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]
2Calculate the size of angle A in a triangle where angle B is 50 degrees and angle C is 60 degrees.
Angle A = 180 - (50 + 60) = 70 degrees
[2 marks]
1In a right-angled triangle, one angle is 30 degrees. What is the size of the other non-right angle?
The other non-right angle = 90 - 30 = 60 degrees
[1 mark]
2Two angles are supplementary. One angle measures 120 degrees. Find the measure of the other angle.
The other angle = 180 - 120 = 60 degrees
[2 marks]
3In a quadrilateral, three angles are 85 degrees, 95 degrees, and 70 degrees. Calculate the fourth angle.
Fourth angle = 360 - (85 + 95 + 70) = 110 degrees
[3 marks]
4Prove that the internal angles of a triangle always sum to 180 degrees.
Let angles be A, B, and C. By the triangle angle sum property, A + B + C = 180 degrees. This holds true for all triangles.
[4 marks]
4In a circle, angle A is inscribed in a semicircle. If angle A measures x degrees, show that angle A = 90 degrees.
By the inscribed angle theorem, an angle inscribed in a semicircle is always 90 degrees. Therefore, x = 90 degrees.
[4 marks]
3In a parallelogram, one angle is 70 degrees. Calculate the other three angles.
Opposite angle = 70 degrees. Adjacent angles = 180 - 70 = 110 degrees. So, angles are 70 degrees, 110 degrees, 70 degrees, 110 degrees.
[3 marks]
5A transversal crosses two parallel lines creating alternate angles. If one alternate angle is (3x + 15) degrees and the other is (2x + 45) degrees, find x and hence find the size of the angles.
Set (3x + 15) = (2x + 45). Solving gives x = 30. Substitute x back to find angles: (3(30) + 15) = 105 degrees and (2(30) + 45) = 105 degrees.
[5 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]
1Calculate the area of a rectangle with a length of 8 cm and a width of 5 cm.
Area = length × width = 8 cm × 5 cm = 40 cm²
[1 mark]
1Find the perimeter of a triangle with side lengths 6 cm, 4 cm, and 5 cm.
Perimeter = 6 cm + 4 cm + 5 cm = 15 cm
[1 mark]
2A circle has a radius of 7 cm. Calculate its circumference. Use π ≈ 3.14.
Circumference = 2 × π × radius = 2 × 3.14 × 7 cm ≈ 43.96 cm
[2 marks]
2Show that the area of a triangle with a base of 10 cm and a height of 6 cm is 30 cm².
Area = 1/2 × base × height = 1/2 × 10 cm × 6 cm = 30 cm²
[2 marks]
3A rectangular garden has a length of 12 m and a width of 9 m. If a path of width 1 m is added around the garden, calculate the new area of the garden including the path.
New length = 12 m + 2(1 m) = 14 m; New width = 9 m + 2(1 m) = 11 m. New area = 14 m × 11 m = 154 m².
[3 marks]
3A square has a perimeter of 48 cm. Calculate its area.
Side length = Perimeter / 4 = 48 cm / 4 = 12 cm; Area = side length² = 12 cm × 12 cm = 144 cm².
[3 marks]
4A rectangular swimming pool measures 25 m by 10 m. A deck surrounds the pool, and the total area of the deck is 180 m². Find the width of the deck.
Let the width of the deck be x. The dimensions of the outer rectangle = (25 + 2x) by (10 + 2x). Area of outer rectangle = (25 + 2x)(10 + 2x) = 250 + 50x + 20x + 4x² = 250 + 70x + 4x². Set up the equation: 250 + 70x + 4x² - 250 = 180; 4x² + 70x - 180 = 0. Use the quadratic formula: x = [−b ± √(b² - 4ac)] / 2a. Calculate to find x = 2.5 m.
[4 marks]
5Prove that the area of a trapezium with parallel sides of length a and b, and height h, is given by the formula A = (1/2)(a + b)h.
To prove this, consider dividing the trapezium into a rectangle and two right triangles. The area of the rectangle is base × height = b × h, and the area of the triangles is (1/2)(a - b) × h. Adding these areas shows that the total area is A = (1/2)(a + b)h.
[5 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]
1Calculate the volume of a cube with a side length of 5 cm.
The volume of a cube is calculated using the formula V = s³, where s is the side length. V = 5³ = 125 cm³.
[1 mark]
2A cylinder has a radius of 3 cm and a height of 10 cm. Calculate its volume.
The volume of a cylinder is calculated using the formula V = πr²h. V = π(3)²(10) = 90π cm³, which is approximately 282.74 cm³.
[2 marks]
2The surface area of a sphere is given as 113.04 cm². Show that the radius of the sphere is approximately 3 cm.
The surface area of a sphere is given by the formula A = 4πr². Setting 4πr² = 113.04, we solve for r: r² = 113.04 / (4π) ≈ 9, thus r ≈ 3 cm.
[2 marks]
3A rectangular prism has dimensions 4 cm, 3 cm, and 10 cm. Calculate its surface area.
The surface area of a rectangular prism is given by the formula SA = 2(lw + lh + wh). SA = 2(4*3 + 4*10 + 3*10) = 2(12 + 40 + 30) = 2(82) = 164 cm².
[3 marks]
4A cone has a radius of 4 cm and a height of 9 cm. Calculate the volume of the cone. Give your answer in terms of π.
The volume of a cone is V = (1/3)πr²h. V = (1/3)π(4)²(9) = (1/3)π(16)(9) = 48π cm³.
[4 marks]
4A cylinder and a cone have the same base radius of 5 cm and the same height of 12 cm. Find the ratio of their volumes.
Volume of cylinder = πr²h = π(5)²(12) = 300π cm³; Volume of cone = (1/3)πr²h = (1/3)π(5)²(12) = 100π cm³. Ratio of volumes = 300π : 100π = 3:1.
[4 marks]
5A hemisphere has a radius of 6 cm. Calculate the total surface area of the hemisphere including its base.
Total surface area of a hemisphere = 3πr². SA = 3π(6)² = 3π(36) = 108π cm². Therefore, total surface area = approximately 339.29 cm².
[5 marks]
6A cylinder has a height of 15 cm and a volume of 300π cm³. Calculate the radius of the cylinder. Show all your workings.
Volume of a cylinder is V = πr²h. Setting 300π = πr²(15), we divide both sides by π: 300 = 15r². Thus, r² = 300/15 = 20, so r = √20 = 2√5 cm.
[6 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]
2A right-angled triangle has one side of length 3 cm and another side of length 4 cm. Calculate the length of the hypotenuse.
Using Pythagoras' Theorem: c² = a² + b² c² = 3² + 4² c² = 9 + 16 c² = 25 c = √25 = 5 cm.
[2 marks]
3A ladder leans against a wall forming a right triangle with the ground. If the foot of the ladder is 6 m away from the wall and the ladder is 10 m long, how high up the wall does the ladder reach? Calculate the height.
Using Pythagoras' Theorem: c² = a² + b² 10² = 6² + h² 100 = 36 + h² h² = 100 - 36 h² = 64 h = √64 = 8 m.
[3 marks]
3Prove that a triangle with sides of lengths 5 cm, 12 cm, and 13 cm is a right-angled triangle.
Check if 5² + 12² = 13². 5² = 25, 12² = 144, 13² = 169. 25 + 144 = 169. Since 5² + 12² = 13², the triangle is right-angled.
[3 marks]
2A rectangular garden measures 8 m by 15 m. Find the length of the diagonal of the garden.
Using Pythagoras' Theorem: c² = a² + b² c² = 8² + 15² c² = 64 + 225 c² = 289 c = √289 = 17 m.
[2 marks]
3A right-angled triangle has one leg measuring 9 cm and a hypotenuse measuring 15 cm. Find the length of the other leg.
Using Pythagoras' Theorem: c² = a² + b² 15² = 9² + b² 225 = 81 + b² b² = 225 - 81 b² = 144 b = √144 = 12 cm.
[3 marks]
4A triangular park has sides measuring 7 m, 24 m, and 25 m. Show that this is a right-angled triangle.
Check if 7² + 24² = 25². 7² = 49, 24² = 576, 25² = 625. 49 + 576 = 625. Since 7² + 24² = 25², the triangle is right-angled.
[4 marks]
5In a right-angled triangle, the lengths of the legs are in the ratio 3:4. If the hypotenuse is 10 cm, find the lengths of the legs.
Let the legs be 3x and 4x. Using Pythagoras' Theorem: (3x)² + (4x)² = 10² 9x² + 16x² = 100 25x² = 100 x² = 4 x = 2. Thus, the lengths of the legs are 3(2) = 6 cm and 4(2) = 8 cm.
[5 marks]
5A right triangle has an area of 30 cm². If one leg is 10 cm long, calculate the length of the other leg. Also, find the length of the hypotenuse.
Area = (1/2) * base * height = 30 (1/2) * 10 * h = 30 5h = 30 h = 6 cm. Now, use Pythagoras' Theorem: c² = 10² + 6² c² = 100 + 36 c² = 136 c = √136 = 2√34 cm.
[5 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)
The graphs of sin, cos and tan are shown below. Key features: - sin x and cos x oscillate between -1 and 1 with period 360° - cos x is the same shape as sin x but shifted 90° to the left - tan x has asymptotes at 90° and 270° and period 180°
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]
2A right-angled triangle has one angle measuring 30 degrees and the side opposite this angle is 5 cm long. Calculate the length of the hypotenuse.
Using the sine function, sin(30) = opposite/hypotenuse. Therefore, 1/2 = 5/hypotenuse. Hypotenuse = 5 / (1/2) = 10 cm.
[2 marks]
2A right triangle has angles A, B, and the right angle. If angle A is 45 degrees and the length of the adjacent side to angle A is 8 cm, find the length of the opposite side.
Using tan(A) = opposite/adjacent, tan(45) = 1. Therefore, 1 = opposite/8, so opposite = 8 cm.
[2 marks]
3In triangle ABC, angle A = 60 degrees and side a (opposite angle A) = 12 cm. Calculate the length of side b (opposite angle B) if angle B = 30 degrees.
Using the sine rule: a/sin(A) = b/sin(B). Therefore, 12/sin(60) = b/sin(30). b = (12 * sin(30)) / sin(60) = (12 * 1/2) / (√3/2) = 12/√3 = 4√3 cm.
[3 marks]
3A ladder leans against a wall, forming an angle of 70 degrees with the ground. If the base of the ladder is 3 m away from the wall, how long is the ladder?
Using cosine: cos(70) = adjacent/hypotenuse. Therefore, hypotenuse = adjacent/cos(70) = 3/cos(70) ≈ 10.68 m.
[3 marks]
4Show that in a right triangle, if one angle is 45 degrees, both the other two sides are equal.
In a right triangle with angle A = 45 degrees, angle B = 45 degrees (since the angles sum to 180 degrees). Using tan(45) = opposite/adjacent, both sides opposite and adjacent to 45 degrees must be equal, hence showing that the two sides are equal.
[4 marks]
4A ship is sailing 10 km north and then turns 60 degrees east of north for another 5 km. Calculate the distance from the starting point.
Using the cosine rule: distance = √(10² + 5² - 2*10*5*cos(60)) = √(100 + 25 - 50) = √75 = 5√3 km ≈ 8.66 km from the starting point.
[4 marks]
5Prove that for any right-angled triangle, the relationship a² + b² = c² holds true using the definition of sine, cosine, and tangent.
Consider a right triangle with sides a, b, and hypotenuse c. By definition, sin² + cos² = 1. From Pythagoras' theorem, a² + b² = c² can be derived through geometric interpretation of the sine and cosine relationships as segments of the triangle formed in a unit circle. The proof follows using triangle decomposition and area comparisons.
[5 marks]
5A 12 m high building casts a shadow of 8 m. Calculate the angle of elevation of the sun from the tip of the shadow.
Using tan(angle) = opposite/adjacent, tan(angle) = 12/8 = 3/2. Therefore, angle = arctan(3/2) ≈ 56.31 degrees.
[5 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]
2A circle has a radius of 5 cm. Calculate the circumference of the circle. Use π ≈ 3.14.
Circumference = 2 × π × radius = 2 × 3.14 × 5 = 31.4 cm.
[2 marks]
3A circle has a diameter of 12 cm. Calculate the area of the circle. Use π ≈ 3.14.
Radius = diameter / 2 = 12 / 2 = 6 cm. Area = π × radius² = 3.14 × 6² = 3.14 × 36 = 113.04 cm².
[3 marks]
4The radius of a circle is increased from 3 cm to 7 cm. Calculate the increase in the area of the circle.
Area before = π × 3² = 3.14 × 9 = 28.26 cm². Area after = π × 7² = 3.14 × 49 = 153.86 cm². Increase = 153.86 - 28.26 = 125.6 cm².
[4 marks]
4Prove that the angle subtended at the circumference by a diameter is a right angle.
Let O be the center and A and B the endpoints of the diameter. Any point C on the circumference creates triangle ABC. By the inscribed angle theorem, angle ACB is half of angle AOB. Since AOB is 180°, angle ACB = 90°. Therefore, angle ACB is a right angle.
[4 marks]
3A circle is inscribed in a square. If the area of the square is 64 cm², find the radius of the circle.
Side of the square = √64 = 8 cm. Diameter of the circle = side of the square. Radius = diameter / 2 = 8 / 2 = 4 cm.
[3 marks]
4Find the length of an arc of a circle with a radius of 10 cm that subtends an angle of 60 degrees at the center.
Arc length = (θ/360) × 2πr = (60/360) × 2 × π × 10 = (1/6) × 20π = (10/3)π ≈ 10.47 cm.
[4 marks]
5A circle has a radius of 8 cm. A chord is drawn that is 10 cm long. Find the distance from the center of the circle to the chord.
Let d be the distance from the center to the chord. Using Pythagoras in the triangle formed: (8² = d² + (10/2)²) => 64 = d² + 25 => d² = 39 => d = √39 ≈ 6.24 cm.
[5 marks]
5The circle with equation x² + y² = 49 is reflected in the y-axis. Write the equation of the new circle and find its center and radius.
The new circle's equation is x² + y² = 49 (reflection does not change the equation). Center = (0, 0), radius = √49 = 7.
[5 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]
1In a circle, the angle subtended at the center is 80 degrees. What is the angle subtended at the circumference on the same arc?
The angle at the circumference is half the angle at the center. Therefore, the angle is 80/2 = 40 degrees.
[1 mark]
2Prove that the opposite angles of a cyclic quadrilateral are supplementary (add up to 180 degrees).
Let ABCD be a cyclic quadrilateral. Angles A and C are subtended by the same arc BD, so angle A + angle C = 180 degrees. Similarly, angles B and D are subtended by the same arc AC, so angle B + angle D = 180 degrees. Therefore, opposite angles are supplementary.
[2 marks]
3In a circle, angle AOB is 60 degrees and point C lies on the circumference. Find the value of angles ACB and ABC.
Angle ACB = 30 degrees (half of angle AOB). Since triangle ABC is isosceles (AB = AC), angle ABC = (180 - 30) / 2 = 75 degrees. Therefore, angle ACB = 30 degrees and angle ABC = 75 degrees.
[3 marks]
4A circle has a radius of 10 cm. A chord in the circle is 12 cm long. Find the distance from the center of the circle to the chord.
Let O be the center and AB be the chord. Draw OM perpendicular to AB, where M is the midpoint of AB. Then AM = 6 cm. Using Pythagoras' theorem: OM² + AM² = OA². OM² + 6² = 10². OM² + 36 = 100. Therefore, OM² = 64, so OM = √64 = 8 cm.
[4 marks]
5A circle has a diameter of 14 cm. Point P lies on the circumference. A tangent is drawn from point P to point A on the circle. If angle APB is 50 degrees, find the angle PAB.
Since AP is a tangent and AB is a radius, angle PAB = 90 degrees. Using angle properties, angle APB + angle PAB = 90 degrees. Therefore, angle PAB = 90 - 50 = 40 degrees.
[5 marks]
6In a circle, points A, B, C, and D lie on the circumference. Angle ACB = 45 degrees and angle ADB = 75 degrees. Prove that points A, B, C, and D are concyclic.
To prove A, B, C, D are concyclic, we need to show that angles ACB and ADB are opposite angles. Since angle ADB = 75 degrees, and the angle ACB = 45 degrees, notice that angle ACB + angle ADB = 45 + 75 = 120 degrees. The angles subtended from points A and B on the opposite sides of the circle must add up to 180 degrees. Thus, A, B, C, and D are cyclic.
[6 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]
2A circle has a radius of 7 cm. Calculate the length of an arc that subtends a central angle of 60 degrees.
The formula for the length of an arc is (θ/360) * 2πr. Here, θ = 60 and r = 7. Length of arc = (60/360) * 2 * π * 7 = (1/6) * 14π = 14π/6 = 7π/3 cm. Therefore, the length of the arc is approximately 7.33 cm.
[2 marks]
2A sector of a circle has a radius of 10 cm and an angle of 90 degrees. Find the area of the sector.
The area of a sector is given by (θ/360) * πr². Here, θ = 90 and r = 10. Area = (90/360) * π * 10² = (1/4) * π * 100 = 25π cm². Therefore, the area of the sector is approximately 78.54 cm².
[2 marks]
4A circle has a diameter of 12 cm. Find the area of the circle. Then find the area of a sector with a central angle of 180 degrees.
The radius r = 12/2 = 6 cm. Area of the circle = πr² = π * 6² = 36π cm². Area of the sector with 180 degrees = (180/360) * π * 6² = (1/2) * 36π = 18π cm². Therefore, the area of the sector is approximately 56.55 cm².
[4 marks]
3Explain how to find the area of a segment of a circle with a radius of 8 cm and a central angle of 120 degrees.
To find the area of the segment, first find the area of the sector using the formula (θ/360) * πr². Then, find the area of the triangle formed by the two radii and the chord using 1/2 * r² * sin(θ). Subtract the area of the triangle from the area of the sector to get the area of the segment.
[3 marks]
4A sector has a radius of 5 m and an angle of 270 degrees. Calculate the area of the sector and the length of the arc.
Area of the sector = (270/360) * π * 5² = (3/4) * π * 25 = 18.75π m². Length of the arc = (270/360) * 2π * 5 = (3/4) * 10π = 7.5π m. Therefore, area ≈ 58.90 m² and arc length ≈ 23.56 m.
[4 marks]
4A circle has a radius of 10 cm. Show that the area of a segment with a central angle of 60 degrees is 25π/3 cm².
Area of the sector = (60/360) * π * 10² = (1/6) * 100π = 50π/3 cm². Area of the triangle = 1/2 * r² * sin(θ) = 1/2 * 10² * sin(60) = 50√3/4 cm². Area of the segment = Area of sector - Area of triangle = (50π/3 - 50√3/4) cm². Check for correctness.
[4 marks]
5A circle has a radius of 15 cm. Calculate the area of the segment formed when the central angle is 150 degrees.
Area of the sector = (150/360) * π * 15² = (5/12) * 225π = 93.75π cm². Area of triangle = 1/2 * 15² * sin(150) = 1/2 * 225 * 1/2 = 56.25 cm². Area of segment = Area of sector - Area of triangle = 93.75π - 56.25 cm². Approximate area ≈ 235.62 cm².
[5 marks]
6A circle has a radius of 12 cm. Find the area of the segment when the central angle is 150 degrees. Also, explain the steps taken to find the answer.
Area of the sector = (150/360) * π * 12² = (5/12) * 144π = 60π cm². Area of triangle = 1/2 * 12² * sin(150) = 1/2 * 144 * 1/2 = 36 cm². Area of segment = Area of sector - Area of triangle = 60π - 36 cm². The area of the segment is thus approximately 60π - 36 cm². Explain that you first calculate the area of the sector using the formula, then the area of the triangle, and subtract to find the segment area.
[6 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]
2What is the standard form of the equation of a circle with center at (3, -2) and a radius of 5?
(x - 3)² + (y + 2)² = 25
[2 marks]
1A circle has the equation x² + y² = 36. Find the radius of the circle.
The radius is √36 = 6.
[1 mark]
3Show that the point (4, 3) lies on the circle with the equation (x - 2)² + (y + 1)² = 25.
Substituting (4, 3): (4 - 2)² + (3 + 1)² = 2² + 4² = 4 + 16 = 20, which is not equal to 25, so the point does not lie on the circle.
[3 marks]
4Find the center and radius of the circle given by the equation x² + y² - 6x + 8y + 9 = 0.
Rearranging gives (x - 3)² + (y + 4)² = 4. The center is (3, -4) and the radius is 2.
[4 marks]
3The circle in standard form has the equation (x + 1)² + (y - 2)² = 16. Find the coordinates of the center and the length of the diameter.
The center is (-1, 2) and the diameter is 16 (2 * radius 4).
[3 marks]
5Prove that the points (2, 1), (2, 5), and (6, 3) are all on the circle defined by the equation (x - 4)² + (y - 3)² = 4.
For (2, 1): (2 - 4)² + (1 - 3)² = 4 + 4 = 8 (not on circle). For (2, 5): (2 - 4)² + (5 - 3)² = 4 + 4 = 8 (not on circle). For (6, 3): (6 - 4)² + (3 - 3)² = 4 + 0 = 4 (on circle). Thus, only (6, 3) is on the circle.
[5 marks]
4Find the equation of a circle that passes through the origin and has its center at (3, -1).
Using the center (3, -1) and the radius to the origin (distance = √(3² + (-1)²) = √10), the equation is (x - 3)² + (y + 1)² = 10.
[4 marks]
2Given the circle (x - a)² + (y - b)² = r², explain how you would find the coordinates of the center and the radius.
The center is (a, b) and the radius is r, from the standard form of the circle's equation.
[2 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]
1A triangle has vertices at A(1, 2), B(3, 4), and C(5, 2). Translate the triangle 4 units to the right and 2 units up. What are the new coordinates of vertex A?
A' (5, 4)
[1 mark]
2A rectangle has corners at (2, 3), (2, 7), (6, 3), and (6, 7). Reflect the rectangle in the line y = 5. What are the coordinates of the new corners?
The new corners are (2, 3), (2, 7), (6, 3), and (6, 7) reflect to (2, 7), (2, 3), (6, 7), (6, 3).
[2 marks]
3A quadrilateral is defined by the points A(2, 1), B(4, 3), C(6, 1), and D(4, -1). Perform a rotation of 90 degrees clockwise about the origin. What are the new coordinates of all vertices?
A'(-1, 2), B'(-3, 4), C'(-1, 6), D'(1, 4)
[3 marks]
4Show that the shape formed by the points (1, 1), (1, 4), (3, 4), and (3, 1) remains congruent after a translation of 2 units left and 3 units down.
New points: (1-2, 1-3) = (-1, -2), (1-2, 4-3) = (-1, 1), (3-2, 4-3) = (1, 1), (3-2, 1-3) = (1, -2). All distances between points remain the same, hence the shape is congruent.
[4 marks]
3An isosceles triangle has vertices at A(2, 1), B(4, 5), and C(6, 1). Scale the triangle by a factor of 2 from the origin. What are the new coordinates of vertices A, B, and C?
A'(4, 2), B'(8, 10), C'(12, 2)
[3 marks]
1A shape is reflected over the line y = x. If one vertex of the shape is at (4, 2), what are the coordinates of the reflected vertex?
(2, 4)
[1 mark]
4A square has vertices at A(1, 1), B(1, 5), C(5, 5), and D(5, 1). Perform a rotation of 180 degrees about the center of the square. Find the coordinates of the new vertices.
New coordinates: A'(5, 5), B'(5, 1), C'(1, 1), D'(1, 5)
[4 marks]
4A triangle has vertices at A(1, 1), B(3, 1), and C(2, 4). Explain how you would perform a dilation with a scale factor of 3 from the origin and provide the coordinates of the triangle after the transformation.
To perform a dilation with scale factor 3, multiply each coordinate by 3. A'(3, 3), B'(9, 3), C'(6, 12).
[4 marks]
5Prove that transforming a triangle by a translation followed by a reflection results in a shape that is congruent to the original triangle. Use the vertices A(0, 0), B(2, 0), C(1, 1) as an example, translating it by (3, 2) then reflecting in the line x = 2.
After translation: A'(3, 2), B'(5, 2), C'(4, 3). Reflecting these points over x = 2 gives A''(1, 2), B''(-1, 2), C''(0, 3). Distances between the points remain the same as all transformations preserve distance, thus the new triangle is congruent to the original.
[5 marks]