Printable · GCSE Higher · ages 14-16
Geometry and measures worksheet — GCSE Higher
Fifteen questions across the geometry and measures statements at Higher tier. Choose the non-calculator filter to rehearse Paper 1, which counts for a third of the marks.
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Geometry and measures worksheet — GCSE Higher
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- 1.Triangle ABC has AB = 8 cm, BC = 10 cm and CA = 6 cm. Triangle LMN has LM = 10 cm, MN = 6 cm and NL = 8 cm. The two triangles are congruent by SSS. Which of the following correctly matches the corresponding vertices?
- 2.A ship sails 12 km on a bearing of 070° from a harbour to a buoy. It then sails a further 9 km, still on a bearing of 070°, from the buoy to a lighthouse. How far is the lighthouse from the harbour, in kilometres?
- 3.Triangle ABC is extended so that side BC is extended beyond C to a point D, forming exterior angle ACD. Angle BAC = 48° and angle ABC = 67°. Which single angle fact gives angle ACD = 115° in one step, without first working out the interior angle at C?
- 4.Lines JK and LM are parallel. A straight line crosses JK at point P and crosses LM at point Q. Angle KPQ = 118°. Angle KPQ and angle MQP are co-interior (allied) angles. Work out angle MQP.
- 5.A written description of a diagram states: 'AB is marked with a single arrow, CD is marked with a single arrow, and EF is marked with a double arrow.' Using the standard convention for marking parallel lines, what can be concluded?
- 6.A landscaper marks out two triangular flower beds using stakes and a tape measure. Bed 1 has two edges of 5 m and 7 m, meeting at a corner with a marked angle of 60° between them. Bed 2 has edges of 5 m and 7 m, meeting at a corner also marked at 60° between them. The landscaper wants to check the two beds will be exactly the same size and shape using only these three measurements. Which condition proves the two beds are congruent?
- 7.The point R(−6, 9) is translated by the vector . Write down the coordinates of the image of R.
- 8.A carpenter measures two triangular braces. Brace 1 has a 55° angle, a 65° angle, and, opposite the 55° angle, a side of 8 cm. Brace 2 has matching 55° and 65° angles and, opposite its 55° angle, a side of 8 cm. The carpenter says that because the equal 8 cm side is not between the two equal angles in either brace, he cannot be sure the two braces are the same size. Is the carpenter correct?
- 9.Triangle ABC and triangle DEF are similar. The area of triangle ABC is 36 cm² and the area of triangle DEF is 100 cm². Work out the ratio of the lengths of triangle ABC to the lengths of triangle DEF, in its simplest form.
- 10.Three of these are rational numbers and one is irrational. Write down the one that is irrational.
- 11.A quadrilateral has vertices A(1, 1), B(5, 1), C(6, 4) and D(2, 4). By comparing the y-coordinates of A, B and of D, C, write down whether the sides AB and DC are parallel, and give a reason for your answer.
- 12.A shape is translated by the vector and then by the vector . Write down the single column vector that has the same effect as those two translations together.
- 13.A, B, C and D are points on a circle, and ABCD is a cyclic quadrilateral with the vertices in that order around the circle. Angle ABC = 108° and angle ADC = 72°. Which circle theorem is the reason that angle ABC + angle ADC = 180°?
- 14.A garden water trough is a prism whose cross-section is a right-angled triangle with base 40 cm and height 30 cm. The trough is 120 cm long. Work out the volume of water needed to fill the trough completely.
- 15.Two straight roads cross at a junction. The angle between them, measured with a protractor, is 118°. What is the size of the angle vertically opposite to it?
Answer key
- (c) A↔N, B↔L, C↔M (ABC≅NLM) — Matching equal side lengths: AB (8 cm) equals NL (8 cm), BC (10 cm) equals LM (10 cm), and CA (6 cm) equals MN (6 cm). This gives the correspondence A with N, B with L, and C with M, so triangle ABC is congruent to triangle NLM, making 'A↔N, B↔L, C↔M (ABC≅NLM)' correct. 'A↔L, B↔M, C↔N (ABC≅LMN)' simply matches the vertices in the order they are written without checking the side lengths: AB (8 cm) would need to equal LM (10 cm), which is false. 'A↔M, B↔N, C↔L (ABC≅MNL)' also fails this check, since AB (8 cm) would need to equal MN (6 cm), which is false. 'A↔N, B↔M, C↔L (ABC≅NML)' gets A correct but swaps B and C, so AB (8 cm) would need to equal NM (6 cm), which is also false.
- (b) 21 — Both legs of the journey are on the same bearing, 070°, so the ship travels in one straight line the whole way and the distances simply add: 12 + 9 = 21 km. A candidate who subtracts instead of adding gets 12 − 9 = 3 km. A candidate who multiplies the two distances gets 12 × 9 = 108. A candidate who assumes the ship changed direction and treats the two legs as the sides of a right-angled triangle works out √(12² + 9²) = √225 = 15 km — but the bearing does not change, so there is no triangle and no hypotenuse to find. Because the ship stays on one straight line, the total distance is 21 km.
- (c) exterior angle = sum of the two interior opposite angles — Angle ACD is the exterior angle at C, and the exterior angle of a triangle is always equal to the sum of the two interior angles at the other two vertices — here, angle BAC and angle ABC — so angle ACD = 48° + 67° = 115°. "angles in a triangle add up to 180°" is a true fact, but on its own it only gives the INTERIOR angle at C (180 − 48 − 67 = 65°), not the exterior angle 115° — reaching 115° that way still needs a further step. "angles on a straight line add up to 180°" is the fact that links the interior and exterior angles at C to each other, not the fact that gives 115° directly from the two OTHER angles. "exterior angle is always double the smallest angle" is not a genuine angle fact: double the smallest angle here is 48° × 2 = 96°, not 115°.
- (a) 62° — Angle KPQ and angle MQP are co-interior (allied) angles between parallel lines, so they add up to 180°: angle MQP = 180° − 118° = 62°. 118° comes from treating the angles as equal, as if this were a corresponding or alternate angle pair, instead of using the co-interior rule. 90° comes from wrongly assuming the crossing line meets JK and LM at right angles. 59° comes from halving 118° instead of subtracting it from 180°.
- (c) AB ∥ CD only; EF not confirmed — By convention, lines marked with the same number of arrows are parallel to each other, but lines marked with a different number of arrows belong to a different, unrelated family of parallel lines. AB and CD both have a single arrow, so AB is parallel to CD. EF has a double arrow, showing it is not part of the same family as AB and CD; it may be parallel to some other line marked with a double arrow, but nothing here confirms it is parallel to AB or CD, so 'AB ∥ CD only; EF not confirmed' is correct. 'AB, CD and EF are all parallel' and 'EF is parallel to AB' both wrongly treat every arrow mark as showing the same relationship. 'None of the lines are parallel' wrongly assumes a different arrow count rules out any parallel relationship at all, when it actually just signals a different pairing.
- (b) SAS — Method: check which condition matches two sides and the angle between them, since that is what has been measured for each bed. Working: the 60° angle is marked at the corner where the 5 m and 7 m edges meet, so it is the included angle — this is two Sides and the included Angle, SAS. Options: SSS would need a third side measured, but only two edges are known; ASA would need two angles and the side between them, but only one angle is measured; RHS needs a right angle, and 60° is not a right angle. Answer: SAS.
- (b) (−1, −2) — To translate R(−6, 9) by $\binom{5}{−11}$, add 5 to the x-coordinate and −11 to the y-coordinate: (−6 + 5, 9 + (−11)) = (−1, −2). (−1, 9) applies only the x-component and leaves the y-coordinate unchanged. (−6, −2) applies only the y-component and leaves the x-coordinate unchanged. (−11, 20) comes from subtracting the vector instead of adding it.
- (d) No — third angle is also fixed — Since both braces have angles of 55° and 65°, their third angles must both be 60°, because angles in a triangle sum to 180°. All three angles now match, so the braces have the same shape. Both 8 cm sides lie in the same position relative to those angles — opposite the 55° angle in each brace — so one matching pair of corresponding sides fixes the size as well, exactly as ASA or AAS would. The braces are therefore guaranteed to be congruent and the carpenter is incorrect: 'No — third angle is also fixed' is correct. 'Yes — side must be included' is wrong because the side does not have to lie physically between the two named angles; once the third angle is fixed, a corresponding equal side anywhere is enough. 'No — any two angles enough alone' is wrong because two equal angles with no side length at all would only show the triangles are similar, not congruent. 'Yes — third angle may differ' is wrong because the third angle is fixed at 60° by the angle sum and cannot vary.
- (a) 3 : 5 — Method: for similar figures the ratio of the areas is the square of the ratio of the lengths, so the length ratio is found by taking the square root of each part of the area ratio. Working: the areas are in the ratio 36 : 100, and taking square roots gives √36 : √100 = 6 : 10, which cancels to 3 : 5. Answer: 3 : 5. The distractors: 9 : 25 is 36 : 100 in its simplest form, so it is still the ratio of the areas and the square roots were never taken; 5 : 3 takes the roots correctly but writes the parts the wrong way round, giving DEF to ABC when ABC to DEF was asked for; 25 : 9 makes both mistakes at once, leaving the area ratio and reversing it.
- (c) cos 45° — cos 45° = √2/2, and √2 cannot be written as an exact fraction, so this value is irrational. tan 0° = 0, sin 90° = 1 and sin 30° = 1/2 are all rational, since each can be written as an exact whole number or fraction.
- (b) Parallel, since AB and DC are both horizontal lines — A and B both have y-coordinate 1, so AB is horizontal; D and C both have y-coordinate 4, so DC is horizontal too. Two horizontal lines are always parallel, so AB and DC are parallel. "Not parallel, since AB and DC have different lengths" is wrong twice over: AB runs from x = 1 to x = 5 and DC from x = 2 to x = 6, so both are in fact 4 units long, and in any case length has no bearing on whether two lines are parallel — only direction does. "Not parallel, since AB is horizontal and DC is vertical" misreads the coordinates of D and C, which share the y-coordinate 4 and so give a horizontal line, not a vertical one. "Parallel, since A and D have the same x-coordinate" rests on a false claim: A has x-coordinate 1 and D has x-coordinate 2, so they do not share an x-coordinate — and even if they did, it would say nothing about whether AB is parallel to DC.
- (a) $\binom{-5}{-4}$ — Method: one translation followed by another is a single translation, and the two vectors are added: top to top, bottom to bottom. Working: across, 4 − 9 = −5; up, −7 + 3 = −4. Answer: $\binom{-5}{-4}$. Subtracting the second vector instead of adding it gives 13 on top and −7 − 3 = −10 underneath. Adding the top numbers correctly but subtracting the bottom ones gives −10 underneath with −5 on top. Adding 4 and 9 as though both were positive and then keeping the minus sign of the larger gives −13 on top.
- (d) Opposite angles in a cyclic quadrilateral sum to 180° — ABCD is a cyclic quadrilateral, and angle ABC and angle ADC are a pair of opposite angles in it — they don't sit next to each other around the quadrilateral. The theorem that fixes the sum of a cyclic quadrilateral's opposite angles at 180° is exactly the one needed here. The other three theorems don't apply to this figure at all: the angle-in-a-semicircle theorem gives a 90° angle only when one side is a diameter, and no diameter is mentioned; the angle-at-the-centre theorem needs a centre point and an inscribed angle on the same arc, and no centre is given here; the tangent–radius theorem gives a 90° angle only where a tangent line touches the circle, and there is no tangent in this figure. Only the cyclic-quadrilateral theorem fits what is actually drawn.
- (b) 72 000 cm³ — Cross-sectional area = 1/2 × 40 × 30 = 600 cm². Volume = cross-sectional area × length = 600 × 120 = 72 000 cm³. (144 000 cm³ comes from forgetting the 1/2 in the triangle's area, using 40 × 30 as the cross-section; 36 000 cm³ comes from halving the correct volume again, as if the 1/2 applied a second time; 720 cm³ comes from adding the cross-sectional area and the length, 600 + 120, instead of multiplying them.)
- (a) 118° — When two straight lines cross, the angles that are vertically opposite each other are always equal. So the angle vertically opposite 118° is also 118°. A candidate who instead finds the angle next to it on the straight line, using 180° − 118° = 62°, has found the adjacent angle, not the vertically opposite one. A candidate who answers 180° has confused the rule with angles on a straight line. A candidate who doubles the angle, giving 236°, has applied no valid angle rule at all. The vertically opposite angle is 118°.
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