17 Geometry and measurement

Reading the shape of the world

Your bedroom floor has a certain area. You know this even without measuring it — you can picture whether a rug would fit, whether the furniture leaves enough space, whether it’s bigger or smaller than a friend’s room.

Your phone screen is described by a diagonal measurement in inches. But the area that matters — the bit you actually look at — is the width times the height, not the diagonal. The diagonal is a shortcut that hides the real shape.

The skateboard ramp at the park is a triangle in cross-section. If you built one, you’d need to know the length of the slope, the height it reaches, and the distance it runs along the ground — and how those three numbers relate to each other.

These are all geometry. This chapter gives you the tools to read the shape of any object you encounter — and to find the measurement you need from the ones you have.

17.1 What the notation is saying

Every formula in this chapter is shorthand for a relationship between measurements. Before using any of them, here is what each symbol means.

$A$ — area. The amount of flat space a shape covers. Measured in square units: square centimetres (cm²), square metres (m²), square kilometres (km²). If a length is in centimetres, the area will be in square centimetres.

$P$ — perimeter. The total distance around the outside edge of a flat shape. Measured in the same units as the lengths themselves — not squared.

$r$ — radius. The distance from the centre of a circle to any point on its edge. The diameter $d$ is twice the radius: $d = 2r$.

$\pi$ — pi. Read it as “pie”. It’s approximately $3.14159$, but it never terminates or repeats — it’s an irrational number. It appears in every circle formula because it captures the relationship between a circle’s diameter and the distance around it. Your calculator has a $\pi$ button; use it rather than the approximation unless you’re told otherwise.

$V$ — volume. The amount of three-dimensional space a solid object fills. Measured in cubic units: cm³, m³, litres (1 litre = 1000 cm³).

$h$ — height. In a prism or cylinder, this is the perpendicular distance between the two bases — how tall the solid is, measured straight through, at a right angle to the base.

$a^2 + b^2 = c^2$ — the Pythagorean theorem. Read it as: the square of $a$ plus the square of $b$ equals the square of $c$. Here $a$ and $b$ are the two shorter sides of a right-angled triangle and $c$ is the longest side — the one opposite the right angle, called the hypotenuse. This only works for right triangles. The right angle is marked with a small square in diagrams.

17.2 The method

17.2.1 Rectangles

A rectangle has two pairs of equal sides. Call the longer side the length $l$ and the shorter side the width $w$.

Area — multiply length by width:

\[A = l \times w\]

A bedroom that is 4 metres long and 3 metres wide has area $4 \times 3 = 12$ m².

Perimeter — add up all four sides. Since opposite sides are equal, that’s two lengths plus two widths:

\[P = 2l + 2w\]

The same bedroom has perimeter $2(4) + 2(3) = 8 + 6 = 14$ m.

17.2.2 Triangles

A triangle has three sides and three angles. For area, you need a base and a height. The base is any side you choose. The height is the perpendicular distance from that base to the opposite corner — straight up, at a right angle to the base.

Area:

\[A = \frac{1}{2} \times b \times h\]

Read it as: half of base times height. The halving is there because a triangle is half a rectangle — if you put two identical triangles together, they make a parallelogram with the same base and height.

A triangle with base 6 cm and height 4 cm has area $\frac{1}{2} \times 6 \times 4 = 12$ cm².

Perimeter — just add all three sides. There’s no shortcut unless the triangle has special properties.

17.2.3 Circles

Circumference — the distance around the edge of a circle:

\[C = 2\pi r\]

Read it as: two times pi times radius. Since $d = 2r$, this is the same as $\pi d$ — pi times diameter. That’s actually where $\pi$ comes from: the ratio of any circle’s circumference to its diameter. Every circle, everywhere, has $C = \pi d$.

Area — the amount of flat space a circle covers:

\[A = \pi r^2\]

Read it as: pi times radius squared. A circle with radius 5 cm has area $\pi \times 5^2 = 25\pi \approx 78.5$ cm².

Note: leave the answer as $25\pi$ when exact form is needed. Use the decimal approximation when you need a number to compare or measure with.

17.2.4 The Pythagorean theorem

For any right-angled triangle with legs $a$ and $b$ and hypotenuse $c$:

\[a^2 + b^2 = c^2\]

Finding the hypotenuse when you know the two legs:

\[c = \sqrt{a^2 + b^2}\]

A right triangle with legs 3 cm and 4 cm:

\[c = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \text{ cm}\]

Finding a leg when you know the hypotenuse and the other leg:

\[a = \sqrt{c^2 - b^2}\]

A right triangle with hypotenuse 13 cm and one leg 5 cm:

\[a = \sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12 \text{ cm}\]

The next step looks slightly mechanical. Square both known values, add (or subtract), then take the square root. In that order, every time.

Code

{
  const a = legA, b = legB;
  const c = Math.sqrt(a * a + b * b);
  const scale = 28;
  const margin = 60;
  const svgW = 520, svgH = 340;

  // Triangle vertices: right angle at bottom-left
  const ox = margin + 20, oy = svgH - margin - 20;
  const ax = ox + b * scale, ay = oy;       // bottom-right
  const bx = ox, by = oy - a * scale;       // top-left

  const svg = d3.create("svg")
    .attr("width", svgW)
    .attr("height", svgH)
    .style("font-family", "inherit");

  // Square on side b (bottom, along x-axis)
  svg.append("rect")
    .attr("x", ox).attr("y", oy)
    .attr("width", b * scale).attr("height", b * scale)
    .attr("fill", "#fef9c3").attr("stroke", "#ca8a04").attr("stroke-width", 1.5);
  svg.append("text")
    .attr("x", ox + b * scale / 2).attr("y", oy + b * scale / 2 + 5)
    .attr("text-anchor", "middle").attr("font-size", "13px").attr("fill", "#92400e")
    .text(`b² = ${b*b}`);

  // Square on side a (left, along y-axis) — goes to the left
  svg.append("rect")
    .attr("x", ox - a * scale).attr("y", by)
    .attr("width", a * scale).attr("height", a * scale)
    .attr("fill", "#dcfce7").attr("stroke", "#16a34a").attr("stroke-width", 1.5);
  svg.append("text")
    .attr("x", ox - a * scale / 2).attr("y", by + a * scale / 2 + 5)
    .attr("text-anchor", "middle").attr("font-size", "13px").attr("fill", "#166534")
    .text(`a² = ${a*a}`);

  // Square on hypotenuse c — rotated square drawn as polygon
  // hypotenuse goes from (ax,ay) to (bx,by)
  // perpendicular direction (outward): normalise then rotate 90°
  const dx = bx - ax, dy = by - ay;
  const len = Math.sqrt(dx*dx + dy*dy);
  const nx = -dy / len, ny = dx / len; // outward normal
  const p1 = {x: ax, y: ay};
  const p2 = {x: bx, y: by};
  const p3 = {x: bx + nx * len, y: by + ny * len};
  const p4 = {x: ax + nx * len, y: ay + ny * len};
  const pts = [p1,p2,p3,p4].map(p => `${p.x},${p.y}`).join(" ");

  svg.append("polygon")
    .attr("points", pts)
    .attr("fill", "#dbeafe").attr("stroke", "#2563eb").attr("stroke-width", 1.5);

  const midCx = (p1.x + p2.x + p3.x + p4.x) / 4;
  const midCy = (p1.y + p2.y + p3.y + p4.y) / 4;
  svg.append("text")
    .attr("x", midCx).attr("y", midCy + 5)
    .attr("text-anchor", "middle").attr("font-size", "13px").attr("fill", "#1e3a8a")
    .text(`c² = ${(c*c).toFixed(2)}`);

  // Triangle fill
  svg.append("polygon")
    .attr("points", `${ox},${oy} ${ax},${ay} ${bx},${by}`)
    .attr("fill", "#f1f5f9").attr("stroke", "#475569").attr("stroke-width", 2);

  // Right-angle marker
  const sq = 8;
  svg.append("rect")
    .attr("x", ox).attr("y", oy - sq)
    .attr("width", sq).attr("height", sq)
    .attr("fill", "none").attr("stroke", "#475569").attr("stroke-width", 1.5);

  // Side labels
  svg.append("text")
    .attr("x", ox + b * scale / 2).attr("y", oy - 8)
    .attr("text-anchor", "middle").attr("font-size", "13px").attr("fill", "#374151")
    .text(`b = ${b}`);

  svg.append("text")
    .attr("x", ox - 14).attr("y", oy - a * scale / 2)
    .attr("text-anchor", "middle").attr("font-size", "13px").attr("fill", "#374151")
    .text(`a = ${a}`);

  // Equation below
  svg.append("text")
    .attr("x", svgW / 2).attr("y", svgH - 16)
    .attr("text-anchor", "middle").attr("font-size", "14px").attr("fill", "#111827")
    .text(`${a}² + ${b}² = ${a*a} + ${b*b} = ${(a*a+b*b).toFixed(2)}   →   c = ${c.toFixed(3)}`);

  return svg.node();
}

Why the Pythagorean theorem works

Draw a right triangle. Now draw a square on each side — one square on leg $a$, one on leg $b$, one on the hypotenuse $c$. The theorem says the area of the big square exactly equals the total area of the two smaller squares: $a^2 + b^2 = c^2$.

You can show this is true by cutting up the two smaller squares and rearranging the pieces to exactly fill the big one — no gaps, no overlaps. That’s the core of most visual proofs of this result. The algebra follows from the geometry.

This only holds when the angle between sides $a$ and $b$ is exactly 90°. Change that angle and the relationship changes — that’s what trigonometry (the next chapter) is about.

17.2.5 Volume of a rectangular prism

A rectangular prism (a box shape) has length $l$, width $w$, and height $h$.

\[V = l \times w \times h\]

Read it as: length times width times height. A box 10 cm long, 5 cm wide, and 3 cm tall has volume $10 \times 5 \times 3 = 150$ cm³.

A useful way to think about it: the base area is $l \times w$, and you’re stacking that base up through height $h$. Volume is always base area times height for any prism.

17.2.6 Volume of a cylinder

A cylinder is a circle stacked up through a height. The base is a circle of radius $r$, and the height is $h$.

\[V = \pi r^2 h\]

Read it as: pi times radius squared times height. The $\pi r^2$ part is just the area of the circular base; multiplying by $h$ stacks it up.

A can of drink is a cylinder. A standard 330 ml can has a radius of about 3.3 cm and a height of about 9.6 cm:

\[V = \pi \times 3.3^2 \times 9.6 = \pi \times 10.89 \times 9.6 \approx 328 \text{ cm}^3\]

Close to 330 ml (330 cm³). The small difference is rounding on the actual dimensions.

17.2.7 Units and unit conversion

Measurements only make sense with their units attached. $15$ is not an area — $15 \text{ m}^2$ is.

When you multiply lengths, the units multiply too:

\[3 \text{ m} \times 4 \text{ m} = 12 \text{ m}^2\]

\[2 \text{ cm} \times 5 \text{ cm} \times 3 \text{ cm} = 30 \text{ cm}^3\]

Converting between units: multiply by a conversion factor equal to 1. For example, $1 \text{ m} = 100 \text{ cm}$, so $\frac{100 \text{ cm}}{1 \text{ m}} = 1$.

\[5 \text{ m} = 5 \times 100 \text{ cm} = 500 \text{ cm}\]

For areas, the conversion factor gets squared:

\[1 \text{ m}^2 = (100 \text{ cm})^2 = 10{,}000 \text{ cm}^2\]

So $3 \text{ m}^2 = 3 \times 10{,}000 = 30{,}000 \text{ cm}^2$.

Why squared? A square metre is a square that is 100 cm wide and 100 cm tall — that’s $100 \times 100 = 10{,}000$ square centimetres. The conversion factor gets multiplied by itself once for each dimension you’re measuring.

For volumes, it gets cubed:

\[1 \text{ m}^3 = (100)^3 \text{ cm}^3 = 1{,}000{,}000 \text{ cm}^3\]

A cubic metre is a cube 100 cm × 100 cm × 100 cm, so the conversion factor multiplies by itself three times. This trips people up. One square metre is not a hundred square centimetres — it’s ten thousand. The number gets big fast because you’re covering a two-dimensional (or three-dimensional) space.

17.3 Worked examples

Example 1 (sci) — Area of a circular clock face.

A clock has a circular face with a diameter of 20 cm. What is the area of the face?

Diameter = 20 cm, so radius = 10 cm.

\[A = \pi r^2 = \pi \times 10^2 = 100\pi \approx 314 \text{ cm}^2\]

The clock face covers approximately 314 cm² — about the same as a standard sheet of A4 paper turned on its side. Doubling the diameter would quadruple the area, because radius is squared in the formula.

Example 2 (eng) — Length of a ramp.

A skateboard ramp rises 1.2 m vertically and runs 1.6 m horizontally along the ground. The ramp surface is the hypotenuse of a right triangle. How long is the ramp surface?

\[c = \sqrt{a^2 + b^2} = \sqrt{1.2^2 + 1.6^2} = \sqrt{1.44 + 2.56} = \sqrt{4.00} = 2.0 \text{ m}\]

The ramp surface is exactly 2 m long.

Note: 1.2, 1.6, 2.0 is a scaled version of the 3-4-5 right triangle (multiply each by 0.4). When you see a clean answer, check whether it’s a Pythagorean triple.

Example 3 (comp) — Screen area from diagonal and aspect ratio.

A phone screen has a 6.1-inch diagonal. The screen is 16:9 aspect ratio (16 units wide for every 9 units tall). What is the actual screen area?

One approach: call the width $16k$ and height $9k$, where $k$ is a scaling factor that sets the actual size. The Pythagorean theorem then gives an equation in $k$:

\[(16k)^2 + (9k)^2 = 6.1^2\]

\[256k^2 + 81k^2 = 37.21\]

\[337k^2 = 37.21\]

\[k^2 = \frac{37.21}{337} \approx 0.11041\]

\[k \approx 0.3323 \text{ inches}\]

Width $= 16 \times 0.3323 \approx 5.32$ inches. Height $= 9 \times 0.3323 \approx 2.99$ inches.

Area $= 5.32 \times 2.99 \approx 15.9$ square inches.

The 6.1-inch measurement is the diagonal — the actual viewing area is about 15.9 square inches, which is roughly the size of a playing card.

Example 4 (geo) — Comparing two screen sizes.

You’re deciding between a phone screen and a tablet screen. The phone screen measures 6 cm by 12 cm on the poster. The tablet screen is shown on a second poster at double the linear scale — so every centimetre on the first poster represents 2 cm on the second. How does the tablet screen area compare to the phone screen area, and what is each real area on the posters?

Phone screen area: $6 \times 12 = 72$ cm² on the poster.

Tablet screen on the second poster: doubling the linear scale means each dimension doubles.

Width $= 2 \times 6 = 12$ cm. Height $= 2 \times 12 = 24$ cm.

Tablet screen area $= 12 \times 24 = 288$ cm² on the second poster.

$288 \div 72 = 4$ — the tablet poster area is four times the phone poster area, even though only the scale doubled. That’s because area scales as the square of the linear scale: $2^2 = 4$.

17.4 Where this goes

The next chapter in this volume is trigonometry (Vol 3, Ch 2). Trigonometry extends the geometry here by handling triangles where you don’t know the right angle — or where you need to find angles as well as lengths. The Pythagorean theorem handles one specific case (right triangles, lengths only); trigonometry handles the general case using the sine, cosine, and tangent ratios. Every oscillating system — sound, light, signals, tides — is described using trigonometry.

Further out, the measurement ideas here feed directly into functions and relations (Vol 3, Ch 4). Area and volume formulas are functions: plug in the radius, get out the area. Once you’re working with functions, you can ask questions like “how fast does the area change as the radius grows?” — and that question leads to calculus.

Where this shows up

A structural engineer calculating how much steel plate is needed for a bridge deck is computing surface area.
A physicist calculating fluid pressure on a pipe cross-section is computing circular area.
A GIS analyst measuring a protected forest region from a satellite image is computing polygon areas at scale.
A game developer checking whether two sprites have collided is comparing rectangular bounding boxes — area and distance calculations happening thousands of times per second.
A packaging designer minimising the amount of cardboard in a cereal box is minimising surface area for a fixed volume.

The formulas are always the same. What changes is what you’re measuring and why it matters.

17.5 Exercises

These are puzzles. Each one has a clean answer, but the interesting part is deciding which formula to use and what counts as the base, height, radius, or leg.

A rectangular phone screen is 13.8 cm wide and 7.0 cm tall. What is its area? What is its diagonal length (to one decimal place)?

Code

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      done.textContent = "✓ Solution complete";
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Code

makeStepperHTML(1, [
  { op: "Find the area", eq: "A = l \\times w = 13.8 \\times 7.0 = 96.6 \\text{ cm}^2", note: null },
  { op: "Write the Pythagorean theorem for the diagonal", eq: "d = \\sqrt{l^2 + w^2}", note: "The diagonal of a rectangle is the hypotenuse of a right triangle whose legs are the width and height." },
  { op: "Substitute values", eq: "d = \\sqrt{13.8^2 + 7.0^2} = \\sqrt{190.44 + 49.00} = \\sqrt{239.44}", note: null },
  { op: "Calculate", eq: "d = 15.5 \\text{ cm}", note: null },
  { op: "Check units", eq: "\\text{Area: } 96.6 \\text{ cm}^2 \\quad \\text{Diagonal: } 15.5 \\text{ cm}", note: null }
])

A circular pizza has a diameter of 30 cm. What is its area (leave your answer in terms of $\pi$, then give a decimal approximation to the nearest whole number)?

Code

makeStepperHTML(2, [
  { op: "Find the radius from the diameter", eq: "r = \\frac{d}{2} = \\frac{30}{2} = 15 \\text{ cm}", note: null },
  { op: "Write the area formula", eq: "A = \\pi r^2", note: null },
  { op: "Substitute", eq: "A = \\pi \\times 15^2 = 225\\pi \\text{ cm}^2", note: null },
  { op: "Decimal approximation", eq: "A \\approx 225 \\times 3.14159 \\approx 707 \\text{ cm}^2", note: null },
  { op: "Check units", eq: "A = 225\\pi \\approx 707 \\text{ cm}^2", note: null }
])

A ramp rises 0.9 m vertically over a horizontal distance of 1.2 m. How long is the sloped surface? (Give your answer to one decimal place.)

Code

makeStepperHTML(3, [
  { op: "Identify the right triangle: legs are the vertical rise and horizontal run, hypotenuse is the slope", eq: "c = \\sqrt{a^2 + b^2}", note: null },
  { op: "Substitute values", eq: "c = \\sqrt{0.9^2 + 1.2^2} = \\sqrt{0.81 + 1.44} = \\sqrt{2.25}", note: null },
  { op: "Calculate", eq: "c = 1.5 \\text{ m}", note: "0.9, 1.2, 1.5 is a scaled 3-4-5 triple (each side multiplied by 0.3)." },
  { op: "Check units", eq: "\\text{Slope length} = 1.5 \\text{ m}", note: null }
])

A cylindrical water bottle has a radius of 4 cm and a height of 20 cm. What is its volume in cm³? What is its capacity in millilitres? (1 cm³ = 1 ml.)

Code

makeStepperHTML(4, [
  { op: "Write the volume formula for a cylinder", eq: "V = \\pi r^2 h", note: null },
  { op: "Substitute values", eq: "V = \\pi \\times 4^2 \\times 20 = \\pi \\times 16 \\times 20 = 320\\pi \\text{ cm}^3", note: null },
  { op: "Decimal approximation", eq: "V \\approx 320 \\times 3.14159 \\approx 1005 \\text{ cm}^3", note: null },
  { op: "Convert to millilitres", eq: "1005 \\text{ cm}^3 = 1005 \\text{ ml} \\approx 1 \\text{ litre}", note: "1 cm³ = 1 ml, so the numeric value stays the same." },
  { op: "Check units", eq: "V = 320\\pi \\approx 1005 \\text{ ml}", note: null }
])

A triangular garden bed has a base of 5 m and a perpendicular height of 3.6 m. What area of soil does it cover? If soil costs $12 per m², what is the total cost?

Code

makeStepperHTML(5, [
  { op: "Write the area formula for a triangle", eq: "A = \\frac{1}{2} \\times b \\times h", note: null },
  { op: "Substitute values", eq: "A = \\frac{1}{2} \\times 5 \\times 3.6 = \\frac{18}{2} = 9 \\text{ m}^2", note: null },
  { op: "Calculate the cost", eq: "\\text{Cost} = 9 \\times 12 = \\$108", note: null },
  { op: "Check units", eq: "\\text{Area} = 9 \\text{ m}^2, \\quad \\text{Cost} = \\$108", note: null }
])

A rectangular swimming pool is 25 m long and 10 m wide. It has a uniform depth of 1.8 m. How many cubic metres of water does it hold? Give your answer in litres. (1 m³ = 1000 litres.)

Code

makeStepperHTML(6, [
  { op: "Write the volume formula for a rectangular prism", eq: "V = l \\times w \\times h", note: null },
  { op: "Substitute values", eq: "V = 25 \\times 10 \\times 1.8 = 450 \\text{ m}^3", note: null },
  { op: "Convert to litres", eq: "450 \\text{ m}^3 \\times 1000 \\text{ litres/m}^3 = 450{,}000 \\text{ litres}", note: "1 m³ = 1000 litres, so multiply by 1000." },
  { op: "Check units", eq: "V = 450 \\text{ m}^3 = 450{,}000 \\text{ litres}", note: null }
])