Geometry and Measures
Shapes, angles, area, volume, transformations, vectors, and constructions.
In this topic
Angles
Angles are measured in degrees. Key angle facts: - Angles on a straight line sum to 180° - Angles at a point sum to 360° - Vertically opposite angles are equal - Angles in a triangle sum to 180° - Angles in a quadrilateral sum to 360°
Parallel line angle rules: - Alternate angles are equal (Z-angles) - Corresponding angles are equal (F-angles) - Co-interior angles sum to 180° (C-angles)
Key Points
- Learn all angle rules — they're frequently tested
- Always state the reason for each angle you find
- Interior angle of a regular n-gon = (n-2) × 180° / n
- Exterior angles of any polygon sum to 360°
Example Questions
2Calculate the interior angle of a regular octagon.
(8-2) × 180° / 8 = 1080° / 8 = 135°
[2 marks]
2Two angles on a straight line are (3x + 10)° and (2x + 20)°. Find x.
3x + 10 + 2x + 20 = 180 → 5x + 30 = 180 → 5x = 150 → x = 30
[2 marks]
Area and Perimeter
Perimeter is the total distance around a shape. Area is the space inside.
Key formulae: - Rectangle: A = l × w, P = 2(l + w) - Triangle: A = ½ × base × height - Parallelogram: A = base × perpendicular height - Trapezium: A = ½(a + b) × h - Circle: A = πr², C = 2πr = πd
Key Points
- Height must be perpendicular to the base
- For compound shapes, split into simpler shapes
- Semicircle area = ½πr², perimeter = πr + 2r
- Units: area in cm², m² etc.
Example Questions
2A trapezium has parallel sides 8cm and 12cm, with height 5cm. Find its area.
A = ½(8 + 12) × 5 = ½ × 20 × 5 = 50 cm²
[2 marks]
3Find the area and circumference of a circle with diameter 14cm. Give answers to 1 d.p.
r = 7cm. Area = π × 7² = 153.9 cm². Circumference = π × 14 = 44.0 cm
[3 marks]
Volume and Surface Area
Volume is the space inside a 3D shape. Surface area is the total area of all faces.
Key formulae: - Cuboid: V = lwh, SA = 2(lw + lh + wh) - Cylinder: V = πr²h, SA = 2πrh + 2πr² - Cone: V = ⅓πr²h, SA = πrl + πr² (l = slant height) - Sphere: V = ⁴⁄₃πr³, SA = 4πr² - Prism: V = cross-sectional area × length
Key Points
- Prism volume = area of cross-section × length
- Learn the sphere and cone formulae (given on formula sheet but know how to use them)
- Units: volume in cm³, m³; surface area in cm², m²
Example Questions
2A cylinder has radius 4cm and height 10cm. Find its volume to 1 d.p.
V = π × 4² × 10 = 160π = 502.7 cm³
[2 marks]
3A sphere has volume 288π cm³. Find its radius.
⁴⁄₃πr³ = 288π → r³ = 216 → r = 6 cm
[3 marks]
Pythagoras' Theorem
In a right-angled triangle, the square of the hypotenuse equals the sum of the squares of the other two sides:
a² + b² = c² (where c is the hypotenuse — the longest side, opposite the right angle)
To find the hypotenuse: c = √(a² + b²) To find a shorter side: a = √(c² - b²)
Key Points
- Only works in RIGHT-ANGLED triangles
- The hypotenuse is always opposite the right angle
- Can be used in 3D by applying it twice
- Pythagorean triples: 3,4,5 and 5,12,13 and 8,15,17
Example Questions
2A right-angled triangle has legs of 6cm and 8cm. Find the hypotenuse.
c² = 6² + 8² = 36 + 64 = 100. c = √100 = 10 cm
[2 marks]
2The hypotenuse of a right-angled triangle is 13cm. One leg is 5cm. Find the other leg.
a² = 13² - 5² = 169 - 25 = 144. a = 12 cm
[2 marks]
Trigonometry
Trigonometry connects angles and sides in right-angled triangles.
SOH CAH TOA: - sin θ = Opposite / Hypotenuse - cos θ = Adjacent / Hypotenuse - tan θ = Opposite / Adjacent
To find a side: rearrange the appropriate formula To find an angle: use the inverse function (sin⁻¹, cos⁻¹, tan⁻¹)
Key Points
- Label the triangle: Hypotenuse, Opposite, Adjacent relative to the angle
- Choose the ratio that uses the two sides you know (or need)
- For non-right-angled triangles: use sine rule or cosine rule
- Sine rule: a/sinA = b/sinB = c/sinC
Example Questions
2In a right-angled triangle, the side opposite 35° is 7cm. Find the hypotenuse.
sin 35° = 7/h → h = 7/sin 35° = 7/0.5736 = 12.2 cm (1 d.p.)
[2 marks]
2Find angle θ if the adjacent side is 8cm and the hypotenuse is 10cm.
cos θ = 8/10 = 0.8 → θ = cos⁻¹(0.8) = 36.9°
[2 marks]
Circles
A circle is a set of points equidistant from a centre. You need to know the key parts, formulae, and circle theorems.
Key parts: radius, diameter, circumference, chord, tangent, arc, sector, segment.
Formulae: - Circumference = πd = 2πr - Area = πr² - Arc length = (θ/360) × 2πr - Sector area = (θ/360) × πr²
Key Points
- A tangent is perpendicular to the radius at the point of contact
- The angle in a semicircle is always 90°
- Angles in the same segment are equal
- Opposite angles in a cyclic quadrilateral sum to 180°
- The angle at the centre is twice the angle at the circumference
- Two tangents from an external point are equal in length
Example Questions
4A sector has radius 10 cm and angle 72°. Calculate the arc length and the area of the sector.
Arc length = (72/360) × 2π × 10 = (1/5) × 20π = 4π = 12.6 cm (3 s.f.). Area = (72/360) × π × 100 = (1/5) × 100π = 20π = 62.8 cm² (3 s.f.)
[4 marks]
2A, B and C are points on a circle. The angle at the centre O for arc BC is 130°. Find the angle BAC.
The angle at the centre is twice the angle at the circumference. Angle BAC = 130° ÷ 2 = 65°
[2 marks]
Circle Theorems
Circle theorems are rules about angles formed by chords, tangents, and radii. You must learn all of them and be able to state the theorem as a reason.
Theorem 1: The angle at the centre is twice the angle at the circumference (when subtended by the same arc).
Theorem 2: The angle in a semicircle is 90°. Any angle inscribed in a semicircle (where the hypotenuse is the diameter) is a right angle.
Theorem 3: Angles in the same segment are equal. If two angles are subtended by the same chord and are on the same side, they are equal.
Theorem 4: Opposite angles of a cyclic quadrilateral sum to 180°. A cyclic quadrilateral has all four vertices on the circumference.
Theorem 5: A tangent to a circle is perpendicular to the radius at the point of contact.
Theorem 6: Two tangents from an external point are equal in length.
Theorem 7: The alternate segment theorem — the angle between a tangent and a chord equals the angle in the alternate segment.
Key Points
- Always state the theorem name when giving a reason
- Look for isosceles triangles formed by two radii
- The perpendicular from the centre to a chord bisects the chord
- These theorems appear in almost every GCSE Higher paper
Example Questions
2In a circle with centre O, points A, B and C lie on the circumference. Angle AOB = 104°. Find angle ACB. State the circle theorem you use.
Angle ACB = 104° ÷ 2 = 52°. The angle at the centre is twice the angle at the circumference (subtended by the same arc AB).
[2 marks]
2ABCD is a cyclic quadrilateral. Angle A = 85° and angle B = 110°. Find angles C and D.
Opposite angles in a cyclic quadrilateral sum to 180°. Angle C = 180° - 85° = 95°. Angle D = 180° - 110° = 70°.
[2 marks]
2A tangent at point P on a circle meets a chord PQ. The angle between the tangent and chord PQ is 55°. Find the angle PRQ, where R is a point on the major arc. State the theorem used.
Angle PRQ = 55°. By the alternate segment theorem, the angle between a tangent and a chord equals the angle in the alternate segment.
[2 marks]
4Prove that the angle in a semicircle is 90°.
Let the diameter be AB and P be any point on the circumference. Let O be the centre. OA = OB = OP = radius. Triangle OAP is isosceles: let angle OAP = angle OPA = α. Triangle OBP is isosceles: let angle OBP = angle OPB = β. In triangle APB: α + β + (α + β) = 180°. So 2(α + β) = 180° → α + β = 90°. Angle APB = α + β = 90°. ∎
[4 marks]
Arcs, Sectors and Segments
An arc is a portion of the circumference. A sector is the region between two radii and an arc (like a pizza slice). A segment is the region between a chord and an arc.
Arc length = (θ/360) × 2πr
Sector area = (θ/360) × πr²
Perimeter of a sector = arc length + 2 × radius
Segment area = sector area − triangle area For the triangle area, use: ½r²sinθ
Key Points
- θ is the angle at the centre in degrees
- For the perimeter of a sector, don't forget to add the two straight edges (radii)
- Segment = sector minus triangle
- Use ½r²sinθ for the triangle area when you know the angle and two sides (radii)
Example Questions
5A sector of a circle has radius 8 cm and angle 150°. (a) Calculate the arc length. (b) Calculate the area of the sector. (c) Calculate the perimeter of the sector.
(a) Arc = (150/360) × 2π × 8 = (5/12) × 16π = 20π/3 = 20.9 cm (3 s.f.) (b) Area = (150/360) × π × 64 = (5/12) × 64π = 320π/12 = 83.8 cm² (3 s.f.) (c) Perimeter = 20.9 + 8 + 8 = 36.9 cm (3 s.f.)
[5 marks]
4A chord AB divides a circle of radius 6 cm into two segments. The angle AOB at the centre is 120°. Calculate the area of the minor segment.
Sector area = (120/360) × π × 36 = 12π = 37.70 cm². Triangle area = ½ × 6² × sin 120° = 18 × (√3/2) = 9√3 = 15.59 cm². Segment area = 37.70 − 15.59 = 22.1 cm² (3 s.f.)
[4 marks]
Equation of a Circle (Higher)
The equation of a circle with centre (0, 0) and radius r is:
x² + y² = r²
The equation of a circle with centre (a, b) and radius r is:
(x − a)² + (y − b)² = r²
To find if a point lies on a circle, substitute into the equation. To find where a line meets a circle, solve simultaneously.
Key Points
- x² + y² = 25 is a circle centre (0,0), radius 5
- (x−3)² + (y+1)² = 16 is centre (3,−1), radius 4
- The tangent at a point is perpendicular to the radius at that point
- To find gradient of tangent: find gradient of radius, then use negative reciprocal
Example Questions
5A circle has equation x² + y² = 50. (a) Write down the radius of the circle. (b) Show that the point (5, 5) lies on the circle. (c) Find the equation of the tangent to the circle at (5, 5).
(a) r = √50 = 5√2 (b) 5² + 5² = 25 + 25 = 50 ✓ (c) Gradient of radius from (0,0) to (5,5) = 5/5 = 1. Tangent is perpendicular: gradient = −1. y − 5 = −1(x − 5) → y = −x + 10
[5 marks]
3A circle has centre (2, 3) and passes through the point (6, 6). Find the equation of the circle.
Radius = distance from (2,3) to (6,6) = √((6−2)² + (6−3)²) = √(16+9) = √25 = 5. Equation: (x−2)² + (y−3)² = 25
[3 marks]
Transformations
The four transformations at GCSE are:
1. Translation: sliding a shape by a vector (x, y) 2. Reflection: flipping a shape over a mirror line 3. Rotation: turning a shape about a centre point by an angle 4. Enlargement: making a shape bigger or smaller using a scale factor from a centre
Translations, reflections, and rotations are isometries (preserve size and shape). Enlargement changes the size but preserves the shape.
Key Points
- Translation: describe with a column vector
- Reflection: give the mirror line equation
- Rotation: state centre, angle, and direction
- Enlargement: state centre and scale factor
Example Questions
3Describe fully the single transformation that maps triangle A to triangle B (B is twice the size, 3 units right and 2 up from the same centre).
Enlargement, scale factor 2, centre of enlargement (specify from the diagram).
[3 marks]