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Admissions tests / ESAT / Maths 1 / Geometry and measures

Test standard. 15 questions, 15 marks, about 22 minutes.

ESAT Mathematics 1: Geometry and measures, set 1

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.
  1. 11 mark

    Three angles meeting at a single point are 3x degrees, 5x degrees and 4x degrees. Together they make a complete turn about that point. Find the value of x.

    1. A x = 15
    2. B x = 30
    3. C x = 45
    4. D x = 90
  2. 21 mark

    A solid pyramid has a hexagonal base and six triangular faces that all meet at a single apex above the base. How many edges does the pyramid have in total?

    1. A 12
    2. B 6
    3. C 14
    4. D 7
  3. 31 mark

    In kite ABCD, AB = AD and CB = CD, so the diagonal AC is a line of symmetry and angle ABC = angle ADC. Angle BAD = 80 degrees and angle BCD = 120 degrees. Find angle ABC.

    1. A 90 degrees
    2. B 160 degrees
    3. C 80 degrees
    4. D 120 degrees
  4. 41 mark

    Triangle ABC has AB = 8 cm, angle A = 50 degrees and angle B = 70 degrees. Triangle DEF has DE = 8 cm, angle D = 50 degrees and angle E = 70 degrees. Which congruence criterion proves that triangle ABC is congruent to triangle DEF?

    1. A ASA
    2. B SSS
    3. C SAS
    4. D RHS
  5. 51 mark

    Triangle ABC is similar to triangle PQR, with side AB corresponding to side PQ. AB = 6 cm, PQ = 9 cm, and the area of triangle ABC is 24 cm^2. Find the area of triangle PQR.

    1. A 36 cm^2
    2. B 81 cm^2
    3. C 16 cm^2
    4. D 54 cm^2
  6. 61 mark

    Triangle ABC has vertices A(2,1), B(4,1) and C(2,5). The triangle is enlarged by scale factor -2 about the origin O. Find the coordinates of the image of point C.

    1. A (4, 10)
    2. B (-4, -10)
    3. C (0, 3)
    4. D (-4, 5)
  7. 71 mark

    OABC is a parallelogram, with O at the origin. Vector OA = a and vector OC = c. M is the midpoint of side AB. Find vector OM in terms of a and c.

    1. A (1/2)a + (1/2)c
    2. B a + c
    3. C a + (1/2)c
    4. D (1/2)a + c
  8. 81 mark

    Isosceles triangle ABC has AB = AC = 13 cm and base BC = 10 cm. Find the height from vertex A to the base BC.

    1. A sqrt(69) cm
    2. B 8 cm
    3. C sqrt(194) cm
    4. D 12 cm
  9. 91 mark

    A cuboid has edges of length 3 cm, 4 cm and 12 cm. Find the length of the longest diagonal inside the cuboid, connecting two opposite corners.

    1. A 13 cm
    2. B 5 cm
    3. C 19 cm
    4. D sqrt(19) cm
  10. 101 mark

    Points A, B and C lie on the circumference of a circle with centre O. The angle at the centre standing on arc AC, angle AOC, is 130 degrees, and B lies on the major arc. Find angle ABC, the angle subtended by the same arc AC at the circumference.

    1. A 130 degrees
    2. B 65 degrees
    3. C 260 degrees
    4. D 50 degrees
  11. 111 mark

    AB is a diameter of a circle, and C is a point on the circumference such that AC = 6 cm and BC = 8 cm. Find the length of the diameter AB.

    1. A 14 cm
    2. B 48 cm
    3. C 10 cm
    4. D 7 cm
  12. 121 mark

    A ship sails from port A on a bearing of 065 degrees to reach port B. Find the bearing of port A from port B (the back bearing).

    1. A 065 degrees
    2. B 295 degrees
    3. C 155 degrees
    4. D 245 degrees
  13. 131 mark

    A sector of a circle has radius 9 cm and an angle of 40 degrees at the centre. Find the exact length of the arc, in terms of pi.

    1. A 2pi cm
    2. B 9pi cm
    3. C pi cm
    4. D 4pi cm
  14. 141 mark

    A cylinder has radius 4 cm and height 10 cm. Find its volume in terms of pi.

    1. A 80pi cm^3
    2. B 160pi cm^3
    3. C 40pi cm^3
    4. D 640pi cm^3
  15. 151 mark

    A right-angled triangle has a hypotenuse of length 12 cm and one angle of 30 degrees. Find the exact length of the side opposite the 30-degree angle.

    1. A 6 sqrt(3) cm
    2. B 24 cm
    3. C 6 cm
    4. D 4 sqrt(3) cm

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 three angles lie all the way around a single point, so they must add up to 360 degrees, not 180 degrees (that rule is for a straight line).
    2. So 3x + 5x + 4x = 360, which simplifies to 12x = 360.
    3. Dividing both sides by 12 gives x = 30.
    • Why not A: This comes from setting 3x + 5x + 4x equal to 180 degrees, the rule for angles on a straight line, rather than 360 degrees for angles at a point.
    • Why not C: This comes from adding only two of the three terms, 3x + 5x = 8x, and forgetting the 4x term before dividing 360 by 8.
    • Why not D: This comes from dividing 360 by 4 instead of by 12, miscounting the number of x-terms in 3x + 5x + 4x as four rather than three (3 + 5 + 4 = 12).

  2. Question 2Answer: A

    1. The base is a hexagon, so it has 6 sides, and each side is an edge of the solid.
    2. Each of the 6 base vertices is also joined to the apex by a slant edge, giving 6 more edges.
    3. In total the pyramid has 6 + 6 = 12 edges.
    • Why not B: This counts only the six edges running from the apex down to each base vertex, forgetting the six edges that form the hexagonal base itself.
    • Why not C: This applies Euler's formula for a solid, faces + vertices - edges = 2, but adds faces and vertices together (7 + 7 = 14) instead of subtracting 2 from that total to isolate the edges.
    • Why not D: This gives the number of faces (six triangles plus one hexagonal base = 7), mistaking a face count for an edge count.

  3. Question 3Answer: C

    1. The angles of any quadrilateral add up to 360 degrees.
    2. So angle BAD + angle BCD + angle ABC + angle ADC = 360, giving 80 + 120 + angle ABC + angle ADC = 360.
    3. Because AC is a line of symmetry, angle ABC = angle ADC, so the remaining 160 degrees is split equally between them: angle ABC = 160 / 2 = 80 degrees.
    • Why not A: This assumes all four angles of the kite are equal and divides 360 by 4, but a kite only has one pair of equal angles (at B and D), not four equal angles.
    • Why not B: This finds the combined size of angles ABC and ADC together (360 - 80 - 120 = 160) but forgets to divide that total between the two equal angles.
    • Why not D: This wrongly assumes angle ABC equals angle BCD, the adjacent angle at C, instead of angle ADC, the angle at D on the other side of the line of symmetry.

  4. Question 4Answer: A

    1. The two given angles and the side between them are known in each triangle: angle A, side AB and angle B correspond to angle D, side DE and angle E.
    2. A known side lying between two known angles is the ASA condition: two angles and the included side.
    3. Since AB = DE, angle A = angle D and angle B = angle E, the ASA condition is satisfied, so the triangles are congruent by ASA.
    • Why not B: SSS requires all three pairs of sides to be known equal, but only one pair of corresponding sides (AB and DE) is given here.
    • Why not C: SAS requires two sides and the angle included between them, but only one side is given in each triangle, not two.
    • Why not D: RHS requires a right angle and the hypotenuse, but no right angle is stated in either triangle.

  5. Question 5Answer: D

    1. The linear scale factor from triangle ABC to triangle PQR is PQ / AB = 9 / 6 = 3/2.
    2. Areas of similar shapes scale with the SQUARE of the linear scale factor, so the area scale factor is (3/2)^2 = 9/4.
    3. The area of triangle PQR is therefore 24 x 9/4 = 54 cm^2.
    • Why not A: This multiplies the area by the linear scale factor 9/6 = 3/2 directly (24 x 3/2 = 36), forgetting that area scales with the SQUARE of the linear scale factor, not the scale factor itself.
    • Why not B: This cubes the linear scale factor instead of squaring it (24 x (3/2)^3 = 81), applying the rule for how VOLUME scales between similar solids rather than how area scales between similar shapes.
    • Why not C: This uses the reciprocal of the scale factor without squaring it (24 x 2/3 = 16), the wrong direction and the wrong power for going from the smaller triangle to the larger one.

  6. Question 6Answer: B

    1. An enlargement with centre the origin and scale factor k maps a point (x, y) to (kx, ky).
    2. Here k = -2 and C = (2, 5), so the image has coordinates (-2 x 2, -2 x 5).
    3. That gives the image of C as (-4, -10).
    • Why not A: This uses scale factor +2 instead of -2, forgetting the negative sign, so the image lands on the same side of the centre as C rather than the opposite side.
    • Why not C: This adds the scale factor -2 to each coordinate of C (2 + (-2), 5 + (-2)) as if the enlargement were a translation, rather than multiplying each coordinate by -2.
    • Why not D: This multiplies only the x-coordinate of C by the scale factor and leaves the y-coordinate unchanged, instead of multiplying both coordinates.

  7. Question 7Answer: C

    1. Since OABC is a parallelogram, side AB is equal and parallel to side OC, so vector AB = vector OC = c.
    2. The position vector of B is therefore OB = OA + AB = a + c.
    3. M is the midpoint of AB, so OM = OA + (1/2) x AB = a + (1/2)c.
    • Why not A: This finds the midpoint of the whole diagonal OB (giving (a + c)/2) instead of the midpoint of side AB.
    • Why not B: This uses the full vector OB = a + c, forgetting that M is the MIDPOINT of AB and not the vertex B itself.
    • Why not D: This halves vector a but keeps vector c whole, the reverse of what is needed: it is the c-part of AB that should be halved, since AB itself equals vector c.

  8. Question 8Answer: D

    1. The height from A meets BC at its midpoint, splitting the isosceles triangle into two congruent right-angled triangles, each with hypotenuse 13 cm and base 10 / 2 = 5 cm.
    2. By Pythagoras' theorem, height^2 = 13^2 - 5^2 = 169 - 25 = 144.
    3. So the height is sqrt(144) = 12 cm.
    • Why not A: This applies Pythagoras' theorem using the full base of 10 cm as one side of the right-angled triangle (sqrt(13^2 - 10^2)), forgetting that the height splits the isosceles triangle into two right-angled triangles each with a base of only half of BC.
    • Why not B: This subtracts half the base directly from the equal side (13 - 5 = 8) instead of using Pythagoras' theorem to combine them.
    • Why not C: This adds the squares of 13 and 5 instead of subtracting them (sqrt(13^2 + 5^2)), the opposite of what Pythagoras' theorem requires for finding a shorter side from the hypotenuse.

  9. Question 9Answer: A

    1. For a cuboid with edges p, q and r, the space diagonal has length sqrt(p^2 + q^2 + r^2), applying Pythagoras' theorem twice: once across a face, once through the solid.
    2. Here p = 3, q = 4 and r = 12, so the diagonal squared is 3^2 + 4^2 + 12^2 = 9 + 16 + 144 = 169.
    3. So the diagonal has length sqrt(169) = 13 cm.
    • Why not B: This applies Pythagoras' theorem to only two of the three edges (sqrt(3^2 + 4^2) = 5), treating the problem as a 2-dimensional right-angled triangle and forgetting the third edge entirely.
    • Why not C: This simply adds the three edge lengths together (3 + 4 + 12 = 19) instead of using the 3-dimensional form of Pythagoras' theorem.
    • Why not D: This takes the square root of the SUM of the edge lengths (sqrt(3 + 4 + 12) = sqrt(19)) instead of the square root of the sum of their SQUARES.

  10. Question 10Answer: B

    1. The angle subtended at the centre of a circle by an arc is always twice the angle subtended at the circumference by the same arc.
    2. Here angle AOC = 130 degrees is the angle at the centre standing on arc AC.
    3. So the angle at the circumference, angle ABC, is half of this: 130 / 2 = 65 degrees.
    • Why not A: This gives the angle at the CENTRE itself (130 degrees) rather than halving it to find the angle at the circumference.
    • Why not C: This doubles the centre angle instead of halving it (130 x 2 = 260), inverting the theorem's relationship between the two angles.
    • Why not D: This subtracts the centre angle from 180 (180 - 130 = 50), applying the rule for opposite angles of a cyclic quadrilateral rather than the angle-at-the-centre theorem.

  11. Question 11Answer: C

    1. The angle in a semicircle is always a right angle, so angle ACB = 90 degrees because AB is a diameter.
    2. Triangle ACB is therefore right-angled at C, with AB as the hypotenuse.
    3. By Pythagoras' theorem, AB^2 = AC^2 + BC^2 = 6^2 + 8^2 = 36 + 64 = 100, so AB = sqrt(100) = 10 cm.
    • Why not A: This simply adds the two chord lengths together (6 + 8 = 14) instead of recognising that angle ACB = 90 degrees (the angle in a semicircle) and applying Pythagoras' theorem.
    • Why not B: This multiplies the two chord lengths together (6 x 8 = 48), which is not a step in Pythagoras' theorem or any circle rule that applies here.
    • Why not D: This takes half of the sum of the two given lengths (6 + 8 = 14, then 14 / 2 = 7), an average with no basis in Pythagoras' theorem or any circle theorem.

  12. Question 12Answer: D

    1. A back bearing, the bearing measured in the opposite direction along the same line, differs from the original bearing by 180 degrees.
    2. Since the original bearing, 065 degrees, is less than 180 degrees, the back bearing is found by adding 180.
    3. So the bearing of A from B is 065 + 180 = 245 degrees.
    • Why not A: This gives the original bearing unchanged (065 degrees), as if the bearing of A from B were simply the same as the bearing of B from A, forgetting that a back bearing points in roughly the opposite direction.
    • Why not B: This subtracts the bearing from 360 (360 - 65 = 295), a rule for a different situation entirely, not for finding a back bearing.
    • Why not C: This adds 90 degrees instead of 180 degrees (65 + 90 = 155), confusing the back-bearing rule with a right-angle turn.

  13. Question 13Answer: A

    1. The arc length is the fraction of the full circumference that the sector's angle represents: (angle / 360) x 2 x pi x r.
    2. Here angle = 40 degrees and r = 9 cm, so the fraction is 40/360 = 1/9.
    3. The arc length is therefore (1/9) x 2 x pi x 9 = 2pi cm.
    • Why not B: This uses the formula for the AREA of the sector (40/360 x pi x 9^2 = 40/360 x pi x 81 = 9pi) instead of the formula for the arc LENGTH, mixing up the two different formulae.
    • Why not C: This treats 9 cm as the DIAMETER rather than the radius, halving it to 4.5 cm before applying the arc length formula (40/360 x 2 x pi x 4.5 = pi).
    • Why not D: This uses 180 degrees as the full turn instead of 360 (40/180 x 2 x pi x 9 = 4pi), as if the sector were being compared to a semicircle rather than a whole circle.

  14. Question 14Answer: B

    1. The volume of a cylinder is given by V = pi x r^2 x h, where r is the radius and h is the height.
    2. Here r = 4 cm and h = 10 cm, so r^2 = 16.
    3. The volume is therefore pi x 16 x 10 = 160pi cm^3.
    • Why not A: This uses the formula for the CURVED SURFACE AREA of a cylinder (2 x pi x r x h = 2 x pi x 4 x 10 = 80pi) instead of the volume formula, mixing up area with volume.
    • Why not C: This forgets to square the radius (pi x r x h = pi x 4 x 10 = 40pi) instead of using pi x r^2 x h.
    • Why not D: This uses the DIAMETER squared in place of the radius squared (pi x 8^2 x 10 = 640pi), doubling the radius before squaring it instead of squaring the radius itself.

  15. Question 15Answer: C

    1. For a right-angled triangle, sin(angle) = opposite / hypotenuse, so opposite = hypotenuse x sin(angle).
    2. The exact value of sin 30 degrees is 1/2, which candidates should know without a calculator.
    3. So the opposite side is 12 x 1/2 = 6 cm.
    • Why not A: This uses cos 30 (= sqrt(3)/2) instead of sin 30, finding the ADJACENT side (12 x sqrt(3)/2 = 6 sqrt(3)) rather than the side opposite the 30-degree angle.
    • Why not B: This divides 12 by sin 30 instead of multiplying (12 / (1/2) = 24), inverting the relationship between the hypotenuse and the opposite side.
    • Why not D: This uses tan 30 (= 1/sqrt(3)) instead of sin 30, treating the hypotenuse as if it were the adjacent side in a tangent ratio (12 x 1/sqrt(3) = 4 sqrt(3)).

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