Home/ Mathematics/ Unit 04 – Geometry & Measures/ Pythagoras' Theorem
Mathematics · AQA 8300 §G6

Pythagoras' Theorem

📖 Spec: AQA 8300 §G6 ⭐⭐⭐ Difficulty 3/5 ⏱ 40 mins 🎓 AQA · Edexcel · OCR Grade 9 Target
  • Apply Pythagoras' theorem to find missing sides in right-angled triangles
  • Use Pythagorean triples to identify right-angled triangles and check answers
  • Find the distance between two coordinate points using the distance formula
  • Apply Pythagoras' theorem in 3D problems involving cuboids, pyramids and cones
  • Leave answers in exact surd form and combine Pythagoras with trigonometry

🔑 Core Concepts

2.1 The Theorem and Why It Works

Pythagoras' theorem is a fundamental relationship between the three sides of any right-angled triangle. It was proven by ancient Greek mathematicians and has thousands of known proofs — the most intuitive is the "square on each side" proof: if you draw squares on all three sides, the area of the square on the hypotenuse exactly equals the combined area of the two smaller squares.

📖
Definition — Pythagoras' Theorem
In a right-angled triangle, the square of the hypotenuse equals the sum of the squares of the two shorter sides. If $a$ and $b$ are the shorter sides and $c$ is the hypotenuse (the side opposite the right angle), then: $$a^2 + b^2 = c^2$$
Core Formula
$$a^2 + b^2 = c^2$$
$c$ = hypotenuse (longest side, opposite right angle) $a, b$ = the two shorter sides
⚠️
Important
The hypotenuse is always the side opposite the right angle (90°). It is always the longest side. You must identify it correctly before applying the formula.
🎯
Exam Tip
Label the sides before writing any equation. Identify the right angle, label the hypotenuse $c$, then label the other two sides $a$ and $b$. This prevents the single most common error in Pythagoras questions.

2.2 Finding the Hypotenuse

When both shorter sides are known and you need the hypotenuse, substitute into $a^2 + b^2 = c^2$ and solve for $c$.

Identify hypotenuse $c$
→
Write $a^2 + b^2 = c^2$
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Substitute known values
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Solve: $c = \sqrt{a^2 + b^2}$
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Round or leave as surd

2.3 Finding a Shorter Side

When the hypotenuse and one shorter side are known, rearrange the formula to find the unknown shorter side. This requires subtraction, not addition.

Finding a Shorter Side
$$a = \sqrt{c^2 - b^2}$$
Subtract the known shorter side squared from hypotenuse squared
✗
Common Mistake
When finding a shorter side, students frequently add instead of subtract: writing $a^2 = c^2 + b^2$. Remember: only when finding the hypotenuse do you add. Rearranging $a^2 + b^2 = c^2$ gives $a^2 = c^2 - b^2$.

2.4 Pythagorean Triples

A Pythagorean triple is a set of three positive integers $(a, b, c)$ that exactly satisfies $a^2 + b^2 = c^2$. Recognising these saves calculation time and allows you to verify that a triangle has a right angle without a calculator.

TripleVerificationScaled version (×2)Scaled version (×3)
(3, 4, 5)$9 + 16 = 25$ ✓(6, 8, 10)(9, 12, 15)
(5, 12, 13)$25 + 144 = 169$ ✓(10, 24, 26)(15, 36, 39)
(8, 15, 17)$64 + 225 = 289$ ✓(16, 30, 34)(24, 45, 51)
(7, 24, 25)$49 + 576 = 625$ ✓(14, 48, 50)(21, 72, 75)
🧠
Memory Trick
"Three-Four-Five — Drive to Stay Alive" — the most common triple is (3,4,5). Any multiple works: (6,8,10), (9,12,15), (15,20,25) etc. Always check if the given sides are a scaled triple before calculating.

2.5 Converse of Pythagoras' Theorem

The converse states: if $a^2 + b^2 = c^2$ for the three sides of a triangle (where $c$ is the longest), then the triangle has a right angle opposite side $c$. This is used to prove that a triangle is right-angled.

📖
Converse of Pythagoras
If the sides of a triangle satisfy $a^2 + b^2 = c^2$, then the triangle is right-angled with the right angle opposite side $c$. Used to prove right angles — a Grade 7+ skill requiring a clear written justification.
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Exam Tip — Converse Proof
In converse problems, you must calculate $a^2 + b^2$ and $c^2$ separately, then state whether they are equal and give a conclusion. A correct answer without a conclusion loses the final mark.

2.6 Distance Between Two Points

The distance formula is derived directly from Pythagoras' theorem. When you plot two points, you can form a right-angled triangle where the horizontal distance is one leg, the vertical distance is the other leg, and the line segment between the points is the hypotenuse.

Distance Formula (2D Coordinates)
$$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$
$(x_1, y_1)$ = first point $(x_2, y_2)$ = second point $d$ = straight-line distance
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Exam Tip — Coordinate Geometry
It does not matter which point you label $(x_1, y_1)$ — since differences are squared, a negative difference gives the same result as a positive one. However, always show your working to earn method marks.

2.7 Pythagoras in 3D Problems

Grade 8–9 questions extend Pythagoras into three dimensions. The key skill is identifying the correct right-angled triangle within the 3D shape, applying Pythagoras in stages — often using an intermediate result as the new leg.

Space Diagonal of a Cuboid

The space diagonal runs from one corner to the diagonally opposite corner through the interior of the cuboid. It requires two applications of Pythagoras:

  1. Find the diagonal of the base rectangle: $d_{base} = \sqrt{l^2 + w^2}$
  2. Use $d_{base}$ as one leg and the height $h$ as the other: $d_{space} = \sqrt{d_{base}^2 + h^2}$
  3. Combining: $d_{space} = \sqrt{l^2 + w^2 + h^2}$
Space Diagonal of Cuboid
$$d = \sqrt{l^2 + w^2 + h^2}$$
$l$ = length $w$ = width $h$ = height

Slant Height of a Cone

The slant height $l$ of a cone runs along the curved surface from apex to base circumference. Combined with the radius $r$ and perpendicular height $h$, these form a right-angled triangle.

Slant Height of a Cone
$$l = \sqrt{r^2 + h^2}$$
$l$ = slant height $r$ = base radius $h$ = perpendicular height

Height of a Square-Based Pyramid

The perpendicular height of a pyramid runs from apex to the centre of the base. The slant edge runs from apex to a base corner. The distance from the centre of the base to a corner equals half the diagonal of the base square.

Find base diagonal $= \sqrt{2}\,s$ for square base of side $s$
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Half-diagonal $= \frac{s\sqrt{2}}{2}$
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Height $h = \sqrt{e^2 - \left(\frac{s\sqrt{2}}{2}\right)^2}$
🎯
Exam Tip — 3D Pythagoras
Always draw the cross-sectional right-angled triangle separately. Label all three sides with what they represent in the 3D shape. This makes the calculation clear and earns method marks even if the final answer is wrong.

2.8 Exact Surd Answers

At Grade 7+, questions often require answers in exact form — leaving the answer as a surd rather than rounding. This is more precise and avoids accumulation of rounding errors in multi-step problems.

📖
Surd
A surd is an irrational root that cannot be simplified to a rational number. For example, $\sqrt{2}$, $\sqrt{3}$, $3\sqrt{5}$ are surds. $\sqrt{4} = 2$ is not a surd because it simplifies to a whole number.
✏️
Worked Example — Surd Simplification
$\sqrt{75} = \sqrt{25 \times 3} = 5\sqrt{3}$. Always look for perfect square factors to simplify surds. In a right triangle with legs 3 and 6: $c = \sqrt{9 + 36} = \sqrt{45} = \sqrt{9 \times 5} = 3\sqrt{5}$.

🗺️ Visual Notes

Pythagoras' Theorem
The Formula
  • $a^2 + b^2 = c^2$
  • $c$ = hypotenuse (opposite 90°)
  • Hypotenuse: always longest
  • Rearrange to find shorter side
Pythagorean Triples
  • 3, 4, 5 and multiples
  • 5, 12, 13 and multiples
  • 8, 15, 17 and multiples
  • Use to prove right angles
2D Applications
  • Composite shapes
  • Coordinate distances
  • $d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$
  • Isosceles triangle height
3D Applications
  • Cuboid space diagonal
  • Slant height of cone
  • Height of pyramid
  • Two-stage calculation
Exact Answers
  • Leave as $\sqrt{\phantom{x}}$ where needed
  • Simplify surds (find sq. factors)
  • Use intermediate exact values
  • Avoids rounding errors
Converse
  • Test if $a^2 + b^2 = c^2$
  • $c$ must be the longest side
  • State the conclusion clearly
  • Used in coordinate proofs

Finding the Hypotenuse vs Finding a Shorter Side

SituationWhat you knowWhat you findFormula usedOperation
Find hypotenuseBoth shorter sides $a$, $b$$c$$c = \sqrt{a^2 + b^2}$ADD, then square root
Find shorter sideHypotenuse $c$ and one side $b$$a$$a = \sqrt{c^2 - b^2}$SUBTRACT, then square root
Converse checkAll three sidesIs it right-angled?Check $a^2 + b^2 = c^2$Test equality
Coordinate distanceTwo points $(x_1,y_1)$, $(x_2,y_2)$$d$$d = \sqrt{\Delta x^2 + \Delta y^2}$ADD squares, then root
3D space diagonal$l$, $w$, $h$ of cuboid$d$$d = \sqrt{l^2 + w^2 + h^2}$ADD three squares, then root

Decision Tree — Which Formula to Use

Is it a right-angled triangle?
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Yes: use Pythagoras
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Do you need hypotenuse?
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Yes → $c=\sqrt{a^2+b^2}$
spacer
→
spacer
→
No (shorter side) → $a=\sqrt{c^2-b^2}$

Quick-Recognition: Is it a Pythagorean Triple?

Given sidesScale factorBase tripleRight-angled?
6, 8, 10×23, 4, 5Yes ✓
9, 12, 15×33, 4, 5Yes ✓
10, 24, 26×25, 12, 13Yes ✓
20, 48, 52×45, 12, 13Yes ✓
16, 30, 34×28, 15, 17Yes ✓
5, 6, 8—Not a tripleNo ✗ ($25+36=61 \neq 64$)

✏️ Worked Examples

Grade 4–5 | Finding the Hypotenuse
A right-angled triangle has shorter sides of length 9 cm and 12 cm. Find the length of the hypotenuse.
1
Identify the hypotenuse
The hypotenuse $c$ is opposite the right angle — it is what we need to find. The two shorter sides are $a = 9$ cm and $b = 12$ cm.
2
Write the formula
$$a^2 + b^2 = c^2$$
3
Substitute known values
$$9^2 + 12^2 = c^2$$ $$81 + 144 = c^2$$ $$225 = c^2$$
4
Solve for $c$
$$c = \sqrt{225} = 15 \text{ cm}$$ Note: $(9, 12, 15)$ is a Pythagorean triple — it is $3 \times (3, 4, 5)$.
Answer: The hypotenuse is 15 cm.
Grade 6–7 | Distance Between Two Points
Point $A$ has coordinates $(2, 5)$ and point $B$ has coordinates $(−3, 17)$. Find the exact distance $AB$, giving your answer in the form $k\sqrt{n}$.
1
Write the distance formula
$$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$
2
Calculate the differences
$x_2 - x_1 = -3 - 2 = -5$ and $y_2 - y_1 = 17 - 5 = 12$
3
Square each difference
$$(-5)^2 = 25 \qquad 12^2 = 144$$
4
Add and take the square root
$$d = \sqrt{25 + 144} = \sqrt{169} = 13$$
5
Interpret the result
Since $\sqrt{169} = 13$ exactly, $(5, 12, 13)$ is a Pythagorean triple. No surd needed here — the answer is already exact.
Answer: $AB = \mathbf{13}$ units.
Grade 9 | 3D Pythagoras — Pyramid
A square-based pyramid has a base of side length 10 cm and slant edges (from apex to each base corner) of length 13 cm. Find the perpendicular height of the pyramid, giving your answer in the form $k\sqrt{n}$ where $k$ and $n$ are integers.
1
Find the diagonal of the square base
The base is a $10 \times 10$ square. Its diagonal is: $$d_{base} = \sqrt{10^2 + 10^2} = \sqrt{200} = 10\sqrt{2} \text{ cm}$$
2
Find the distance from centre to corner
The perpendicular from the apex meets the centre of the base. The centre is at the intersection of the diagonals: $$d_{centre-to-corner} = \frac{10\sqrt{2}}{2} = 5\sqrt{2} \text{ cm}$$
3
Identify the right-angled triangle
From centre of base to corner = $5\sqrt{2}$ cm (horizontal leg), perpendicular height = $h$ (vertical leg), slant edge = 13 cm (hypotenuse). These form a right-angled triangle.
4
Apply Pythagoras — shorter side formula
$$h^2 = 13^2 - (5\sqrt{2})^2$$ $$h^2 = 169 - 25 \times 2 = 169 - 50 = 119$$ $$h = \sqrt{119} \text{ cm}$$
5
Check if the surd simplifies
$119 = 7 \times 17$ — no perfect square factors, so $\sqrt{119}$ is already in simplest form. Here $k = 1$, $n = 119$.
Answer: Perpendicular height $= \mathbf{\sqrt{119}}$ cm $\approx 10.9$ cm.

❓ Exam Questions

Q1 [2 marks]

A right-angled triangle has legs of length 5 cm and 7 cm. Calculate the length of the hypotenuse, giving your answer correct to 1 decimal place.

Mark scheme:
M1: Correct substitution $c^2 = 5^2 + 7^2 = 25 + 49 = 74$
A1: $c = \sqrt{74} = 8.6$ cm (to 1 d.p.)

Full solution: $c = \sqrt{5^2 + 7^2} = \sqrt{74} \approx 8.6022... = \mathbf{8.6}$ cm
Q2 [3 marks]

Show that the triangle with sides 7 cm, 24 cm and 25 cm is right-angled. State which angle is 90°.

Mark scheme:
M1: Test $a^2 + b^2$: $7^2 + 24^2 = 49 + 576 = 625$
M1: Test $c^2$: $25^2 = 625$
A1: Since $7^2 + 24^2 = 25^2$, by the converse of Pythagoras the triangle is right-angled with the 90° angle opposite the 25 cm side.

Note: You must state "by the converse of Pythagoras" or equivalent to earn full marks. Simply writing numbers is insufficient.
Q3 [4 marks]

Points $P(−2, 3)$ and $Q(4, k)$ are such that $PQ = 10$. Find the two possible values of $k$.

Mark scheme:
M1: Write $PQ^2 = (4-(-2))^2 + (k-3)^2 = 36 + (k-3)^2$
M1: Set equal to $10^2 = 100$: $36 + (k-3)^2 = 100 \Rightarrow (k-3)^2 = 64$
A1: $k - 3 = \pm 8$
A1: $k = 11$ or $k = -5$

Full working: $\sqrt{36 + (k-3)^2} = 10 \Rightarrow (k-3)^2 = 64 \Rightarrow k = 3 \pm 8$, so $k = 11$ or $k = -5$.
Q4 [4 marks]

A cuboid has dimensions $6$ cm by $8$ cm by $h$ cm. The space diagonal of the cuboid is $\sqrt{149}$ cm. Find the value of $h$.

Mark scheme:
M1: Use $d^2 = l^2 + w^2 + h^2$: $(\sqrt{149})^2 = 6^2 + 8^2 + h^2$
M1: $149 = 36 + 64 + h^2 \Rightarrow 149 = 100 + h^2$
A1: $h^2 = 49$
A1: $h = 7$ cm

Note: $h$ must be positive since it is a length, so $h = 7$ cm only.
Q5 [6 marks]

A right pyramid has a rectangular base measuring 12 cm by 16 cm. The apex is directly above the centre of the base, and each slant edge (apex to corner) has length 18 cm.
(a) Find the perpendicular height of the pyramid, giving your answer in the form $\sqrt{n}$.
(b) Hence find the angle the slant edge makes with the base, correct to 1 decimal place.

Mark scheme:
(a)
M1: Diagonal of rectangular base: $d = \sqrt{12^2 + 16^2} = \sqrt{144+256} = \sqrt{400} = 20$ cm
M1: Distance from centre to corner $= 20 \div 2 = 10$ cm
M1: $h^2 = 18^2 - 10^2 = 324 - 100 = 224$
A1: $h = \sqrt{224}$ cm (or $4\sqrt{14}$ cm)

(b)
M1: Right-angled triangle with adjacent = 10 cm, hypotenuse = 18 cm
A1: $\cos\theta = \frac{10}{18} \Rightarrow \theta = \cos^{-1}\left(\frac{10}{18}\right) = 56.3°$ (to 1 d.p.)
Q6 [1 mark]

A right-angled isosceles triangle has a hypotenuse of 10 cm. Write down the exact length of each shorter side.

Mark scheme:
A1: Let each shorter side be $a$. Then $2a^2 = 100 \Rightarrow a^2 = 50 \Rightarrow a = \sqrt{50} = 5\sqrt{2}$ cm.

Answer: Each shorter side is $5\sqrt{2}$ cm.

⭐ Grade 9 Model Answers

Full annotated solution to Q5 — the highest-demand question, combining 3D Pythagoras with trigonometry.

Q5 Full Model Answer

Part (a): Find the perpendicular height

Step 1 — Draw and label the 3D diagram Visualisation

Label the rectangular base $ABCD$ with $AB = 12$ cm, $BC = 16$ cm. Let $O$ be the centre of the base. The apex $V$ is directly above $O$. Slant edge $VA = 18$ cm.

Step 2 — Find diagonal of base M1

$$AC = \sqrt{AB^2 + BC^2} = \sqrt{12^2 + 16^2} = \sqrt{144 + 256} = \sqrt{400} = 20 \text{ cm}$$

Note: $(12, 16, 20) = 4 \times (3, 4, 5)$ — a scaled Pythagorean triple, giving an exact integer result. Recognising this speeds up the calculation.

Step 3 — Find distance from centre to corner M1

$$OA = \frac{AC}{2} = \frac{20}{2} = 10 \text{ cm}$$

This is because the diagonals of a rectangle bisect each other, so $O$ is the midpoint of $AC$.

Step 4 — Apply Pythagoras in triangle $VOA$ M1 A1

In right-angled triangle $VOA$: $VO$ = height $h$, $OA = 10$ cm, $VA = 18$ cm (hypotenuse).

$$h^2 = VA^2 - OA^2 = 18^2 - 10^2 = 324 - 100 = 224$$

$$h = \sqrt{224} = \sqrt{16 \times 14} = 4\sqrt{14} \text{ cm}$$

Both forms are acceptable; $4\sqrt{14}$ is fully simplified.

Part (b): Find the angle

Step 5 — Identify the correct triangle and angle M1

The angle the slant edge $VA$ makes with the base is angle $\angle VAO$ in triangle $VOA$. We know: adjacent side $OA = 10$ cm, hypotenuse $VA = 18$ cm.

Step 6 — Apply trigonometry A1

$$\cos(\angle VAO) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{10}{18} = \frac{5}{9}$$

$$\angle VAO = \cos^{-1}\!\left(\frac{5}{9}\right) = 56.251...° \approx 56.3°$$

Why this earns Grade 9: The answer requires two separate applications of Pythagoras (base diagonal, then height), plus a trigonometric calculation. At each stage, exact intermediate values are carried through to avoid rounding errors. The student identifies the correct triangle for the angle (triangle $VOA$, not the slant face triangle), and gives a justified conclusion for part (a) in surd form.

🎯
Exam Tip — Multi-Step Grade 9 Problems
In problems combining Pythagoras and trigonometry: (1) always find all lengths first using Pythagoras; (2) then apply trig. Never mix them in one equation. Carry exact surd values between steps to prevent rounding errors accumulating.

📋 Revision Sheet

Key Definitions
TermMeaning
HypotenuseLongest side; opposite 90°
Pythagorean tripleInteger set satisfying $a^2+b^2=c^2$
SurdIrrational root in exact form
ConverseUsing $a^2+b^2=c^2$ to prove a right angle
Space diagonalLongest diagonal inside a 3D solid
Slant heightFrom apex to midpoint of base edge
Essential Formulae

$$a^2 + b^2 = c^2$$

$$c = \sqrt{a^2 + b^2} \qquad a = \sqrt{c^2 - b^2}$$

$$d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$$

$$d_{cuboid} = \sqrt{l^2 + w^2 + h^2}$$

$$l_{cone} = \sqrt{r^2 + h^2}$$

Memory Hooks
  • 3-4-5: the simplest triple
  • Hypotenuse = PLUS, Shorter = MINUS
  • Converse needs a conclusion sentence
  • 3D = two-stage Pythagoras
  • Exact answer → leave as $\sqrt{\phantom{x}}$
  • Surd simplify: find biggest square factor
Exam Tips
  • Always label $a$, $b$, $c$ before calculating
  • Check if sides form a known triple first
  • 3D: draw the cross-section triangle separately
  • State the converse conclusion explicitly
  • Carry exact values through multi-step problems
  • Negative coordinates: square negatives carefully
  • "Exact form" = leave as surd, do not round

🔄 Flashcards

Click a card to reveal the answer. Work through all 15 before moving on.

✗ Common Mistakes

✗
Mistake 1 — Adding when finding a shorter side
What students do wrong: Writing $a^2 = c^2 + b^2$ instead of $a^2 = c^2 - b^2$ when the hypotenuse is given.
Why marks are lost: The final answer is larger than the hypotenuse, which is impossible — but students do not check this.
How to avoid it: Always ask "am I finding the hypotenuse (add) or a shorter side (subtract)?" before writing the equation.
✗
Mistake 2 — Confusing the hypotenuse
What students do wrong: Labelling the longest side as $c$ without checking it is opposite the right angle (e.g. in obtuse triangles, Pythagoras does not apply at all).
Why marks are lost: Applying Pythagoras to a non-right-angled triangle gives an incorrect equation from step 1.
How to avoid it: Always confirm the 90° angle is marked or stated before using $a^2 + b^2 = c^2$.
✗
Mistake 3 — Forgetting the converse conclusion
What students do wrong: Calculating $a^2 + b^2$ and $c^2$, showing they are equal, but not writing a conclusion sentence.
Why marks are lost: The final mark in a "show that / prove" question always requires a clear written conclusion: "Therefore the triangle is right-angled by the converse of Pythagoras."
How to avoid it: Practise writing the conclusion sentence every time — even if it feels obvious.
✗
Mistake 4 — Rounding intermediate values in 3D
What students do wrong: Finding the base diagonal (e.g. $\sqrt{200} \approx 14.1$), rounding it, then using the rounded value in the next Pythagoras step.
Why marks are lost: Final answer is inaccurate due to accumulated rounding error. The accuracy mark is withheld if the answer differs from exact by more than the permitted tolerance.
How to avoid it: Store the exact value ($\sqrt{200}$ or $10\sqrt{2}$) in your calculator's memory (STO) and use that in step 2.
✗
Mistake 5 — Squaring a negative coordinate difference incorrectly
What students do wrong: Computing $(x_2 - x_1)^2$ where one coordinate is negative: e.g. $(-3-2)^2 = -5^2 = -25$ (forgetting that squaring a negative gives a positive).
Why marks are lost: A negative value under a square root produces an error or an imaginary number — an obvious clue that something is wrong, but students miss it.
How to avoid it: Always square the bracket as a whole: $(-5)^2 = +25$.
✗
Mistake 6 — Using slant height instead of perpendicular height (and vice versa)
What students do wrong: In cone and pyramid problems, confusing the perpendicular height $h$ (from apex perpendicular to base) with the slant height $l$ (along the curved surface).
Why marks are lost: Using the wrong length in a volume, surface area, or further Pythagoras calculation propagates an error through all subsequent steps.
How to avoid it: Draw a cross-section triangle. Label: perpendicular height (vertical), radius/half-base (horizontal), slant height (hypotenuse). Identify which one the question gives and which it asks for.

✅ Final Checklist

Tick each item as you become confident. Aim for 100% before your exam.

  • I can state Pythagoras' theorem: $a^2 + b^2 = c^2$
  • I can identify the hypotenuse as the side opposite the right angle
  • I can find the hypotenuse given two shorter sides
  • I can find a shorter side given the hypotenuse and one side
  • I know the three main Pythagorean triples: (3,4,5), (5,12,13), (8,15,17)
  • I can use the converse of Pythagoras and write a proper conclusion
  • I can apply the distance formula to two coordinate points
  • I can find the space diagonal of a cuboid using $\sqrt{l^2+w^2+h^2}$
  • I can find the slant height of a cone using $\sqrt{r^2+h^2}$
  • I can find the perpendicular height of a square-based pyramid (two-stage)
  • I can leave answers in exact surd form and simplify surds
  • I can carry exact intermediate values through multi-step problems
  • I can combine Pythagoras with trigonometry in 3D problems
  • I recognise when a negative coordinate difference becomes positive when squared
  • I can prove a triangle is right-angled using the converse and explain which angle is 90°
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