The practical math behind tech and STEM work: numbers, algebra, functions, geometry, logic and counting, with short code exercises.

Math for Tech & STEM

## Why this matters

How much cable runs around a room? How much paint covers a wall? How many litres does a tank hold? Those three questions are perimeter, area and volume, and they show up in civil and mechanical engineering, lab work, electrical installation, game levels and even data-centre planning. The formulas are short; the skill is choosing the right one and keeping the units straight.

## Three kinds of measurement

| Measures | Units | Think of it as |
|---|---|---|
| **Perimeter** | length (m, ft) | the distance around the edge |
| **Area** | length² (m², ft²) | how many unit squares cover it |
| **Volume** | length³ (m³, ft³) | how many unit cubes fill it |

The unit power tells you what you have computed. If an "area" comes out in metres, something went wrong.

## Formulas for flat shapes

| Shape | Perimeter | Area |
|---|---|---|
| Rectangle (length l, width w) | 2l + 2w | l × w |
| Square (side s) | 4s | s² |
| Triangle (base b, height h) | sum of the three sides | ½ × b × h |
| Circle (radius r) | 2πr (the circumference) | πr² |

The triangle's height is measured **perpendicular** to the base, not along a slanted side. The circle's radius is half the diameter; mixing the two is a very common slip.

## Formulas for solids

| Solid | Volume |
|---|---|
| Rectangular box (l, w, h) | l × w × h |
| Cylinder (radius r, height h) | πr²h |
| Cone | ⅓ πr²h |
| Sphere | (4/3) πr³ |

A pattern: a box and a cylinder are both "area of the base × height". If you remember that idea, you can work out the volume of any prism.

## Worked example 1: flooring

A room is 4.2 m by 3.5 m. Flooring is sold in boxes covering 2.2 m².

- Area: 4.2 × 3.5 = 14.7 m²
- Boxes: 14.7 ÷ 2.2 ≈ 6.68, so buy **7 boxes** (round up, and installers usually add about 10% for cuts and waste).
- Skirting board around the edge: perimeter = 2(4.2) + 2(3.5) = **15.4 m**.

## Worked example 2: a cylindrical tank

A tank has radius 0.5 m and height 1.2 m. How many litres does it hold?

- Volume: π × 0.5² × 1.2 = π × 0.25 × 1.2 ≈ 0.942 m³
- 1 m³ = 1000 L, so about **942 litres**.

If the radius had been given in centimetres (50 cm), working in cm gives π × 50² × 120 ≈ 942,478 cm³, and 1 L = 1000 cm³ gives the same 942 L. Pick one unit system and convert at the end.

## Worked example 3: composite shapes

An L-shaped patio can be split into two rectangles: 6 m × 2 m and 3 m × 2 m. Total area = 12 + 6 = **18 m²**. Most irregular shapes can be broken into rectangles, triangles and circles, then added (or subtracted for holes).

## Scaling: why doubling is not just double

Double every length of a shape and:

- perimeter doubles (×2)
- area becomes **4 times** larger (×2²)
- volume becomes **8 times** larger (×2³)

This matters in practice. A pipe with twice the diameter has four times the cross-sectional area, and a scale model's weight grows with the cube of its size.

## In code

```python
import math

def circle_area(r):
    return math.pi * r ** 2

def box_volume(l, w, h):
    return l * w * h

print(round(circle_area(3), 2))    # 28.27
print(box_volume(2, 3, 4))         # 24
```

Use `math.pi` rather than 3.14 when precision matters; 3.14 is already wrong in the third digit.

## Common mistakes

- **Using diameter where the formula needs radius.**
- **Using a slanted side as a triangle's height.**
- **Forgetting the ½ in the triangle area** or the ⅓ in a cone.
- **Mixing units** (one side in cm, another in m).
- **Giving area in m or volume in m²**. Check the unit power.

## Quick recap

- Perimeter is length, area is length², volume is length³.
- Rectangle area l × w, triangle ½bh, circle πr²; prism volume = base area × height.
- Break odd shapes into simple ones.
- Scaling lengths by k scales area by k² and volume by k³.

## Try it

In the **Do** tab, add the cylinder volume formula to a small geometry toolkit, then use it to check the tank from example 2.


Compute the perimeter, area and volume of common shapes and solids, with the right units.

Area, volume, and perimeter

Math Antics - Area

Math Antics - Volume

## Why this matters

Right angles are everywhere people build things: walls meet floors, screens are rectangles, city blocks form grids. Whenever you need a diagonal or a straight-line distance across a right angle, the Pythagorean theorem gives it. Carpenters use it to check that corners are square, electricians use it to estimate conduit runs, game developers use it to measure distance between characters, and data scientists use its big brother, Euclidean distance.

## The core idea

In a **right triangle** (one angle exactly 90°):

- The two sides that form the right angle are the **legs**, a and b.
- The longest side, opposite the right angle, is the **hypotenuse**, c.

The **Pythagorean theorem** says:

**a² + b² = c²**

The squares on the two legs add up to the square on the hypotenuse. It works only for right triangles.

## Finding the hypotenuse

c = √(a² + b²)

Legs 3 and 4: c = √(9 + 16) = √25 = **5**.

## Finding a missing leg

Rearrange: a = √(c² − b²)

A 13 m ladder reaches 12 m up a wall. How far is its foot from the wall? a = √(169 − 144) = √25 = **5 m**.

Notice you **subtract** when finding a leg. If you add, the "leg" comes out longer than the hypotenuse, which is impossible and a useful warning sign.

## Worked example 1: screen size

A monitor is 60 cm wide and 34 cm tall. Screens are sold by diagonal:

c = √(60² + 34²) = √(3600 + 1156) = √4756 ≈ 68.96 cm, about **27.1 inches** (÷ 2.54).

## Worked example 2: is the corner square?

The **3-4-5 rule**: measure 3 units along one wall and 4 along the other. If the diagonal between the marks is exactly 5, the corner is 90°. This works because the converse of the theorem is also true: if a² + b² = c², the angle is a right angle. Builders often scale it up to 6-8-10 or 9-12-15 for better accuracy.

Common **Pythagorean triples** (whole-number right triangles): 3-4-5, 5-12-13, 8-15-17, 7-24-25, and any multiple of these.

## Worked example 3: distance between two points

The distance between points (x₁, y₁) and (x₂, y₂) is a hypotenuse whose legs are the x and y differences:

d = √((x₂ − x₁)² + (y₂ − y₁)²)

From (1, 2) to (7, 10): d = √(6² + 8²) = √100 = **10**.

In three dimensions, add a third squared term: d = √(Δx² + Δy² + Δz²). The diagonal of a 3 × 4 × 12 box is √(9 + 16 + 144) = √169 = 13.

## In code

Python's `math` module has the square root, and even a ready-made hypotenuse function:

```python
import math

print(math.sqrt(3**2 + 4**2))    # 5.0
print(math.hypot(60, 34))        # about 68.96
print(math.dist((1, 2), (7, 10)))  # 10.0
```

Writing the formula yourself once is the best way to understand what `hypot` and `dist` are doing.

## Common mistakes

- **Using it on a triangle without a right angle.** Use the law of cosines instead (you will meet it later).
- **Labelling the wrong side as c.** The hypotenuse is always opposite the right angle and is always the longest.
- **Adding instead of subtracting** when finding a leg.
- **Forgetting the square root**: a² + b² gives c², not c.
- **Writing √(a² + b²) as a + b.** √(9 + 16) is 5, not 7.

## Quick recap

- For a right triangle: a² + b² = c², with c the hypotenuse.
- Hypotenuse: c = √(a² + b²). Leg: a = √(c² − b²).
- If the sides satisfy the equation, the triangle has a right angle (3-4-5 check).
- Distance between points is Pythagoras on the coordinate differences.

## Try it

In the **Do** tab, write the hypotenuse formula. A missing-leg function and a distance function are already written, and they depend on yours.


Use the Pythagorean theorem to find missing sides, distances and diagonals in right triangles.

Right triangles and Pythagoras

The Pythagorean theorem intro | Right triangles and trigonometry | Geometry | Khan Academy

## Why this matters

Pythagoras finds a side when you know two sides. Trigonometry goes further: it connects **angles** to sides. That is how surveyors measure a building's height from the ground, how engineers resolve a force into horizontal and vertical parts, how a roof pitch becomes a rafter length, and how games rotate a sprite to face a target.

## The core idea

In a right triangle, pick one of the two acute angles and call it θ (theta). Relative to that angle, name the sides:

- **Hypotenuse (H):** the longest side, opposite the right angle.
- **Opposite (O):** the side across from θ.
- **Adjacent (A):** the side next to θ that is not the hypotenuse.

The labels depend on which angle you choose: the opposite side of one acute angle is the adjacent side of the other.

The three ratios are remembered as **SOH-CAH-TOA**:

| Ratio | Definition |
|---|---|
| **S**ine | sin θ = **O**pposite / **H**ypotenuse |
| **C**osine | cos θ = **A**djacent / **H**ypotenuse |
| **T**angent | tan θ = **O**pposite / **A**djacent |

For a given angle these ratios are fixed, however big the triangle is. sin 30° is always 0.5.

## Finding a side

**Step 1:** label the sides relative to the known angle. **Step 2:** choose the ratio that uses the side you know and the side you want. **Step 3:** rearrange and compute.

### Worked example 1: a ramp

A 10 m ramp rises at 30°. How high does it go?

- Known: hypotenuse = 10. Want: opposite (the height).
- O and H: use **sine**. sin 30° = O / 10.
- O = 10 × sin 30° = 10 × 0.5 = **5 m**.

### Worked example 2: a building's height

You stand 40 m from a building and the angle up to its roof is 52°.

- Known: adjacent = 40. Want: opposite.
- O and A: use **tangent**. O = 40 × tan 52° ≈ 40 × 1.280 = **51.2 m** (plus your eye height, if you measured from your eyes).

## Finding an angle

To go from a ratio back to an angle, use the **inverse** functions: sin⁻¹ (arcsin), cos⁻¹ (arccos) and tan⁻¹ (arctan).

### Worked example 3: roof pitch

A roof rises 4 m over a horizontal run of 6 m. What angle does it make?

- O = 4, A = 6, so tan θ = 4 / 6 ≈ 0.667.
- θ = tan⁻¹(0.667) ≈ **33.7°**.

## Degrees and radians

Angles can be measured in degrees (a full turn is 360°) or **radians** (a full turn is 2π). Most programming languages, including Python and JavaScript, use **radians** in their trig functions. Convert first:

```python
import math

print(math.sin(math.radians(30)))      # 0.49999999999999994, i.e. 0.5
print(40 * math.tan(math.radians(52)))  # about 51.2
print(math.degrees(math.atan(4 / 6)))  # about 33.69
```

The tiny error in the first line is normal floating-point rounding. Compare with a tolerance, not `==`.

`math.atan2(y, x)` is a safer angle function in code: it handles all four directions and x = 0 without dividing by zero. Games use it to point an object towards a target.

## Resolving a vector

A force of 100 N pulls at 30° above horizontal. Its parts are:

- horizontal = 100 × cos 30° ≈ 86.6 N
- vertical = 100 × sin 30° = 50 N

Splitting a quantity into x and y parts this way is the foundation of physics and engineering mechanics.

## Common mistakes

- **Calculator or code in the wrong angle mode.** sin(30) in radians is about −0.988, not 0.5.
- **Mislabelling opposite and adjacent.** Always label from the angle you are using.
- **Using trig ratios on a non-right triangle.** SOH-CAH-TOA needs a 90° angle.
- **Dividing when you should multiply.** If the unknown is on top of the ratio, multiply; if it is on the bottom, divide (A = O / tan θ).

## Quick recap

- Label O, A and H relative to the chosen angle.
- SOH-CAH-TOA: sin = O/H, cos = A/H, tan = O/A.
- Inverse functions turn a ratio back into an angle.
- Code uses radians; convert with `math.radians` and `math.degrees`.

## Try it

In the **Do** tab, finish a function that finds the opposite side from an angle and the hypotenuse. An angle-finding function is already written as a model.


Use SOH-CAH-TOA to find unknown sides and angles in right triangles.

Sine, cosine, and tangent

Trigonometry | SOH CAH TOA | Sin, Cos, Tan

Perimeter, area and volume, the Pythagorean theorem, and sine, cosine and tangent for right triangles.