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.A right-angled triangle has a hypotenuse of 14 cm. One of the other angles is 30°. Work out the exact length of the side opposite the 30° angle.
- 2.A fan-shaped rug is in the shape of a sector of a circle with radius 4 m and angle 180°. It costs £4 per square metre to make. Using π = 3.14, work out the total cost of the rug.
- 3.Two similar triangles have lengths in the ratio 5 : 8. A side of the smaller triangle is 6.5 cm. Work out the length of the corresponding side of the larger triangle.
- 4.Rectangle ABCD has vertices A(−1, 2), B(4, 2), C(4, 5) and D(−1, 5). The rectangle is translated so that C moves to (0, −1). Write down the coordinates of the image of A.
- 5.Point A(1, 2) is enlarged to give image point A′(7, −4). The centre of enlargement lies on the x-axis. Work out the scale factor of the enlargement.
- 6.Triangle PQR has vertices P(0, 0), Q(6, 0) and R(0, 8). Work out the perimeter of the triangle.
- 7.State whether the line segment joining A(−4, 3) and B(2, 6) is horizontal, vertical or neither, and give a reason for your answer.
- 8.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?
- 9.A skateboard ramp is designed so that a straight rail RT is tangent to a circular curve at the point T, and TC is a chord of the curve from T to another point C on the curve. A second rail continues straight through T on the other side, so that the two rails together form a straight line, and the angle between that second rail and the chord TC is 109°. A support point D is placed on the curve, on the major arc TC. Work out the size of angle TDC, the angle subtended by the chord TC at D.
- 10.A solid has 5 faces: one square face and four triangular faces, which all meet at a single point above the square. What is the name of this solid?
- 11.A right-angled triangle has its two shorter sides equal to 8 cm and 15 cm. Work out the value of cos θ, where θ is the angle opposite the 8 cm side. Give your answer as a fraction.
- 12.A pinhole camera is set up at a pitch-side stand, with its aperture at the point O = (2, 1) on a grid. Light from the corner F of a triangular flag, at the point (4, 2), passes through the aperture and forms an inverted image on the film behind it. The projection is an enlargement of scale factor −1.5 centred at O. Work out the coordinates of the image of corner F.
- 13.In triangle ABC the angle at A is 90° and BC = 10 cm. For the angle at B, sin B = 3/5. Work out the length of AC.
- 14.A shape is made by joining two rectangles edge to edge, with no overlap. One rectangle measures 12 cm by 4 cm and the other measures 4 cm by 8 cm. Work out the total area of the shape.
- 15.A cube has edges of length 5 cm. Work out the length of the diagonal that runs from one corner of the cube to the opposite corner. Give your answer as a surd in its simplest form.
Answer key
- (b) 7 cm — The side opposite the 30° angle is found using sin 30° = opposite/hypotenuse, so opposite = 14 × sin 30° = 14 × 1/2 = 7 cm. 7√3 cm comes from using cos 30° = √3/2 instead of sin 30° (mixing up the opposite and adjacent sides). 14/√3 cm comes from using tan 30° = 1/√3 instead of sin 30°. 28 cm comes from dividing 14 by sin 30° instead of multiplying by it.
- (d) £100.48 — Area of the rug = (angle ÷ 360) × π × r² = (180 ÷ 360) × 3.14 × 16 = 0.5 × 50.24 = 25.12 m². Cost = 25.12 × £4 = £100.48. (£200.96 comes from finding the area of a full circle, 3.14 × 16 = 50.24 m², and forgetting the angle fraction before costing it; £25.12 comes from forgetting to square the radius, using 0.5 × 3.14 × 4 = 6.28 m², and then costing that; £50.24 comes from using the arc length formula, 0.5 × 2 × 3.14 × 4 = 12.56, in place of the area, and costing that.)
- (a) 10.4 cm — The scale factor from the smaller triangle to the larger triangle is 8 ÷ 5 = 1.6, so the larger side is 6.5 × 1.6 = 10.4 cm. The distractor 4.0625 cm comes from using the ratio the wrong way round, 6.5 × 5 ÷ 8 = 4.0625. The distractor 9.5 cm comes from adding the difference between the ratio numbers (8 − 5 = 3) to the given length, 6.5 + 3 = 9.5. The distractor 13 cm comes from doubling the given length, treating the scale factor as 2 instead of 1.6.
- (d) (−5, −4) — Method: every vertex of a translated shape moves by the same vector, so find that vector from the one vertex whose image is given, then apply it to A. Working: C(4, 5) moves to (0, −1), so across 0 − 4 = −4 and up −1 − 5 = −6, giving the vector $\binom{-4}{-6}$. Applying it to A(−1, 2): −1 − 4 = −5 and 2 − 6 = −4. Answer: the image of A is (−5, −4). Working the vector out as object minus image gives 4 to the right and 6 up, which applied to A gives (3, 8). Getting the horizontal movement right but reversing the vertical one gives (−5, 8). Treating (0, −1) as the image of every vertex ignores that a translation carries each vertex to a different place.
- (a) −2 — Let the centre be C = (c, 0). The vector from the centre to the image equals the scale factor k times the vector from the centre to the object: (7 − c, −4) = k(1 − c, 2). The y-coordinate gives −4 = 2k, so k = −2 — this doesn't depend on knowing c. (Checking: substituting k = −2 into the x-equation gives c = 3, consistent with a centre on the x-axis.) '2' comes from taking the magnitude of the ratio without noticing the image is on the opposite side of the centre from the object, so the sign should be negative. '−1/2' comes from inverting the scale factor, dividing the object's coordinate by the image's instead of the other way round. '3' is the x-coordinate of the centre, mistaken for the scale factor.
- (c) 24 — PQ lies along the x-axis with length 6, and PR lies along the y-axis with length 8, meeting at a right angle at P, so QR = √(6² + 8²) = √100 = 10. The perimeter is 6 + 8 + 10 = 24. "14" adds only the two shorter sides, PQ and PR, and leaves out the hypotenuse QR completely. "28" comes from finding QR incorrectly as 6 + 8 = 14 instead of using Pythagoras' theorem, then adding 6 + 8 + 14. "48" comes from multiplying the two shorter sides, 6 × 8, instead of finding and adding all three sides of the triangle.
- (b) Neither, because both coordinates differ — A line segment is horizontal only when both points share the same y-coordinate, and vertical only when both points share the same x-coordinate. Here A has x-coordinate −4 and B has x-coordinate 2, which differ, and A has y-coordinate 3 and B has y-coordinate 6, which also differ, so the segment is neither horizontal nor vertical. Every point has a y-coordinate and an x-coordinate, so simply having one is not a reason for the line to be horizontal or vertical — both of those wrong reasons ignore that the coordinates must match, not just exist. The x-coordinates do increase from A to B, but an increasing x-coordinate on its own describes a slope, not a horizontal line.
- (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.
- (a) 71° — Method: first use the straight line through T to find the tangent-chord angle RTC, then apply the alternate segment theorem, which states that this angle equals the angle subtended by the chord in the alternate segment. Working: the two rails form a straight line through T, so angle RTC = 180 − 109 = 71 degrees. D lies in the alternate segment of the chord TC, so angle TDC = angle RTC = 71°. Answer: 71°. Find angle RTC FIRST from the straight line before applying the theorem: using the given 109° directly, doubling the tangent-chord angle, or taking its complement from 90° all give the wrong angle at D.
- (a) Square-based pyramid — A solid with one square base and four triangular faces meeting at a single apex above the base is a square-based pyramid. A triangular prism has two triangular faces and three rectangular faces, not a square base with four triangles, so that is a different solid. A cube has six square faces, and a cuboid has six rectangular faces — neither has any triangular faces at all. The solid described is a square-based pyramid.
- (c) 15/17 — Method: cos θ = adjacent ÷ hypotenuse, so find the hypotenuse with Pythagoras' theorem first and then decide which short side is next to θ. Working: the hypotenuse is √(8² + 15²) = √(64 + 225) = √289 = 17 cm. The angle θ is opposite the 8 cm side, so the side next to it is the 15 cm side, and cos θ = 15 ÷ 17. Answer: 15/17. The distractors: 8/17 is sin θ, opposite over hypotenuse, used in place of the cosine; 8/15 is tan θ, opposite over adjacent; 17/15 comes from writing the cosine ratio upside down, as hypotenuse over adjacent.
- (a) (−1, −0.5) — Method: for an enlargement centred at a point O that is not the origin, first find the point's position relative to O, multiply that by the scale factor, then add O's coordinates back on — image = O + k × (F − O). Working: F relative to O is (4, 2) − (2, 1) = (2, 1). Multiplying by the scale factor, −1.5 × (2, 1) = (−3, −1.5). Adding O back on: (2, 1) + (−3, −1.5) = (−1, −0.5). Answer: (−1, −0.5). Measure F's position from the aperture O, not from the origin, apply the full signed scale factor −1.5 to that displacement, and add O's own coordinates back on afterwards: measuring from the origin instead of O, treating the factor as +1.5, or adding O onto the wrong displacement all place the image mark somewhere else.
- (b) 6 cm — Method: the right angle is at A, so BC is the hypotenuse and AC is the side opposite the angle at B; sin B = opposite ÷ hypotenuse therefore gives AC ÷ BC = 3/5. Working: AC ÷ 10 = 3/5, so AC = 10 × 3 ÷ 5 = 6. Answer: 6 cm. The distractors: 8 cm is AB, the side next to the angle at B, which is what cos B = 4/5 produces — the right method used on the wrong side; 3 cm comes from reading the 3 in the ratio as a length and never scaling it up to the 10 cm hypotenuse; 30 cm comes from multiplying by 3 and forgetting to divide by 5.
- (c) 80 cm² — Method: a composite shape made of two rectangles that do not overlap has an area equal to the sum of the two rectangle areas, so work out each area and add them. Working: the first rectangle has area 12 × 4 = 48 cm² and the second has area 4 × 8 = 32 cm², so the total is 48 + 32 = 80. Answer: 80 cm². The distractors: 48 cm² comes from working out the larger rectangle only and treating it as the whole shape; 32 cm² comes from working out the smaller rectangle only; 28 cm comes from adding the four given lengths, 12 + 4 + 4 + 8, which is a distance rather than an area and so carries a plain centimetre unit.
- (c) 5√3 cm — The space diagonal of a cube with edge a satisfies d² = a² + a² + a² = 3a², using Pythagoras' theorem in three dimensions. With a = 5, d² = 3 × 5² = 3 × 25 = 75, so d = √75 = √(25 × 3) = 5√3 cm. Finding the diagonal of one face instead, using only two of the three edges, gives d = √(5² + 5²) = √50 = 5√2 cm, which leaves out the third dimension. Adding the three edges directly, 5 + 5 + 5 = 15 cm, ignores that Pythagoras' theorem works with squares of lengths, not the lengths themselves. Squaring the edge and multiplying by 3 correctly, 3 × 5² = 75, but then forgetting to take the square root, leaves 75 cm — the squared length, not the diagonal itself.
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