10 free IGCSE Mathematics Geometry questions written in the style of Cambridge Mathematics papers, each with its answer and mark scheme. 5 of them are A* challenge questions, built to catch the mistakes strong students still make.
This topic covers Angles & parallel lines, Polygons, Similarity, Circle theorems I, Circle theorems II. Work through the questions below, then generate a full Geometry paper with a fresh set every time.
The n vertices of a regular polygon lie on a circle with centre O. Three of the vertices, P, Q and R, are joined to form triangle PQR.
(a)
Explain why each angle of triangle PQR is a multiple of 180n degrees.
[3]
(b)
Angle PQR = 99°. Find the smallest possible value of n.
[3]
(c)
Angle PQR = 99° and triangle PQR is isosceles. Find the smallest possible value of n, and the other two angles of the triangle.
[3]
(d)
n = 20 and angle PQR = 99°. Find the number of non-congruent triangles PQR that are possible.
[2]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
each side of the polygon subtends 360n° at O; an angle of the triangle is half the angle at O subtended by k sides
each side of the polygon subtends 360n° at the centre B1 angle at the circumference is half the angle at the centre (same arc) M1 arc opposite the angle contains k sides, so angle = k × 180nA1
3
(b)
n = 20
99 = 180knM1 n = 20k11, so k is a multiple of 11 M1 n = 20 (k = 11, the other arcs total 9 sides) A1
3
(c)
n = 40; 40.5° and 40.5°
the other two angles must be equal: 40.5° each B1 40.5 = 180jn needs n a multiple of 40, and 99° needs n a multiple of 20 M1 n = 40 A1
3
(d)
4
arc PR (not containing Q) has 11 sides, so arcs PQ and QR together contain 9 sides M1 splits 1 + 8, 2 + 7, 3 + 6, 4 + 5: 4 triangles A1
2
Question 2Structured · 14 marksA* challenge
Some regular polygons fit exactly around a point, with no gaps or overlaps, so that the point is a vertex of each polygon.
(a)
Three regular polygons fit exactly around a point. One of them is an equilateral triangle. Find all the possible pairs of polygons that complete the arrangement. Give the number of sides of each.
[6]
(b)
A regular polygon X and two identical regular polygons fit exactly around a point. Find every possible polygon X, and the two polygons that go with it.
[4]
(d)
Four regular polygons, which are not all the same, fit exactly around a point. Find all the possible sets of four polygons.
[4]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
(7, 42), (8, 24), (9, 18), (10, 15), (12, 12)
interior angle of an n-sided regular polygon = 180 − 360nB1 60 + (180 − 360b) + (180 − 360c) = 360 M1 1b + 1c = 16A1 with b ≤ c, 16 < 2b, so 7 ≤ b ≤ 12 M1 at least three correct pairs A1 all five pairs and no others A1
6
(b)
(3; 12, 12), (4; 8, 8), (6; 6, 6), (10; 5, 5)
1x + 2y = 12M1 y = 4xx − 2 = 4 + 8x − 2M1 x − 2 is a factor of 8: x = 3, 4, 6, 10 A1 3 with 12, 12; 4 with 8, 8; 6 with 6, 6; 10 with 5, 5 A1
4
(d)
{3, 3, 4, 12}, {3, 3, 6, 6}, {3, 4, 4, 6}
1a + 1b + 1c + 1d = 1 B1 the smallest polygon must be a triangle (four squares are all the same) M1 systematic search on the second polygon M1 all three sets and no others A1
4
Question 3Structured · 7 marksA* challenge
In triangle ABC, D is the point on AB with AD : DB = 1 : 2 and E is the point on AC with AE : EC = 1 : 2. The lines BE and CD meet at X.
(a)
Explain why DE is parallel to BC and find the ratio DE : BC.
[2]
(b)
Find the fraction of the area of triangle ABC that lies inside triangle XBC.
[5]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
triangle ADE is an enlargement of ABC, scale factor 1/3; DE : BC = 1 : 3
ADE similar to ABC (common angle, sides in ratio 1 : 3) M1 so DE ∥ BC and DE : BC = 1 : 3 A1
2
(b)
12
triangles XDE and XCB are similar, ratio 1 : 3 M1 so EX : XB = 1 : 3 A1 area EBC = 23 area ABC (same height, base EC = 23AC) M1 area XBC = 34 area EBC M1 12 (allow any equivalent route, e.g. DX : XC = 1 : 3) A1
5
Question 4Structured · 7 marksA* challenge
A, B, C and D lie on a circle, centre O. Angle AOC = 124°.
Fig. 4.1
(a)
Find angle ABC. Give a reason for your answer.
[2]
(b)
Find angle ADC. Give a reason for your answer.
[2]
(c)
The tangent to the circle at C is drawn. Find the angle between this tangent and the chord CA, on the same side of CA as D. Give a reason.
[2]
(d)
Find angle OAC.
[1]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
62° (angle at centre is twice the angle at the circumference)
62 B1 angle at centre = 2 × angle at circumference B1
2
(b)
118° (opposite angles of a cyclic quadrilateral add up to 180°)
118 B1 opposite angles of a cyclic quadrilateral sum to 180° B1
2
(c)
62° (alternate segment theorem)
62 B1 alternate segment theorem B1
2
(d)
28°
28 (isosceles triangle OAC) B1
1
Question 5Structured · 4 marks
The diagram shows two parallel lines crossed by a straight line.
Fig. 5.1
Find the value of y and the value of z. Give a geometrical reason for each answer.
[4]
Show answers and mark scheme
Part
Answer and marking points
Marks
y = 62, z = 118
y = 62 B1 Corresponding angles are equal B1 z = 118 B1 Angles on a straight line add up to 180° B1
4
Question 6Structured · 3 marks
Each interior angle of a regular polygon is 135°. Work out the number of sides of the polygon.
[3]
Show answers and mark scheme
Part
Answer and marking points
Marks
8
180 − 135 = 45 M1 360 ÷ 45 M1 8 A1
3
Question 7Structured · 4 marks
Triangle PQR is similar to triangle XYZ. PQ = 8 cm, QR = 8 cm and XY = 16 cm.
(a)
Find YZ.
[2]
(b)
The area of triangle PQR is 32 cm². Find the area of triangle XYZ.
[2]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
16 cm
scale factor 2 M1 16 A1
2
(b)
128 cm²
× 22M1 128 A1
2
Question 8Structured · 3 marks
Two parallel lines are crossed by a transversal. One angle is (3x + 14)° and the corresponding angle on the other line is (5x − 34)°. Find x and the size of the angle.
[3]
Show answers and mark scheme
Part
Answer and marking points
Marks
x = 24; 86°
equate the angles (corresponding angles are equal) M1 x = 24 A1 86° A1
3
Question 9Structured · 2 marks
A, B, C and D are points on the circle.
Fig. 9.1
Find the value of y. Give a reason for your answer.
[2]
Show answers and mark scheme
Part
Answer and marking points
Marks
y = 25
25 B1 Angles in the same segment are equal B1
2
Question 10Structured · 9 marksA* challenge
A solid cone has base radius 8 cm and height 36 cm. A smaller cone of height 12 cm is cut off the top by a cut parallel to the base, leaving a frustum. [Volume of a cone = ⅓πr²h]
(a)
Calculate the volume of the frustum.
[4]
(b)
Show that the volume of the frustum is 2627 of the volume of the original cone.
[2]
(c)
A mathematically similar frustum has a total surface area 4 times as large. It is made of a material with density 1.2 g/cm³. Calculate its mass in kilograms.
[3]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
2320 cm³
small cone is similar: radius 2.67 cm M1 large cone 2413 M1 small cone 89.36 M1 2320 cm³ A1
In the 0580 syllabus, Geometry covers Angles & parallel lines, Polygons, Similarity, Circle theorems I, Circle theorems II.
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Original Cambridge-style practice questions written for mentorriyaz.in. These are not Cambridge past paper questions, and this site is not affiliated with or endorsed by Cambridge University Press & Assessment. Updated 2026-11-04. Paper generator · mentorriyaz.in