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Admissions tests / ESAT / Maths 1 / Geometry and measures

Demanding. 15 questions, 15 marks, about 27 minutes.

ESAT Mathematics 1: Geometry and measures, set 3

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 demanding set: expect multi-step working and less signposting than the real exam.
  1. 11 mark

    A convex hexagon has exterior angles, taken in order around the hexagon, of x degrees, (x + 15) degrees, (x + 30) degrees, (x + 45) degrees, (x + 60) degrees and (x + 75) degrees. Find x.

    1. A x = 60
    2. B x = 22.5
    3. C x = 82.5
    4. D x = 27
  2. 21 mark

    A quadrilateral has diagonals that bisect each other, and the diagonals cross at right angles, but the two diagonals are not equal in length. Which of the following must this quadrilateral be?

    1. A Square
    2. B Rectangle
    3. C Kite
    4. D Rhombus
  3. 31 mark

    In triangle ABC, D is the midpoint of side BC, and AD is perpendicular to BC. AB = (4n + 3) cm, AC = (7n - 15) cm, and BD = (n + 2) cm. Find the length BC.

    1. A 8 cm
    2. B 12 cm
    3. C 16 cm
    4. D 27 cm
  4. 41 mark

    Shape T is reflected in the line y = x to give shape T'. Shape T' is then reflected in the x-axis to give shape T''. Describe the single transformation that maps T directly onto T''.

    1. A A rotation of 90 degrees clockwise about the origin
    2. B A rotation of 90 degrees anticlockwise about the origin
    3. C A reflection in the line y = -x
    4. D A reflection in the x-axis
  5. 51 mark

    A cuboid box has a rectangular base measuring 7 cm by 24 cm. The space diagonal running from a bottom corner to the opposite top corner has length 65 cm. Find the height of the box.

    1. A 25 cm
    2. B 34 cm
    3. C 3600 cm
    4. D 60 cm
  6. 61 mark

    From an external point T, two tangents TA and TB touch a circle with centre O at points A and B. The angle between the tangents, angle ATB, is 40 degrees. Find angle AOB, the angle between the radii OA and OB.

    1. A 40 degrees
    2. B 140 degrees
    3. C 50 degrees
    4. D 220 degrees
  7. 71 mark

    ABCD is a cyclic quadrilateral. Angle A = (2x + 15) degrees and angle C = (3x - 25) degrees, and A and C are opposite angles of the quadrilateral. Find angle A.

    1. A 91 degrees
    2. B 38 degrees
    3. C 89 degrees
    4. D 95 degrees
  8. 81 mark

    Points A(-3, 1) and B(5, 7) are the endpoints of a diameter of a circle. Find the radius of the circle.

    1. A 10
    2. B 7
    3. C 5
    4. D 25
  9. 91 mark

    A convex polyhedron has 30 edges and 20 vertices. Using Euler's formula, find the number of faces it has.

    1. A 10
    2. B 48
    3. C 8
    4. D 12
  10. 101 mark

    The bearing of point B from point A is 065 degrees. The bearing of point C from point B is 155 degrees. AB = BC. Find the bearing of C from A.

    1. A 155 degrees
    2. B 110 degrees
    3. C 20 degrees
    4. D 90 degrees
  11. 111 mark

    A trapezium has parallel sides of length (3n + 1) cm and (n + 7) cm, and the perpendicular distance between them is 6 cm. Its area is 72 cm^2. Find n.

    1. A n = 1
    2. B n = 15
    3. C n = 4
    4. D n = 8
  12. 121 mark

    Two similar solid cones have volumes in the ratio 8 : 27. Find the ratio of their heights, in its simplest form.

    1. A 2 : 3
    2. B 8 : 27
    3. C 4 : 9
    4. D 3 : 2
  13. 131 mark

    A sector of a circle has radius 6 cm and area 6pi cm^2. Find the angle at the centre of the sector.

    1. A 30 degrees
    2. B 180 degrees
    3. C 360 degrees
    4. D 60 degrees
  14. 141 mark

    Triangle ABC has AB = 10 cm, BC = 14 cm, and angle B = 60 degrees. Find the exact area of triangle ABC.

    1. A 35 sqrt(3) cm^2
    2. B 70 sqrt(3) cm^2
    3. C 70 cm^2
    4. D 35 cm^2
  15. 151 mark

    OABC is a parallelogram, with O at the origin. Vector OA = a and vector OC = c. P is the point on AB such that AP : PB = 1 : 2. Q is the point on OC such that OQ : QC = 1 : 2. Find vector PQ in terms of a and c.

    1. A -a - (1/3)c
    2. B -a
    3. C -a + (1/3)c
    4. D a

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. The exterior angles of any convex polygon always sum to 360 degrees, however many sides it has and however unequal the individual angles are.
    2. Adding the six given expressions gives 6x + (15 + 30 + 45 + 60 + 75) = 6x + 225.
    3. Setting this equal to 360 gives 6x + 225 = 360, so 6x = 135, and x = 135 / 6 = 22.5.
    • Why not A: This assumes all six exterior angles are equal in size and simply divides 360 by 6, ignoring that the angles actually increase by 15 degrees each time.
    • Why not C: This confuses the exterior angle sum with the INTERIOR angle sum of a hexagon, (6 - 2) x 180 = 720 degrees, setting 6x + 225 = 720 instead of 6x + 225 = 360.
    • Why not D: This correctly finds that the five added amounts (15, 30, 45, 60, 75) sum to 225, and that 360 - 225 = 135, but then divides 135 by 5 (the number of ADDED amounts) instead of by 6 (the total number of angles).

  2. Question 2Answer: D

    1. A quadrilateral whose diagonals bisect each other is always a parallelogram; this rules out a plain kite, whose diagonals do not both bisect each other.
    2. A parallelogram whose diagonals also cross at right angles is, by definition, a rhombus; this is the property that distinguishes a rhombus from a general parallelogram or a rectangle.
    3. Since the diagonals here are also stated to be unequal, the quadrilateral cannot be the special case where a rhombus is also a square, so it must be a (non-square) rhombus.
    • Why not A: A square's diagonals do bisect each other at right angles, but a square's diagonals are always EQUAL in length, which contradicts the given condition that the diagonals are unequal.
    • Why not B: A rectangle's diagonals bisect each other and are equal in length, but they are only at right angles in the special case of a square, so a general rectangle fails the right-angle condition.
    • Why not C: A kite's diagonals do cross at right angles, but only ONE diagonal is bisected by the other; the diagonals of a kite do not bisect each OTHER, so the mutual-bisection condition fails.

  3. Question 3Answer: C

    1. Since D is the midpoint of BC, BD = DC, and since AD is perpendicular to BC, angle ADB = angle ADC = 90 degrees; with AD common to both triangles, triangle ABD is congruent to triangle ACD by SAS.
    2. Corresponding sides of congruent triangles are equal, so AB = AC, giving 4n + 3 = 7n - 15.
    3. Solving gives 18 = 3n, so n = 6, and BD = n + 2 = 8 cm. Since D is the midpoint, BC = 2 x BD = 16 cm.
    • Why not A: This is the length of BD alone; it forgets that D is the MIDPOINT of BC, so BC is twice BD, not BD itself.
    • Why not B: This comes from a sign error when solving 4n + 3 = 7n - 15: writing 4n - 7n = -15 + 3 instead of 4n - 7n = -15 - 3 gives n = 4 instead of n = 6, and then doubling the resulting (wrong) BD of 6 gives 12.
    • Why not D: This is the length of AB (and AC), the two equal sides of the isosceles triangle, not the base BC that the question asks for.

  4. Question 4Answer: A

    1. Reflecting a point (x, y) in the line y = x swaps its coordinates, giving the image (y, x).
    2. Reflecting that image (y, x) in the x-axis negates its second coordinate, giving the final image (y, -x).
    3. The overall rule (x, y) maps to (y, -x) is exactly what a rotation of 90 degrees clockwise about the origin does, so the combined transformation is a rotation of 90 degrees clockwise about the origin.
    • Why not B: This reverses the order the two reflections are applied in: reflecting first in the x-axis and then in y = x would give this rotation instead, but the question applies y = x first.
    • Why not C: This assumes that combining two reflections must always give another single reflection, but reflecting in two lines that cross (here at the origin) actually gives a ROTATION about that crossing point, through twice the angle between the lines.
    • Why not D: This only accounts for the SECOND reflection (in the x-axis) and ignores the effect of the first reflection in y = x entirely.

  5. Question 5Answer: D

    1. For a cuboid with edges p, q and r, the space diagonal d satisfies d^2 = p^2 + q^2 + r^2, applying Pythagoras' theorem across the base and then through the solid.
    2. Here d = 65, p = 7 and q = 24, so the height h satisfies 65^2 = 7^2 + 24^2 + h^2, giving 4225 = 49 + 576 + h^2, so h^2 = 4225 - 625 = 3600.
    3. So h = sqrt(3600) = 60 cm.
    • Why not A: This is the length of the diagonal of the rectangular BASE only, sqrt(7^2 + 24^2) = 25, which stops after a genuine but incomplete calculation and never uses the 65 cm space diagonal at all.
    • Why not B: This treats the space diagonal as though lengths simply add up along the three edges, computing 65 - 7 - 24 = 34, instead of combining the lengths using Pythagoras' theorem.
    • Why not C: This is height^2 = 65^2 - 7^2 - 24^2 = 3600, the value reached just before the final step; forgetting to take the square root leaves this squared value instead of the height itself.

  6. Question 6Answer: B

    1. A tangent to a circle is always perpendicular to the radius drawn to the point of contact, so angle OAT = angle OBT = 90 degrees.
    2. O, A, T and B form a quadrilateral whose four interior angles sum to 360 degrees: angle AOB + angle OAT + angle ATB + angle TBO = 360.
    3. So angle AOB = 360 - 90 - 90 - 40 = 140 degrees.
    • Why not A: This wrongly assumes the angle between the two radii equals the angle between the two tangents, ignoring the two right angles where the tangents meet the radii.
    • Why not C: This subtracts only ONE of the two right angles from 90 (90 - 40 = 50), instead of accounting for both right angles at A and B when finding the fourth angle of the quadrilateral OATB.
    • Why not D: This comes from a sign slip when finding the fourth angle of quadrilateral OATB, computing 360 - 90 - 90 + 40 instead of 360 - 90 - 90 - 40.

  7. Question 7Answer: A

    1. Opposite angles of a cyclic quadrilateral are supplementary: they always sum to 180 degrees.
    2. So angle A + angle C = 180 gives (2x + 15) + (3x - 25) = 180, which simplifies to 5x - 10 = 180.
    3. Solving gives 5x = 190, so x = 38, and angle A = 2(38) + 15 = 91 degrees.
    • Why not B: This is the value of x, not of angle A; the question asks for the angle itself, so this stops one step short of substituting x back into the expression for angle A.
    • Why not C: This is the value of angle C, not angle A; substituting x = 38 into the expression 3x - 25 gives 89, which answers the wrong one of the two angles.
    • Why not D: This wrongly assumes that opposite angles of a cyclic quadrilateral are EQUAL, solving 2x + 15 = 3x - 25 (giving x = 40) instead of using the correct rule that they are supplementary and sum to 180 degrees.

  8. Question 8Answer: C

    1. The centre of the circle is the midpoint of the diameter AB: midpoint = ((-3 + 5) / 2, (1 + 7) / 2) = (1, 4).
    2. The radius is the distance from this centre (1, 4) to either endpoint, for example A(-3, 1): the differences are 1 - (-3) = 4 and 4 - 1 = 3.
    3. By Pythagoras' theorem, radius = sqrt(4^2 + 3^2) = sqrt(25) = 5.
    • Why not A: This is the length of the whole diameter AB, sqrt(8^2 + 6^2) = 10, forgetting to halve it to get the radius.
    • Why not B: This averages the horizontal and vertical distances between A and B directly, (8 + 6) / 2 = 7, instead of combining them with Pythagoras' theorem before halving.
    • Why not D: This is 4^2 + 3^2 = 25, the value under the square root when finding the radius from the centre to A; forgetting to take the square root leaves this squared value instead of the radius itself.

  9. Question 9Answer: D

    1. Euler's formula for any convex polyhedron states V - E + F = 2, relating its vertices, edges and faces.
    2. Rearranging to make F the subject gives F = 2 - V + E.
    3. Substituting V = 20 and E = 30 gives F = 2 - 20 + 30 = 12.
    • Why not A: This drops the constant term from Euler's formula, computing F = E - V = 30 - 20 = 10 instead of F = E - V + 2.
    • Why not B: This mis-signs Euler's formula, using F = V + E - 2 = 20 + 30 - 2 = 48, as though F were being SUBTRACTED in the original formula V - E + F = 2, rather than correctly isolating F.
    • Why not C: This keeps the correct E and V terms but gets the sign of the constant wrong, computing F = E - V - 2 = 30 - 20 - 2 = 8 instead of F = E - V + 2.

  10. Question 10Answer: B

    1. The bearing of A from B (the back bearing of AB) is 065 + 180 = 245 degrees, so the interior angle ABC of the triangle is 245 - 155 = 90 degrees.
    2. Since AB = BC, triangle ABC is isosceles with equal base angles at A and C: angle BAC = angle BCA = (180 - 90) / 2 = 45 degrees.
    3. Because C lies further clockwise than B as seen from A, the bearing of C from A is the bearing of B from A plus this base angle: 065 + 45 = 110 degrees.
    • Why not A: This simply repeats the given bearing of C from B, as though a bearing measured from one point applies unchanged when measured from a different point.
    • Why not C: This subtracts the triangle's base angle from the bearing of B from A (65 - 45) instead of adding it, taking the base angle on the wrong side of line AB.
    • Why not D: This is the triangle's own interior angle at B (found from 245 - 155), mistaking that interior angle for the bearing of C measured from A.

  11. Question 11Answer: C

    1. The area of a trapezium is half the sum of the parallel sides multiplied by the perpendicular height: Area = (1/2)(a + b)h.
    2. Substituting a = 3n + 1, b = n + 7, h = 6 and Area = 72 gives (1/2)((3n + 1) + (n + 7)) x 6 = 72, which simplifies to 3(4n + 8) = 72, so 12n + 24 = 72.
    3. Solving gives 12n = 48, so n = 4.
    • Why not A: This drops the factor of 1/2 from the trapezium area formula, solving ((3n + 1) + (n + 7)) x 6 = 72 instead of (1/2) x ((3n + 1) + (n + 7)) x 6 = 72.
    • Why not B: This uses the DIFFERENCE of the two parallel sides, (3n + 1) - (n + 7), instead of their sum, inside the area formula.
    • Why not D: This comes from a sign slip when isolating the n-term, computing 12n = 72 + 24 = 96 instead of 12n = 72 - 24 = 48.

  12. Question 12Answer: A

    1. For similar solids, the ratio of volumes equals the CUBE of the ratio of corresponding lengths, here the heights.
    2. The volumes are in ratio 8 : 27, and since 8 = 2^3 and 27 = 3^3, this is the cube of the ratio 2 : 3.
    3. So the ratio of the heights is 2 : 3.
    • Why not B: This treats the given volume ratio as though it were already the ratio of heights, forgetting that volume scales with the CUBE of the linear ratio, so 8 : 27 must be cube-rooted rather than used directly.
    • Why not C: This correctly finds the underlying length ratio 2 : 3, but then squares it, 2^2 : 3^2 = 4 : 9, as though the question asked for the ratio of SURFACE AREAS (which scales with the square of the linear ratio) rather than the ratio of heights.
    • Why not D: This finds the right two numbers 2 and 3 but states them in the wrong order, giving the larger cone's height compared to the smaller's rather than smaller to larger, the order the volume ratio 8 : 27 (smaller : larger) was given in.

  13. Question 13Answer: D

    1. The area of a sector is the fraction of the full circle's area that its angle represents: Area = (theta / 360) x pi x r^2.
    2. Here Area = 6pi cm^2 and r = 6 cm, so 6pi = (theta / 360) x pi x 36; dividing both sides by pi gives 6 = (theta / 360) x 36.
    3. So theta / 360 = 6 / 36 = 1/6, giving theta = 360 / 6 = 60 degrees.
    • Why not A: This uses 180 degrees as the angle of a full turn (as for angles on a straight line) instead of the correct 360 degrees for a full circle.
    • Why not B: This uses the sector ARC LENGTH formula, (theta / 360) x 2 x pi x r, in place of the sector AREA formula, mixing up the two different formulae.
    • Why not C: This uses the radius itself in place of the radius SQUARED in the area formula, forgetting to square r before multiplying by pi.

  14. Question 14Answer: A

    1. The area of a triangle can be found from two sides and the included angle: Area = (1/2) x AB x BC x sin(B), since angle B lies between sides AB and BC.
    2. The exact value of sin 60 degrees is sqrt(3)/2, which candidates should know without a calculator.
    3. So Area = (1/2) x 10 x 14 x (sqrt(3)/2) = 70 x (sqrt(3)/2) = 35 sqrt(3) cm^2.
    • Why not B: This drops the 1/2 factor from the area formula, computing 10 x 14 x sin 60 instead of (1/2) x 10 x 14 x sin 60.
    • Why not C: This omits the sine term altogether, computing (1/2) x 10 x 14 as though angle B were 90 degrees and AB, BC were the two legs of a right-angled triangle.
    • Why not D: This uses cos 60 (= 1/2) in place of sin 60 (= sqrt(3)/2), the wrong trigonometric ratio for this area formula.

  15. Question 15Answer: B

    1. Since OABC is a parallelogram, vector AB = vector OC = c, so the position vector of B is OB = OA + AB = a + c.
    2. P divides AB with AP : PB = 1 : 2, so P is one-third of the way from A to B: vector OP = a + (1/3)c. Q divides OC with OQ : QC = 1 : 2, so vector OQ = (1/3)c.
    3. Vector PQ = OQ - OP = (1/3)c - (a + (1/3)c) = -a, since the two (1/3)c terms cancel exactly.
    • Why not A: This reverses the ratio AP : PB when locating P, placing P two-thirds of the way from A to B instead of one-third, so the c-terms fail to cancel and leave -a - (1/3)c.
    • Why not C: This reverses the ratio OQ : QC when locating Q, placing Q two-thirds of the way from O to C instead of one-third, so the c-terms fail to cancel and leave -a + (1/3)c.
    • Why not D: This finds vector QP (from Q to P) instead of vector PQ (from P to Q), the opposite direction.

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