135 free, high-quality past-paper-style questions on Geometry for Cambridge IGCSE Maths, each with a full mark scheme. Covers: Similarity, congruence and symmetry; Angle properties of lines, triangles and polygons; Circle theorems. For when you have finished the past papers, or want to practise one topic at a time.Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.753 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · How questions are checked
1IGMC-044-01-01Similarity, congruence and symmetryPaper CEasy3 marks
No calculator
Answer the following questions about symmetry.
(a)
Write down the number of lines of symmetry of a regular pentagon.
(1)
(b)
Write down the order of rotational symmetry of a rhombus that is not a square.
(1)
(c)
A regular polygon has rotational symmetry of order 9. Write down the number of lines of symmetry of this polygon.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 5
✓ 1
Part (b)
B1 2
✓ 1
Part (c)
B1 9
✓ 1
Common error: A rhombus has no lines of symmetry along its sides. It has two lines of symmetry, along its diagonals, and rotational symmetry of order 2. Do not mix up the number of lines with the order of rotation.
2IGMC-044-01-02Similarity, congruence and symmetryPaper CEasy4 marks
Calculator
Triangle ABC is mathematically similar to triangle PQR. A corresponds to P, B corresponds to Q and C corresponds to R. AB = 6 cm, BC = 8 cm and AC = 9 cm. PQ = 15 cm.
Diagram not to scale
(a)
Work out the scale factor of the enlargement from triangle ABC to triangle PQR.
(1)
(b)
Calculate the length of QR.
(2)
(c)
Calculate the length of PR.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 15 ÷ 6 = 2.5
✓ 1
Part (b)
M1 8 × 2.5 or 8 × 15 ÷ 6
✓ 1
A1 20 cm
✓ 1
Part (c)
B1 22.5 cm
✓ 1
Exam tip: Match the corresponding sides carefully using the order of the letters. Find the scale factor by dividing a length in the new shape by the matching length in the original shape.
3IGMC-044-01-03Similarity, congruence and symmetryPaper CEasy5 marks
No calculator
For each pair of triangles, decide whether the triangles must be congruent. Give a reason for each answer.
(a)
Triangle PQR has PQ = 5 cm, QR = 7 cm and PR = 9 cm. Triangle XYZ has XY = 9 cm, YZ = 5 cm and XZ = 7 cm.
(2)
(b)
Triangle DEF and triangle KLM each have angles of 40°, 60° and 80°.
(2)
(c)
Two right-angled triangles have hypotenuses of equal length, and one other pair of sides of equal length. Write down the condition of congruence that applies.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 Yes, congruent
✓ 1
B1 SSS: all three pairs of sides are equal
✓ 1
Part (b)
B1 Not necessarily congruent
✓ 1
B1 Equal angles only show the triangles are similar; the sides could be different lengths
✓ 1
Part (c)
B1 RHS (right angle, hypotenuse, side)
✓ 1
Common error: AAA is not a test for congruence. Three equal angles show that two triangles are similar. The triangles could still be different sizes.
4IGMC-044-01-04Similarity, congruence and symmetryPaper CEasy4 marks
Calculator
At a certain time of day, a vertical pole of height 1.8 m casts a shadow 2.4 m long on horizontal ground. At the same time, a nearby vertical tree casts a shadow 14 m long. The triangles formed by each object and its shadow are similar.
Diagram not to scale
(a)
Calculate the height of the tree.
(2)
(b)
At the same time, a person of height 1.6 m stands on the same ground. Calculate the length of the person's shadow.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 1.8 × 14 ÷ 2.4 or 14 × 0.75
✓ 1
A1 10.5 m
✓ 1
Part (b)
M1 1.6 × 2.4 ÷ 1.8
✓ 1
A1 2.13 m (2.1333…)
✓ 1
Exam tip: Write the ratio as height ÷ shadow = 0.75 for the pole. The same ratio applies to every object at that time, so this gives a quick check on both answers.
5IGMC-044-01-05Similarity, congruence and symmetryPaper CEasy4 marks
No calculator
Triangle ABC is isosceles with AB = AC. It is not equilateral. Angle ABC = 52° and BC = 10 cm. The line of symmetry of the triangle meets BC at M.
Diagram not to scale
(a)
Write down the number of lines of symmetry of triangle ABC.
(1)
(b)
Write down the length of BM.
(1)
(c)
Work out angle BAM.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 1
✓ 1
Part (b)
B1 5 cm
✓ 1
Part (c)
M1 angle BAC = 180 − 2 × 52 = 76°, then halved; OR angle AMB = 90° used, giving 180 − 90 − 52
✓ 1
A1 38°
✓ 1
Exam tip: The line of symmetry of an isosceles triangle bisects the apex angle. It is also the perpendicular bisector of the base, so angle AMB = 90°. Either fact leads to the answer, and using both lets you check it.
6IGMC-044-00-01Angle properties of lines, triangles and polygonsPaper CEasy3 marks
No calculator
Two straight lines PQ and RS intersect at the point O. Angle POR = (3x + 10)° and angle ROQ = (2x − 5)°.
Diagram not to scale
(a)
Find the value of x.
(2)
(b)
Find the size of angle QOS.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
3x + 10 + 2x − 5 = 180 (angles on a straight line) M1
✓ 1
x = 35 A1
✓ 1
Part (b)
115° B1 [vertically opposite to angle POR = 3 × 35 + 10]
✓ 1
Exam tip: Angle POR and angle QOS are vertically opposite because P is opposite Q and R is opposite S. Find x first, then substitute it back into the expression.
7IGMC-044-00-02Angle properties of lines, triangles and polygonsPaper CEasy4 marks
No calculator
This question is about the angles of polygons.
(a)
Find the size of one exterior angle of a regular 12-sided polygon.
(1)
(b)
Find the size of one interior angle of a regular 12-sided polygon.
(1)
(c)
Find the sum of the interior angles of a 9-sided polygon.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
360 ÷ 12 = 30° B1
✓ 1
Part (b)
180 − 30 = 150° B1
✓ 1
Part (c)
(9 − 2) × 180 M1
✓ 1
1260° A1
✓ 1
Common error: Using 9 × 180 instead of (9 − 2) × 180. A polygon with n sides splits into (n − 2) triangles from one vertex.
8IGMC-044-00-03Angle properties of lines, triangles and polygonsPaper CEasy3 marks
No calculator
In triangle ABC, AB = AC and angle BAC = 48°. The side BC is extended beyond C to the point D.
Diagram not to scale
(a)
Find angle ABC.
(2)
(b)
Find angle ACD.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
(180 − 48) ÷ 2 M1
✓ 1
66° A1
✓ 1
Part (b)
180 − 66 = 114° B1
✓ 1
Exam tip: In an isosceles triangle the two equal angles are opposite the two equal sides. Here AB = AC, so the equal angles are at B and C, not at A.
9IGMC-044-00-04Angle properties of lines, triangles and polygonsPaper CEasy3 marks
No calculator
The straight lines AB and CD are parallel. A and C are on the same side of the straight line EF. The straight line EF crosses AB at G and CD at H, with G between E and H. Angle AGH = 68°.
Diagram not to scale
(a)
Find angle GHD. Give a geometrical reason for your answer.
(2)
(b)
Find angle CHG.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
68° B1
✓ 1
Alternate angles B1
✓ 1
Part (b)
180 − 68 = 112° B1 [co-interior angles]
✓ 1
Common error: Writing 'Z angles' or 'F angles' as a reason. Use the correct terms: alternate angles, corresponding angles and co-interior angles.
10IGMC-044-00-05Angle properties of lines, triangles and polygonsPaper CEasy4 marks
No calculator
The four interior angles of a quadrilateral are x°, 2x°, (3x − 20)° and 110°.
Diagram not to scale
(a)
Find the value of x.
(3)
(b)
Find the size of the largest angle in the quadrilateral.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Angles in a quadrilateral sum to 360° B1
✓ 1
x + 2x + 3x − 20 + 110 = 360 M1
✓ 1
x = 45 A1
✓ 1
Part (b)
115° B1
✓ 1
Exam tip: Once you have found x, work out all four angles. The 110° angle is not the largest one here.
11IGMC-044-02-02Circle theoremsPaper CEasy6 marks
No calculator
AB is a diameter of a circle and C is a point on the circumference. Angle CAB = x° and angle CBA = (2x + 15)°.
Diagram not to scale
(a)
Write down the size of angle ACB. Give a reason for your answer.
(2)
(b)
Form an equation in x and solve it.
(3)
(c)
Find angle CBA.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1: 90°
✓ 1
B1: the angle in a semicircle is 90°
✓ 1
Part (b)
M1: x + 2x + 15 = 90, or x + 2x + 15 + 90 = 180
✓ 1
M1: 3x = 75
✓ 1
A1: x = 25
✓ 1
Part (c)
B1: 65°
✓ 1
Common error: Do not stop once you have x. Read what the question asks for. Here part (c) needs 2x + 15, not x.
12IGMC-044-02-04Circle theoremsPaper CEasy5 marks
No calculator
AB is a diameter of a circle. The line BD is the tangent to the circle at B. C is a point on the circle, and the straight line AC is extended to meet the tangent at D. Angle BAC = 35°.
Diagram not to scale
(a)
Write down angle ABD. Give a reason for your answer.
(2)
(b)
Find angle ADB.
(1)
(c)
Find angle CBD.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1: 90°
✓ 1
B1: a tangent is perpendicular to the radius (diameter AB) at the point of contact
✓ 1
Part (b)
B1: 180 − 90 − 35 = 55° (angles in triangle ABD)
✓ 1
Part (c)
M1: any correct method, e.g. angle ACB = 90° (angle in a semicircle) so angle BCD = 90° and 180 − 90 − 55; or angle ABC = 180 − 90 − 35 = 55° so 90 − 55; or angle CBD = angle BAC (alternate segment theorem)
✓ 1
A1: angle CBD = 35°
✓ 1
Exam tip: A diagram with a diameter and a tangent at one end often has two right angles. Find both of them before you start the angle chase.
13IGMC3-4-01d-01Circle theoremsPaper CEasy3 marks
No calculator
A roof truss is an isosceles triangle ABC with AB = AC. Angle BAC = 38°.
Diagram not to scale
(a)
Calculate angle ABC.
(2)
(b)
The beam BC is extended beyond C to a point D. Calculate angle ACD.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 (180 − 38) ÷ 2
✓ 1
A1 71°
✓ 1
Part (b)
B1 109° (ECF from (a): 180 − angle ACB)
✓ 1
Exam tip: In an isosceles triangle the two equal sides sit opposite the two equal angles, so find the equal pair before you subtract.
14IGMC3-4-01d-02Angle properties of lines, triangles and polygonsPaper CEasy4 marks
No calculator
A paving pattern is built from tiles that are regular polygons.
(a)
One tile is a regular polygon with 12 sides. Calculate the size of one exterior angle.
(1)
(b)
Calculate the size of one interior angle of this 12-sided tile.
(1)
(c)
A different regular polygon has exterior angles of 18°. Calculate its number of sides.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 30° (360 ÷ 12)
✓ 1
Part (b)
A1 150° (180 − 30)
✓ 1
Part (c)
M1 360 ÷ 18
✓ 1
A1 20
✓ 1
Exam tip: Exterior angles of any polygon add up to 360°, which makes a regular polygon quick to handle without the interior-angle formula.
15IGMC3-4-01d-03Angle properties of lines, triangles and polygonsPaper CEasy4 marks
No calculator
Two straight steel rails AB and CD of a solar-panel frame are parallel. A and C are on the left and B and D on the right, with CD below AB. A straight strut PQ joins P on AB to Q on CD. Angle BPQ = 54°.
Diagram not to scale
(a)
Write down the size of angle PQC and give a reason for your answer.
(2)
(b)
Calculate angle PQD.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 54°
✓ 1
B1 alternate angles (are equal)
✓ 1
Part (b)
M1 180 − 54 (co-interior angles add to 180°)
✓ 1
A1 126°
✓ 1
Common error: Alternate angles form a Z shape and co-interior angles form a C shape; mixing them up gives 126° where 54° is needed, or the other way round.
16IGMC3-4-01d-04Angle properties of lines, triangles and polygonsPaper CEasy6 marks
No calculator
A ceiling fan has five identical blades equally spaced around the central motor, so the fan has rotational symmetry.
(a)
Write down the order of rotational symmetry of the fan.
(1)
(b)
Calculate the smallest angle through which the fan can turn so that it looks unchanged.
(2)
(c)
The motor is held on by a nut shaped like a regular hexagon. Write down the number of lines of symmetry of the nut.
(1)
(d)
Calculate the size of one interior angle of the regular hexagon.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 5
✓ 1
Part (b)
M1 360 ÷ 5
✓ 1
A1 72°
✓ 1
Part (c)
B1 6
✓ 1
Part (d)
M1 360 ÷ 6 = 60 for the exterior angle (or (6 − 2) × 180 ÷ 6)
✓ 1
A1 120°
✓ 1
Exam tip: A regular polygon with n sides has n lines of symmetry and rotational symmetry of order n.
17IGMC3-4-01d-05Similarity, congruence and symmetryPaper CEasy5 marks
Calculator
A flagpole stands on level ground. At the same moment, a vertical stick of height 1.5 m casts a shadow 0.8 m long and the flagpole casts a shadow 4.8 m long.
Diagram not to scale
(a)
Calculate the height of the flagpole.
(3)
(b)
At the same moment a vertical post of height 6 m stands nearby. Calculate the length of its shadow.
Exam tip: Find the scale factor from a pair of corresponding lengths, then multiply every other length by it.
21IGMC3-4-08r8-07Circle theoremsPaper CEasy4 marks
No calculator
A circular lake has centre O. The points A and B are at opposite ends of a diameter, and the point C is on the shore of the lake. Angle CAB = 29°.
Diagram not to scale
(a)
Find angle ACB and give a reason.
(1)
(b)
Find angle CBA.
(2)
(c)
A straight path is tangent to the lake at B and meets the line AC extended at the point T. Find angle ATB.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 90° because the angle in a semicircle is a right angle
✓ 1
Part (b)
M1 180 − 90 − 29
✓ 1
A1 61°
✓ 1
Part (c)
A1 61° (angle ABT = 90° as the tangent is perpendicular to the diameter, so 180 − 90 − 29)
✓ 1
Common error: Do not assume an angle is 90° unless you can name the circle theorem that makes it so.
22IGMC3-4-11r30-01Angle properties of lines, triangles and polygonsPaper CEasy6 marks
No calculator
Honeybees build wax cells in the shape of a regular hexagon. A beekeeper studies one cell.
(a)
Calculate the size of one exterior angle of a regular hexagon.
(2)
(b)
Calculate the size of one interior angle of a regular hexagon.
(2)
(c)
Explain why three of these cells can meet at a point with no gaps.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 360 ÷ 6
✓ 1
A1 60°
✓ 1
Part (b)
M1 180 − 60 (ECF from (a))
✓ 1
A1 120°
✓ 1
Part (c)
B1 3 × 120° = 360°
✓ 1
B1 angles around a point add up to 360°
✓ 1
Exam tip: Interior angle plus exterior angle on a straight line always equals 180°, so you can move between them quickly.
23IGMC3-4-11r30-02Angle properties of lines, triangles and polygonsPaper CEasy4 marks
No calculator
A hospital ward bay has a floor in the shape of a pentagon. Four of its interior angles are 100°, 115°, 108° and 122°. The fifth interior angle is x°.
Diagram not to scale
(a)
Work out the sum of the interior angles of a pentagon.
(2)
(b)
Find the value of x.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 (5 − 2) × 180
✓ 1
A1 540°
✓ 1
Part (b)
M1 540 − (100 + 115 + 108 + 122) (ECF from (a))
✓ 1
A1 95
✓ 1
Common error: Using 360° (the exterior angle total) instead of the interior angle total of (n − 2) × 180°.
24IGMC3-4-11r30-03Angle properties of lines, triangles and polygonsPaper CEasy6 marks
No calculator
In a Formula 1 garage, two straight rails AB and CD are parallel. A third straight rail crosses AB at P and CD at Q. Points B and D are on the same side of PQ, and angle BPQ = 112°.
Diagram not to scale
(a)
Find angle PQD. Give a reason for your answer.
(3)
(b)
Find angle APQ. Give a reason for your answer.
(2)
(c)
Write down the name of the relationship between angle APQ and angle PQD.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 180 − 112
✓ 1
A1 68°
✓ 1
B1 co-interior (allied) angles between parallel lines add up to 180°
✓ 1
Part (b)
B1 68°
✓ 1
B1 angles on a straight line add up to 180°
✓ 1
Part (c)
B1 alternate angles (equal)
✓ 1
Exam tip: Name the angle property in words (for example co-interior or alternate) whenever the question asks for a reason.
25IGMC3-4-11r30-04Similarity, congruence and symmetryPaper CEasy4 marks
No calculator
A support strut on a Formula 1 front wing is a triangle ABC with AB = AC = 7.0 cm and angle BAC = 48°.
Diagram not to scale
(a)
Calculate angle ABC.
(2)
(b)
A second strut PQR has PQ = PR = 7.0 cm and angle QPR = 48°. State whether triangles ABC and PQR are congruent and give the reason.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 (180 − 48) ÷ 2
✓ 1
A1 66°
✓ 1
Part (b)
B1 congruent
✓ 1
B1 two sides and the included angle are equal (SAS)
✓ 1
Common error: In an isosceles triangle the equal angles are opposite the equal sides, so the angle between the equal sides is not one of the equal angles.
26IGMC3-4-12r33-01Angle properties of lines, triangles and polygonsPaper CEasy5 marks
No calculator
A triangular bracket supporting a solar panel has interior angles x°, (2x + 10)° and (3x − 10)°.
Diagram not to scale
(a)
Form and solve an equation to find x.
(3)
(b)
Write down the size of the largest interior angle.
(1)
(c)
Find the exterior angle at the vertex with the largest interior angle.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 x + (2x + 10) + (3x − 10) = 180
✓ 1
M1 6x = 180
✓ 1
A1 x = 30
✓ 1
Part (b)
B1 80°
✓ 1
Part (c)
B1 180° − 80° = 100° (ECF from (b))
✓ 1
Exam tip: Interior angles of a triangle sum to 180°; always substitute back to check that your three angles add up.
27IGMC3-4-12r33-02Angle properties of lines, triangles and polygonsPaper CEasy5 marks
No calculator
The floor pattern in a retail store uses tiles shaped as a regular polygon. Each exterior angle of the polygon is 24°.
(a)
Find the number of sides of the polygon.
(2)
(b)
Find the size of one interior angle.
(1)
(c)
Find the sum of the interior angles of the polygon.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 360 ÷ 24
✓ 1
A1 15
✓ 1
Part (b)
B1 180° − 24° = 156°
✓ 1
Part (c)
M1 (15 − 2) × 180 (or 15 × 156)
✓ 1
A1 2340°
✓ 1
Exam tip: Exterior angles of any polygon sum to 360°, so a regular polygon has 360 ÷ n degrees at each vertex.
28IGMC3-4-12r33-10Angle properties of lines, triangles and polygonsPaper CEasy4 marks
No calculator
A solar farm site is a hexagon. Five of its interior angles are 120°, 135°, 110°, 125° and 140°. The sixth interior angle is x°.
Common error: The interior angle sum is (n − 2) × 180°, not n × 180°.
29IGMC3-4-12r33-11Similarity, congruence and symmetryPaper CEasy4 marks
No calculator
A wind-turbine rotor has identical blades equally spaced around the hub.
(a)
A rotor has 3 blades. Find the angle between two adjacent blades.
(1)
(b)
Write down the order of rotational symmetry of a rotor with 5 blades.
(1)
(c)
Find the angle between two adjacent blades of a rotor with 5 blades.
(1)
(d)
Another rotor has an angle of 15° between adjacent blades. Find the number of blades.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 120°
✓ 1
Part (b)
B1 5
✓ 1
Part (c)
B1 72°
✓ 1
Part (d)
B1 360 ÷ 15 = 24
✓ 1
Exam tip: For n equally spaced blades, the angle between neighbours is 360° ÷ n and the order of rotational symmetry is n.
30IGMC3-4-10r37-01Angle properties of lines, triangles and polygonsPaper CEasy5 marks
No calculator
A railway station has a floor tile in the shape of a regular polygon. Each exterior angle of the polygon is 24°.
(a)
Work out the number of sides of the polygon.
(2)
(b)
Work out the size of each interior angle.
(1)
(c)
Work out the sum of the interior angles of the polygon.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 360 ÷ 24
✓ 1
A1 15
✓ 1
Part (b)
B1 180 − 24 = 156°
✓ 1
Part (c)
M1 (15 − 2) × 180 or 15 × 156
✓ 1
A1 2340°
✓ 1
Exam tip: Exterior angles of any polygon add up to 360°, so for a regular polygon divide 360° by the exterior angle to find the number of sides.
31IGMC3-4-10r37-02Angle properties of lines, triangles and polygonsPaper CEasy4 marks
No calculator
A triangular steel bracket on a railway signal gantry is shaped like triangle ABC with AB = AC. Angle ABC = (2x + 10)° and angle ACB = (3x − 20)°.
Diagram not to scale
(a)
Write down an equation in x and solve it to find x.
(2)
(b)
Work out angle BAC.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 2x + 10 = 3x − 20 (base angles of an isosceles triangle are equal)
✓ 1
A1 x = 30
✓ 1
Part (b)
M1 base angle = 2 × 30 + 10 = 70° and 180 − 2 × 70
✓ 1
A1 40°
✓ 1
Common error: In an isosceles triangle the equal angles are opposite the equal sides, so with AB = AC the equal angles are at B and C, not at A.
32IGMC3-4-10r37-08Angle properties of lines, triangles and polygonsPaper CEasy5 marks
Calculator
A sports pitch is in the shape of a pentagon. Four of its interior angles are 118°, 104°, 97° and 126°.
Diagram not to scale
(a)
Work out the sum of the interior angles of a pentagon.
(2)
(b)
Work out the size of the fifth interior angle.
(2)
(c)
Work out the exterior angle at the vertex where the interior angle is 126°.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 (5 − 2) × 180
✓ 1
A1 540°
✓ 1
Part (b)
M1 540 − (118 + 104 + 97 + 126) = 540 − 445
✓ 1
A1 95° (ECF from (a))
✓ 1
Part (c)
B1 54°
✓ 1
Exam tip: Check your answer by adding all five interior angles; the total must equal 540°.
33IGMC3-4-10r37-10Similarity, congruence and symmetryPaper CEasy5 marks
No calculator
Triangle ABC is part of a railway bridge truss, with AB = 10 m, angle BAC = 48° and angle ABC = 75°. Triangle PQR is a second truss with PQ = 10 m, angle QPR = 48° and angle PQR = 75°.
Diagram not to scale
(a)
Show that triangles ABC and PQR are congruent, stating the condition for congruence.
(2)
(b)
Work out angle PRQ.
(2)
(c)
Triangle XYZ has angles of 48°, 75° and 57°. A student says that triangle XYZ must be congruent to triangle ABC. Explain why the student may be wrong.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 angle BAC = angle QPR, angle ABC = angle PQR and AB = PQ
✓ 1
B1 two angles and the included (corresponding) side are equal, so congruent by ASA
✓ 1
Part (b)
M1 180 − 48 − 75
✓ 1
A1 57°
✓ 1
Part (c)
B1 equal angles only show the triangles are similar; no side length is given, so XYZ could be larger or smaller
✓ 1
Common error: Three equal angles (AAA) show similarity but do not prove congruence; at least one side must also match.
34IGMC3-4-10r41-01Angle properties of lines, triangles and polygonsPaper CEasy4 marks
No calculator
A triangular field has angles of (2x + 10)°, (3x − 5)° and (x + 25)°.
Diagram not to scale
(a)
Write down and solve an equation to find the value of x.
(3)
(b)
Work out the size of the smallest angle of the field.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 sum of the three angles equals 180
✓ 1
A1 6x + 30 = 180
✓ 1
A1 x = 25
✓ 1
Part (b)
B1 50°
✓ 1
Exam tip: Collect like terms carefully before solving: 2x + 3x + x = 6x and 10 − 5 + 25 = 30.
35IGMC3-4-10r41-02Angle properties of lines, triangles and polygonsPaper CEasy5 marks
Calculator
A sound-absorbing panel in a recording studio is a regular decagon (a polygon with 10 equal sides and 10 equal angles).
(a)
Calculate the size of one exterior angle of the decagon.
(2)
(b)
Calculate the size of one interior angle.
(1)
(c)
Calculate the sum of all the interior angles of the decagon.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 360 ÷ 10
✓ 1
A1 36°
✓ 1
Part (b)
B1 144° (ft from (a))
✓ 1
Part (c)
M1 (10 − 2) × 180 or 10 × 144
✓ 1
A1 1440°
✓ 1
Exam tip: Interior and exterior angles at a vertex of a polygon add up to 180°.
36IGMC3-4-10r41-03Similarity, congruence and symmetryPaper CEasy4 marks
No calculator
The wing of a paper aeroplane is a kite ABCD with AB = AD and CB = CD. Angle ABC is 110° and angle BAD is 80°.
Diagram not to scale
(a)
Write down the number of lines of symmetry of a kite.
(1)
(b)
Write down the order of rotational symmetry of a rhombus.
(1)
(c)
Work out the size of angle BCD.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 1
✓ 1
Part (b)
B1 2
✓ 1
Part (c)
M1 angle ADC = angle ABC = 110° (symmetry) and angles in a quadrilateral sum to 360
✓ 1
A1 360 − 110 − 110 − 80 = 60°
✓ 1
Common error: In a kite, the two equal angles are the ones between a short side and a long side, not the angles on the line of symmetry.
37IGMC3-4-10r41-12Angle properties of lines, triangles and polygonsPaper CEasy6 marks
No calculator
A circular irrigation sprinkler waters a field in five sectors. The angles of the sectors are 60°, 90°, x°, 2x° and 3x°.
Diagram not to scale
(a)
Write down and solve an equation to find x.
(3)
(b)
Write down the size of the largest sector angle.
(1)
(c)
Write the largest sector as a fraction of the full circle. Give your answer in its simplest form.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 angles at a point add up to 360, so 60 + 90 + x + 2x + 3x = 360
✓ 1
A1 6x + 150 = 360
✓ 1
A1 x = 35
✓ 1
Part (b)
B1 105°
✓ 1
Part (c)
M1 105 ÷ 360
✓ 1
A1 7/24
✓ 1
Exam tip: Angles around a point add up to 360°, and a fraction should be cancelled fully before you give it as the answer.
38IGMC3-4-12r44-01Angle properties of lines, triangles and polygonsPaper CEasy5 marks
No calculator
Three wind turbines A, B and C stand at the corners of a triangular site. Angle A = 48°. Angle B is three times as large as angle C.
(a)
Show that angle B + angle C = 132°.
(2)
(b)
Find the size of angle B.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 angles in a triangle add up to 180°
✓ 1
A1 180 − 48 = 132 shown
✓ 1
Part (b)
M1 4 × angle C = 132
✓ 1
A1 angle C = 33°
✓ 1
A1 angle B = 99°
✓ 1
Common error: Write the equation before solving. Here 4 × C = 132 comes from C + 3C, so do not divide by 3 or by 2.
39IGMC3-4-12r44-02Angle properties of lines, triangles and polygonsPaper CEasy6 marks
Calculator
A paddling pool in a swimming complex is in the shape of a regular polygon. Each exterior angle is 24°.
(a)
Work out the number of sides of the pool.
(2)
(b)
Work out the size of one interior angle.
(2)
(c)
Work out the sum of the interior angles of the pool.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 360 ÷ 24
✓ 1
A1 15
✓ 1
Part (b)
M1 180 − 24
✓ 1
A1 156°
✓ 1
Part (c)
M1 (15 − 2) × 180 or 15 × 156
✓ 1
A1 2340°
✓ 1
Exam tip: Exterior angles of any polygon add up to 360°. Interior and exterior angles at the same vertex add up to 180°.
40IGMC3-4-12r44-03Angle properties of lines, triangles and polygonsPaper CEasy4 marks
No calculator
Two lane lines AB and CD on a pool are parallel straight lines. A straight rope crosses them, meeting AB at G and CD at H. Points A and C are on the same side of the rope, and points B and D are on the other side. Angle AGH = 68°.
Diagram not to scale
(a)
Find angle GHC. Give a reason for your answer.
(2)
(b)
Find angle GHD. Give a reason for your answer.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 co-interior (allied) angles between parallel lines add up to 180°
✓ 1
A1 112°
✓ 1
Part (b)
B1 68°
✓ 1
B1 reason: alternate angles are equal (or angles CHD on a straight line add up to 180°)
✓ 1
Common error: Name the angle relationship in words, such as alternate or co-interior. Do not just write "parallel lines" as the reason.
41IGMC3-4-12r44-04Similarity, congruence and symmetryPaper CEasy4 marks
Calculator
Two library posters are mathematically similar rectangles. The small poster is 12 cm wide and 18 cm tall. The large poster is 30 cm wide.
Diagram not to scale
(a)
Find the scale factor of enlargement from the small poster to the large poster.
(2)
(b)
Find the height of the large poster.
(2)
Show mark scheme
Marking point
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Notes
Part (a)
M1 30 ÷ 12
✓ 1
A1 2.5
✓ 1
Part (b)
M1 18 × 2.5
✓ 1
A1 45 cm
✓ 1
Common error: Match corresponding sides: width with width, height with height. Do not divide a width by a height.
42IGMC-044-01-07Similarity, congruence and symmetryPaper CMedium7 marks
Calculator
The straight lines AE and BD cross at C. AB is parallel to DE. AB = 7.2 cm, DE = 4.8 cm, AC = 6 cm and BD = 11 cm.
Diagram not to scale
(a)
Give reasons why triangle ABC is similar to triangle EDC.
(2)
(b)
Calculate CE.
(2)
(c)
Calculate BC.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 angle BAC = angle DEC (alternate angles, since AB is parallel to DE)
Exam tip: In a crossed (bow-tie) diagram the sides that correspond are on opposite sides of C. BD is not a side of either triangle. Split it as BC + CD in the ratio 1.5 : 1.
43IGMC-044-01-11Similarity, congruence and symmetryPaper CMedium6 marks
Calculator
A rectangular photograph measures 15 cm by 10 cm. It is enlarged to make a mathematically similar photograph with longer side 36 cm.
(a)
Calculate the length of the shorter side of the enlarged photograph.
(2)
(b)
A frame measures 40 cm by 28 cm. Show that the frame is not mathematically similar to the original photograph.
(2)
(c)
Work out how many times larger the area of the enlarged photograph is than the area of the original.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 10 × 36 ÷ 15 or 10 × 2.4
✓ 1
A1 24 cm
✓ 1
Part (b)
M1 40 ÷ 15 = 2.67 and 28 ÷ 10 = 2.8, or 40 : 28 = 1.43 and 15 : 10 = 1.5
✓ 1
A1 the ratios are different, so not similar
✓ 1
Part (c)
M1 (36 × 24) ÷ (15 × 10) or 2.42
✓ 1
A1 5.76
✓ 1
Exam tip: To test for similarity, compare the scale factors for both pairs of sides. If the two scale factors are not equal, the rectangles are not similar.
44IGMC-044-00-06Angle properties of lines, triangles and polygonsPaper CMedium5 marks
No calculator
Each interior angle of a regular polygon is 8 times as large as each of its exterior angles.
(a)
Find the number of sides of the polygon.
(3)
(b)
Explain why there is no regular polygon with an interior angle of 155°.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
e + 8e = 180, or equivalent such as \(\frac{180(n-2)}{n} = 8 \times \frac{360}{n}\) M1
✓ 1
e = 20 (interior = 160), or n − 2 = 16 A1
✓ 1
n = 360 ÷ 20 = 18 A1
✓ 1
Part (b)
Exterior angle = 180 − 155 = 25°, or forms \(\frac{180(n-2)}{n} = 155\) M1
✓ 1
360 ÷ 25 = 14.4 (or n = 14.4), which is not a whole number, so there is no such polygon A1
✓ 1
Exam tip: An interior angle and its exterior angle sum to 180°. For a regular polygon, the number of sides equals 360 ÷ exterior angle, and this must be a whole number.
45IGMC-044-00-07Angle properties of lines, triangles and polygonsPaper CMedium5 marks
Calculator
The interior angles of a pentagon are 100°, 2x°, 2x°, 3x° and (3x + 20)°.
Diagram not to scale
(a)
Show that the interior angles of a pentagon add up to 540°.
(2)
(b)
Find the value of x.
(2)
(c)
Find the size of the largest angle in the pentagon.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
(5 − 2) × 180, or diagonals from one vertex split the pentagon into 3 triangles, giving 3 × 180 M1
Common error: Collecting the x terms wrongly: 2x + 2x + 3x + 3x = 10x. Check your answer by confirming that all five angles add up to 540°.
46IGMC-044-00-08Angle properties of lines, triangles and polygonsPaper CMedium5 marks
No calculator
ABCDEF is a regular hexagon. ABPQ is a square drawn on the side AB, outside the hexagon. The points F, A and Q are joined to form triangle FAQ.
Diagram not to scale
(a)
Write down the size of one interior angle of a regular hexagon.
(1)
(b)
Find angle FAQ.
(2)
(c)
Find angle AQF.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
120° B1
✓ 1
Part (b)
360 − 120 − 90 (angles at a point) M1
✓ 1
150° A1
✓ 1
Part (c)
AF = AQ, so triangle FAQ is isosceles: (180 − 150) ÷ 2 M1
✓ 1
15° A1
✓ 1
Exam tip: The square and the hexagon share the side AB, so AF = AB = AQ. Spotting this equal pair of sides shows that triangle FAQ is isosceles.
47IGMC-044-00-09Angle properties of lines, triangles and polygonsPaper CMedium5 marks
No calculator
In triangle PQR, angle QPR = (x + 15)° and angle PQR = (2x − 10)°. The side QR is extended beyond R to the point S, and angle PRS = (4x − 25)°.
Diagram not to scale
(a)
Find the value of x.
(3)
(b)
Find angle PRQ.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Correct equation, e.g. x + 15 + 2x − 10 = 4x − 25 (exterior angle) or x + 15 + 2x − 10 + 180 − (4x − 25) = 180 (angle sum) M1
✓ 1
Correct simplification, e.g. 3x + 5 = 4x − 25 M1
✓ 1
x = 30 A1
✓ 1
Part (b)
180 − (4 × 30 − 25) or 180 − 45 − 50 M1
✓ 1
85° A1
✓ 1
Exam tip: Check your answer: with x = 30 the angles of the triangle are 45°, 50° and 85°, which add up to 180°.
48IGMC-044-00-11Angle properties of lines, triangles and polygonsPaper CMedium4 marks
Calculator
The bearing of town B from town A is 064°. Town C is due south of town B. The bearing of town C from town A is 118°.
Diagram not to scale
(a)
Find the bearing of A from B.
(1)
(b)
Find angle BAC.
(1)
(c)
Find angle ACB.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
064 + 180 = 244°
✓ 1
Part (b)
118 − 64 = 54°
✓ 1
Part (c)
Either: angle ABC = 64° (alternate angles between the parallel north lines), so ACB = 180 − 64 − 54. Or: angle ACB = 180 − 118 (co-interior angles between the parallel north lines). Any one complete correct method
✓ 1
62°
✓ 1
Exam tip: The north lines through A and B are parallel, and BC lies along the north line through B. You can find angle ACB with alternate angles plus the angle sum of a triangle, or directly with co-interior angles. Give the reason as well as the working.
49IGMC-044-02-07Circle theoremsPaper CMedium6 marks
Calculator
A, B and C are points on a circle. AB is a diameter. AC = 7.2 cm and BC = 9.6 cm. The region inside the circle but outside triangle ABC is shaded.
Diagram not to scale. The shaded region is inside the circle and outside the triangle.
(a)
Explain why angle ACB = 90°.
(1)
(b)
Calculate the diameter AB.
(2)
(c)
Calculate the area of the shaded region.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1: it is the angle in a semicircle, because AB is a diameter
✓ 1
Part (b)
M1: AB = √(7.22 + 9.62)
✓ 1
A1: 12 cm
✓ 1
Part (c)
M1: circle area = π × 62 (= 113.09...)
✓ 1
M1: subtract triangle area ½ × 7.2 × 9.6 (= 34.56)
✓ 1
A1: 78.5 cm2
✓ 1
Common error: In the circle area, use the radius 6 cm and not the diameter 12 cm. Using 12 gives an area four times too big.
50IGMC-044-02-11Circle theoremsPaper CMedium5 marks
Calculator
A circle has centre O and radius 8 cm. A point P is 17 cm from O. A tangent from P touches the circle at T.
Diagram not to scale
(a)
Calculate the length PT.
(3)
(b)
Calculate angle OPT.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1: angle OTP = 90° (tangent perpendicular to radius)
✓ 1
M1: PT = √(172 − 82)
✓ 1
A1: 15 cm
✓ 1
Part (b)
M1: sin OPT = 8/17, or tan OPT = 8/15, or cos OPT = 15/17
✓ 1
A1: 28.1°
✓ 1
Common error: OP is the hypotenuse because it is opposite the right angle at T. Subtract the squares here, do not add them.
51IGMC3-4-01d-06Angle properties of lines, triangles and polygonsPaper CMedium6 marks
No calculator
A garden bed is an irregular pentagon. Its interior angles are 95°, 118°, 2x°, (2x + 10)° and (x + 7)°.
Diagram not to scale
(a)
Calculate the sum of the interior angles of a pentagon.
(2)
(b)
Form an equation in x and solve it.
(3)
(c)
Calculate the size of the largest angle of the pentagon.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 (5 − 2) × 180
✓ 1
A1 540°
✓ 1
Part (b)
M1 95 + 118 + 2x + (2x + 10) + (x + 7) = 540
✓ 1
M1 5x + 230 = 540
✓ 1
A1 x = 62
✓ 1
Part (c)
B1 134° (ECF from (b): 2x + 10)
✓ 1
Common error: Collect the numbers and the x terms separately before solving; a slip in the constant 230 is the usual reason this equation goes wrong.
52IGMC3-4-01d-08Angle properties of lines, triangles and polygonsPaper CMedium5 marks
No calculator
A metal gate panel is a rhombus ABCD. Angle ABC = 70°. The diagonal BD is drawn.
Diagram not to scale
(a)
Write down the size of angle BCD and give a reason.
(2)
(b)
Calculate angle DBC.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 110°
✓ 1
B1 co-interior angles between parallel sides add up to 180° (180 − 70)
✓ 1
Part (b)
M1 BC = CD (all sides of a rhombus are equal), so triangle BCD is isosceles
✓ 1
M1 (180 − 110) ÷ 2
✓ 1
A1 35°
✓ 1
Exam tip: State the geometric fact you are using, such as 'sides of a rhombus are equal', as well as the number you calculate.
53IGMC3-4-01d-10Angle properties of lines, triangles and polygonsPaper CMedium7 marks
No calculator
Two triangular steel plates ABC and PQR are cut for a bridge. In plate ABC, angle BAC = 48°, angle ABC = 64° and AB = 7.5 m. In plate PQR, angle QPR = 48°, angle PQR = 64° and PQ = 7.5 m. The length AC is 7.3 m.
Diagram not to scale
(a)
Calculate angle ACB.
(2)
(b)
Explain why the two plates are congruent.
(2)
(c)
Write down the length of PR.
(1)
(d)
Write down the size of angle PRQ and give a reason.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 180 − 48 − 64
✓ 1
A1 68°
✓ 1
Part (b)
B1 two pairs of equal angles (48° and 64°)
✓ 1
B1 and the corresponding sides AB and PQ are equal, so the triangles are congruent (ASA / AAS)
✓ 1
Part (c)
B1 7.3 m
✓ 1
Part (d)
A1 68°
✓ 1
B1 corresponding angles in congruent triangles are equal
✓ 1
Exam tip: When you name a congruence condition, check that the equal sides and angles are in matching positions in both triangles.
54IGMC3-4-08r8-03Circle theoremsPaper CMedium6 marks
Calculator
In a rope-bridge frame, triangle ABC has AB = AC and angle BAC = 38°. The side BC is extended beyond C to a point D such that CD = CA.
Diagram not to scale
(a)
Find angle ABC.
(2)
(b)
Find angle ACD.
(2)
(c)
Find angle CAD.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 (180 − 38) ÷ 2
✓ 1
A1 71°
✓ 1
Part (b)
M1 angles on a straight line (or exterior angle): 180 − 71
✓ 1
A1 109°
✓ 1
Part (c)
M1 triangle CAD is isosceles with CA = CD, so 180 − 109 shared between two equal angles
✓ 1
A1 35.5°
✓ 1
Common error: In an isosceles triangle, the equal angles are opposite the equal sides; check which two sides are equal before halving.
55IGMC3-4-08r8-05Similarity, congruence and symmetryPaper CMedium5 marks
Calculator
A vertical telecoms mast stands on level ground. At the same moment, the mast casts a shadow 15 m long and a vertical pole 1.8 m tall casts a shadow 2.4 m long.
Diagram not to scale
(a)
Give a reason why the triangle formed by the mast and its shadow is similar to the triangle formed by the pole and its shadow.
(1)
(b)
Calculate the height of the mast.
(2)
(c)
A support cable runs in a straight line from the top of the mast to the end of its shadow. Calculate the length of the cable.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
R1 the sun's rays are parallel, so the angles are equal (both have a right angle and the same angle of elevation)
✓ 1
Part (b)
M1 height ÷ 15 = 1.8 ÷ 2.4
✓ 1
A1 11.25 m
✓ 1
Part (c)
M1 sqrt(11.252 + 152)
✓ 1
A1 18.75 m
✓ 1
Common error: Make sure each ratio compares corresponding sides: height to shadow in both triangles.
56IGMC3-4-08r8-06Circle theoremsPaper CMedium5 marks
Calculator
A circular mountain lake has centre O and radius 6 km. A straight path is tangent to the lake at the point A. The point T is on the path, with AT = 8 km.
Diagram not to scale
(a)
Write down the size of angle OAT and give a reason.
(1)
(b)
Calculate the distance OT.
(2)
(c)
Calculate angle AOT, giving your answer to 1 decimal place.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 90° because a tangent is perpendicular to the radius at the point of contact
✓ 1
Part (b)
M1 OT2 = 62 + 82
✓ 1
A1 10 km
✓ 1
Part (c)
M1 tan(AOT) = 8 ÷ 6
✓ 1
A1 53.1°
✓ 1
Exam tip: Whenever a tangent meets a radius, mark the right angle straight away; it gives you a right-angled triangle to work in.
57IGMC3-4-08r8-11Similarity, congruence and symmetryPaper CMedium4 marks
No calculator
A telecoms company arranges antennas at the corners of a regular hexagon.
(a)
State the number of lines of symmetry of a regular hexagon.
(1)
(b)
State the order of rotational symmetry of a regular hexagon.
(1)
(c)
Triangles ABC and PQR have AB = PQ = 7 cm, angle ABC = angle PQR = 50° and BC = QR = 9 cm. Give the condition that shows the triangles are congruent.
(1)
(d)
Write down the side of triangle PQR that equals AC.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 6
✓ 1
Part (b)
B1 6
✓ 1
Part (c)
B1 SAS (two sides and the included angle)
✓ 1
Part (d)
B1 PR
✓ 1
Common error: SAS needs the equal angle to lie between the two equal sides; an equal angle elsewhere does not prove congruence.
58IGMC3-4-08r8-12Angle properties of lines, triangles and polygonsPaper CMedium5 marks
No calculator
Each interior angle of a regular polygon used in an antenna array is 132° larger than each exterior angle.
(a)
Find the size of an exterior angle.
(2)
(b)
Find the number of sides of the polygon.
(1)
(c)
Find the sum of the interior angles of the polygon.
Exam tip: Interior and exterior angles at the same vertex always add up to 180°; use this to set up an equation.
59IGMC3-4-11r30-05Similarity, congruence and symmetryPaper CMedium5 marks
Calculator
A beekeeper wants to know the height of a tree next to an apiary. At the same time of day, a hive stand 1.2 m tall casts a shadow 0.9 m long and the tree casts a shadow 7.5 m long. The stand and tree are both vertical and the ground is level.
Diagram not to scale
(a)
Calculate the height of the tree.
(3)
(b)
The beekeeper is 1.8 m tall. Calculate the length of the beekeeper's shadow at that time.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 scale factor 7.5 ÷ 0.9 (or 1.2 ÷ 0.9 = height per metre of shadow)
✓ 1
M1 1.2 × 7.5 ÷ 0.9
✓ 1
A1 10 m
✓ 1
Part (b)
M1 1.8 × 0.9 ÷ 1.2
✓ 1
A1 1.35 m
✓ 1
Exam tip: Write which lengths correspond before calculating, and check that the larger object has the larger scale factor.
60IGMC3-4-11r30-06Circle theoremsPaper CMedium7 marks
Calculator
A circular hospital helipad has centre O and radius 12 m. A straight path touches the edge of the helipad at T. A point P is on the path with PT = 16 m.
Diagram not to scale
(a)
State the size of angle OTP and give a reason.
(2)
(b)
Calculate the distance OP.
(3)
(c)
Calculate angle TOP. Give your answer correct to 1 decimal place.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 90°
✓ 1
B1 a tangent is perpendicular to the radius at the point of contact
✓ 1
Part (b)
M1 OP2 = 122 + 162
✓ 1
A1 400
✓ 1
A1 20 m
✓ 1
Part (c)
M1 cos(TOP) = 12 ÷ 20 (or tan = 16 ÷ 12, or sin = 16 ÷ 20)
✓ 1
A1 53.1°
✓ 1
Common error: Treating the tangent as if it passed through the centre; the radius to the point of contact meets it at a right angle.
61IGMC3-4-11r30-07Angle properties of lines, triangles and polygonsPaper CMedium7 marks
No calculator
The corridor floor in a hospital is tiled with regular dodecagons (12 equal sides) and equilateral triangles.
(a)
Calculate one interior angle of a regular dodecagon.
(2)
(b)
Show that two dodecagons and one equilateral triangle fit together at a point with no gaps.
(2)
(c)
Each interior angle of a different regular polygon is 162°. Calculate the number of sides of this polygon.
Exam tip: The exterior angles of any polygon add up to 360°, so for a regular polygon the number of sides is 360 divided by one exterior angle.
62IGMC3-4-11r30-09Angle properties of lines, triangles and polygonsPaper CMedium6 marks
Calculator
A roof garden at a hospital is a kite ABCD with AB = AD and CB = CD. Angle BAD = 80° and angle BCD = 110°.
Diagram not to scale
(a)
Write down the number of lines of symmetry of the kite.
(1)
(b)
Calculate angle ABC.
(3)
(c)
Calculate angle DBC.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 1
✓ 1
Part (b)
M1 (360 − 80 − 110) ÷ 2 (the angles at B and D are equal)
✓ 1
M1 angles in a quadrilateral add up to 360°
✓ 1
A1 85°
✓ 1
Part (c)
M1 angle ABD = (180 − 80) ÷ 2 = 50° (triangle ABD is isosceles)
✓ 1
A1 85 − 50 = 35° (ECF from (b))
✓ 1
Exam tip: A kite has one pair of equal angles, so subtract the two known angles from 360° and halve the remainder.
63IGMC3-4-12r33-03Angle properties of lines, triangles and polygonsPaper CMedium6 marks
No calculator
Lines AB and CD are parallel, with AB above CD, A to the left of B and C to the left of D. A straight line EF crosses AB at G and CD at H, with E above AB and F below CD. Angle EGB = 112°.
Diagram not to scale
(a)
Find angle GHD and give a reason for your answer.
(2)
(b)
Find angle DHF and give a reason for your answer.
(2)
(c)
The line HK bisects angle GHD, with K between HG and HD. Find angle KHF.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 112°
✓ 1
B1 corresponding angles (parallel lines) are equal
✓ 1
Part (b)
B1 68° (ECF from (a))
✓ 1
B1 angles on a straight line sum to 180° (G, H, F are collinear)
✓ 1
Part (c)
M1 angle GHK = 112° ÷ 2 = 56°
✓ 1
A1 180° − 56° = 124°
✓ 1
Common error: Always give the named angle property as a reason, such as corresponding or co-interior, and not just the angle size.
64IGMC3-4-12r33-04Similarity, congruence and symmetryPaper CMedium7 marks
Calculator
A health district map is drawn to a scale of 1 : 25 000.
(a)
Two clinics are 7.2 cm apart on the map. Find the actual distance between them in kilometres.
(2)
(b)
A flood-risk zone has area 14 cm² on the map. Find its actual area in square kilometres.
(3)
(c)
A real park has area 2.4 km². Find its area on the map in square centimetres.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 7.2 × 25 000 = 180 000 cm
✓ 1
A1 1.8 km
✓ 1
Part (b)
M1 1 cm represents 0.25 km
✓ 1
M1 14 × 0.25²
✓ 1
A1 0.875 km²
✓ 1
Part (c)
M1 2.4 ÷ 0.25²
✓ 1
A1 38.4 cm²
✓ 1
Common error: Areas scale with the square of the length scale factor, so square the factor before multiplying or dividing.
65IGMC3-4-12r33-05Circle theoremsPaper CMedium7 marks
No calculator
The points A and B are the ends of a diameter of a circular shop display with centre O. The point C lies on the circumference and angle CAB = 34°. The tangent to the circle at C meets AB extended beyond B at T.
Diagram not to scale
(a)
Find angle ACB and give a reason for your answer.
(2)
(b)
Find angle COB and give a reason for your answer.
(2)
(c)
Find angle CTO.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 90°
✓ 1
B1 angle in a semicircle is a right angle
✓ 1
Part (b)
B1 68°
✓ 1
B1 angle at the centre is twice the angle at the circumference
✓ 1
Part (c)
B1 angle OCT = 90° (tangent is perpendicular to radius)
✓ 1
M1 180° − 90° − 68°
✓ 1
A1 22°
✓ 1
Exam tip: Mark every right angle you can justify first (semicircle, tangent and radius) and then use angle sums.
66IGMC3-4-12r33-12Angle properties of lines, triangles and polygonsPaper CMedium7 marks
No calculator
A retail store floor is tiled with regular polygons.
(a)
Calculate the interior angle of a regular octagon.
(2)
(b)
Show that two regular octagons and one square fit together exactly around a point.
(2)
(c)
A regular 12-sided tile has interior angle 150°. One of these tiles and one square meet at a point with a third regular polygon so that they fit exactly. Find the number of sides of the third polygon.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 exterior angle = 360 ÷ 8 = 45°
✓ 1
A1 135°
✓ 1
Part (b)
M1 135 + 135 + 90
✓ 1
A1 = 360°, so they fit with no gap or overlap
✓ 1
Part (c)
M1 360 − 150 − 90 = 120° interior angle
✓ 1
M1 exterior angle = 180 − 120 = 60°
✓ 1
A1 360 ÷ 60 = 6 sides
✓ 1
Exam tip: To fit regular polygons around a point, their interior angles must sum to exactly 360°.
67IGMC3-4-12r33-14Angle properties of lines, triangles and polygonsPaper CMedium5 marks
No calculator
In a vaccination-site sign, triangle ABC has AB = AC and angle BAC = 48°. The point D lies on BC extended beyond C.
Diagram not to scale
(a)
Find angle ABC and give a reason for your answer.
(2)
(b)
Find angle ACD and give a reason for your answer.
(1)
(c)
The line CE bisects angle ACD, with E on the same side of BD as A. Find angle BCE.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 (180° − 48°) ÷ 2
✓ 1
A1 66° (base angles of an isosceles triangle are equal)
✓ 1
Part (b)
A1 114° (angles on a straight line, or exterior angle = 48° + 66°)
✓ 1
Part (c)
M1 angle ECD = 114° ÷ 2 = 57°
✓ 1
A1 180° − 57° = 123°
✓ 1
Exam tip: Write each reason beside the step it justifies, for example isosceles triangle, angles on a line or exterior angle.
68IGMC3-4-10r37-03Similarity, congruence and symmetryPaper CMedium5 marks
Calculator
On a sunny day, a vertical signal post of height 3.2 m beside a railway line casts a shadow of length 2.0 m on level ground. At the same moment, a vertical gantry pole beside the line casts a shadow of length 7.5 m.
Diagram not to scale
(a)
Give a reason why the triangle formed by the post and its shadow is similar to the triangle formed by the pole and its shadow.
(1)
(b)
Work out the height of the gantry pole.
(2)
(c)
Later, the shadow of the gantry pole is 9.0 m long. Work out the length of the shadow of the 3.2 m signal post at this time.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 the sun's rays are parallel, so the angles of elevation are equal and each triangle has a right angle (the corresponding angles are equal)
Exam tip: For similar shapes, set up the scale factor between matching lengths first and check that each ratio compares the same two objects.
69IGMC3-4-10r37-05Circle theoremsPaper CMedium5 marks
No calculator
The circular floor of an observatory has centre O. Points A and B are at opposite ends of a diameter and point C lies on the circumference. Angle CAB = 34°.
Diagram not to scale
(a)
Find angle ACB and give a geometrical reason for your answer.
(2)
(b)
Find angle ABC.
(1)
(c)
A straight path touches the circle at B. The point T lies on the path, on the line AC extended beyond C. Find angle ATB.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 90°
✓ 1
B1 the angle in a semicircle is a right angle
✓ 1
Part (b)
B1 56° (ECF from (a))
✓ 1
Part (c)
M1 angle ABT = 90° because a tangent is perpendicular to the radius (diameter) at the point of contact
✓ 1
A1 56°
✓ 1
Exam tip: State the circle theorem by name or in words next to each angle you find; the reason is often worth a mark of its own.
70IGMC3-4-10r37-12Angle properties of lines, triangles and polygonsPaper CMedium5 marks
No calculator
Two straight gymnastics bars AB and CD are parallel. A straight cable passes through the points E, G, H and F in that order, crossing AB at G and CD at H. The points B and D are on the same side of the cable. Angle EGB = 112°.
Diagram not to scale
(a)
Find angle AGH and give a geometrical reason.
(2)
(b)
Find angle GHD and give a geometrical reason.
(2)
(c)
Find angle FHD.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 112°
✓ 1
B1 vertically opposite angles are equal
✓ 1
Part (b)
B1 112°
✓ 1
B1 corresponding angles between parallel lines are equal
✓ 1
Part (c)
B1 68° (angles on a straight line add up to 180°)
✓ 1
Common error: Always name the angle relationship you use (corresponding, alternate or co-interior); stating only the angle value does not earn the reason mark.
71IGMC3-4-10r41-04Similarity, congruence and symmetryPaper CMedium5 marks
Calculator
Triangle ABC is a plan of a field with AB = 60 m, BC = 80 m and CA = 100 m. Triangle PQR is a plan of a smaller similar field, with PQ = 45 m, QR corresponding to BC and PR corresponding to CA.
Diagram not to scale
(a)
Write down the scale factor of enlargement from triangle ABC to triangle PQR.
(1)
(b)
Work out the lengths QR and PR.
(2)
(c)
Fencing for field PQR costs £4.20 per metre. Work out the cost of fencing all the way round the field.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 0.75 or 3/4
✓ 1
Part (b)
B1 QR = 60 m (ft from (a))
✓ 1
B1 PR = 75 m (ft from (a))
✓ 1
Part (c)
M1 perimeter 45 + 60 + 75 = 180 m
✓ 1
A1 4.20 × 180 = £756.00
✓ 1
Exam tip: The scale factor from the larger shape to the smaller one is less than 1.
72IGMC3-4-10r41-06Angle properties of lines, triangles and polygonsPaper CMedium5 marks
No calculator
A plot of land beside a runway is a trapezium ABCD in which AB is parallel to DC. Angle DAB = (3x + 12)° and angle ADC = (2x + 18)°.
Diagram not to scale. Arrows mark the parallel sides.
(a)
Give a reason why angle DAB + angle ADC = 180°.
(1)
(b)
Write down and solve an equation to find x.
(3)
(c)
Work out the size of angle DAB.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 co-interior (allied) angles between parallel lines add up to 180
✓ 1
Part (b)
M1 (3x + 12) + (2x + 18) = 180
✓ 1
A1 5x + 30 = 180
✓ 1
A1 x = 30
✓ 1
Part (c)
B1 3 × 30 + 12 = 102°
✓ 1
Exam tip: Always give the geometric reason for an angle fact in full, for example 'co-interior angles add up to 180°'.
73IGMC3-4-10r41-07Circle theoremsPaper CMedium7 marks
Calculator
AB is a diameter of the circular skin of a drum, with AB = 24 cm. Point C lies on the circumference and angle CAB = 38°. The tangent to the circle at B meets the line AC extended beyond C at point T.
Diagram not to scale
(a)
Write down the size of angle ACB, giving a reason.
(2)
(b)
Work out the size of angle CBA.
(1)
(c)
Work out the size of angle ATB.
(2)
(d)
Calculate the length of AC. Give your answer correct to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 90°
✓ 1
B1 angle in a semicircle is a right angle
✓ 1
Part (b)
B1 180 − 90 − 38 = 52°
✓ 1
Part (c)
B1 angle ABT = 90° (tangent is perpendicular to the radius or diameter)
✓ 1
A1 angle ATB = 180 − 90 − 38 = 52°
✓ 1
Part (d)
M1 AC = 24 × cos 38°
✓ 1
A1 18.9 cm
✓ 1
Exam tip: Spot a diameter first: it gives a right angle at the circumference and, with a tangent, another right angle at the end of the diameter.
74IGMC3-4-10r41-09Similarity, congruence and symmetryPaper CMedium7 marks
No calculator
Triangle ABC is the tail fin of a model aeroplane with AB = AC = 17 cm and BC = 16 cm. M is the midpoint of BC.
Diagram not to scale
(a)
Show that triangles ABM and ACM are congruent.
(3)
(b)
Hence show that angle AMB = 90°.
(2)
(c)
Work out the length of AM.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 AB = AC (given)
✓ 1
B1 BM = MC (M is the midpoint of BC)
✓ 1
B1 AM is common to both triangles, so they are congruent (SSS)
✓ 1
Part (b)
M1 angle AMB = angle AMC (congruent triangles)
✓ 1
R1 these angles lie on a straight line and add up to 180°, so each is 90°
✓ 1
Part (c)
M1 AM2 = 172 − 82
✓ 1
A1 AM = 15 cm
✓ 1
Exam tip: To prove congruence, state the three facts and the condition used (SSS, SAS, ASA or RHS).
75IGMC3-4-10r41-13Angle properties of lines, triangles and polygonsPaper CMedium7 marks
Calculator
A pentagonal stage ABCDE in a concert hall has interior angles 2x°, (x + 20)°, 130°, (3x − 10)° and 100°.
Diagram not to scale
(a)
Calculate the sum of the interior angles of a pentagon.
(2)
(b)
Write down and solve an equation to find x.
(3)
(c)
Write down the size of the largest interior angle.
(1)
(d)
Work out the size of the exterior angle at the vertex with the largest interior angle.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 (5 − 2) × 180
✓ 1
A1 540°
✓ 1
Part (b)
M1 2x + (x + 20) + 130 + (3x − 10) + 100 = 540
✓ 1
A1 6x + 240 = 540
✓ 1
A1 x = 50
✓ 1
Part (c)
B1 3 × 50 − 10 = 140°
✓ 1
Part (d)
B1 180 − 140 = 40° (ft from (c))
✓ 1
Exam tip: Check your value of x by adding all five angles back up to 540°.
76IGMC3-4-12r44-05Angle properties of lines, triangles and polygonsPaper CMedium6 marks
No calculator
A garden at a library has the shape of a pentagon ABCDE. Its interior angles are (x + 20)°, 2x°, (x + 40)°, (3x − 10)° and 70°.
Diagram not to scale
(a)
Show that 7x + 120 = 540.
(2)
(b)
Find the value of x.
(2)
(c)
Find the size of the largest interior angle.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 interior angles of a pentagon add up to (5 − 2) × 180 = 540°
✓ 1
A1 adds all five angles to obtain 7x + 120
✓ 1
Part (b)
M1 7x = 420
✓ 1
A1 x = 60
✓ 1
Part (c)
M1 substitutes x = 60 into the angles
✓ 1
A1 170°
✓ 1
Common error: After solving for x, substitute back into every angle expression. The value of x is rarely the angle itself.
77IGMC3-4-12r44-06Circle theoremsPaper CMedium5 marks
Calculator
A circular exclusion zone around a wind farm has centre O and radius 7 km. A village P is 25 km from O. A straight path from P just touches the edge of the zone at T (the path is a tangent to the circle).
Diagram not to scale
(a)
State the size of angle OTP.
(1)
(b)
Calculate the length of the path PT.
(2)
(c)
Calculate angle TOP. Give your answer correct to 1 decimal place.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 90°
✓ 1
Part (b)
M1 PT² = 25² − 7²
✓ 1
A1 24 km
✓ 1
Part (c)
M1 cos(TOP) = 7/25 (or tan = 24/7, sin = 24/25)
✓ 1
A1 73.7°
✓ 1
Common error: The right angle is between the tangent and the radius at the point of contact. The hypotenuse is OP, not OT.
78IGMC3-4-12r44-08Similarity, congruence and symmetryPaper CMedium5 marks
Calculator
A vertical pole of height 1.8 m casts a shadow 2.4 m long on level ground. At the same moment a vertical wind turbine tower casts a shadow 36 m long on the same level ground.
Diagram not to scale
(a)
Give one reason why the triangles formed by the pole and its shadow, and by the tower and its shadow, are similar.
(1)
(b)
Calculate the height of the tower.
(2)
(c)
Later, the pole's shadow is 3.2 m long. Calculate the length of the tower's shadow at that time.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 the Sun's rays are parallel so the angles of elevation are equal, and both objects make a right angle with the ground
✓ 1
Part (b)
M1 1.8 × 36 ÷ 2.4 (or scale factor 36 ÷ 2.4 = 15)
✓ 1
A1 27 m
✓ 1
Part (c)
M1 27 ÷ 1.8 × 3.2 (or 15 × 3.2)
✓ 1
A1 48 m
✓ 1
Exam tip: In shadow problems, compare height with shadow length for each object. The ratio height ÷ shadow is the same for both at the same moment.
79IGMC3-4-12r44-11Angle properties of lines, triangles and polygonsPaper CMedium6 marks
Calculator
A library courtyard plan shows a regular hexagonal garden and a regular octagonal pond that share one side AB. The two shapes do not overlap. At vertex A, the other side of the hexagon and the other side of the octagon form an angle that lies outside both shapes.
Diagram not to scale
(a)
Work out the size of an interior angle of the hexagon.
(2)
(b)
Work out the size of an interior angle of the octagon.
(2)
(c)
Work out the size of the angle outside both shapes at A.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 360 ÷ 6 = 60 for the exterior angle, or (6 − 2) × 180 ÷ 6
✓ 1
A1 120°
✓ 1
Part (b)
M1 360 ÷ 8 = 45 for the exterior angle, or (8 − 2) × 180 ÷ 8
✓ 1
A1 135°
✓ 1
Part (c)
M1 360 − 120 − 135
✓ 1
A1 105°
✓ 1
Common error: Angles around a point add up to 360°. Subtract every interior angle that meets at the point, not just one of them.
80IGMC-044-00-12Angle properties of lines, triangles and polygonsPaper CHard5 marks
No calculator
A tiling design has three regular polygons meeting at a single point with no gaps and no overlaps. The three polygons are a regular pentagon, a square and a regular polygon with n sides.
(a)
Find the size of one interior angle of a regular pentagon.
(1)
(b)
Show that three regular pentagons cannot fit together around a point without a gap.
(1)
(c)
Find the value of n.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
108° (from 540 ÷ 5 or 180 − 72) B1
✓ 1
Part (b)
3 × 108 = 324, which is less than 360, so a gap (of 36°) is left B1
✓ 1
Part (c)
Interior angle of the n-sided polygon = 360 − 108 − 90 = 162° M1
✓ 1
Exterior angle = 180 − 162 = 18°, or form (n − 2) × 180 = 162n M1
✓ 1
n = 360 ÷ 18 = 20, or 18n = 360 giving n = 20 A1
✓ 1
Exam tip: It is quicker to find n from the exterior angle (360 ÷ exterior angle) than to solve (n − 2) × 180 ÷ n = 162, but both methods earn full marks.
81IGMC3-4-01d-13Similarity, congruence and symmetryPaper CHard7 marks
No calculator
In a model roof frame, a horizontal beam DE is parallel to the base BC of triangle ABC, with D on AB and E on AC. AD = x cm, AB = (x + 4) cm, DE = 6 cm, BC = 9 cm and AC = 15 cm.
Diagram not to scale
(a)
Explain why triangles ADE and ABC are similar.
(2)
(b)
Write down an equation in x and solve it.
(3)
(c)
Calculate the length AE.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 angle A is common to both triangles
✓ 1
B1 angle ADE = angle ABC (corresponding angles, DE parallel to BC), so the triangles are similar
✓ 1
Part (b)
M1 x ÷ (x + 4) = 6 ÷ 9
✓ 1
M1 9x = 6x + 24
✓ 1
A1 x = 8
✓ 1
Part (c)
M1 AE ÷ 15 = 8 ÷ 12
✓ 1
A1 10 cm
✓ 1
Exam tip: Write the ratio of corresponding sides with the small triangle on top throughout, so each fraction compares like with like.
82IGMC3-4-10r37-14Angle properties of lines, triangles and polygonsPaper CHard6 marks
No calculator
A station floor pattern uses a regular hexagon ABCDEF and a regular pentagon ABGHI. They share the side AB and lie on opposite sides of AB. Every side of both shapes is 4 m long.
Diagram not to scale
(a)
Write down the size of an interior angle of the regular hexagon and of the regular pentagon.
(2)
(b)
Work out angle FAI, the angle at A that lies inside neither shape.
(2)
(c)
Work out angle AFI.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 120°
✓ 1
B1 108°
✓ 1
Part (b)
M1 360 − 120 − 108
✓ 1
A1 132° (ECF from (a))
✓ 1
Part (c)
M1 AF = AI = 4 m, so triangle AFI is isosceles and angle AFI = (180 − 132) ÷ 2
✓ 1
A1 24° (ECF from (b))
✓ 1
Common error: Angles around a point add up to 360°; subtract every angle that is already used before finding the missing one.
83IGMC3-4-12r44-14Angle properties of lines, triangles and polygonsPaper CHard8 marks
Calculator
Two fences surround a wind-farm compound. Fence P is a regular polygon with n sides. Fence Q is a regular polygon with 2n sides. Each exterior angle of P is 15° larger than each exterior angle of Q.
(a)
Write down an equation in n using this information.
(2)
(b)
Solve your equation to find n.
(2)
(c)
Find the size of each interior angle of Q.
(2)
(d)
Find the sum of the interior angles of Q.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 exterior angles are 360/n and 360/(2n)
✓ 1
A1 360/n − 180/n = 15
✓ 1
Part (b)
M1 180/n = 15
✓ 1
A1 n = 12
✓ 1
Part (c)
M1 Q has 24 sides, so exterior angle is 360 ÷ 24 = 15°
✓ 1
A1 165°
✓ 1
Part (d)
M1 (24 − 2) × 180
✓ 1
A1 3960°
✓ 1
Exam tip: Exterior angles are often the quickest route for regular polygons, because each one is 360° divided by the number of sides.
84IGMC-044-02-01Circle theoremsPaper EEasy4 marks
No calculator
Points A, B and C lie on a circle with centre O. C lies on the major arc AB, and angle ACB = 37°.
Diagram not to scale
(a)
Find angle AOB. Give a reason for your answer.
(2)
(b)
Find angle OAB.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1: angle AOB = 74°
✓ 1
B1: the angle at the centre is twice the angle at the circumference
✓ 1
Part (b)
M1: triangle OAB is isosceles (OA = OB are radii), so angle OAB = (180 − 74) ÷ 2
✓ 1
A1: 53°
✓ 1
Exam tip: When O is the centre, look for radii. Two radii always make an isosceles triangle, which gives you the base angles.
85IGMC-044-02-03Circle theoremsPaper EEasy3 marks
No calculator
PQRS is a cyclic quadrilateral. Angle PQR = 104° and angle QRS = 71°.
Diagram not to scale
(a)
Find angle PSR. Give a reason for your answer.
(2)
(b)
Find angle QPS.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1: 76°
✓ 1
B1: opposite angles of a cyclic quadrilateral add up to 180°
✓ 1
Part (b)
B1: 180 − 71 = 109°
✓ 1
Common error: Only opposite angles of a cyclic quadrilateral add up to 180°. Angles PQR and QRS are next to each other, so they are not supplementary.
86IGMC-044-02-05Circle theoremsPaper EEasy4 marks
No calculator
P is a point outside a circle with centre O. The two tangents from P touch the circle at A and B. Angle APB = 46°.
Diagram not to scale
(a)
Find angle PAB.
(2)
(b)
Find angle AOB.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1: PA = PB (tangents from an external point are equal), so angle PAB = (180 − 46) ÷ 2
Exam tip: In the kite OAPB, the angle at O and the angle at P always add up to 180°, because the two other angles are both 90°.
87IGMC-044-01-06Similarity, congruence and symmetryPaper EMedium4 marks
Calculator
Two bottles are mathematically similar. The smaller bottle has height 12 cm. The larger bottle has height 18 cm.
(a)
The surface area of the smaller bottle is 150 cm2. Calculate the surface area of the larger bottle.
(2)
(b)
The capacity of the larger bottle is 810 ml. Calculate the capacity of the smaller bottle.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 150 × (18/12)2 or 150 × 2.25
✓ 1
A1 337.5 cm2
✓ 1
Part (b)
M1 810 ÷ (18/12)3 or 810 × (12/18)3
✓ 1
A1 240 ml
✓ 1
Common error: Using the length scale factor 1.5 for an area or a volume gives the wrong answer. Square it for area (2.25) and cube it for volume or capacity (3.375).
88IGMC-044-01-08Similarity, congruence and symmetryPaper EMedium6 marks
No calculator
ABCD is a parallelogram. The diagonal AC is drawn.
Diagram not to scale
(a)
Prove that triangle ABC is congruent to triangle CDA.
(3)
(b)
Hence write down which angle in triangle CDA is equal to angle ABC.
(1)
(c)
Angle BAC = 35° and angle ACB = 47°. Work out angle ADC.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 AB = CD and BC = DA (opposite sides of a parallelogram are equal)
✓ 1
B1 AC is common to both triangles
✓ 1
B1 congruent by SSS
✓ 1
Part (b)
B1 angle CDA
✓ 1
Part (c)
M1 angle ABC = 180 − 35 − 47 (angle sum of a triangle), or angle DAC = 47° and angle DCA = 35° (alternate angles) so angle ADC = 180 − 47 − 35
✓ 1
A1 angle ADC = 98°, since angle ADC = angle ABC (from congruence) or by the direct method
✓ 1
Exam tip: A congruence proof needs three facts, each with a reason, and then the condition (SSS, SAS, ASA or RHS). The word 'common' is enough of a reason for a shared side. When using a congruence later, state which angles correspond.
89IGMC-044-01-09Similarity, congruence and symmetryPaper EMedium7 marks
Calculator
In triangle ABC, D is a point on AB and E is a point on AC. DE is parallel to BC. AD = 5 cm, DB = 3 cm, DE = 3 cm and AE = 4 cm.
Diagram not to scale
(a)
Calculate BC.
(2)
(b)
Calculate EC.
(2)
(c)
The area of triangle ADE is 6 cm2. Calculate the area of the trapezium DBCE.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 3 × 8 ÷ 5 or 3 × \(\frac{8}{5}\)
✓ 1
A1 4.8 cm
✓ 1
Part (b)
M1 4 × 8 ÷ 5 (= 6.4) then 6.4 − 4, or directly 4 × 3 ÷ 5
✓ 1
A1 2.4 cm
✓ 1
Part (c)
M1 6 × (8/5)2 (= 15.36)
✓ 1
M1 15.36 − 6
✓ 1
A1 9.36 cm2
✓ 1
Common error: Using 5 : 3 as the scale factor is wrong. The similar triangles are ADE and ABC, so compare AD with AB = 8 cm, not with DB. For areas, square the length scale factor.
90IGMC-044-01-10Similarity, congruence and symmetryPaper EMedium5 marks
No calculator
Answer the following questions about symmetry in three-dimensional shapes.
(a)
Write down the number of planes of symmetry of a cuboid measuring 3 cm by 4 cm by 5 cm.
(1)
(b)
Write down the number of planes of symmetry of a right pyramid with a square base.
(1)
(c)
A right prism has a regular hexagon as its cross-section. Write down the order of rotational symmetry of the prism about the axis through the centres of its two hexagonal faces.
(1)
(d)
Work out the total number of planes of symmetry of this right hexagonal prism.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 3
✓ 1
Part (b)
B1 4
✓ 1
Part (c)
B1 6
✓ 1
Part (d)
B1 6 planes containing the axis through the hexagon centres (3 through opposite vertices, 3 through midpoints of opposite edges)
✓ 1
B1 total 7, including the plane parallel to the hexagonal faces, halfway along the prism
✓ 1
Common error: A right prism also has a plane of symmetry halfway along its length, parallel to the end faces. Count this plane as well as the planes that contain the central axis.
91IGMC-044-02-06Circle theoremsPaper EMedium4 marks
No calculator
Points A, B and C lie on a circle. The straight line TAS is the tangent to the circle at A. B is on the same side of A as T, and C is on the same side of A as S. Angle TAB = 67° and angle SAC = 52°.
Diagram not to scale
(a)
Find angle ACB. Give a reason for your answer.
(2)
(b)
Find angle BAC.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1: 67°
✓ 1
B1: alternate segment theorem (the angle between a tangent and a chord equals the angle in the alternate segment)
✓ 1
Part (b)
M1: 180 − 67 − 52 (angles on a straight line), or angle ABC = 52° (alternate segment) then 180 − 67 − 52
✓ 1
A1: 61°
✓ 1
Common error: The angle between the tangent and chord AB is equal to the angle at C, the vertex opposite AB. It is not equal to the angle at B. Match each chord with the angle that stands on it.
92IGMC-044-02-08Circle theoremsPaper EMedium7 marks
No calculator
ABCD is a cyclic quadrilateral. Angle ABC = (3x + 10)° and angle ADC = (2x − 5)°.
Diagram not to scale
(a)
Find the value of x.
(3)
(b)
Find angle ADC.
(1)
(c)
The diagonal AC is drawn. Angle BAC = 38°. Find angle ACB.
(1)
(d)
The diagonal BD is drawn. Find angle ADB. Give a reason for your answer.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1: angle ABC + angle ADC = 180° (opposite angles of a cyclic quadrilateral)
✓ 1
M1: 5x + 5 = 180
✓ 1
A1: x = 35
✓ 1
Part (b)
B1: 65°
✓ 1
Part (c)
B1: 180 − 115 − 38 = 27°
✓ 1
Part (d)
B1: 27°
✓ 1
B1: angles in the same segment are equal (both stand on chord AB)
✓ 1
Exam tip: To use 'angles in the same segment', find the chord that both angles stand on. Angles ACB and ADB both stand on AB.
93IGMC-044-02-09Circle theoremsPaper EMedium6 marks
Calculator
A circle has centre O and radius 13 cm. AB is a chord of length 24 cm. M is the midpoint of AB.
Diagram not to scale
(a)
Calculate the distance OM.
(3)
(b)
Calculate angle AOB.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1: OM is perpendicular to AB and AM = 12 cm
✓ 1
M1: OM = √(132 − 122)
✓ 1
A1: 5 cm
✓ 1
Part (b)
M1: sin AOM = 12/13, or tan AOM = 12/5, or cos AOM = 5/13
✓ 1
A1: angle AOM = 67.38...°
✓ 1
A1: angle AOB = 134.8°
✓ 1
Exam tip: The perpendicular from the centre bisects the chord. This splits triangle AOB into two right-angled triangles, so you can use Pythagoras and right-angled trigonometry.
94IGMC-044-02-10Circle theoremsPaper EMedium5 marks
No calculator
P, Q and R are points on a circle. The tangent to the circle at P meets the line RQ extended at T, so that Q lies between T and R. Angle QPT = 34° and angle PTQ = 28°.
Diagram not to scale
(a)
Find angle PRQ. Give a reason for your answer.
(2)
(b)
Find angle PQR.
(2)
(c)
Find angle QPR.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1: 34°
✓ 1
B1: alternate segment theorem
✓ 1
Part (b)
M1: exterior angle of triangle PQT = 34 + 28, or 180 − (180 − 34 − 28)
✓ 1
A1: 62°
✓ 1
Part (c)
B1: 180 − 34 − 62 = 84°
✓ 1
Exam tip: The exterior angle of a triangle equals the sum of the two opposite interior angles. This is quicker than finding angle PQT first.
95IGMC3-4-01d-07Circle theoremsPaper EMedium4 marks
No calculator
A, B and C are points on the circumference of a circular pond with centre O. Angle ACB = 37°, and C lies on the major arc AB.
Diagram not to scale
(a)
Calculate angle AOB and give a reason for your answer.
(2)
(b)
Calculate angle OAB.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 74°
✓ 1
B1 the angle at the centre is twice the angle at the circumference
✓ 1
Part (b)
M1 triangle OAB is isosceles (OA = OB), so angle OAB = (180 − 74) ÷ 2
✓ 1
A1 53°
✓ 1
Common error: Do not stop at angle AOB: triangle OAB has two equal radii, which is the key to every angle at its base.
96IGMC3-4-01d-09Similarity, congruence and symmetryPaper EMedium6 marks
Calculator
Two fuel tanks are mathematically similar. The smaller tank has height 1.8 m and the larger tank has height 2.7 m. The surface area of the smaller tank is 14 m2.
(a)
Calculate the surface area of the larger tank.
(3)
(b)
The volume of the larger tank is 40.5 m3. Calculate the volume of the smaller tank.
Common error: Lengths scale by k, areas by k² and volumes by k³. Applying the length factor to an area or volume is the most frequent slip.
97IGMC3-4-01d-11Circle theoremsPaper EMedium6 marks
No calculator
Two straight ropes PA and PB are tangents to a circular reservoir with centre O. The ropes touch the reservoir at A and B. Angle APB = 50°.
Diagram not to scale
(a)
Calculate angle AOB.
(3)
(b)
Calculate angle OAB.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 angles OAP and OBP are each 90° (a tangent is perpendicular to the radius)
✓ 1
M1 360 − 90 − 90 − 50 (angles in a quadrilateral)
✓ 1
A1 130°
✓ 1
Part (b)
M1 OA = OB, so triangle OAB is isosceles
✓ 1
M1 (180 − 130) ÷ 2
✓ 1
A1 25°
✓ 1
Exam tip: Whenever a tangent appears, draw in the radius to the point of contact: it gives a right angle you can use straight away.
98IGMC3-4-08r8-08Circle theoremsPaper EMedium6 marks
No calculator
The points A, B, C and D lie on the circumference of a circle, in that order, forming the cyclic quadrilateral ABCD used to model a microwave dish rim.
Diagram not to scale
(a)
Angle ABC = 100°. Find angle ADC and give a reason.
(2)
(b)
In a different quadrilateral of this type, angle DAB = (2x + 10)° and angle BCD = (3x − 20)°. Find the value of x.
(3)
(c)
Find angle BCD for this quadrilateral.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 80°
✓ 1
R1 opposite angles of a cyclic quadrilateral add up to 180°
✓ 1
Part (b)
M1 (2x + 10) + (3x − 20) = 180
✓ 1
A1 5x − 10 = 180
✓ 1
A1 x = 38
✓ 1
Part (c)
A1 94°
✓ 1
Exam tip: State the theorem in words, 'opposite angles of a cyclic quadrilateral sum to 180°', as part of your working.
99IGMC3-4-08r8-10Circle theoremsPaper EMedium6 marks
Calculator
A circular flatbread has centre O and radius 13 cm. A chord AB of length 24 cm is cut across it with a knife.
Diagram not to scale
(a)
Calculate the perpendicular distance from O to the chord AB.
(3)
(b)
Calculate angle AOB to 1 decimal place.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 the perpendicular from the centre bisects the chord, giving a right-angled triangle with sides 12 and 13
✓ 1
M1 sqrt(132 − 122)
✓ 1
A1 5 cm
✓ 1
Part (b)
M1 sin(half the angle) = 12 ÷ 13
✓ 1
M1 angle = 2 × 67.38...
✓ 1
A1 134.8°
✓ 1
Exam tip: Draw the perpendicular from the centre to the chord in your own sketch; it splits the problem into two right-angled triangles.
100IGMC3-4-11r30-10Circle theoremsPaper EMedium6 marks
Calculator
A circular garden at a hospital has centre O. Points A, B and C lie on its boundary, with C on the major arc AB. Angle ACB = 37°.
Diagram not to scale
(a)
Find angle AOB. Give a reason for your answer.
(2)
(b)
Find angle OAB. Give a reason for your answer.
(2)
(c)
A fourth point D lies on the minor arc AB. Find angle ADB. Give a reason for your answer.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 74°
✓ 1
B1 the angle at the centre is twice the angle at the circumference
✓ 1
Part (b)
M1 (180 − 74) ÷ 2 (ECF from (a))
✓ 1
A1 53° because OA = OB (radii) so triangle OAB is isosceles
✓ 1
Part (c)
M1 180 − 37
✓ 1
A1 143° because opposite angles of a cyclic quadrilateral ADBC add up to 180°
✓ 1
Common error: The angle at the centre is twice the angle at the circumference only when both angles stand on the same arc.
101IGMC3-4-11r30-11Circle theoremsPaper EMedium6 marks
Calculator
Four hive stands A, B, C and D, in that order, lie on the circumference of a circular clearing. Angle ABC = (3x + 15)° and angle ADC = (2x + 35)°.
Diagram not to scale
(a)
Show that 5x + 50 = 180 and solve it to find x.
(3)
(b)
Find angle ABC.
(1)
(c)
Find angle ADC.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 opposite angles of a cyclic quadrilateral add up to 180°
✓ 1
M1 3x + 15 + 2x + 35 = 180, so 5x + 50 = 180
✓ 1
A1 x = 26
✓ 1
Part (b)
A1 93° (ECF from (a))
✓ 1
Part (c)
M1 2 × 26 + 35 (ECF from (a))
✓ 1
A1 87°
✓ 1
Exam tip: After solving for x, check that your two angles add up to 180°.
102IGMC3-4-12r33-06Similarity, congruence and symmetryPaper EMedium5 marks
Calculator
Two solar thermal hot-water tanks are mathematically similar. The smaller tank has height 1.5 m, surface area 4.5 m² and volume 1.35 m³. The larger tank has height 2.4 m.
(a)
Calculate the surface area of the larger tank.
(2)
(b)
Calculate the volume of the larger tank. Give your answer correct to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 (2.4 ÷ 1.5)² = 2.56
✓ 1
A1 4.5 × 2.56 = 11.52 m²
✓ 1
Part (b)
M1 (2.4 ÷ 1.5)³ = 4.096
✓ 1
M1 1.35 × 4.096
✓ 1
A1 5.53 m³
✓ 1
Common error: Lengths scale by k, areas by k² and volumes by k³; do not use the length factor for areas or volumes.
103IGMC3-4-12r33-09Similarity, congruence and symmetryPaper EMedium7 marks
No calculator
A triangular tent frame at a vaccination site is modelled by triangle ABC, with AB = AC = 13 cm and BC = 10 cm. M is the midpoint of BC.
Diagram not to scale
(a)
Prove that triangles ABM and ACM are congruent.
(3)
(b)
Hence show that AM is perpendicular to BC.
(2)
(c)
Calculate the length AM.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 AB = AC (given)
✓ 1
B1 BM = MC (M is the midpoint)
✓ 1
B1 AM is common, so the triangles are congruent (SSS)
✓ 1
Part (b)
B1 angle AMB = angle AMC (congruent triangles)
✓ 1
B1 the two angles lie on a straight line so sum to 180°, so each is 90°
✓ 1
Part (c)
M1 AM² = 13² − 5²
✓ 1
A1 12 cm
✓ 1
Exam tip: State the congruence condition (SSS, SAS, ASA or RHS) in full and make sure each of the three facts has a reason.
104IGMC3-4-10r37-06Circle theoremsPaper EMedium6 marks
No calculator
Four sensors A, B, C and D are placed in that order around the edge of a circular running track with centre O, so that ABCD is a cyclic quadrilateral. Angle ABC = 112°.
Diagram not to scale
(a)
Find angle ADC and give a geometrical reason.
(2)
(b)
Find angle AOC, where angle AOC is less than 180°.
(2)
(c)
Find angle OAC.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 68°
✓ 1
B1 opposite angles of a cyclic quadrilateral add up to 180°
✓ 1
Part (b)
M1 the angle at the centre is twice the angle at the circumference, 2 × 68
✓ 1
A1 136° (ECF from (a))
✓ 1
Part (c)
M1 triangle OAC is isosceles because OA = OC (radii), so (180 − 136) ÷ 2
✓ 1
A1 22° (ECF from (b))
✓ 1
Common error: The angle at the centre is twice the angle at the circumference only when both angles stand on the same arc; check which arc each angle subtends.
105IGMC3-4-10r37-07Similarity, congruence and symmetryPaper EMedium5 marks
Calculator
A sports science lab uses two mathematically similar drinks bottles. The smaller bottle has height 20 cm. The larger bottle has height 30 cm and holds 1.35 litres.
(a)
Work out the capacity of the smaller bottle.
(3)
(b)
The label on the smaller bottle has an area of 240 cm2. Work out the area of the corresponding label on the larger bottle.
Common error: Lengths scale by k, areas by k² and volumes by k³; do not use the length scale factor for areas or volumes.
106IGMC3-4-10r37-13Circle theoremsPaper EMedium5 marks
No calculator
A circular crater on the Moon has radius 10 km and centre O. A straight survey line crosses the crater from A to B on the rim, with AB = 16 km. The point M is the midpoint of AB.
Diagram not to scale
(a)
Give a reason why angle OMA = 90°.
(1)
(b)
Work out the length OM.
(2)
(c)
Work out the area of triangle OAB.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 a line from the centre of a circle to the midpoint of a chord is perpendicular to the chord
✓ 1
Part (b)
M1 OM2 = 102 − 82
✓ 1
A1 6 km
✓ 1
Part (c)
M1 ½ × 16 × 6
✓ 1
A1 48 km2 (ECF from (b))
✓ 1
Exam tip: Whenever a chord and the centre of a circle appear together, draw the perpendicular from the centre to the chord and look for a right-angled triangle.
107IGMC3-4-10r41-05Similarity, congruence and symmetryPaper EMedium7 marks
Calculator
Two water tanks on a farm are mathematically similar. The smaller tank has height 1.2 m, holds 800 litres and its outer surface area is 6.4 m2. The larger tank has height 1.8 m.
(a)
Write down the length scale factor from the smaller tank to the larger tank.
(1)
(b)
Work out the volume of water that the larger tank holds.
(2)
(c)
Work out the outer surface area of the larger tank.
(2)
(d)
One litre of weatherproof paint covers 4.5 m2. The paint is sold in 1 litre tins at £11.50 each. Work out the cost of the tins needed to give the larger tank one coat.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 1.8 ÷ 1.2 = 1.5
✓ 1
Part (b)
M1 800 × 1.53
✓ 1
A1 2700 litres
✓ 1
Part (c)
M1 6.4 × 1.52
✓ 1
A1 14.4 m2
✓ 1
Part (d)
M1 14.4 ÷ 4.5 = 3.2 litres, so 4 tins are needed
✓ 1
A1 4 × 11.50 = £46.00
✓ 1
Common error: Area scale factor is the length scale factor squared and volume scale factor is cubed; do not just multiply by 1.5.
108IGMC3-4-10r41-08Circle theoremsPaper EMedium6 marks
No calculator
Points A, B, C and D lie in that order on the circle that marks the edge of a radar display, so ABCD is a cyclic quadrilateral. Angle DAB = (2x + 10)° and angle BCD = (3x − 20)°. The diagonal BD is drawn and angle ABD = 40°.
Diagram not to scale
(a)
Write down and solve an equation to find x.
(3)
(b)
Write down the size of angle BCD.
(1)
(c)
The diagonal AC is also drawn. Work out the size of angle ACB, giving a reason.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 opposite angles of a cyclic quadrilateral add up to 180, so (2x + 10) + (3x − 20) = 180
A1 angle ACB = 54° because angles in the same segment are equal
✓ 1
Common error: Opposite angles of a cyclic quadrilateral add up to 180°; adjacent angles do not.
109IGMC3-4-12r44-07Circle theoremsPaper EMedium6 marks
No calculator
Four marker buoys A, B, C and D lie on the circumference of a circular swimming area with centre O. They are in the order A, B, C, D round the circle. Angle ABC = 74°.
Diagram not to scale
(a)
Find angle ADC. Give a reason for your answer.
(2)
(b)
Find angle AOC, the angle at the centre that is less than 180°. Give a reason for your answer.
(2)
(c)
Find angle OAC.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 opposite angles of a cyclic quadrilateral add up to 180°
✓ 1
A1 106°
✓ 1
Part (b)
M1 angle at the centre is twice the angle at the circumference
✓ 1
A1 148°
✓ 1
Part (c)
M1 triangle OAC is isosceles because OA = OC (radii), so angle OAC = (180 − 148) ÷ 2
✓ 1
A1 16°
✓ 1
Common error: Use the right theorem for the right angle: opposite angles in a cyclic quadrilateral add to 180°, but the angle at the centre is double, not supplementary.
110IGMC3-4-12r44-09Similarity, congruence and symmetryPaper EMedium6 marks
Calculator
Two swimming pools are mathematically similar in shape. The small pool has a surface area of 40 m², a depth of 1.2 m and a volume of 48 m³. The large pool has a surface area of 250 m².
(a)
Find the length scale factor from the small pool to the large pool.
(2)
(b)
Find the depth of the large pool.
(2)
(c)
Find the volume of the large pool.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 √(250 ÷ 40)
✓ 1
A1 2.5
✓ 1
Part (b)
M1 1.2 × 2.5
✓ 1
A1 3 m
✓ 1
Part (c)
M1 48 × 2.5³
✓ 1
A1 750 m³
✓ 1
Common error: Area scales with the square of the length scale factor and volume with the cube. Do not multiply the volume by the area ratio.
111IGMC3-4-12r44-12Circle theoremsPaper EMedium5 marks
No calculator
In a circle with centre O, AB is a diameter. C is a point on the circumference with angle BAC = (3y + 1)° and angle ABC = (2y + 4)°. The line BT is the tangent to the circle at B, with T on the same side of AB as C.
Diagram not to scale
(a)
State the size of angle ACB and give a reason.
(1)
(b)
Find the value of y.
(2)
(c)
Find angle TBC. Give a reason for your answer.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 90° (angle in a semicircle)
✓ 1
Part (b)
M1 (3y + 1) + (2y + 4) + 90 = 180
✓ 1
A1 y = 17
✓ 1
Part (c)
M1 angle between tangent and chord equals the angle in the alternate segment (or 90° − angle ABC)
✓ 1
A1 52°
✓ 1
Common error: A diameter always gives a right angle at the circumference. Do not forget to use it as the third angle in the triangle.
112IGMC-044-01-12Similarity, congruence and symmetryPaper EHard7 marks
Calculator
A solid cone has height 20 cm and base radius 9 cm. A small cone of height 8 cm is cut from the top of the cone by a cut parallel to the base. This leaves a frustum. The small cone is mathematically similar to the original cone. [The volume of a cone is \(\frac{1}{3}\pi r^2 h\).]
Diagram not to scale
(a)
Calculate the base radius of the small cone.
(2)
(b)
Show that the volume of the original cone is 540π cm3.
(2)
(c)
Calculate the volume of the frustum.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 9 × 8 ÷ 20
✓ 1
A1 3.6 cm
✓ 1
Part (b)
M1 \(\frac{1}{3}\) × π × 92 × 20
✓ 1
A1 = 540π with working seen
✓ 1
Part (c)
M1 small cone = 540π × (8/20)3 (= 34.56π) or \(\frac{1}{3}\)π × 3.62 × 8
✓ 1
M1 540π − 34.56π
✓ 1
A1 1590 cm3 (1587.9…)
✓ 1
Exam tip: Keep π in your working until the last line. This avoids rounding errors when you subtract the volume of the small cone.
113IGMC-044-01-13Similarity, congruence and symmetryPaper EHard8 marks
Calculator
Triangle ABC has a right angle at A. AB = 9 cm, AC = 12 cm and BC = 15 cm. D is the point on BC such that AD is perpendicular to BC.
Diagram not to scale
(a)
Explain why triangle DBA is similar to triangle ABC.
(2)
(b)
Calculate BD.
(2)
(c)
Calculate AD.
(2)
(d)
Calculate the area of triangle ADC.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 angle ADB = angle CAB = 90°
✓ 1
B1 angle ABD = angle CBA (common angle), so the third angles are also equal
✓ 1
Part (b)
M1 BD/9 = 9/15
✓ 1
A1 5.4 cm
✓ 1
Part (c)
M1 AD/12 = 9/15 or \(\sqrt{9^2 - 5.4^2}\)
✓ 1
A1 7.2 cm
✓ 1
Part (d)
M1 \(\frac{1}{2}\) × 7.2 × (15 − 5.4)
✓ 1
A1 34.56 cm2
✓ 1
Common error: The matching vertices are D↔A, B↔B and A↔C. If you list the vertices in the wrong order, you will pair the wrong sides and get a wrong value for BD.
114IGMC-044-01-14Similarity, congruence and symmetryPaper EHard9 marks
Calculator
ABCD is a square. E is a point on BC and F is a point on CD such that BE = CF. The lines AE and BF meet at P.
Diagram not to scale
(a)
Prove that triangle ABE is congruent to triangle BCF.
(3)
(b)
Hence prove that AE is perpendicular to BF.
(3)
(c)
AB = 10 cm and BE = 4 cm. Calculate BP.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 AB = BC (sides of a square)
✓ 1
B1 angle ABE = angle BCF = 90° (angles of a square) and BE = CF (given)
✓ 1
B1 congruent by SAS
✓ 1
Part (b)
M1 let angle BAE = x; then angle CBF = x (corresponding angles in the congruent triangles) and so angle PBE = x (P lies on BF and E lies on BC)
✓ 1
M1 angle AEB = 90° − x (angle sum of triangle ABE), so angle PEB = 90° − x (P lies on AE)
✓ 1
A1 in triangle BPE, angle BPE = 180° − x − (90° − x) = 90°, so AE is perpendicular to BF
✓ 1
Part (c)
M1 AE = \(\sqrt{10^2 + 4^2}\) = \(\sqrt{116}\)
✓ 1
M1 BP × AE = AB × BE (area of triangle ABE), or use similar triangles BPE and ABE, or BP = 10 sin(angle BAE) with tan(angle BAE) = \(\frac{4}{10}\)
✓ 1
A1 3.71 cm (3.713…)
✓ 1
Exam tip: When you calculate BP, notice that BP is the height of triangle ABE drawn to the side AE. The area is ½ × 10 × 4 = 20 cm², and it is also ½ × AE × BP. Equate the two expressions.
115IGMC-044-00-13Angle properties of lines, triangles and polygonsPaper EHard5 marks
No calculator
Regular polygon A has n sides. Regular polygon B has 2n sides. Each interior angle of B is 15° larger than each interior angle of A.
(a)
Show that \(\frac{180}{n} = 15\).
(3)
(b)
Find the value of n and the size of one interior angle of polygon A.
(2)
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Part (a)
Correct expressions for both interior angles, e.g. \(180 - \frac{360}{n}\) and \(180 - \frac{360}{2n}\), or \(\frac{(n-2)\times180}{n}\) and \(\frac{(2n-2)\times180}{2n}\)
✓ 1
Equation formed with B's angle minus A's angle, e.g. \(\left(180 - \frac{360}{2n}\right) - \left(180 - \frac{360}{n}\right) = 15\)
✓ 1
Correct simplification with no errors, e.g. \(\frac{360}{n} - \frac{180}{n} = \frac{180}{n} = 15\)
✓ 1
Part (b)
n = 12
✓ 1
Interior angle of A = 180 − 360 ÷ 12 = 150°
✓ 1
Common error: Subtracting the interior angles the wrong way round. Polygon B has more sides, so its interior angle is the larger one and its exterior angle is the smaller one.
116IGMC-044-00-14Angle properties of lines, triangles and polygonsPaper EHard6 marks
No calculator
In triangle ABC, D is a point on BC such that AD = BD and AD = AC. Angle ABC = x°.
Diagram not to scale
(a)
Show that angle ADC = 2x°.
(2)
(b)
Find an expression, in terms of x, for angle DAC.
(2)
(c)
Given that angle BAC = 105°, find the value of x.
(2)
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Part (a)
Angle BAD = x (triangle ABD is isosceles because AD = BD) B1
✓ 1
Angle ADC = x + x = 2x (exterior angle of triangle ABD) B1
✓ 1
Part (b)
Angle ACD = 2x (triangle ADC is isosceles because AD = AC) M1
✓ 1
Angle DAC = 180 − 4x A1
✓ 1
Part (c)
x + (180 − 4x) = 105, or 180 − 3x = 105 M1
✓ 1
x = 25 A1
✓ 1
Exam tip: In a 'show that' part, state the reason for each step: isosceles triangle, exterior angle of a triangle. Then use the given result in the later parts even if you could not prove it.
117IGMC-044-02-12Circle theoremsPaper EHard6 marks
No calculator
A, B, C and D are points on a circle. The chords AC and BD meet at X, inside the circle.
Diagram not to scale
(a)
Show that triangle AXB is similar to triangle DXC.
(3)
(b)
AX = 6 cm, XC = 4 cm and BX = 8 cm. Find XD.
(2)
(c)
AB = 7.5 cm. Find DC.
(1)
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Part (a)
B1: angle BAC = angle BDC (angles in the same segment, on chord BC)
✓ 1
B1: angle ABD = angle ACD (angles in the same segment, on chord AD), or angle AXB = angle DXC (vertically opposite angles)
✓ 1
B1: two pairs of equal angles, so the third pair is also equal (angle sum of a triangle) and the triangles are similar, with A↔D, X↔X, B↔C
B1: the scale factor is BX/CX = 8/4 = 2, so DC = 7.5 ÷ 2 = 3.75 cm
✓ 1
Common error: Match the vertices in order: A goes with D, X with X, and B with C. Then AX corresponds to DX, not to XC. Matching the wrong sides gives the wrong ratio.
118IGMC-044-02-13Circle theoremsPaper EHard7 marks
No calculator
T is a point outside a circle with centre O. The tangents from T touch the circle at A and B. C is a point on the major arc AB. Angle ATB = 2x° and angle ACB = (3x − 10)°.
B1: the angles in quadrilateral OATB add up to 360°, so angle AOB = 360 − 90 − 90 − 2x = 180 − 2x
✓ 1
Part (b)
M1: angle ACB = ½(180 − 2x) = 90 − x (angle at centre is twice angle at circumference)
✓ 1
M1: 3x − 10 = 90 − x
✓ 1
A1: x = 25
✓ 1
Part (c)
M1: TA = TB so angle TAB = (180 − 50) ÷ 2, or angle TAB = angle ACB (alternate segment theorem)
✓ 1
A1: 65°
✓ 1
Exam tip: After you find x, check it. Here angle ACB = 65° and angle TAB = 65°, which agrees with the alternate segment theorem.
119IGMC-044-02-14Circle theoremsPaper EHard7 marks
Calculator
A circle has centre O and radius 9 cm. A, B and C are points on the circle, with C on the major arc AB. Angle ACB = 40°.
Diagram not to scale
(a)
Find angle AOB. Give a reason for your answer.
(2)
(b)
Calculate the length of the chord AB.
(2)
(c)
Calculate the area of the minor segment cut off by the chord AB.
(3)
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Part (a)
B1: 80°
✓ 1
B1: the angle at the centre is twice the angle at the circumference
✓ 1
Part (b)
M1: AB = 2 × 9 sin 40°, or AB2 = 92 + 92 − 2 × 9 × 9 cos 80°
✓ 1
A1: 11.6 cm
✓ 1
Part (c)
M1: sector area = (80/360) × π × 92 (= 56.54...)
✓ 1
M1: subtract triangle area ½ × 9 × 9 × sin 80° (= 39.88...)
✓ 1
A1: 16.7 cm2
✓ 1
Common error: The sector and the triangle use the angle at the centre, 80°. Do not use the 40° angle at the circumference.
120IGMC3-4-01d-12Circle theoremsPaper EHard9 marks
No calculator
A, B, C and D are points on a circle, in that order, so ABCD is a cyclic quadrilateral. Angle DAB = (3x + 5)° and angle BCD = (2x + 15)°.
Diagram not to scale
(a)
Form an equation and find the value of x.
(3)
(b)
Find angle DAB.
(1)
(c)
The side BC is extended beyond C to a point E. Calculate angle DCE.
(2)
(d)
Angle ADB = 38°. Calculate angle ACD and give a reason.
(3)
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Part (a)
M1 opposite angles of a cyclic quadrilateral add up to 180°
✓ 1
M1 (3x + 5) + (2x + 15) = 180, so 5x + 20 = 180
✓ 1
A1 x = 32
✓ 1
Part (b)
A1 101° (ECF from (a))
✓ 1
Part (c)
M1 angle BCD = 79° and angles on a straight line add up to 180°
✓ 1
A1 101°
✓ 1
Part (d)
M1 angle ABD = 180 − 101 − 38
✓ 1
A1 41°
✓ 1
B1 angle ACD = angle ABD (angles in the same segment are equal)
✓ 1
Common error: Opposite angles of a cyclic quadrilateral are supplementary, not equal. Equal angles come from the same segment, which is a different theorem.
121IGMC3-4-01d-14Angle properties of lines, triangles and polygonsPaper EHard9 marks
No calculator
A decorative window is a regular pentagon ABCDE with all five vertices on a circle with centre O. The diagonals AC and BD cross at X.
Diagram not to scale
(a)
Calculate angle AOB.
(2)
(b)
Calculate the interior angle ABC of the pentagon.
(2)
(c)
Calculate angle ACB and give a reason.
(2)
(d)
Calculate angle AXB.
(3)
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Part (a)
M1 360 ÷ 5
✓ 1
A1 72°
✓ 1
Part (b)
M1 (5 − 2) × 180 ÷ 5
✓ 1
A1 108°
✓ 1
Part (c)
A1 36°
✓ 1
B1 the angle at the circumference is half the angle at the centre on the same arc (72 ÷ 2)
✓ 1
Part (d)
M1 angle BAX = 36° (same arc BC as in part (c), equal chords)
✓ 1
M1 angle ABX = 108 − 36 = 72°
✓ 1
A1 180 − 36 − 72 = 72°
✓ 1
Exam tip: In a regular polygon inscribed in a circle, equal sides cut off equal arcs, so equal inscribed angles are available all round the shape.
122IGMC3-4-08r8-09Circle theoremsPaper EHard7 marks
Calculator
Two tangents are drawn from a point T outside a circle with centre O, touching the circle at A and B. Angle ATB = 56°. The point C lies on the major arc AB.
Diagram not to scale
(a)
Find angle AOB, giving reasons.
(3)
(b)
Find angle ACB, giving a reason.
(2)
(c)
Triangle TAB is isosceles because TA = TB. Use this to find angle TAB and state how your answer agrees with the alternate segment theorem.
(2)
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Part (a)
B1 angles OAT and OBT are 90° (tangent perpendicular to radius)
✓ 1
M1 angles in quadrilateral OATB sum to 360°: 360 − 90 − 90 − 56
✓ 1
A1 124°
✓ 1
Part (b)
M1 angle at the centre is twice the angle at the circumference: 124 ÷ 2
✓ 1
A1 62°
✓ 1
Part (c)
M1 (180 − 56) ÷ 2 = 62°
✓ 1
R1 it equals angle ACB, the angle in the alternate segment
✓ 1
Common error: Equal tangent lengths from an external point make triangle TAB isosceles; the equal angles are at A and B, not at T.
123IGMC3-4-08r8-13Similarity, congruence and symmetryPaper EHard7 marks
Calculator
A food company sells two similar tins. The small tin has height 10 cm, holds 0.5 litres and uses metal costing €1.40. The large tin has height 15 cm. The cost of the metal is proportional to the surface area of the tin. Use unrounded values in your working.
(a)
Calculate the volume of the large tin, giving your answer to 3 significant figures.
(2)
(b)
Calculate the cost of the metal for the large tin.
(2)
(c)
Calculate the percentage decrease in metal cost per litre of volume from the small tin to the large tin, to 3 significant figures.
(3)
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Part (a)
M1 volume scale factor 1.53 = 3.375
✓ 1
A1 1.6875 litres, so 1.69 litres
✓ 1
Part (b)
M1 area scale factor 1.52 = 2.25 and 1.40 × 2.25
✓ 1
A1 €3.15
✓ 1
Part (c)
M1 small tin: 1.40 ÷ 0.5 = €2.80 per litre
✓ 1
M1 large tin: 3.15 ÷ 1.6875 = €1.8667 per litre
✓ 1
A1 (2.80 − 1.8667) ÷ 2.80 × 100 = 33.3%
✓ 1
Common error: Lengths scale by k, areas by k² and volumes by k³; do not apply the length scale factor to areas or volumes.
124IGMC3-4-08r8-14Circle theoremsPaper EHard6 marks
No calculator
The points A, B, C and D lie on a circular crater rim with centre O. The point B is on the major arc AC and the point D is on the minor arc AC. Angle ABC = (3x + 4)° and angle AOC = (8x − 4)°, where AOC is the angle at the centre, less than 180°.
Diagram not to scale
(a)
Form an equation in x and show that x = 6.
(3)
(b)
Find angle ADC, giving a reason.
(2)
(c)
Find the size of the reflex angle AOC.
(1)
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Part (a)
M1 angle at centre = 2 × angle at circumference: 8x − 4 = 2(3x + 4)
✓ 1
M1 8x − 4 = 6x + 8
✓ 1
A1 2x = 12, so x = 6
✓ 1
Part (b)
M1 ABCD is a cyclic quadrilateral so ADC = 180 − (3 × 6 + 4)
✓ 1
A1 158°
✓ 1
Part (c)
A1 360 − 44 = 316°
✓ 1
Exam tip: Check which arc each point lies on before choosing between 'angle at the centre' and 'cyclic quadrilateral'.
125IGMC3-4-11r30-12Circle theoremsPaper EHard7 marks
Calculator
On a circular skid pad with centre O, two straight barriers meet at a point P outside the circle. They are tangents that touch the circle at A and B. Angle APB = 48°. The point C lies on the major arc AB.
Diagram not to scale
(a)
Calculate angle AOB. Give reasons for your working.
(3)
(b)
Calculate angle OAB.
(2)
(c)
Calculate angle ACB and give a reason.
(2)
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Notes
Part (a)
B1 angles OAP and OBP are 90° (tangent perpendicular to radius)
A1 66° because the angle at the circumference is half the angle at the centre
✓ 1
Common error: Forgetting the two right angles at the points of contact when finding the angle at the centre.
126IGMC3-4-11r30-13Circle theoremsPaper EHard8 marks
Calculator
A circular go-kart track at a hospital charity day is marked on a field. The circle has radius 25 m. A straight rope AB of length 40 m is a chord of the circle and the centre is O.
Diagram not to scale
(a)
Calculate the shortest distance from O to the rope AB.
(3)
(b)
Calculate angle AOB, correct to 1 decimal place.
(3)
(c)
Calculate the length of the minor arc AB. Use the unrounded value of angle AOB and give your answer correct to 3 significant figures.
(2)
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Part (a)
M1 the perpendicular from O bisects the chord, so half the chord is 20 m
✓ 1
M1 252 − 202 = 225
✓ 1
A1 15 m
✓ 1
Part (b)
M1 sin(½AOB) = 20 ÷ 25 (or cos = 15 ÷ 25)
✓ 1
M1 double the angle found (53.13...°)
✓ 1
A1 106.3°
✓ 1
Part (c)
M1 (angle AOB ÷ 360) × 2 × π × 25
✓ 1
A1 46.4 m
✓ 1
Exam tip: Drop a perpendicular from the centre to a chord: it bisects the chord and creates a right-angled triangle.
127IGMC3-4-11r30-14Similarity, congruence and symmetryPaper EHard8 marks
Calculator
A triangular roof panel on a beehive is triangle ABC. The point D is on AB and the point E is on AC so that DE is parallel to BC. AD = 6 cm, DB = 4 cm and DE = 9 cm. The area of triangle ADE is 27 cm2.
Diagram not to scale
(a)
Show that triangles ADE and ABC are similar.
(2)
(b)
Calculate the length of BC.
(2)
(c)
Calculate the area of triangle ABC.
(3)
(d)
Calculate the area of quadrilateral DBCE.
(1)
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Part (a)
B1 angle A is common
✓ 1
B1 angle ADE = angle ABC (corresponding angles, DE parallel to BC), so the triangles are similar (AA)
✓ 1
Part (b)
M1 AB = 10 and scale factor 10 ÷ 6
✓ 1
A1 15 cm
✓ 1
Part (c)
M1 area scale factor (10 ÷ 6)2 (or 25 ÷ 9)
✓ 1
M1 27 × 25 ÷ 9
✓ 1
A1 75 cm2
✓ 1
Part (d)
A1 75 − 27 = 48 cm2 (ECF from (c))
✓ 1
Common error: Use the area scale factor (the square of the length factor) for areas; the length factor alone gives a wrong area.
128IGMC3-4-12r33-07Circle theoremsPaper EHard8 marks
No calculator
The points A, B, C and D lie in that order on the circular boundary of an outdoor testing site, so that ABCD is a cyclic quadrilateral. Angle ABC = 104°, angle CAD = 41° and angle BDC = 29°.
Diagram not to scale
(a)
Find angle ADC and give a reason for your answer.
(2)
(b)
Find angle ACD.
(2)
(c)
Find angle BAC and give a reason for your answer.
(2)
(d)
Find angle BCA.
(2)
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Notes
Part (a)
B1 76°
✓ 1
B1 opposite angles of a cyclic quadrilateral sum to 180°
✓ 1
Part (b)
M1 180° − 41° − 76°
✓ 1
A1 63°
✓ 1
Part (c)
B1 29°
✓ 1
B1 angles in the same segment are equal (angles BDC and BAC stand on chord BC)
✓ 1
Part (d)
M1 180° − 104° − 29°
✓ 1
A1 47° (ECF from (c))
✓ 1
Exam tip: For same-segment angles, check that both angles stand on the same chord and sit on the same side of it.
129IGMC3-4-12r33-08Circle theoremsPaper EHard10 marks
Calculator
A circular fountain with centre O and radius 5 m stands in a retail park. Two straight rails, PA and PB, touch the edge of the fountain at A and B and meet at P, where PA = PB = 12 m.
Diagram not to scale
(a)
Calculate the distance OP.
(3)
(b)
Calculate angle APB, correct to 1 decimal place.
(3)
(c)
Calculate the area bounded by the rails PA and PB and the minor arc AB, correct to 3 significant figures.
(4)
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Part (a)
B1 angle OAP = 90° (tangent is perpendicular to radius)
✓ 1
M1 OP² = 5² + 12²
✓ 1
A1 13 m
✓ 1
Part (b)
M1 angle APO = tan⁻¹(5/12) = 22.62...°
✓ 1
M1 angle APB = 2 × angle APO
✓ 1
A1 45.2°
✓ 1
Part (c)
M1 area of kite OAPB = 2 × ½ × 5 × 12 = 60 m²
✓ 1
M1 angle AOB = 180° − 45.24...° = 134.76...°
✓ 1
M1 sector area = (134.76/360) × π × 5²
✓ 1
A1 30.6 m²
✓ 1
Exam tip: Tangents from an external point are equal in length, so the kite splits into two congruent right-angled triangles.
130IGMC3-4-10r37-09Circle theoremsPaper EHard6 marks
No calculator
A, B and T are points on the circle drawn on the plan of an observatory dome. The straight line PT touches the circle at T. The points P and B lie on opposite sides of the chord TA. Angle PTA = 52° and BT = BA.
Diagram not to scale
(a)
Find angle TBA and give a geometrical reason.
(2)
(b)
Find angle BTA.
(2)
(c)
Find angle PTB.
(2)
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Part (a)
B1 52°
✓ 1
B1 alternate segment theorem: the angle between a tangent and a chord equals the angle in the alternate segment
✓ 1
Part (b)
M1 triangle BTA is isosceles with BT = BA, so angles at T and A are equal, (180 − 52) ÷ 2
✓ 1
A1 64° (ECF from (a))
✓ 1
Part (c)
M1 angle PTB = angle PTA + angle ATB = 52 + 64
✓ 1
A1 116° (ECF from (b))
✓ 1
Common error: For the alternate segment theorem, the angle must be in the segment on the opposite side of the chord from the angle between tangent and chord.
131IGMC3-4-10r37-11Similarity, congruence and symmetryPaper EHard5 marks
Calculator
Two solid glass lenses used in an astronomy experiment are mathematically similar. The smaller lens has a surface area of 45 cm2 and the larger lens has a surface area of 80 cm2. The smaller lens has a mass of 135 g. Both lenses are made from the same glass.
(a)
Show that the ratio of corresponding lengths, smaller : larger, is 3 : 4.
Exam tip: Mass is proportional to volume for the same material, so use the cube of the length scale factor.
132IGMC3-4-10r41-10Circle theoremsPaper EHard8 marks
Calculator
A circular pond has centre O and radius 6.5 m. The point P is outside the pond with OP = 16 m. The straight lines from P that touch the pond are tangents, touching the pond at A and B. Give lengths and angles correct to 3 significant figures.
Diagram not to scale
(a)
Give a reason why angle OAP = 90°.
(1)
(b)
Calculate the length PA.
(2)
(c)
Calculate angle APO.
(2)
(d)
A fence is built round the part of the pond's edge between A and B that is nearer to P (the minor arc). Calculate the length of the fence.
(3)
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Part (a)
B1 a tangent is perpendicular to the radius at the point of contact
Exam tip: Two tangents from the same point are equal in length, so line OP splits the figure into two congruent right-angled triangles.
133IGMC3-4-10r41-14Circle theoremsPaper EHard8 marks
Calculator
A guitar-maker marks a triangular inlay ABC with all three corners on the circular edge of a sound-hole of radius 9 cm. The straight line PAQ is the tangent to the circle at A, where P lies on the opposite side of AB from C, and Q lies on the opposite side of AC from B. Angle PAB = 48° and angle QAC = 67°.
Diagram not to scale
(a)
Write down the size of angle ACB, giving a reason.
(2)
(b)
Write down the size of angle ABC, giving a reason.
(2)
(c)
Work out the size of angle BAC.
(2)
(d)
The chord BC has length 2R sin A, where R is the radius of the circle and A is the angle BAC. Calculate the length of BC. Give your answer correct to 3 significant figures.
(2)
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Notes
Part (a)
B1 48°
✓ 1
B1 alternate segment theorem (angle between tangent and chord equals the angle in the alternate segment)
✓ 1
Part (b)
B1 67°
✓ 1
B1 alternate segment theorem (angle between tangent AQ and chord AC equals the angle in the alternate segment)
✓ 1
Part (c)
M1 angles in triangle ABC add up to 180 (or angles on straight line PAQ add up to 180)
✓ 1
A1 180 − 48 − 67 = 65°
✓ 1
Part (d)
M1 2 × 9 × sin 65°
✓ 1
A1 16.3 cm
✓ 1
Exam tip: The alternate segment is on the opposite side of the chord from the angle formed with the tangent.
134IGMC3-4-12r44-10Circle theoremsPaper EHard6 marks
No calculator
A circle has centre O. The line PT touches the circle at T. A is another point on the circle and angle PTA = 62°. B is a point on the major arc TA.
Diagram not to scale
(a)
Find angle OTA. Give a reason for your answer.
(2)
(b)
Find angle TOA.
(2)
(c)
Find angle TBA. Give a reason for your answer.
(2)
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Notes
Part (a)
B1 angle OTP = 90° (tangent is perpendicular to radius)
✓ 1
A1 28°
✓ 1
Part (b)
M1 triangle OTA is isosceles (OT = OA), so TOA = 180 − 2 × 28
✓ 1
A1 124°
✓ 1
Part (c)
M1 angle at the centre is twice the angle at the circumference
✓ 1
A1 62° (this equals angle PTA, the angle in the alternate segment)
✓ 1
Exam tip: The angle between a tangent and a chord equals the angle in the alternate segment. It is a quick way to check an answer found by other routes.
135IGMC3-4-12r44-13Circle theoremsPaper EHard10 marks
Calculator
A circular wind-farm site has centre O and radius 6 km. Straight roads PA and PB from a village P touch the boundary at A and B (they are tangents). Angle APB = 50°.
Diagram not to scale
(a)
Show that angle AOB = 130°.
(3)
(b)
Calculate the length of the chord AB. Give your answer correct to 3 significant figures.
(3)
(c)
Calculate the length of the road PA.
(2)
(d)
Calculate the area of quadrilateral OAPB. Use the unrounded value of PA. Give your answer correct to 3 significant figures.
(2)
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Notes
Part (a)
B1 angles OAP and OBP are both 90° (tangent perpendicular to radius)
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