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ESAT Maths 1 Geometry questions

12 original ESAT Maths 1 geometry questions in the real exam style, each with a step-by-step worked solution and an explanation of why the tempting wrong answers are wrong. Try each one before opening the solution.

Methods on this page: Similar shapes: length k, area k², volume k³Circle theorems: angle at the centre, cyclic quadrilaterals, alternate segmentVectors: walk a route of known vectorsMelt or pour: the volume stays the samePolygon angles: exterior angles sum to 360°Pythagoras in 3D: use it twice (or a² + b² + c²)Area of an awkward shape: subtract the simple pieces around it

Question 1ESAT Maths 1Warm-up

AA, BB and CC lie on a circle with centre OO. Angle AOB=122∘AOB = 122^\circ, and CC lies on the major arc ABAB. What is angle ACBACB?
  1. A29∘29^\circ
  2. B59∘59^\circ
  3. C60∘60^\circ
  4. D61∘61^\circ
  5. E62∘62^\circ
  6. F63∘63^\circ
  7. G119∘119^\circ
  8. H122∘122^\circ
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Answer: D · Method: Circle theorems: angle at the centre, cyclic quadrilaterals, alternate segment

  1. The angle at the centre is twice the angle at the circumference: angle ACB=122÷2=61∘ACB = 122 \div 2 = 61^\circ.

Why the tempting wrong answers are wrong

  • A: That's angle OABOAB in the isosceles triangle.
  • G: CC is on the major arc.
  • H: The angle at the centre is twice the angle at the circumference.

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Question 2ESAT Maths 1Warm-up

Each interior angle of a regular polygon is 150∘150^\circ. How many sides does it have?
  1. A88
  2. B1010
  3. C1111
  4. D1212
  5. E1313
  6. F1414
  7. G1515
  8. H1616
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Answer: D · Method: Polygon angles: exterior angles sum to 360°

  1. Each exterior angle is 180∘−150∘=30∘180^\circ - 150^\circ = 30^\circ.
  2. So the polygon has 360÷30=12360 \div 30 = 12 sides.

Why the tempting wrong answers are wrong

  • H: Divide 360° by the exterior angle, not the interior one.

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Question 3ESAT Maths 1Warm-up

A cuboid measures 7 cm×7 cm×6 cm7\ \text{cm} \times 7\ \text{cm} \times 6\ \text{cm}. What is the length of a diagonal joining two opposite corners of the cuboid, in cm?
  1. A252\sqrt{5}
  2. B727\sqrt{2}
  3. C134\sqrt{134}
  4. D183\sqrt{183}
  5. E186\sqrt{186}
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Answer: C · Method: Pythagoras in 3D: use it twice (or a² + b² + c²)

  1. The diagonal of the base is 72+72\sqrt{7^2 + 7^2}. Using Pythagoras again with the height: 72+72+62=134\sqrt{7^2 + 7^2 + 6^2} = \sqrt{134}.

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Question 4ESAT Maths 1Standard

Two solids are mathematically similar. Their surface areas are 4 cm24\ \text{cm}^2 and 9 cm29\ \text{cm}^2. The volume of the first solid is 24 cm324\ \text{cm}^3. What is the volume of the second solid, in cm3\text{cm}^3?
  1. A3636
  2. B5454
  3. C8080
  4. D8181
  5. E8282
  6. F2432\dfrac{243}{2}
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Answer: D · Method: Similar shapes: length k, area k², volume k³

  1. Area factor =94=94= \tfrac{9}{4} = \tfrac{9}{4}, so the length factor is 32\tfrac{3}{2}.
  2. Volume factor =(32)3= \left(\tfrac{3}{2}\right)^3, so the volume is 24×278=8124 \times \tfrac{27}{8} = 81.

Why the tempting wrong answers are wrong

  • A: That's the length scale factor. Volumes scale by its cube.
  • B: Volumes scale by the cube of the length factor, not the area factor.

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Question 5ESAT Maths 1Standard

OABCOABC is a parallelogram with OA→=a\overrightarrow{OA} = \mathbf{a} and OC→=c\overrightarrow{OC} = \mathbf{c}. MM is the point on ABAB with AM:MB=3:1AM : MB = 3 : 1, and NN is the point on the diagonal OBOB with ON:NB=1:3ON : NB = 1 : 3. What is MN→\overrightarrow{MN}?
  1. A−34a+12c-\tfrac{3}{4}\mathbf{a} + \tfrac{1}{2}\mathbf{c}
  2. B−34a−12c-\tfrac{3}{4}\mathbf{a} - \tfrac{1}{2}\mathbf{c}
  3. C34a+12c\tfrac{3}{4}\mathbf{a} + \tfrac{1}{2}\mathbf{c}
  4. D54a+c\tfrac{5}{4}\mathbf{a} + \mathbf{c}
  5. E−34a-\tfrac{3}{4}\mathbf{a}
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Answer: B · Method: Vectors: walk a route of known vectors

  1. AB→=OC→=c\overrightarrow{AB} = \overrightarrow{OC} = \mathbf{c}, so OM→=a+34c\overrightarrow{OM} = \mathbf{a} + \tfrac{3}{4}\mathbf{c}.
  2. OB→=a+c\overrightarrow{OB} = \mathbf{a} + \mathbf{c}, so ON→=14(a+c)\overrightarrow{ON} = \tfrac{1}{4}(\mathbf{a} + \mathbf{c}).
  3. MN→=ON→−OM→=−34a−12c\overrightarrow{MN} = \overrightarrow{ON} - \overrightarrow{OM} = -\tfrac{3}{4}\mathbf{a} - \tfrac{1}{2}\mathbf{c}.

Why the tempting wrong answers are wrong

  • C: That's NM→\overrightarrow{NM}. MN→=ON→−OM→\overrightarrow{MN} = \overrightarrow{ON} - \overrightarrow{OM}.
  • E: AM:MB=3:1AM : MB = 3 : 1, so MM is 34\tfrac{3}{4} of the way from AA to BB.

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Question 6ESAT Maths 1Standard

A cone of radius 4 cm4\ \text{cm} and height 5 cm5\ \text{cm} is full of water. The water is poured into an empty cylinder of radius 5 cm5\ \text{cm}. How deep is the water in the cylinder, in cm? (Volume of a cone =13πr2h= \tfrac{1}{3}\pi r^2 h.)
  1. A115\dfrac{1}{15}
  2. B1615\dfrac{16}{15}
  3. C43\dfrac{4}{3}
  4. D3115\dfrac{31}{15}
  5. E165\dfrac{16}{5}
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Answer: B · Method: Melt or pour: the volume stays the same

  1. Cone volume =13π×42×5=803π= \tfrac{1}{3}\pi \times 4^2 \times 5 = \dfrac{80}{3}\pi.
  2. In the cylinder: π×52×d=803π\pi \times 5^2 \times d = \dfrac{80}{3}\pi, so d=1615d = \dfrac{16}{15}.

Why the tempting wrong answers are wrong

  • C: Areas of circles depend on the radius squared.
  • E: A cone holds a third of the cylinder with the same base and height.

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Question 7ESAT Maths 1Standard

WXYZWXYZ is a square of side length 1, labelled anticlockwise. MM lies on WXWX, NN on XYXY and PP on YZYZ, with WM:MX=1:2WM : MX = 1 : 2, XN:NY=1:4XN : NY = 1 : 4 and YP:PZ=3:2YP : PZ = 3 : 2. What is the area of triangle MNPMNP?
  1. A350\dfrac{3}{50}
  2. B2375\dfrac{23}{75}
  3. C49150\dfrac{49}{150}
  4. D103300\dfrac{103}{300}
  5. E12\dfrac{1}{2}
  6. F101150\dfrac{101}{150}
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Answer: C · Method: Area of an awkward shape: subtract the simple pieces around it

  1. Put WW at the origin: M=(13,0)M = (\dfrac{1}{3}, 0), N=(1,15)N = (1, \dfrac{1}{5}), P=(25,1)P = (\dfrac{2}{5}, 1).
  2. Around the triangle are: a triangle at XX of area 115\dfrac{1}{15}, a triangle at YY of area 625\dfrac{6}{25}, and the trapezium WMPZWMPZ of area 1130\dfrac{11}{30}.
  3. So the area of MNPMNP is 1−115−625−1130=491501 - \dfrac{1}{15} - \dfrac{6}{25} - \dfrac{11}{30} = \dfrac{49}{150}.

Why the tempting wrong answers are wrong

  • A: Check which part of each side each ratio describes.
  • E: The triangle's base and height aren't both 1: subtract the pieces around it instead.

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Question 8ESAT Maths 1Standard

AA, BB and CC lie on a circle. The tangent to the circle at AA makes an angle of 59∘59^\circ with the chord ABAB, on the opposite side of ABAB from CC. Angle ABC=55∘ABC = 55^\circ. What is angle BACBAC?
  1. A31∘31^\circ
  2. B62∘62^\circ
  3. C65∘65^\circ
  4. D66∘66^\circ
  5. E67∘67^\circ
  6. F68∘68^\circ
  7. G114∘114^\circ
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Answer: D · Method: Circle theorems: angle at the centre, cyclic quadrilaterals, alternate segment

  1. Alternate segment theorem: angle ACBACB equals the angle between the tangent and chord ABAB, so it's 59∘59^\circ.
  2. Angles in triangle ABCABC: angle BAC=180∘−59∘−55∘=66∘BAC = 180^\circ - 59^\circ - 55^\circ = 66^\circ.

Why the tempting wrong answers are wrong

  • A: The radius to AA is perpendicular to the tangent, but you don't need it here.
  • B: By the alternate segment theorem, angle ACBACB equals the tangent–chord angle.

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Question 9ESAT Maths 1Standard

A solid metal sphere of radius 6 cm6\ \text{cm} is melted down and recast as a solid cylinder of radius 3 cm3\ \text{cm}. What is the height of the cylinder, in cm? (Volume of a sphere =43πr3= \tfrac{4}{3}\pi r^3.)
  1. A1616
  2. B2424
  3. C3131
  4. D3232
  5. E3333
  6. F3434
  7. G6464
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Answer: D · Method: Melt or pour: the volume stays the same

  1. Sphere volume =43π×63=8643π= \tfrac{4}{3}\pi \times 6^3 = \tfrac{864}{3}\pi.
  2. Cylinder: π×32×h=8643π\pi \times 3^2 \times h = \tfrac{864}{3}\pi, so h=32h = 32.

Why the tempting wrong answers are wrong

  • A: Check the powers: r3r^3 for the sphere and R2R^2 for the cylinder.
  • B: Don't forget the 43\tfrac{4}{3} in the sphere's volume.

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Question 10ESAT Maths 1Standard

A polygon has 10 sides. 9 of its interior angles add up to 1143∘1143^\circ. What is the remaining interior angle?
  1. A295∘295^\circ
  2. B296∘296^\circ
  3. C297∘297^\circ
  4. D298∘298^\circ
  5. E299∘299^\circ
  6. F307∘307^\circ
  7. G477∘477^\circ
  8. H657∘657^\circ
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Answer: C · Method: Polygon angles: exterior angles sum to 360°

  1. The interior angles of a 10-sided polygon sum to 180×(10−2)=1440∘180 \times (10 - 2) = 1440^\circ.
  2. So the remaining angle is 1440−1143=297∘1440 - 1143 = 297^\circ.

Why the tempting wrong answers are wrong

  • H: The interior angles of an n-sided polygon sum to 180(n−2)180(n - 2), not 180n180n.

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Question 11ESAT Maths 1Standard

A cuboid has base ABCDABCD with AB=1 cmAB = 1\ \text{cm} and BC=3 cmBC = 3\ \text{cm}, and height 6 cm6\ \text{cm}. MM is the midpoint of the vertical edge above CC. What is the distance AMAM, in cm?
  1. A10\sqrt{10}
  2. B13\sqrt{13}
  3. C44
  4. D19\sqrt{19}
  5. E46\sqrt{46}
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Answer: D · Method: Pythagoras in 3D: use it twice (or a² + b² + c²)

  1. AC2=12+32=10AC^2 = 1^2 + 3^2 = 10, and MM is 33 cm above CC.
  2. So AM=10+9=19AM = \sqrt{10 + 9} = \sqrt{19}.

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Question 12ESAT Maths 1Challenging

Two solid objects XX and YY are mathematically similar. The surface area of YY is 3 times the surface area of XX. The volume of YY is 104 cm3104\ \text{cm}^3 more than the volume of XX. What is the volume of XX, in cm3\text{cm}^3?
  1. A4−1234 - 12\sqrt{3}
  2. B12+4312 + 4\sqrt{3}
  3. C4+1234 + 12\sqrt{3}
  4. D5252
  5. E52+52352 + 52\sqrt{3}
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Answer: C · Method: Similar shapes: length k, area k², volume k³

  1. Area factor 3, so the length factor is 3\sqrt{3} and the volume factor is 333\sqrt{3}.
  2. So VY=33 VXV_Y = 3\sqrt{3}\,V_X and (33−1)VX=104(3\sqrt{3} - 1)V_X = 104.
  3. Rationalise: VX=104(33+1)27−1=4+123V_X = \dfrac{104(3\sqrt{3} + 1)}{27 - 1} = 4 + 12\sqrt{3}.

Why the tempting wrong answers are wrong

  • A: Rationalising 1kk−1\tfrac{1}{k\sqrt{k} - 1} multiplies by kk+1k\sqrt{k} + 1: check the sign.
  • D: The volume factor isn't the area factor 3: it's (3)3(\sqrt{3})^3.
  • E: That uses the length factor 3\sqrt{3} for the volumes.

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