About Geometry in O Level Mathematics
Geometry is one of the 5 main topics in our O Level Mathematics (Cambridge (CAIE) 4024) question bank, and about 10% of its real questions come from it. Tayyari Karo has 39 real questions on this topic from 2021 to 2026, from Paper 1, Paper 2. By difficulty: 22 medium, 9 hard, 8 easy.
Work through the questions below, then open the explanation for each one. Redo the ones you missed a few days later.
Try 8 real questions
These are real past-paper questions. Think of your answer first, then open the explanation.
May/June 2025 · Paper 1 · Geometry
AEC and BED are straight lines, so lines AC and BD cross at E. ED = EC, so triangle CED is isosceles. Angle EDC = 35° (at D) and angle ABE = 48° (at B). Find the value of x, the angle at A in triangle ABE.
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Triangle CED is isosceles with ED = EC, so angle ECD = angle EDC = 35°.
Angle DEC = 180 − 35 − 35 = 110°
Angle AEB = angle DEC = 110° (vertically opposite angles, since AEC and BED are straight lines)
In triangle ABE: x + 110 + 48 = 180
x = 180 − 110 − 48
x = 22
Use the isosceles triangle to find the angle at E, then vertically opposite angles and the angle sum of a triangle.
Oct/Nov 2025 · Paper 1 · Geometry
Points Q, R, S and T lie on a circle, centre O. UV is a tangent to the circle at T. SOQV is a straight line (so SQ is a diameter, extended to meet the tangent at V).
Angle QST = 32°.
(a) Find angle QRT. Give a geometrical reason.
(b) Find angle TQS.
(c) Find angle TVQ.
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(a) Angle QRT and angle QST both stand on the same arc QT, so
Angle QRT = 32° (angles in the same segment are equal)
(b) SQ is a diameter, so angle STQ = 90° (angle in a semicircle).
In triangle STQ: angle QST = 32°, angle STQ = 90°
Angle TQS = 180 − 90 − 32
= 58°
(c) OT = OQ (radii), so triangle OTQ is isosceles.
Angle OQT = angle TQS = 58° (same angle, since O lies on SQ), so angle OTQ = 58° too.
UV is a tangent at T, so angle OTV = 90° (radius ⊥ tangent).
Angle QTV = angle OTV − angle OTQ = 90 − 58 = 32°
In triangle QTV: angle TQV = 180 − 58 = 122° (angles on a straight line SQV, since V is on the opposite side of Q from S)
Angle TVQ = 180 − 122 − 32
= 26°
Use angles in the same segment, the angle in a semicircle, the isosceles triangle formed by two radii, and the tangent–radius right angle.
May/June 2024 · Paper 1 · Geometry
A, B, C and D are points on the circumference of a circle, centre O. Angle BCD = 58° and angle DBC = 72°.
(a) Complete the statement: Angle DAB = ___ because ___.
(b) (i) Find angle DOC.
(ii) Find angle BCO.
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(a) ABCD is a cyclic quadrilateral, and angle BCD is opposite angle DAB.
Angle DAB = 180 − 58 = 122°, because opposite angles in a cyclic quadrilateral sum to 180°.
(b)(i) Angle DBC = 72° is the angle at the circumference standing on arc DC.
Angle DOC (at the centre, same arc) = 2 × 72 = 144°
(b)(ii) OB = OC (both radii), so triangle OCD is isosceles with angle DOC = 144°.
Base angles: angle ODC = angle OCD = (180 − 144)/2 = 18°
Angle BCD = angle BCO + angle OCD
58 = angle BCO + 18
angle BCO = 40°
Use the cyclic quadrilateral rule (opposite angles sum to 180°) and the circle theorem that the angle at the centre is twice the angle at the circumference; radii form isosceles triangles.
May/June 2023 · Paper 1 · Geometry
ABC is a straight line (A, B and C in that order). BCD is an equilateral triangle drawn so that D is not on the line ABC.
The angle y° is the angle ABD, that is, the angle between BA and BD, measured at B.
Work out the value of y.
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BCD is equilateral, so angle DBC = 60°.
ABC is a straight line, so angles on the line at B add to 180°:
angle ABD + angle DBC = 180°
y + 60 = 180
y = 120
Angles on a straight line sum to 180°; every angle of an equilateral triangle is 60°.
Oct/Nov 2023 · Paper 1 · Geometry
(a) Four exterior angles of a pentagon are 150°, 100°, 45° and 35°.
Calculate the size of the remaining exterior angle.
(b) Calculate the interior angle of a regular decagon.
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(a) The exterior angles of any polygon add up to 360°.
Sum of the four given angles = 150 + 100 + 45 + 35 = 330
Remaining exterior angle = 360 − 330 = 30°
(b) A regular decagon has 10 equal exterior angles, so each exterior angle = 360 ÷ 10 = 36°.
Interior angle = 180 − 36 = 144°
Exterior angles of any polygon sum to 360°; interior + exterior = 180° at each vertex.
May/June 2022 · Paper 1 · Geometry
B, D, E, F and G are points on the circumference of a circle, centre O. AC is a tangent to the circle at B.
Angle DFG = 75° and angle ABG = 48°.
(a) Find angle DEG.
(b) Find angle DOG.
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(a) Angles DFG and DEG are both angles in the same segment, standing on the same arc DG.
Angle DEG = angle DFG = 75°
(b) The angle at the centre is twice the angle at the circumference standing on the same arc DG.
Angle DOG = 2 × 75° = 150°
Angles in the same segment on the same arc are equal; the angle at the centre is twice the angle at the circumference.
May/June 2021 · Paper 1 · Geometry
Quadrilateral ABCD has AD extended beyond D to a point E.
Angle BCD = 135°, angle BAD = 83° and angle CDE = 122° (the exterior angle at D, between DC and DE).
Angle ABC = x°.
Find the value of x.
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Angle CDE is the exterior angle at D, so the interior angle ADC = 180 − 122 = 58°
The interior angles of quadrilateral ABCD sum to 360°:
angle BAD + angle ABC + angle BCD + angle CDA = 360
83 + x + 135 + 58 = 360
276 + x = 360
x = 84
Find the missing interior angle at D from its exterior angle, then use angle sum of a quadrilateral = 360°.
May/June 2021 · Paper 2 · Geometry
Triangle KLM is similar to triangle KNJ, with K common to both triangles, L corresponding to N, and M corresponding to J. JKM and NKL are straight lines.
KL = 6 cm, KM = 10.5 cm and KN = 10 cm. The area of triangle JKN is 75 cm².
Find the area, y cm², of triangle KLM.
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Since triangle KLM ~ triangle KNJ, the linear scale factor from KNJ to KLM is
KL/KN = 6/10 = 0.6
Area scale factor = 0.6² = 0.36
Area of KLM = 0.36 × 75
y = 27
Area of triangle KLM = 27 cm²
Find the linear scale factor from a matching pair of sides, then square it to scale the area.
