Original exam-style questions with full mark schemes. Use the filters to choose level and difficulty, and tick questions to build a printable test.
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Fullmark schemes
Original questions in the style of each exam boardFull worked mark schemes with method and accuracy marksCreated by Evan Kotronis with AI assistanceUpdated 15 October 2026 · new questions added continuouslyFree access until 30 November 2026How every question is checked →
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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.
(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.
(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.
(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)°.
(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.
(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°.
(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°.
(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)°.
(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°.
(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.
13IGMC-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.
(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.
14IGMC-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.
15IGMC-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.
16IGMC-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)°.
(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°.
17IGMC-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.
(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.
18IGMC-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)°.
(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°.
19IGMC-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°.
(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.
20IGMC-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.
(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.
21IGMC-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.
(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.
22IGMC-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.
23IGMC-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°.
(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.
24IGMC-044-02-03Circle theoremsPaper EEasy3 marks
No calculator
PQRS is a cyclic quadrilateral. Angle PQR = 104° and angle QRS = 71°.
(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.
25IGMC-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°.
(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°.
26IGMC-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).
27IGMC-044-01-08Similarity, congruence and symmetryPaper EMedium6 marks
No calculator
ABCD is a parallelogram. The diagonal AC is drawn.
(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.
28IGMC-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.
(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.
29IGMC-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
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Part (d)
B1 6 planes containing the axis through the hexagon centres (3 through opposite vertices, 3 through midpoints of opposite edges)
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B1 total 7, including the plane parallel to the hexagonal faces, halfway along the prism
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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.
30IGMC-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°.
(a)
Find angle ACB. Give a reason for your answer.
(2)
(b)
Find angle BAC.
(2)
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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
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A1: 61°
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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.
31IGMC-044-02-08Circle theoremsPaper EMedium7 marks
No calculator
ABCD is a cyclic quadrilateral. Angle ABC = (3x + 10)° and angle ADC = (2x − 5)°.
(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)
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Part (a)
B1: angle ABC + angle ADC = 180° (opposite angles of a cyclic quadrilateral)
✓ 1
M1: 5x + 5 = 180
✓ 1
A1: x = 35
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Part (b)
B1: 65°
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Part (c)
B1: 180 − 115 − 38 = 27°
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Part (d)
B1: 27°
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B1: angles in the same segment are equal (both stand on chord AB)
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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.
32IGMC-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.
(a)
Calculate the distance OM.
(3)
(b)
Calculate angle AOB.
(3)
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Part (a)
B1: OM is perpendicular to AB and AM = 12 cm
✓ 1
M1: OM = √(132 − 122)
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A1: 5 cm
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Part (b)
M1: sin AOM = 12/13, or tan AOM = 12/5, or cos AOM = 5/13
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A1: angle AOM = 67.38...°
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A1: angle AOB = 134.8°
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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.
33IGMC-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°.
(a)
Find angle PRQ. Give a reason for your answer.
(2)
(b)
Find angle PQR.
(2)
(c)
Find angle QPR.
(1)
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Part (a)
B1: 34°
✓ 1
B1: alternate segment theorem
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Part (b)
M1: exterior angle of triangle PQT = 34 + 28, or 180 − (180 − 34 − 28)
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A1: 62°
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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.
34IGMC-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 \(\frac{1}{3}\pi r^2 h\).]
(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)
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Part (a)
M1 9 × 8 ÷ 20
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A1 3.6 cm
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Part (b)
M1 \(\frac{1}{3}\) × π × 92 × 20
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A1 = 540π with working seen
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Part (c)
M1 small cone = 540π × (8/20)3 (= 34.56π) or \(\frac{1}{3}\)π × 3.62 × 8
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M1 540π − 34.56π
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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.
35IGMC-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.
(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)
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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
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A1 5.4 cm
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Part (c)
M1 AD/12 = 9/15 or \(\sqrt{9^2 - 5.4^2}\)
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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.
36IGMC-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.
(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)
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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 = \(\sqrt{10^2 + 4^2}\) = \(\sqrt{116}\)
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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.
37IGMC-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
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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.
38IGMC-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°.
(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.
39IGMC-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.
(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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Notes
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.
40IGMC-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)°.
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