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.Postcard P is similar to postcard Q. The 6 cm side of P corresponds to the 15 cm side of Q, and the 8 cm side of P corresponds to a matching side of Q. Work out the length of that matching side of Q.
- 2.Triangles ABE and CDE share the vertex E, where lines AC and BD cross at E. AE equals CE, and BE equals DE. Angle AEB and angle CED are formed as vertically opposite angles where the lines cross. Which condition proves that triangle ABE is congruent to triangle CDE?
- 3.A market stall sells apples at £2.40 per kilogram. Which of these expressions gives the cost, in pounds, of buying m kilograms of apples?
- 4.A shape is reflected in the line x = 1, and the image is then reflected in the line x = 5. Which single transformation is equivalent to this combination, for every point?
- 5.In triangle ABC and triangle DEF, the angle at A equals the angle at D, and AB ÷ DE = AC ÷ DF. Write down what this proves about the two triangles.
- 6.Point B is at (3, 5). It is reflected in the line y = 2. Work out the coordinates of the image of point B.
- 7.A cyclist's journey from her house to the shop is represented by the column vector with top number 2 and bottom number 5, where each unit is 1 km east and 1 km north. She then cycles from the shop to the park, represented by the column vector with top number 4 and bottom number −3. What single column vector represents her journey from her house directly to the park?
- 8.In triangle ABC, AB = 5 cm, BC = 3 cm and AC = 7 cm. Work out the size of angle ABC.
- 9.The sum of the interior angles of a polygon is 1980°. Work out the number of sides of the polygon.
- 10.A, B and C are points on a circle with centre O. B is on the major arc AC. Angle AOC = (4x + 20)°. Work out the size of angle ABC in terms of x, in its simplest form.
- 11.Two metal window frames are checked before installation. Frame A has edges of 40 cm and 65 cm, with a marked angle of 35° at the far end of the 65 cm edge, away from where the two given edges meet. Frame B has matching edges of 40 cm and 65 cm, with a matching 35° angle in the same position. The fitter wants to know if these measurements alone guarantee the two frames are congruent. What should the fitter be told?
- 12.A rectangular name card has a perimeter of 30 cm. The card is 9 cm long. Work out the width of the card.
- 13.Work out the exact value of tan 30° + tan 30°.
- 14.O is the centre of a circle, and PT is a tangent to the circle at the point T. A is a point on the circle, with OA and OT both radii and angle AOT = 122°. At the point T, the chord TA lies between the radius TO and the tangent TP. Work out the size of angle ATP, the angle between the chord and the tangent.
- 15.In quadrilateral WXYZ, which of the following correctly names the angle at vertex Y using standard notation?
Answer key
- (d) 20 cm — The scale factor from P to Q is 15 ÷ 6 = 2.5. Apply the same scale factor to the other side: 8 × 2.5 = 20 cm. A pupil who divides instead of multiplying by the scale factor gets 8 ÷ 2.5 = 3.2 cm. A pupil who adds the difference between the two known sides, 15 − 6 = 9, to 8 instead of scaling gets 8 + 9 = 17 cm. A pupil who rounds the scale factor 2.5 down to 2 gets 8 × 2 = 16 cm. The correct length is 20 cm.
- (d) SAS, vertically opposite angle included — AE equals CE and BE equals DE give two pairs of equal sides, and angle AEB equals angle CED because they are vertically opposite angles formed where AC and BD cross; vertically opposite angles are always equal without needing to be measured. This included angle sits between the two known sides in each triangle, giving SAS, so 'SAS, vertically opposite angle included' is correct. 'ASA, vertically opposite angle at E' is wrong because ASA needs two pairs of equal angles with the side between them, but only one angle is known in each triangle here, and the two other known facts are sides, not angles. 'SSS, three equal side pairs' is wrong because only two pairs of sides are given; there is no third pair of equal sides. 'Cannot prove — no angle measured' is wrong because vertically opposite angles are always equal automatically when two straight lines cross, so no separate measurement is needed.
- (a) 2.40m — Method: a cost found from a rate is the rate multiplied by the amount bought. Working: the rate is £2.40 per kilogram and the amount is m kilograms, so the cost is 2.40 × m. Answer: 2.40m. A candidate who divides the amount by the rate instead of multiplying writes m/2.40. A candidate who adds the rate to the amount instead of multiplying writes 2.40 + m. A candidate who subtracts the rate from the amount instead of multiplying writes m − 2.40.
- (c) Translation by the vector (8, 0) — Method: reflecting twice in two parallel vertical lines is always equivalent to a single translation, at right angles to the lines, of twice the distance between them. Working: the two lines are 5 − 1 = 4 units apart, so the translation is 2 × 4 = 8 units in the positive x-direction. Answer: translation by the vector (8, 0). Using just the gap itself, without doubling it, gives (4, 0); translating in the negative x-direction, from the second line back towards the first, gives (−8, 0); and describing the combination as a single reflection in the line halfway between them, x = 3, confuses this combination with the effect of a single reflection — two reflections in parallel lines are always equivalent to a translation, never to another reflection. Always double the gap between the lines, and translate in the direction from the first line towards the second.
- (d) Similar, by SAS — Method: read off whether the given sides are equal or only in the same ratio, and where the given angle sits. Sides in proportion point to similarity; sides equal in length would be needed for congruence. Working: AB ÷ DE = AC ÷ DF gives two pairs of corresponding sides in the same ratio, and the equal angle at A and D lies between AB and AC in the first triangle and between DE and DF in the second, so it is the included angle. Two pairs of sides in proportion with the included angle equal is the SAS condition for similarity. Answer: similar, by SAS. The distractors: 'Similar, by AA' quotes a condition that needs two pairs of equal angles, while only one pair is given here; 'Congruent, by SAS' reads AB ÷ DE = AC ÷ DF as AB = DE and AC = DF, turning a statement about proportion into one about equal lengths; 'Congruent, by SSS' makes that same misreading and adds an assumption that the third pair of sides is equal too, which nothing in the question says.
- (b) (3, −1) — Method: find the distance from the point to the mirror line, then place the image the same distance on the other side of the line. Working: B is 5 − 2 = 3 units above the line y = 2, so its image is 3 units below the line, at y = 2 − 3 = −1, giving (3, −1); the x-coordinate does not change, since the mirror line is horizontal. Options: (3, −5) comes from reflecting in the x-axis (y = 0) instead of the line y = 2; (−3, 5) comes from reflecting in the y-axis instead of a horizontal line; (3, −4) comes from doubling the distance from the line instead of reflecting it, giving 2 − 2×3 = −4. Answer: (3, −1).
- (b) (6, 2) — The overall journey from house to park is the sum of the two vectors: top = 2 + 4 = 6, bottom = 5 + (−3) = 2, giving (6, 2). A candidate who subtracts the second vector from the first instead of adding gets (2 − 4, 5 − (−3)) = (−2, 8). A candidate who subtracts the other way round gets (4 − 2, −3 − 5) = (2, −8). A candidate who forgets the negative sign on the second vector's bottom number and adds 3 instead of −3 gets (6, 8). Because the journeys join end to end, the correct resultant vector is (6, 2).
- (a) 120° — Method: three sides are known, so use the cosine rule rearranged as cos A = (b² + c² − a²) ÷ 2bc, with a the side facing the angle wanted. Working: angle ABC lies between AB = 5 cm and BC = 3 cm and faces AC = 7 cm, so cos ABC = (5² + 3² − 7²) ÷ (2 × 5 × 3) = (25 + 9 − 49) ÷ 30 = −15 ÷ 30 = −0.5. The angle between 0° and 180° whose cosine is −0.5 is 180° − 60°. Answer: angle ABC = 120°. The distractors: 60° comes from taking the subtraction the other way round, (49 − 25 − 9) ÷ 30 = 0.5, which loses the minus sign that makes the angle obtuse; 90° comes from the instinct that three known sides always mean Pythagoras, and 5² + 3² = 34 is not 49, so the triangle is not right-angled; 150° comes from knowing the cosine is −0.5 but subtracting 30° from 180°, using the angle whose sine is 0.5 rather than the angle whose cosine is 0.5.
- (b) 13 — Using the sum of interior angles formula, (n − 2) × 180° = 1980°, so n − 2 = 1980 ÷ 180 = 11, and n = 11 + 2 = 13. 11 stops after the division, forgetting to add 2 back to find n. 15 adds 2 twice by mistake, giving 11 + 2 + 2. 22 divides 1980 by 90 instead of 180.
- (b) (2x + 10)° — By the angle at the centre theorem, angle ABC is half of angle AOC, because both stand on the same arc AC: angle ABC = (4x + 20)° ÷ 2 = (2x + 10)°. Writing down the centre angle itself, without halving at all, gives (4x + 20)°. Halving only the constant term and leaving the x-term unchanged gives (4x + 10)°. Doubling the centre angle instead of halving it gives (8x + 40)°. Halve every term in the expression, and (2x + 10)° is what you get.
- (a) No — the angle is not between the two sides — Method: check whether the given angle sits between the two given sides. Working: the 35° angle is marked away from the corner where the 40 cm and 65 cm edges meet, so it is not the included angle — this is SSA, which is not one of the four basic congruence conditions, so it does not guarantee congruence. Options: 'Yes — SAS' wrongly treats any two sides plus any angle as SAS, without checking the angle's position; 'Yes — SSS' wrongly counts the angle as if it were a third side; 'No — only two sides measured' is not the real reason, since SAS itself only needs two sides, so this reasoning is beside the point. Answer: no, the angle is not between the two sides.
- (a) 6 cm — Method: a rectangle is made of two lengths and two widths, so the perimeter is 2 × (length + width). Half the perimeter is therefore one length plus one width, and the width is what is left of that half once the length is taken away. Working: 30 ÷ 2 = 15 cm is one length added to one width, and 15 − 9 = 6. Answer: 6 cm. The distractors: 21 cm comes from taking the length off the whole perimeter, 30 − 9, so three sides are still counted in the figure that is written down; 12 cm comes from removing both lengths correctly, 30 − 2 × 9 = 12, and then recording that remainder as the width instead of halving it, because it is in fact the two widths together; 7.5 cm comes from dividing the perimeter by four, 30 ÷ 4, which is the calculation for the side of a square and throws away the length that was given.
- (a) 2√3/3 — tan 30° = √3/3, so tan 30° + tan 30° = 2 × √3/3 = 2√3/3. √3 comes from wrongly treating tan 30° + tan 30° as tan(30° + 30°) = tan 60° = √3 — adding angles is not the same as adding ratios. √3/3 comes from forgetting to double the value and just writing down tan 30° on its own. 2√3 comes from doubling the numerator of √3/3 but forgetting to keep the denominator of 3.
- (b) 61° — Triangle OAT is isosceles because OA = OT, both radii, so its base angles are equal: angle OTA = angle OAT = (180° − 122°) ÷ 2 = 29°. The tangent PT meets the radius OT at 90°, by the tangent–radius theorem, so angle OTP = 90°. Since TA lies between TO and TP, angle OTP splits into angle OTA and angle ATP: angle ATP = 90° − 29° = 61°. Stopping after the isosceles step and reporting the base angle itself gives 29°. Finding the base angles as 180° − 122° without halving, then subtracting from 90°, gives 90° − 58° = 32°. Adding the base angle to 90° instead of subtracting it gives 90° + 29° = 119°. Subtract the base angle from the right angle, and 61° is what's left.
- (c) ∠XYZ — The angle at a named vertex is written with that vertex's letter in the middle, flanked by its two neighbouring vertices. The angle at Y sits between X and Z, its neighbours in quadrilateral WXYZ, so it is written ∠XYZ. ∠WXY names the angle at X, since X is the middle letter, not Y. ∠YZW names the angle at Z, since Z is the middle letter. ∠ZWX names the angle at W, since W is the middle letter.
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