text{L} = 1{,}000\ \text{cm}^3 = 1\ \text{dm}^3$, and therefore $1\ \text{m}^3 = 1{,}000$ litres.

4. Density and rates

Density is mass per unit volume, $\rho = \dfrac{m}{V}$, rearranged as $m = \rho V$ and $V = \dfrac{m}{\rho}$. Fresh water is $1$ g/cm$^3$, which is the same as $1{,}000$ kg/m$^3$ — the factor between those two units is exactly $1{,}000$.

Rates convert by chaining fractions so unwanted units cancel:

$$72\ \frac{\text{km}}{\text{h}} \times \frac{1{,}000\ \text{m}}{1\ \text{km}} \times \frac{1\ \text{h}}{3{,}600\ \text{s}} = 20\ \frac{\text{m}}{\text{s}}.$$

The shortcut worth memorizing: divide km/h by $3.6$ to get m/s, multiply by $3.6$ to go back. Peso rates behave the same way — diesel at ₱62.50 per litre is $62.50 \times 3.785 \approx$ ₱236.56 per US gallon.

Round only at the very end, and round up whenever the answer counts whole purchased items (bags of cement, cans of paint, jeepney trips).

5. Worked example 1 — barangay water tank

Problem. A cylindrical tank has diameter $2.8$ m and height $3$ m. Using $\pi \approx \frac{22}{7}$: (a) find its volume in cubic metres, (b) convert to litres, (c) if $20$ households each draw $220$ L per day, how long does a full tank last, and (d) how many $1$-litre cans of paint cover the curved outside wall if one can covers $8$ m$^2$?

1. Radius $r = \frac{2.8}{2} = 1.4$ m, so $r^2 = 1.96$ m$^2$.
2. $V = \pi r^2 h = \frac{22}{7}(1.96)(3)$. Since $\frac{1.96}{7} = 0.28$, this is $22 \times 0.28 \times 3 = 18.48$ m$^3$.
3. $18.48\ \text{m}^3 \times 1{,}000 = 18{,}480$ litres.
4. Daily draw $= 20 \times 220 = 4{,}400$ L, so the tank lasts $\frac{18{,}480}{4{,}400} = 4.2$ days.
5. Curved wall $= 2\pi r h = 2 \times \frac{22}{7} \times 1.4 \times 3 = 2 \times 4.4 \times 3 = 26.4$ m$^2$, needing $\frac{26.4}{8} = 3.3$ cans — buy 4, because paint is sold whole.

6. Worked example 2 — concrete footing and the scaling trap

Problem. A rectangular concrete slab measures $4$ m by $3$ m by $150$ mm thick. (a) Find its volume in m$^3$ and cm$^3$. (b) At an assumed $9$ bags of cement per cubic metre, how many bags must be bought? (c) If the client later doubles the length, width and thickness, what volume is poured?

1. Convert the thickness first: $150$ mm $= 0.15$ m. Mixing millimetres with metres is what wrecks this item.
2. $V = 4 \times 3 \times 0.15 = 1.8$ m$^3$.
3. In cubic centimetres, $1.8 \times 1{,}000{,}000 = 1{,}800{,}000$ cm$^3$ (equivalently $1{,}800$ litres).
4. Cement: $1.8 \times 9 = 16.2$ bags, so 17 bags are purchased.
5. Doubling every linear dimension multiplies volume by $2^3 = 8$: $1.8 \times 8 = 14.4$ m$^3$. Answering $3.6$ m$^3$ is the trap.

7. Worked example 3 — density of a steel ball

A solid steel ball has radius $3$ cm and steel has density $7.8$ g/cm$^3$. Its volume is $V = \frac{4}{3}\pi(3)^3 = 36\pi \approx 113.1$ cm$^3$, so its mass is $m = \rho V = 7.8 \times 113.1 \approx 882$ g, or about $0.88$ kg.

8. Common traps

• Using vertical height inside $\pi r \ell$, or slant height inside $\frac{1}{3}\pi r^2 h$.
• Believing $1$ m$^2 = 100$ cm$^2$; it is $10{,}000$ cm$^2$.
• Forgetting the $\frac{1}{3}$ in cone and pyramid volumes, which inflates the answer threefold.
• Taking the total surface area of a hemisphere as $2\pi r^2$ when the flat face is included — it is $3\pi r^2$.
• Being handed a diameter and using it as the radius.
• Assuming doubled dimensions double the volume.
• Leaving one measurement in mm or cm while the rest are in m.