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Mathematics · 4. Geometry and measurement

From diagram to geometric proof

Use definitions, angle relations, similarity and Pythagoras to reason rather than measure a sketch.

40 minutesContent version: 2.1
Start the interactive pathwayOpen the virtual experiment

Driving question

What turns a suggestive diagram into a proof valid in every case?

Curriculum coverage

Vietnam Mathematics 6–10; Cambridge IGCSE Geometry and mensuration.

Model scope and limitations

Diagrams need not be to scale; screen measurement can suggest a conjecture but cannot replace proof.

Learning objectives

  • Use angle relations, triangle properties and Laws of Cosines/Sines.
  • Apply Pythagoras and Heron’s formula for triangle area.
  • Present a clear statement–reason chain.

Prerequisite knowledge

  • Squares, square roots and length units.

Core knowledge

Pythagoras

In a right triangle, the hypotenuse square equals the sum of the leg squares.

Triangle laws and Heron’s formula

Law of Cosines: a² = b² + c² − 2bc·cos(A); Law of Sines: a/sin(A) = 2R. Area: S = (1/2)bc·sin(A) = √(p(p−a)(p−b)(p−c)) with semiperimeter p.

Dihedral angle & Area projection

The angle between two planes is found via lines perpendicular to the intersection line or area projection cos(φ) = S′/S (where S′ is the projection of area S).

Classical landmarks & Lines in plane triangles

Euler line: Orthocenter H, Centroid G and Circumcenter O satisfy vec(HG) = 2 vec(GO). 9-point Feuerbach circle has radius R/2. Simson line: Projections of circumcircle point P onto the three sides are always collinear.

Solid geometry: Distance & Circumscribed sphere radius

1. Distance from height foot A to side face (SBC) uses two-step perpendiculars: 1/h² = 1/SA² + 1/AK². 2. Circumscribed sphere radius: R = √(R_base² + (h/2)²) for edge-perpendicular pyramids; R = SA²/(2h) for regular pyramids; and composite formula for perpendicular-face pyramids.

Worked example

A right triangle has legs 6 and 8: c² = 6²+8² = 100, so c = 10, taking the positive root for a length.

Misconceptions to avoid

  • The hypotenuse is opposite the right angle, not merely the side that looks longest.
  • AAA proves similarity, not congruence.
  • Distance from an arbitrary point to a plane must be ratio-transferred to the height foot before applying the two-step inverse square formula.

Virtual experiment procedure: Pythagoras theorem proof

  1. Mark givens and the target; do not add assumptions from appearance.
  2. Choose a theorem whose conditions hold and state those conditions.
  3. Check units, root signs and plausible lengths/angles.

Safety and cautions

  • In dynamic geometry, drag points through varied and near-degenerate cases before trusting a conjecture.

Evidence to collect

  • A solution with marked diagram, reasoning chain and correctly unitised conclusion.
Open the virtual experiment
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Diagnostic check

1. Where is the hypotenuse?

  1. Opposite the right angle
  2. Adjacent to every acute angle
  3. Always horizontal
Show answer and explanation

Correct answer: Opposite the right angle

The hypotenuse is defined as the side opposite 90°.

2. Angles in a plane triangle sum to?

  1. 180°
  2. 360°
  3. 90°
Show answer and explanation

Correct answer: 180°

In plane Euclidean geometry, the sum is 180°.

Application practice

1. A right triangle has hypotenuse 13 and one leg 5. Other leg?

  1. 12
  2. 18
  3. 8
Show answer and explanation

Correct answer: 12

√(13²−5²) = √144 = 12.

2. A triangle with sides a=13, b=14, c=15 has what Heron area?

  1. 84
  2. 91
  3. 105
Show answer and explanation

Correct answer: 84

Semiperimeter p = 21. S = √(21·8·7·6) = √7056 = 84.

3. A polygon of area S = 40 in (P) projects to area S′ = 20 on (Q). What is the angle between (P) and (Q)?

  1. 60°
  2. 30°
  3. 45°
Show answer and explanation

Correct answer: 60°

Using area projection: cos(φ) = S′/S = 20/40 = 1/2, hence φ = 60°.

4. Pyramid S.ABC has SA perpendicular to base, SA = 4, ABC right at B with AB = 3. What is the distance from A to plane (SBC)?

  1. 2.4
  2. 3.5
  3. 5
Show answer and explanation

Correct answer: 2.4

Since AB ⊥ BC, step 1 is AB. Step 2: AH ⊥ SB. Thus d(A, (SBC)) = (SA · AB) / √(SA² + AB²) = (4 · 3)/5 = 2.4.

5. On the Euler line of a triangle, what is the relation between Orthocenter H, Centroid G, and Circumcenter O?

  1. vec(HG) = 2 vec(GO)
  2. vec(HO) = 2 vec(HG)
  3. vec(OG) = 2 vec(GH)
Show answer and explanation

Correct answer: vec(HG) = 2 vec(GO)

The Centroid G lies on segment HO such that HG = 2 GO, meaning vec(HG) = 2 vec(GO).

Academic references

  1. Cambridge IGCSE Mathematics 0580 — official syllabus and resources
  2. Cambridge AS & A Level Mathematics 9709 — official syllabus
  3. OpenStax Precalculus 2e — functions, geometry and probability

This lesson is maintained against the cited sources below. Simulations are learning models and do not replace supervised physical-laboratory safety procedures.

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