O Level Mathematics D (4024)
Geometry Banker
95 questions · about 10% of the marks we indexed · examined in 9 of 9 sessions.
Practice questions
Original questions written by HomiLabs in the style of the exam, with answers and mark schemes. They are not
Cambridge past paper questions and have not been reviewed by Cambridge. Try each one before opening the answer.
Q1Circle theorems · Core · Paper 1 style · no calculator · 4 marksA, B and C are points on a circle, centre O. B lies on the major arc AC. Angle AOC = 136°.
- (a) Find angle ABC. Give a reason for your answer. [2]
- (b) Find angle OAC. Give a reason for your answer. [2]
Show answer
- (a) 68°
136 ÷ 2 = 68 M1; reason: angle at the centre is twice the angle at the circumference A1.
- (b) 22°
OA = OC (radii) so triangle OAC is isosceles M1; (180 − 136) ÷ 2 = 22 A1.
A, C, B and D lie on a circle in that order. AB is a diameter. Angle CAB = 27°.
- (a) Find angle ACB. Give a reason. [1]
- (b) Find angle ABC. [1]
- (c) Find angle ADC. Give a reason. [2]
- (d) Find angle CDB. Give a reason. [1]
Show answer
- (a) 90°
Angle in a semicircle B1
- (b) 63°
180 − 90 − 27 B1
- (c) 63°
Equals angle ABC B1; angles in the same segment (both subtended by AC from the same arc) B1
- (d) 27°
Equals angle CAB, angles in the same segment B1
P, Q, R and S lie on a circle in that order. PR is a diameter. Angle QPR = 31° and angle PRS = 47°.
- (a) Find angle PQR. Give a reason. [1]
- (b) Find angle QRP. [1]
- (c) Find angle QSR. Give a reason. [2]
- (d) Find angle QPS. Give a reason. [2]
Show answer
- (a) 90°
Angle in a semicircle B1.
- (b) 59°
180 − 90 − 31 = 59 B1.
- (c) 31°
Angle QSR = angle QPR M1; angles in the same segment A1.
- (d) 74°
Angle QRS = 59 + 47 = 106 M1; opposite angles of a cyclic quadrilateral add to 180°, 180 − 106 = 74 A1.
A, B, C and D lie on a circle in that order, so ABCD is a cyclic quadrilateral. Angle ABC = 112°. The line TC is the tangent to the circle at C, with T on the opposite side of the line CD from A. Angle TCD = 35°.
- (a) Find angle ADC. Give a reason. [2]
- (b) Find angle CAD. Give a reason. [2]
- (c) Find angle ACD. [1]
Show answer
- (a) 68°
180 − 112 M1; opposite angles of a cyclic quadrilateral add to 180° A1
- (b) 35°
Angle between tangent and chord equals angle in the alternate segment B2
- (c) 77°
180 − 68 − 35 B1
A, B and C lie on a circle, centre O. C lies on the major arc AB. TA and TB are tangents to the circle at A and B. Angle TAB = 58°.
- (a) Find angle TBA. Give a reason. [2]
- (b) Find angle ATB. [1]
- (c) Find angle AOB. Give a reason. [2]
- (d) Find angle ACB. Give a reason. [1]
- (e) D is a point on the minor arc AB. Find angle ADB. Give a reason. [1]
Show answer
- (a) 58°
TA = TB (tangents from an external point are equal) M1; triangle TAB isosceles so angle TBA = 58 A1.
- (b) 64°
180 − 58 − 58 = 64 B1.
- (c) 116°
Angles OAT = OBT = 90° (tangent ⟂ radius) M1; 360 − 90 − 90 − 64 = 116 A1.
- (d) 58°
Alternate segment theorem (angle TAB = angle ACB), or half of angle AOB B1.
- (e) 122°
180 − 58 = 122; opposite angles of cyclic quadrilateral ACBD B1.
A, B and C lie on a circle, centre O. O lies inside triangle ABC. Angle OAB = x° and angle OCB = 2x°.
- (a) Explain why angle OBA = x° and angle OBC = 2x°. [1]
- (b) Angle AOC (the angle at O facing the minor arc AC) is 132°. Find x. Give a reason. [3]
- (c) Find angle OAC. [1]
- (d) D is a point on the minor arc AC. Find angle ADC. [1]
Show answer
- (a) OA = OB = OC (radii), so triangles OAB and OCB are isosceles and their base angles are equal.
B1
- (b) x = 22
Angle ABC = 3x B1; angle at centre is twice angle at circumference: 6x = 132 M1; x = 22 A1
- (c) 24°
(180 − 132) ÷ 2, isosceles triangle OAC B1
- (d) 114°
ABCD cyclic: 180 − 66 B1
- (a) Find the size of one interior angle of a regular decagon (10 sides). [2]
- (b) Each exterior angle of a regular polygon is 24°. How many sides does the polygon have? [1]
Show answer
- (a) 144°
Exterior angle 360 ÷ 10 = 36 M1; 180 − 36 = 144 A1.
- (b) 15
360 ÷ 24 = 15 B1.
- (a) Find the sum of the interior angles of an octagon. [1]
- (b) Find the size of one interior angle of a regular octagon. [1]
- (c) Each interior angle of a regular polygon is 156°. Find the number of sides. [2]
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- (a) 1080°
(8 − 2) × 180 B1
- (b) 135°
1080 ÷ 8 B1
- (c) 15
Exterior angle 24° M1; 360 ÷ 24 A1
The interior angles of a pentagon are x°, 2x°, (2x + 10)°, (3x − 20)° and 110°.
- (a) Find the value of x. [3]
- (b) Find the size of the largest angle of the pentagon. [1]
Show answer
- (a) x = 55
Angle sum of pentagon = 540 B1; 8x + 100 = 540 M1; x = 55 A1.
- (b) 145°
3(55) − 20 = 145 B1.
ABCD is a trapezium with AB parallel to DC. AD = BD. Angle DAB = 68° and angle BCD = 75°.
- (a) Find angle ADB. [2]
- (b) Find angle BDC. Give a reason. [2]
- (c) Find angle DBC. [1]
Show answer
- (a) 44°
Triangle ABD is isosceles, angle ABD = 68° M1; 180 − 136 A1
- (b) 68°
Equals angle ABD B1; alternate angles, AB parallel to DC B1
- (c) 37°
180 − 68 − 75 B1
- (a) The interior angle of a regular polygon is 4 times its exterior angle. Find the number of sides. [3]
- (b) Regular polygon P has n sides and regular polygon Q has (n + 3) sides. Each exterior angle of P is 6° larger than each exterior angle of Q. Show that n(n + 3) = 180 and hence find n. [4]
Show answer
- (a) 10
e + 4e = 180 M1; e = 36 A1; 360 ÷ 36 = 10 A1.
- (b) n = 12
360/n − 360/(n + 3) = 6 M1; 360(n + 3) − 360n = 6n(n + 3) ⇒ 1080 = 6n(n + 3) ⇒ n(n + 3) = 180 A1; n² + 3n − 180 = 0, (n + 15)(n − 12) = 0 M1; n = 12 A1.
The exterior angles of a hexagon are x°, (x + 10)°, (x + 20)°, (x + 30)°, (x + 40)° and (x + 50)°.
- (a) Find x. [2]
- (b) Find the largest interior angle of the hexagon. [1]
- (c) Show that no regular polygon has an interior angle of 100°. [2]
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- (a) x = 35
6x + 150 = 360 M1; x = 35 A1
- (b) 145°
Smallest exterior angle 35°, 180 − 35 B1
- (c) The exterior angle would be 80°, and 360 ÷ 80 = 4.5 is not a whole number.
Exterior 80° M1; 4.5 not an integer, so impossible A1
- (a) Triangle ABC has AB = AC. M is the midpoint of BC. Show that triangles ABM and ACM are congruent, giving the condition you use. [2]
- (b) Triangles PQR and XYZ are similar, with P corresponding to X, Q to Y and R to Z. PQ = 6 cm, QR = 8 cm, PR = 7 cm and XY = 9 cm. Find YZ and XZ. [3]
Show answer
- (a) Congruent (SSS)
AB = AC (given), BM = CM (M midpoint), AM common M1; congruent, SSS A1.
- (b) YZ = 12 cm, XZ = 10.5 cm
Scale factor 9 ÷ 6 = 1.5 M1; YZ = 12 A1; XZ = 10.5 A1.
At the same time of day, a girl 1.6 m tall casts a shadow 2.4 m long and a minaret casts a shadow 51 m long. The ground is horizontal.
- (a) Find the height of the minaret. [2]
- (b) A photograph 10 cm wide and 15 cm long is enlarged so that it is 25 cm wide. Find its new length. [2]
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- (a) 34 m
Similar triangles: 51 ÷ 2.4 = 21.25 M1; × 1.6 A1
- (b) 37.5 cm
Scale factor 2.5 M1 A1
Two juice bottles are mathematically similar. The smaller bottle is 12 cm tall and the larger bottle is 18 cm tall.
- (a) The volume of the smaller bottle is 320 cm³. Find the volume of the larger bottle. [2]
- (b) The surface area of the larger bottle is 495 cm². Find the surface area of the smaller bottle. [2]
Show answer
- (a) 1080 cm³
Volume factor (18/12)³ = 3.375 M1; 320 × 3.375 = 1080 A1.
- (b) 220 cm²
Area factor (12/18)² = 4/9 M1; 495 × 4/9 = 220 A1.
ABCD is a square. E is a point on BC and F is a point on CD such that BE = CF.
- (a) Prove that triangles ABE and BCF are congruent. [3]
- (b) Hence explain why AE = BF. [1]
Show answer
- (a) AB = BC (sides of square); angle ABE = angle BCF = 90°; BE = CF (given). Congruent by SAS.
B1 for each pair of equal sides/angles with reason, including the condition SAS (max 3)
- (b) They are corresponding sides of congruent triangles.
B1
In triangle ABC, D lies on AB and E lies on AC so that DE is parallel to BC. AD = 4 cm, DB = 6 cm and DE = 5 cm.
- (a) Explain why triangle ADE is similar to triangle ABC. [2]
- (b) Find BC. [2]
- (c) The area of triangle ADE is 12 cm². Find the area of the trapezium DBCE. [3]
Show answer
- (a) Angle A common; angle ADE = angle ABC (corresponding angles, DE ∥ BC), so the triangles are similar (AA).
Angle A common B1; corresponding angles equal with reason B1.
- (b) 12.5 cm
Scale factor AB/AD = 10/4 = 2.5 M1; BC = 5 × 2.5 = 12.5 A1.
- (c) 63 cm²
Area factor 2.5² = 6.25 M1; area ABC = 75 A1; 75 − 12 = 63 A1.
ABCD is a trapezium with AB parallel to DC. The diagonals AC and BD meet at X. AB = 6 cm, DC = 15 cm and AX = 4 cm.
- (a) Explain why triangles ABX and CDX are similar. [2]
- (b) Find CX. [2]
- (c) The area of triangle ABX is 8 cm². Find the area of triangle CDX. [2]
- (d) Find the area of triangle ADX and hence the area of the trapezium ABCD. [2]
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- (a) Angle AXB = angle CXD (vertically opposite); angle BAX = angle DCX (alternate angles, AB ∥ DC). So all angles are equal.
B1 B1
- (b) 10 cm
Scale factor 15/6 = 2.5 M1; 4 × 2.5 A1
- (c) 50 cm²
Area factor 2.5² = 6.25 M1 A1
- (d) 20 cm²; 98 cm²
Same height from A, DX = 2.5 BX: 20 M1; ABCD = 8 + 50 + 20 + 20 A1
A map is drawn to a scale of 1 : 50 000.
- (a) On the map, the distance between two villages is 7.4 cm. Find the actual distance in kilometres. [2]
- (b) A straight road is 2.5 km long. How long is it on the map, in centimetres? [1]
- (c) The bearing of a mosque M from a school S is 064°. Find the bearing of S from M. [1]
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- (a) 3.7 km
7.4 × 50 000 = 370 000 cm M1; = 3.7 km A1.
- (b) 5 cm
250 000 ÷ 50 000 = 5 B1.
- (c) 244°
64 + 180 = 244 B1.
A boat sails 12 km from port P on a bearing of 050° to a buoy Q. It then sails 9 km on a bearing of 140° to an island R.
- (a) Show that angle PQR = 90°. [2]
- (b) Find the distance PR. [2]
- (c) Find the bearing of R from P. Give your answer correct to the nearest 0.1°. [3]
Show answer
- (a) Angle PQR = 90°
Bearing of P from Q is 230° M1; 230 − 140 = 90 A1.
- (b) 15 km
√(12² + 9²) M1; = 15 A1.
- (c) 086.9°
tan QPR = 9/12 M1; angle QPR = 36.87° A1; 50 + 36.87 = 086.9° A1.
Town Q is due east of town P. Town R is on a bearing of 035° from P and on a bearing of 310° from Q.
- (a) Find angle RPQ and angle RQP. [2]
- (b) Find angle PRQ. [1]
- (c) Find the bearing of P from R. [1]
- (d) On a map with scale 1 : 200 000, PQ is 6 cm. Find the actual distance PQ in kilometres. [2]
Show answer
- (a) Angle RPQ = 55°, angle RQP = 40°
90 − 35 B1; 310 − 270 B1
- (b) 85°
180 − 55 − 40 B1
- (c) 215°
035 + 180 B1
- (d) 12 km
6 × 200 000 = 1 200 000 cm M1; 12 km A1
A lighthouse L is 8 km from a harbour H on a bearing of 120°. A boat B is 5 km from H on a bearing of 240°.
- (a) Show that angle LHB = 120°. [1]
- (b) Find the distance LB. Give your answer correct to 3 significant figures. [3]
- (c) Find the bearing of B from L. Give your answer correct to the nearest degree. [3]
- (d) A map has a scale of 1 : 200 000. Find the length of LB on the map, in centimetres, correct to 1 decimal place. [1]
Show answer
- (a) 240 − 120 = 120
240 − 120 = 120 B1.
- (b) 11.4 km
LB² = 8² + 5² − 2 × 8 × 5 × cos 120° M1; = 129 A1; LB = 11.4 A1.
- (c) 278°
sin HLB / 5 = sin 120° / 11.36 M1; angle HLB = 22.4° A1; bearing of H from L is 300°, 300 − 22.4 = 277.6 ≈ 278° A1.
- (d) 5.7 cm
1 135 800 ÷ 200 000 = 5.68 ≈ 5.7 B1.
A is the point (2, 1) and B is the point (8, 1).
- (a) Describe the locus of points that are equidistant from A and B. Give its equation. [2]
- (b) Describe the locus of points that are 3 units from A. [1]
- (c) Find the coordinates of any point that lies on both loci. [1]
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- (a) The perpendicular bisector of AB; x = 5
Perpendicular bisector B1; x = 5 B1.
- (b) A circle, centre A, radius 3 units
B1.
- (c) (5, 1)
Distance from A to the line x = 5 is 3, so the line touches the circle at (5, 1) only B1.
On a farm plan, A is the point (1, 2) and B is the point (7, 6).
- (a) Find the midpoint of AB. [1]
- (b) Find the equation of the perpendicular bisector of AB. Give your answer in the form y = mx + c. [3]
- (c) A well W is to be dug on the y-axis so that it is the same distance from A and from B. Find the coordinates of W. [1]
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- (a) (4, 4)
B1.
- (b) y = −3/2 x + 10
Gradient AB = 4/6 = 2/3 B1; perpendicular gradient −3/2 M1; y − 4 = −3/2(x − 4) ⇒ y = −3/2 x + 10 A1.
- (c) (0, 10)
Put x = 0 in the bisector B1.
ABCD is a rectangular field with AB = 20 m and BC = 12 m. A goat is tied to a post at corner A with a rope 8 m long.
- (a) Describe the region that the goat can reach and find its area in terms of π. [2]
- (b) Describe the locus of points in the field that are the same distance from AB as from AD. [1]
- (c) Find, in terms of π, the area of the part of the field the goat can reach that is nearer to AB than to AD. [2]
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- (a) 16π m² (a quarter circle, centre A, radius 8 m)
Quarter circle B1; ¼ × π × 64 B1
- (b) The bisector of angle DAB (a line from A at 45° to AB).
B1
- (c) 8π m²
Half of the quarter circle M1 A1
A rectangular garden ABCD has A at (0, 0), B at (12, 0), C at (12, 8) and D at (0, 8). Units are metres.
- (a) Describe the locus of points inside the garden that are equidistant from AB and AD, and give its equation. [2]
- (b) Show that the perpendicular bisector of AC has equation y = −1.5x + 13. [2]
- (c) A tree T is planted so that it is equidistant from AB and AD and also equidistant from A and C. Find the coordinates of T. [2]
- (d) Find the distance AT, correct to 3 significant figures. [1]
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- (a) The bisector of angle DAB; y = x
Angle bisector of angle A B1; y = x B1.
- (b) y = −1.5x + 13
Midpoint of AC (6, 4) and gradient of AC 2/3 M1; perpendicular gradient −3/2, y − 4 = −1.5(x − 6) A1.
- (c) (5.2, 5.2)
x = −1.5x + 13 M1; x = 5.2 A1.
- (d) 7.35 m
√(5.2² + 5.2²) = 7.35 B1.
- (a) A regular hexagon: write down the number of lines of symmetry and the order of rotational symmetry. [1]
- (b) A parallelogram that is not a rectangle or a rhombus: write down the number of lines of symmetry and the order of rotational symmetry. [1]
- (c) An isosceles trapezium: write down the number of lines of symmetry and the order of rotational symmetry. [1]
Show answer
- (a) 6 lines; order 6
B1.
- (b) 0 lines; order 2
B1.
- (c) 1 line; order 1
B1.
- (a) A regular polygon has rotational symmetry of order 8. Write down the number of lines of symmetry it has. [1]
- (b) Find the size of one interior angle of this polygon. [2]
- (c) A quadrilateral has exactly 2 lines of symmetry and rotational symmetry of order 2. Name two quadrilaterals that fit this description. [2]
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- (a) 8
A regular n-sided polygon has n lines of symmetry B1.
- (b) 135°
Exterior angle 360 ÷ 8 = 45 M1; 180 − 45 = 135 A1.
- (c) Rectangle (not a square) and rhombus (not a square)
B1 each. A square has 4 lines, so it does not fit.
A circle has centre O and radius 13 cm. AB is a chord of length 24 cm.
- (a) Use the symmetry of a circle to find the perpendicular distance from O to AB. [2]
- (b) CD is another chord, parallel to AB, with length 10 cm. Find the two possible distances between AB and CD. [3]
Show answer
- (a) 5 cm
Perpendicular from centre bisects chord, half-chord 12 M1; √(13² − 12²) = 5 A1.
- (b) 17 cm or 7 cm
Distance O to CD = √(13² − 5²) = 12 M1 A1; 12 + 5 = 17 or 12 − 5 = 7 A1.
- (a) Write down the type of each angle: (i) 137°, (ii) 215°, (iii) 90°. [2]
- (b) An angle is 38°. Find its complement and its supplement. [2]
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- (a) (i) obtuse (ii) reflex (iii) right angle
All three B2; two correct B1.
- (b) Complement 52°, supplement 142°
90 − 38 = 52 B1; 180 − 38 = 142 B1.
PQ and RS are parallel lines. A straight line crosses PQ at X and RS at Y, with P and S on opposite sides of XY. Angle PXY = (3x + 10)° and angle XYS = (5x − 30)°.
- (a) Write down the name of the pair of angles PXY and XYS. [1]
- (b) Find x. [2]
- (c) Find angle XYR. [1]
Show answer
- (a) Alternate angles
B1.
- (b) x = 20
3x + 10 = 5x − 30 M1; x = 20 A1.
- (c) 110°
Angle XYS = 70, 180 − 70 = 110 (angles on a straight line) B1.
In triangle ABC, angle A = (x + 15)°, angle B = 2x° and angle C = (3x − 15)°.
- (a) Find x and the three angles of triangle ABC. [3]
- (b) Describe triangle ABC using two geometrical terms, one for its angles and one for its sides. [1]
- (c) Side BC is extended beyond C to a point E. Find the exterior angle ACE. [1]
- (d) The angles of another triangle are in the ratio 1 : 1 : 2. Name this type of triangle. [1]
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- (a) x = 30; A = 45°, B = 60°, C = 75°
6x = 180 M1; x = 30 A1; angles A1.
- (b) Acute-angled and scalene
B1.
- (c) 105°
180 − 75 = 105 (or 45 + 60) B1.
- (d) Right-angled isosceles
Angles 45°, 45°, 90° B1.
The diagonals of a rhombus are 10 cm and 24 cm long.
- (a) Find the length of a side of the rhombus. [2]
- (b) Find the area of the rhombus. [1]
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- (a) 13 cm
Diagonals bisect at right angles, half-diagonals 5 and 12 M1; √(25 + 144) = 13 A1.
- (b) 120 cm²
½ × 10 × 24 = 120 B1.
ABCD is a kite with AB = AD and CB = CD. Angle BAD = 64° and angle BCD = 118°.
- (a) Find angle ABC. [2]
- (b) Find angle BAC. Give a reason. [1]
- (c) The diagonals AC and BD meet at M. Write down angle AMB and hence find angle ABD. [2]
Show answer
- (a) 89°
Angle ABC = angle ADC (kite symmetry) M1; (360 − 64 − 118) ÷ 2 = 89 A1.
- (b) 32°
The diagonal AC bisects angle BAD (line of symmetry) B1.
- (c) Angle AMB = 90°; angle ABD = 58°
Diagonals of a kite cross at right angles B1; 180 − 90 − 32 = 58 B1.
ABCD is a parallelogram with AB = 9 cm, AD = 6 cm and angle DAB = 50°.
- (a) Find angle ABC. Give a reason. [1]
- (b) Find the area of the parallelogram, correct to 3 significant figures. [2]
- (c) Find the length of the longer diagonal AC, correct to 3 significant figures. [3]
Show answer
- (a) 130°
Co-interior angles of a parallelogram add to 180° B1.
- (b) 41.4 cm²
9 × 6 × sin 50° M1; = 41.4 A1.
- (c) 13.7 cm
AC² = 9² + 6² − 2 × 9 × 6 × cos 130° M1; = 186.4 A1; AC = 13.7 A1.
