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.
Geometry and measures worksheet — GCSE Higher
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- 1.Triangle DEF has a right angle at E. Angle DFE = 60° and EF = 9 cm. Work out the exact length of DE.
- 2.Two similar model cars have a length scale factor of 2 from the small model to the large model. Priya works out that the area scale factor is 2² = 4, and then says the volume scale factor must also be 4, since that is the number she just calculated. Is Priya correct?
- 3.Shape T has a vertex at (5, 3). T is enlarged by scale factor −1, centre (2, 1). Work out the coordinates of the image of the vertex (5, 3).
- 4.A parallelogram has an area of 54 cm² and a base of 9 cm. Work out its perpendicular height.
- 5.A quadrilateral has exactly two lines of symmetry, its two pairs of opposite angles are equal, and its diagonals are not equal in length. Write down the name of this quadrilateral.
- 6.A stained-glass window panel is designed as a kite. Two adjacent sides are each 25 cm, and the other two adjacent sides are each 40 cm. A frame is fitted around the whole panel. The frame costs £2.50 per metre, sold only in whole metres. Work out the total cost of the frame.
- 7.A pop-up canopy has two sloping supports that meet at the top. Each support makes an angle of 30° with the ground and is 6 m long. Using the exact value of cos 30°, work out the total width of the canopy's base.
- 8.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°?
- 9.A solid has 6 faces and 8 vertices. Using the formula F + V − E = 2, work out how many edges it has.
- 10.Triangle DEF is isosceles, with DE = DF. Angle E = 58°. Work out angle F.
- 11.In triangle XYZ, the side opposite vertex X is labelled using the standard lowercase-letter convention. Which label is correct?
- 12.Two triangular offcuts of wood, PQR and STU, are cut for a construction project. PQ = 8 cm, QR = 6 cm and angle PQR = 90°. ST = 8 cm, TU = 6 cm and angle STU = 90°. A carpenter wants to check the two pieces are identical in shape and size before using them as a matching pair. Using only the measurements given, and without working out any further lengths, which condition proves that triangle PQR is congruent to triangle STU?
- 13.A kitchen scale is marked in equal divisions of 20 g. The pointer rests three divisions past the 400 g mark. Work out the reading on the scale.
- 14.A shop sells rice in packs of 250 g. Aisha buys 4 packs. Work out the total mass of rice she buys, in kilograms.
- 15.A model car is built to a scale of 1 : 24. The real car is 4.32 m long. Work out the length of the model, in centimetres.
Answer key
- (d) 9√3 cm — DE is opposite the 60° angle at F, and EF is adjacent to it, so DE = EF × tan 60° = 9 × √3 = 9√3 cm. 9√3/2 cm comes from using sin 60° = √3/2 instead of tan 60°. 3√3 cm comes from using tan 30° = 1/√3 instead of tan 60° (9 × 1/√3 = 9/√3 = 3√3). 18 cm is the hypotenuse DF, not DE: it comes from using cos 60° = 1/2 and working out 9 ÷ 1/2 = 18, which finds the wrong side of the triangle.
- (d) No — the volume factor is 2³ = 8, not the area factor 4. — For similar solids, area scales with the square of the length scale factor and volume scales with its cube — different powers of the same number, so they are not usually equal. The length scale factor here is 2, so the area scale factor is 2² = 4 (Priya's figure) but the volume scale factor is 2³ = 8. Priya is wrong: the correct volume scale factor is 8, not 4. The option claiming area and volume always scale by the same factor treats the two as interchangeable, which is only true for the length factor itself. The option giving 6 comes from multiplying the length factor by the number of dimensions (2 × 3) rather than cubing it. The option repeating '4' for volume simply reuses Priya's own area calculation.
- (b) (−1, −1) — An enlargement by scale factor −1, centre (2, 1), sends a point P to the point on the opposite side of the centre, the same distance away: the image is 2 × centre − P. For the vertex (5, 3), this gives (2 × 2 − 5, 2 × 1 − 3) = (4 − 5, 2 − 3) = (−1, −1). Treating the centre as though it were the origin, and simply negating the point's coordinates, gives (−5, −3) — this ignores that the true centre is (2, 1), not (0, 0). Using scale factor +1 instead of −1 leaves the point exactly where it started, at (5, 3). Adding the point's displacement from the centre instead of subtracting it gives (2 × 2 + 5, 2 × 1 + 3) = (9, 5). Double the centre and subtract the point, and the image is (−1, −1).
- (c) 6 cm — For a parallelogram, area = base × height, so height = area ÷ base = 54 ÷ 9 = 6 cm. 12 cm comes from using the triangle's reverse formula, height = 2 × area ÷ base, which does not apply to a parallelogram. 45 cm comes from subtracting the base from the area, 54 − 9, instead of dividing. 486 cm comes from multiplying the area by the base, 54 × 9, instead of dividing.
- (b) Rhombus — A rhombus has exactly two lines of symmetry, formed by its two diagonals, and both pairs of opposite angles are equal, but its diagonals are unequal in length. A square also has opposite angles equal, but it has four lines of symmetry and its diagonals ARE equal, so it does not fit. A kite normally has only one line of symmetry and only one pair of opposite angles equal, so it does not fit. A general parallelogram has no lines of symmetry at all, so it does not fit. The correct answer is rhombus.
- (b) £5.00 — Perimeter of the kite = 25 + 25 + 40 + 40 = 130 cm = 1.3 m. The frame is sold only in whole metres, so 2 m must be bought. Cost = 2 × £2.50 = £5.00. A student who buys the exact 1.3 m instead of rounding up to whole metres gets 1.3 × £2.50 = £3.25. A student who never converts the perimeter from centimetres to metres and costs 130 × £2.50 gets £325.00.9 m, which rounds up to 3 whole metres, costing 3 × £2.50 = £7.50.
- (b) 6√3 m — The horizontal distance covered by one support is adjacent to the 30° angle, so it equals 6 × cos 30° = 6 × √3/2 = 3√3 m. The total base width is made up of both supports, so it is 2 × 3√3 = 6√3 m. '3√3 m' gives only one support's horizontal distance and forgets to double it for the total width. '6 m' comes from using sin 30° instead of cos 30° for the horizontal distance (6 × sin 30° = 3, doubled to 6). '12 m' comes from doubling the full sloping length of 6 m without using any trigonometry at all.
- (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.
- (d) 12 — Method: rearrange F + V − E = 2 so that E is on its own: E = F + V − 2. Working: E = 6 + 8 − 2 = 12. A student who answers 14 has added F and V but forgotten to subtract 2 at all. A student who answers 16 has added 2 instead of subtracting it. A student who answers 10 has subtracted 2 twice by mistake. Answer: 12 edges.
- (c) 58° — In an isosceles triangle, the base angles opposite the equal sides are equal. Since DE = DF, angle F is the base angle equal to angle E, so angle F = 58°.
- (c) x — By convention, the side opposite a vertex is labelled with the lowercase version of that vertex's letter, so the side opposite X is labelled x. y is the label for the side opposite Y, not X. z is the label for the side opposite Z, not X. X is the vertex's own uppercase letter — the convention specifically switches to lowercase for the side, so the uppercase letter on its own is not correct.
- (b) SAS (two sides and the included angle) — PQ = ST and QR = TU are two pairs of matching sides, and the angle between them (angle Q and angle T) is 90° in both triangles — two sides and the angle between them are equal, so the triangles are congruent by SAS, using only the measurements given. RHS needs the hypotenuses to be known equal; here only the two legs are given, so you would first have to work out each hypotenuse by Pythagoras' theorem — extra working the question rules out. SSS has the same problem: the third side of each triangle is not given, only calculable. ASA needs two angles and the side between them, but here it is two sides and the angle between them that are given, not two angles.
- (c) 460 g — Each division is 20 g, so three divisions past the mark is 3 × 20 = 60 g. Adding this to the 400 g mark gives 400 + 60 = 460 g. Treating each division as worth 1 g instead of 20 g gives 400 + 3 = 403 g. Working out the extra amount correctly but forgetting to add the 400 g mark gives just 60 g. Treating each division as worth 10 g instead of 20 g gives 400 + 30 = 430 g.
- (d) 1 kg — First find the total mass in grams: 4 × 250 = 1000 g. Then convert to kilograms by dividing by 1000: 1000 ÷ 1000 = 1 kg. Finding the correct total in grams but forgetting to divide by 1000 gives 1000 kg. Adding the number of packs to the pack mass instead of multiplying, 4 + 250 = 254 g, gives 0.254 kg. Dividing the pack mass by the number of packs instead of multiplying, 250 ÷ 4 = 62.5 g, gives 0.0625 kg.
- (b) 18 — Method: convert the real length to centimetres, then divide by the scale factor to shrink it down to the model's size. Working: 4.32 m = 432 cm; 432 cm / 24 = 18 cm. A student who answers 432 has converted the units correctly but forgotten to divide by the scale factor at all. A student who answers 10368 has multiplied by the scale factor instead of dividing (432 x 24). A student who answers 1.8 has converted the metres to centimetres by multiplying by 10 instead of 100, getting 43.2 cm, and then divided by 24. Answer: 18 cm.
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