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Admissions tests / ESAT / Maths 1 / Geometry and measures

Stretch. 15 questions, 15 marks, about 30 minutes.

ESAT Mathematics 1: Geometry and measures, set 4

Angles and polygons, congruence and similarity, Pythagoras and trigonometry, circles and circle theorems, mensuration, transformations and vectors.

Download the questions (PDF) Download with worked solutions (PDF)

  • Answer all questions. No calculator allowed.
  • Each question has exactly one correct answer.
  • No diagrams are provided. Read each description carefully and, if it helps, sketch the shape yourself before answering.
  • This is a stretch set: most questions combine two specification points, and several go beyond what a typical exam question would ask. Look for the short route before grinding through algebra.
  1. 11 mark

    A regular polygon has each interior angle equal to 140 degrees. Find the order of rotational symmetry of the polygon.

    1. A 27
    2. B 9
    3. C 1260
    4. D 40
  2. 21 mark

    Triangle ABC is isosceles with AB = AC and angle BAC = 52 degrees. A straight line through A, parallel to BC, is drawn. Find the angle this line makes with AB, measured on the side of AB away from C.

    1. A 26 degrees
    2. B 128 degrees
    3. C 116 degrees
    4. D 64 degrees
  3. 31 mark

    ABCD is a parallelogram. The diagonal AC is drawn, splitting it into triangle ABC and triangle CDA. Which congruence criterion proves that triangle ABC is congruent to triangle CDA?

    1. A SSS
    2. B SAS
    3. C ASA
    4. D RHS
  4. 41 mark

    Point P has coordinates (3, 5). P is first reflected in the line y = x, and the image is then rotated 90 degrees clockwise about the origin O. Find the coordinates of the final image of P.

    1. A (-3, 5)
    2. B (5, -3)
    3. C (3, -5)
    4. D (5, 3)
  5. 51 mark

    A rhombus has diagonals of length 16 cm and 30 cm. Find the side length of the rhombus.

    1. A 34 cm
    2. B 17 cm
    3. C 23 cm
    4. D 289 cm
  6. 61 mark

    A circle has centre O and radius 9 cm. From an external point T, with OT = 15 cm, two tangents TA and TB are drawn, touching the circle at points A and B. Find the area of the quadrilateral OATB formed by the two radii OA, OB and the two tangents TA, TB.

    1. A 135 cm^2
    2. B 216 cm^2
    3. C 108 cm^2
    4. D 54 cm^2
  7. 71 mark

    A solid prism has a regular polygon as its cross-section, and its total number of FACES is 15. Find the number of EDGES the prism has.

    1. A 28
    2. B 39
    3. C 26
    4. D 45
  8. 81 mark

    A solid is built from unit cubes. Viewed from directly above, its plan is a 3 by 3 square with the single centre square removed, so the plan looks like a square picture frame made of 8 unit squares. Every column of the solid, all the way round the frame, is exactly 2 cubes tall, and the centre square has no cubes in it at any height. Find the total surface area of the solid, in square units, counting every face: the top, the bottom, the outer sides and the inside walls of the central hole.

    1. A 48
    2. B 40
    3. C 32
    4. D 16
  9. 91 mark

    A ship sails from port A on a bearing of 050 degrees for 6 km to reach point B. It then changes course and sails on a bearing of 110 degrees for 10 km to reach point C. Find the distance AC.

    1. A 2 sqrt(19) km
    2. B 2 sqrt(34) km
    3. C 16 km
    4. D 14 km
  10. 101 mark

    A prism has length 20 cm. Its cross-section is a trapezium with parallel sides of length 7 cm and 13 cm, and the perpendicular distance between them is 6 cm. Find the volume of the prism.

    1. A 2400 cm^3
    2. B 1560 cm^3
    3. C 1200 cm^3
    4. D 60 cm^3
  11. 111 mark

    TP is a tangent to a circle at point P. PQ is a chord of the circle, and R is a point on the major arc, in the alternate segment relative to the tangent-chord angle at P. The angle between the tangent TP and the chord PQ is 72 degrees. Find angle PRQ, the angle subtended by PQ at R.

    1. A 108 degrees
    2. B 72 degrees
    3. C 36 degrees
    4. D 144 degrees
  12. 121 mark

    A sector OAB of a circle has centre O, radius 10 cm, and angle AOB = 90 degrees. Find the area of the minor segment cut off by the chord AB.

    1. A 25pi - 50 cm^2
    2. B 25pi cm^2
    3. C 25pi + 50 cm^2
    4. D 100pi - 50 cm^2
  13. 131 mark

    Two similar cones have volumes in the ratio 27:8. Find the ratio of their surface areas.

    1. A 27:8
    2. B 3:2
    3. C 729:64
    4. D 9:4
  14. 141 mark

    Triangle ABC is isosceles with AB = AC = 8 cm and angle BAC = 120 degrees. Find the exact length of BC.

    1. A 4 sqrt(3) cm
    2. B 8 cm
    3. C 8 sqrt(3) cm
    4. D 8 sqrt(2) cm
  15. 151 mark

    Points A and B have position vectors a = (2, 1) and b = (22, 11) relative to origin O. Point C lies on line segment AB such that AC:CB = 2:3. Find the position vector of C.

    1. A (14, 7)
    2. B (10, 5)
    3. C (12, 6)
    4. D (8, 4)

Worked solutions

Every question below carries the reasoning, not just the answer. The official material for this test publishes a correct option letter and nothing else.

  1. Question 1Answer: B

    1. Each exterior angle of the polygon is 180 - 140 = 40 degrees.
    2. The exterior angles of any polygon sum to 360 degrees, so the number of sides is n = 360 / 40 = 9.
    3. A regular polygon with n sides has rotational symmetry of order n, so the order of rotational symmetry is 9.
    • Why not A: This is the number of diagonals of the 9-sided polygon (9 x 6 / 2 = 27), found after correctly obtaining n = 9 but then answering with the diagonal count instead of the order of rotational symmetry, which is simply n itself.
    • Why not C: This is the sum of the interior angles, (9-2) x 180 = 1260 degrees, found after correctly obtaining n = 9 but then reporting the angle sum rather than n.
    • Why not D: This is the exterior angle (180 - 140 = 40 degrees), the value used along the way to find n, but not the order of symmetry itself.

  2. Question 2Answer: D

    1. The base angles of the isosceles triangle are equal: angle ABC = angle ACB = (180 - 52) / 2 = 64 degrees.
    2. The line through A is parallel to BC, so with transversal AB, the alternate angle to angle ABC is equal to it, since alternate angles between parallel lines are equal.
    3. So the angle the line makes with AB, on the side away from C, is 64 degrees.
    • Why not A: This halves the apex angle (52 / 2 = 26), as if the line through A were bisecting angle BAC, rather than finding the base angle of the isosceles triangle.
    • Why not B: This gives 180 - 52 = 128, the combined size of the two base angles together, forgetting to split this between angle ABC and angle ACB.
    • Why not C: This finds the CO-INTERIOR angle with AB (180 - 64 = 116) instead of the ALTERNATE angle, using the wrong parallel-line angle rule for the side of the line the question asks about.

  3. Question 3Answer: A

    1. In a parallelogram, opposite sides are equal in length, so AB = CD and BC = DA.
    2. The diagonal AC is a side shared by both triangles, so AC = AC.
    3. All three sides of triangle ABC match all three sides of triangle CDA (AB = CD, BC = DA, AC = AC), so the triangles are congruent by SSS.
    • Why not B: SAS needs two sides and the angle INCLUDED between them. The parallelogram directly gives equal sides (AB = CD and BC = DA), not an equal angle, so SAS is not the criterion these facts support on their own.
    • Why not C: ASA needs two angles and the side between them. The parallelogram's defining property used here is that opposite SIDES are equal, not that any angles are given as equal.
    • Why not D: RHS needs a right angle in each triangle, but a general parallelogram need not have any right angles at all.

  4. Question 4Answer: C

    1. Reflecting P(3, 5) in the line y = x swaps its coordinates, giving the image (5, 3).
    2. Rotating a point 90 degrees clockwise about the origin maps (x, y) to (y, -x).
    3. Applying this to (5, 3) gives (3, -5), so the final image of P is (3, -5).
    • Why not A: This applies the rotation BEFORE the reflection: rotating (3, 5) by 90 degrees clockwise first gives (5, -3), and reflecting that in y = x afterwards (swapping coordinates) gives (-3, 5), the reverse of the stated order.
    • Why not B: This skips the reflection step entirely and rotates the ORIGINAL point (3, 5) by 90 degrees clockwise directly, using (x, y) -> (y, -x) to get (5, -3), instead of rotating the already-reflected point (5, 3).
    • Why not D: This reflects P in the y-axis, (x, y) -> (-x, y), instead of in the line y = x, giving (-3, 5) as the intermediate image instead of (5, 3), before correctly rotating 90 degrees clockwise to reach (5, 3).

  5. Question 5Answer: B

    1. The diagonals of a rhombus bisect each other at right angles, splitting it into four congruent right-angled triangles.
    2. Half of each diagonal gives the two legs of one such triangle: 16 / 2 = 8 cm and 30 / 2 = 15 cm.
    3. By Pythagoras' theorem, side^2 = 8^2 + 15^2 = 64 + 225 = 289, so the side length is sqrt(289) = 17 cm.
    • Why not A: This applies Pythagoras' theorem to the FULL diagonals, sqrt(16^2 + 30^2) = sqrt(1156) = 34, instead of the half-diagonals, forgetting that the diagonals of a rhombus bisect each other, so each right-angled triangle has legs of 8 cm and 15 cm, not 16 cm and 30 cm.
    • Why not C: This simply adds the two half-diagonals together (8 + 15 = 23) instead of applying Pythagoras' theorem to combine them.
    • Why not D: This correctly finds 8^2 + 15^2 = 289 but forgets to take the square root at the end, giving the SQUARE of the side length rather than the side length itself.

  6. Question 6Answer: C

    1. TA and TB are tangents from the external point T, so OA and OB (the radii) are each perpendicular to the tangents at A and B, making triangles OAT and OBT right-angled at A and B respectively.
    2. In right-angled triangle OAT, the hypotenuse is OT = 15 cm and one leg is OA = 9 cm, so by Pythagoras' theorem, AT^2 = OT^2 - OA^2 = 15^2 - 9^2 = 225 - 81 = 144, giving AT = 12 cm.
    3. Triangles OAT and OBT are congruent (both right-angled, with OA = OB and OT common, RHS), so each has area (1/2) x OA x AT = (1/2) x 9 x 12 = 54 cm^2.
    4. The quadrilateral OATB is made up of these two triangles, so its total area is 2 x 54 = 108 cm^2.
    • Why not A: This uses OT, the distance from the centre to the external point, directly as if it were the tangent length AT, skipping the Pythagorean step needed to find the actual tangent length: Area = OA x OT = 9 x 15 = 135, instead of first finding AT = 12 cm.
    • Why not B: This correctly finds the tangent length AT = 12 cm using Pythagoras' theorem, but then forgets the 1/2 in the area-of-a-triangle formula for BOTH of the two right-angled triangles making up the kite, giving 2 x (9 x 12) = 216 instead of 2 x (1/2 x 9 x 12).
    • Why not D: This correctly finds the area of only ONE of the two right-angled triangles, (1/2) x 9 x 12 = 54, but forgets that the quadrilateral OATB is made up of TWO such triangles (OAT and OBT), so this is half the true area.

  7. Question 7Answer: B

    1. An n-sided prism has n rectangular side faces plus 2 polygonal end faces, so its total number of faces is n + 2.
    2. Here the total is 15, so n + 2 = 15, giving n = 13.
    3. Such a prism has 3n edges (n on each end, plus n connecting them), so the number of edges is 3 x 13 = 39.
    • Why not A: This uses Euler's formula F + V - E = 2 but wrongly assumes the prism has the same number of vertices as faces (V = F = 15), giving E = 2(15) - 2 = 28, instead of finding the true vertex count (2n) from the polygon's number of sides.
    • Why not C: This correctly finds the number of sides of the cross-section (n = 13) and the number of VERTICES (2n = 26), but answers with the vertex count rather than the edge count the question asks for.
    • Why not D: This forgets that a prism has TWO polygonal end faces, not one, so it treats the number of sides of the cross-section as equal to the total face count (n = 15) instead of n = 15 - 2 = 13, giving 3 x 15 = 45 edges instead of 3 x 13 = 39.

  8. Question 8Answer: A

    1. The plan shows 8 unit squares in a ring shape, so the top face has area 8 and, since the solid is 2 cubes tall throughout, the bottom face also has area 8.
    2. The outer boundary of the 3 by 3 footprint has perimeter 4 x 3 = 12 units, so the outer side faces have total area 12 x 2 = 24 square units (perimeter x height).
    3. The missing centre square leaves a 1 by 1 hole running through the solid, whose four inner walls contribute a further 4 x 1 x 2 = 8 square units.
    4. Adding these together: 8 + 8 + 24 + 8 = 48 square units.
    • Why not B: This correctly finds the top area (8), the bottom area (8) and the outer side area (12 x 2 = 24), but forgets that the missing centre square creates a hole running through the solid, whose four inner walls are also part of the surface area (4 x 1 x 2 = 8 more square units).
    • Why not C: This uses a height of 1 cube instead of 2 when finding the side areas (outer: 12 x 1 = 12, inner hole: 4 x 1 = 4), forgetting that every column of the solid is 2 cubes tall, not 1.
    • Why not D: This gives the VOLUME of the solid (8 columns x 2 cubes each = 16 cubic units) rather than its surface area, confusing the number of cubes used with the total area of the faces around them.

  9. Question 9Answer: D

    1. The bearing changes from 050 to 110 degrees, a turn of 60 degrees, so the interior angle of the triangle at B, between BA and BC, is 180 - 60 = 120 degrees.
    2. By the cosine rule, AC^2 = AB^2 + BC^2 - 2 x AB x BC x cos(B) = 6^2 + 10^2 - 2(6)(10)cos(120).
    3. The exact value cos(120) = -1/2, so AC^2 = 36 + 100 - 120 x (-1/2) = 136 + 60 = 196, giving AC = sqrt(196) = 14 km.
    • Why not A: This uses the 60-degree TURN in bearing (110 - 050 = 60) as the triangle's interior angle at B directly, instead of the correct interior angle 180 - 60 = 120 degrees, giving AC^2 = 6^2 + 10^2 - 2(6)(10)cos(60) = 76.
    • Why not B: This applies Pythagoras' theorem directly, AC^2 = 6^2 + 10^2 = 136, as though angle B were a right angle, when the interior angle at B is actually 120 degrees and the cosine rule is needed instead.
    • Why not C: This simply adds the two distances travelled, 6 + 10 = 16, treating AC as if the ship's path were a single straight line, and ignoring that the ship changed direction at B.

  10. Question 10Answer: C

    1. The area of a trapezium is (1/2) x (sum of parallel sides) x (perpendicular height) = (1/2)(7 + 13)(6) = 60 cm^2.
    2. The volume of a prism is its cross-sectional area multiplied by its length.
    3. So the volume is 60 x 20 = 1200 cm^3.
    • Why not A: This forgets the 1/2 in the trapezium area formula, computing (7 + 13) x 6 = 120 cm^2 for the cross-section instead of (1/2)(7 + 13)(6) = 60 cm^2, doubling the correct area and therefore the volume.
    • Why not B: This treats the cross-section as a simple rectangle using only the LONGER parallel side, 13 x 6 = 78 cm^2, instead of averaging the two parallel sides as the trapezium area formula requires.
    • Why not D: This correctly finds the cross-sectional area of the trapezium, 60 cm^2, but forgets to multiply by the length of the prism, giving an area rather than a volume.

  11. Question 11Answer: B

    1. The alternate segment theorem states that the angle between a tangent and a chord equals the angle subtended by that chord in the alternate segment, the segment on the other side of the chord.
    2. Here R lies in the alternate segment relative to the tangent-chord angle at P, so angle PRQ equals the tangent-chord angle directly.
    3. Therefore angle PRQ = 72 degrees.
    • Why not A: This assumes R lies in the SAME segment as the tangent-chord angle, opposite to where it is actually stated to be, and applies the cyclic-quadrilateral rule that opposite angles sum to 180, giving 180 - 72 = 108, instead of using the alternate segment theorem directly.
    • Why not C: This halves the given angle (72 / 2 = 36), mistakenly applying the circle theorem that the angle at the centre is twice the angle at the circumference, which does not apply here since no angle at the centre is involved.
    • Why not D: This doubles the given angle (72 x 2 = 144), the same centre-angle confusion in the other direction, treating the tangent-chord angle as if it needed doubling to reach angle PRQ.

  12. Question 12Answer: A

    1. The sector area is (90/360) x pi x r^2 = (1/4) x pi x 100 = 25pi cm^2.
    2. Since angle AOB = 90 degrees and OA = OB = 10 cm (both radii), triangle OAB is right-angled at O with area (1/2)(10)(10) = 50 cm^2.
    3. The minor segment is the sector with the triangle removed, so its area is 25pi - 50 cm^2.
    • Why not B: This gives the area of the SECTOR only, forgetting to subtract the area of triangle OAB to isolate the segment, which lies between the chord and the arc.
    • Why not C: This ADDS the triangle's area to the sector's area instead of subtracting it, the wrong operation for finding a segment, which is the sector with the triangular part removed.
    • Why not D: This uses the area of the WHOLE circle (pi x 10^2 = 100pi) instead of just the 90-degree sector, one quarter of the circle, forgetting to apply the 90/360 fraction before subtracting the triangle.

  13. Question 13Answer: D

    1. For similar solids, volume scales with the CUBE of the linear scale factor, so the linear scale factor is the cube root of the volume ratio: cube root of 27/8 = 3/2.
    2. Surface area scales with the SQUARE of the linear scale factor, so the area ratio is (3/2)^2 = 9/4.
    3. The ratio of surface areas is therefore 9:4.
    • Why not A: This uses the given volume ratio directly as the surface area ratio, without converting through the linear scale factor at all.
    • Why not B: This correctly takes the cube root of the volume ratio to find the linear scale factor, cube root of 27/8 = 3/2, but then forgets that AREA scales with the SQUARE of the linear scale factor, giving the linear ratio itself instead of squaring it.
    • Why not C: This squares the volume ratio directly, (27/8)^2 = 729/64, instead of first taking the cube root to find the linear scale factor and then squaring that.

  14. Question 14Answer: C

    1. Drop a perpendicular from A to the midpoint M of BC, splitting the isosceles triangle into two congruent right-angled triangles, each with hypotenuse AB = 8 cm and an angle of 120 / 2 = 60 degrees at A.
    2. In this right-angled triangle, BM = AB x sin(60) = 8 x (sqrt(3)/2) = 4 sqrt(3) cm.
    3. BC is twice BM, so BC = 2 x 4 sqrt(3) = 8 sqrt(3) cm.
    • Why not A: This correctly drops a perpendicular from A to the midpoint M of BC and finds BM = 8 sin(60) = 4 sqrt(3), but then forgets to DOUBLE this to get the full length BC = 2 x BM, reporting only half of BC.
    • Why not B: This uses cos(120) = +1/2 in the cosine rule instead of the correct value -1/2, giving BC^2 = 8^2 + 8^2 - 2(8)(8)(1/2) = 64, so BC = 8, the wrong sign turning an obtuse-angle calculation into an acute one.
    • Why not D: This applies Pythagoras' theorem directly, BC = sqrt(8^2 + 8^2) = 8 sqrt(2), as though angle BAC were a right angle, when it is actually 120 degrees and Pythagoras' theorem does not apply.

  15. Question 15Answer: B

    1. Since AC:CB = 2:3, point C divides AB so that AC is 2/5 of the way from A to B, i.e. OC = a + (2/5)(b-a).
    2. Here b-a = (22-2, 11-1) = (20, 10), so (2/5)(b-a) = (8, 4).
    3. Adding this to a = (2, 1) gives OC = (2+8, 1+4) = (10, 5).
    • Why not A: This uses the ratio the wrong way round, treating AC:CB as 3:2 instead of the given 2:3, so C = a + (3/5)(b-a) instead of a + (2/5)(b-a), placing C nearer to B than it should be.
    • Why not C: This finds the MIDPOINT of AB, (a+b)/2, ignoring the given ratio 2:3 entirely and assuming C is halfway between A and B.
    • Why not D: This correctly computes the scaled vector (2/5)(b-a) = (8, 4) but forgets to ADD the position vector a of the starting point, giving a displacement rather than C's actual position vector.

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