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
MathsUKwww.geekhero.co.uk
- 1.A line joins the points A(0, 0, 0) and B(9, 12, 8) in a three-dimensional coordinate system in which the z-axis is vertical. Work out the angle this line makes with the horizontal (the xy-plane). Give your answer correct to 1 decimal place.
- 2.Two similar company logos are printed at different sizes. The area of the larger logo is 2.25 times the area of the smaller logo. Work out the length scale factor from the smaller logo to the larger logo.
- 3.A student says the reverse of the translation vector is . Write down the correct reverse translation vector.
- 4.A drone flies from base station D on a bearing of 065° for 14 km to a checkpoint E. At E it changes course and flies on a bearing of 165° for 20 km to a delivery point F. Work out the direct distance from D to F. Give your answer to 1 decimal place.
- 5.PT is a tangent to a circle with centre O, touching the circle at T. C is a point on the circle such that C lies inside angle OTP (the right angle between the radius OT and the tangent PT), and OC is a radius. Angle TOC = 130°. Work out the size of angle PTC, the angle between the tangent PT and the chord TC.
- 6.A parallelogram-shaped tile has two sides of 6 cm and 9 cm, with an angle of 60° between them. Work out the area of the tile. Give your answer to 1 decimal place.
- 7.A photocopier enlarges a document so that every length is multiplied by the same scale factor. A line on the original document is 3.2 cm long, and the same line measures 11.2 cm on the enlarged copy. A second line on the original document is 2.5 cm long. Work out the length of the second line on the enlarged copy, giving your answer to 2 decimal places.
- 8.A cuboid measures 5 cm long (left to right), 3 cm deep (front to back) and 2 cm tall. Its front elevation is 5 cm wide by 2 cm high. Its side elevation is 3 cm wide by 2 cm high. What are the dimensions of its plan view, looking down from above?
- 9.A cuboid has a rectangular base measuring 6 cm by 8 cm, and a height of 4 cm. Work out the angle between the diagonal from a base corner to the opposite top corner, and the base of the cuboid. Give your answer correct to 1 decimal place.
- 10.A rhombus-shaped floor tile has sides of length 12 cm and an interior angle of 55°. Work out the area of the tile. Give your answer to 1 decimal place.
- 11.A shopkeeper builds a display from two cuboid boxes, shown in the diagram. She wants to cover the front of the display with coloured paper. Work out the total area of the front elevation, in square centimetres.
- 12.A shape is translated by the vector , and one vertex of the image is at the point (1, 6). The original shape is instead translated by the vector . Work out the coordinates of the image of that same vertex under this second translation.
- 13.A shop sells a scale model of a van. The model and the real van are similar shapes, with lengths in the ratio 1 : 24. The model is 18.5 cm long. Work out the length of the real van, giving your answer in metres.
- 14.A window is made from a rectangle with a semicircle on top. The rectangle is 40 cm wide and 60 cm tall. The semicircle has the same width as the rectangle. Work out the total area of the window. Use π = 3.14. Give your answer to the nearest whole number.
- 15.Point P has coordinates (4, 2). P is rotated 90° clockwise about the point (1, 1), and the image is then reflected in the line x = 1. Work out the coordinates of the final image of P.
Answer key
- (a) 28.1° — The angle a line makes with the horizontal plane lies in the right-angled triangle formed by the vertical rise, the horizontal distance travelled, and the line itself. The horizontal distance from A to B is √(9² + 12²) = √(81 + 144) = √225 = 15, using only the x- and y-coordinates. The vertical rise is the z-coordinate, 8, so tan(angle) = 8 ÷ 15, giving angle = 28.1° (1 d.p.). Inverting the ratio, tan(angle) = 15 ÷ 8, gives 61.9° instead — the complement of the angle, not the angle with the horizontal. Using only the x-coordinate as if it were the whole horizontal distance, tan(angle) = 8 ÷ 9, gives 41.6°. Using the y-coordinate alone in the same way, tan(angle) = 8 ÷ 12, gives 33.7°.
- (c) 1.5 — The area scale factor is the length scale factor squared, so if n is the length factor, n² = 2.25. Taking the positive square root gives n = 1.5 (check: 1.5² = 2.25). 2.25 is just the area factor restated, with no root taken. 1.125 comes from halving 2.25 instead of taking its square root. 5.0625 comes from squaring 2.25 instead of rooting it.
- (c) $\binom{-4}{9}$ — The reverse of a translation negates both components, so the reverse of $\binom{4}{-9}$ is $\binom{-4}{9}$. $\binom{4}{9}$ is the student's mistake — only the bottom number has been negated, and the top number was left unchanged. $\binom{-4}{-9}$ makes the opposite error, negating only the top number. $\binom{4}{-9}$ is simply the original vector, unchanged.
- (c) 22.3 km — The bearing of E from D is 065°, so the bearing of D from E (the back bearing) is 065° + 180° = 245°. The bearing of F from E is 165°, so the angle at E between ED and EF is 245° − 165° = 80°. Using the cosine rule, DF² = DE² + EF² − 2 × DE × EF × cos(80°) = 14² + 20² − 2 × 14 × 20 × cos(80°) = 196 + 400 − 560 × cos(80°). Since cos(80°) ≈ 0.17365, 560 × cos(80°) ≈ 97.24, so DF² ≈ 596 − 97.24 = 498.76. Taking the square root, DF ≈ 22.333, which rounds to 22.3 km. 26.3 km comes from using 100° as the angle at E — the difference between the two given bearings taken directly (165° − 65°) without converting to the back bearing first. 498.8 km is DF² itself, rounded, with the square root never taken. 23.4 km comes from leaving out the factor of 2 in the cosine rule formula, computing DF² = 196 + 400 − 14 × 20 × cos(80°) ≈ 547.38 instead.
- (a) 65° — OT and OC are both radii, so triangle OTC is isosceles with OT = OC, and its base angles are equal: angle OTC = angle OCT = (180° − 130°) ÷ 2 = 25°. A tangent is perpendicular to the radius at the point of contact, so angle OTP = 90°. Since C lies inside angle OTP, the chord TC splits this right angle into angle OTC and angle PTC, so angle PTC = angle OTP − angle OTC = 90° − 25° = 65°. Stopping after finding the base angle of the isosceles triangle, without subtracting it from the right angle at T, leaves 25° instead of the angle actually asked for. Adding the base angle to the right angle instead of subtracting it, 90° + 25° = 115°, reverses the direction the two angles combine in. Finding the sum of the two base angles of the isosceles triangle, 180° − 130° = 50°, and stopping there, gives another wrong value entirely.
- (b) 46.8 cm² — Method: for a parallelogram, not a triangle, the two sides and the angle between them give Area = ab sin C — there is no 1/2. Working: Area = 6 × 9 × sin 60° = 46.8 cm² (1 d.p.). Answer: 46.8 cm². Using the triangle formula, (1/2)ab sin C, on a parallelogram by mistake gives half the true area, 23.4 cm²; using cos 60° instead of sin 60° gives 27.0 cm²; and adding the two sides before multiplying by sin 60° gives 13.0 cm². A parallelogram is exactly two of the triangles this formula was built for, so never carry the 1/2 across from the triangle version.
- (d) 8.75 cm — The scale factor of the enlargement is 11.2 ÷ 3.2 = 3.5. Applying this to the second line, 2.5 × 3.5 = 8.75 cm. The distractor 10.50 cm comes from adding the difference between the first line's two lengths (11.2 − 3.2 = 8) to the second line's original length, 2.5 + 8 = 10.5. The distractor 0.71 cm comes from using the scale factor the wrong way round, 2.5 × (3.2 ÷ 11.2) = 0.71 (to 2 d.p.). The distractor 8.70 cm comes from directly subtracting 11.2 − 2.5 = 8.7, muddling the two different lines instead of scaling the second one.
- (b) 5 cm by 3 cm — The plan view looks straight down on the cuboid's footprint, so it shows the length (5 cm, left to right) and the depth (3 cm, front to back) — the two dimensions that do not involve height. "5 cm by 2 cm" repeats the front elevation's dimensions, pairing the length with the height instead of the depth. "3 cm by 2 cm" repeats the side elevation's dimensions, again pairing the depth with the height rather than with the length. "5 cm by 5 cm" comes from mistakenly assuming the plan must be a square, pairing the length with itself instead of with the depth.
- (a) 21.8° — The base diagonal has length √(6² + 8²) = √(36 + 64) = √100 = 10 cm, using Pythagoras' theorem on the rectangular base. The angle between the space diagonal and the base lies in the right-angled triangle formed by the height (4 cm, opposite), the base diagonal (10 cm, adjacent) and the space diagonal (hypotenuse), so tan(angle) = 4 ÷ 10 = 0.4, giving angle = 21.8° (1 d.p.). Using the height and one base edge instead of the full base diagonal, tan(angle) = 4 ÷ 8 = 0.5, gives 26.6° instead. Inverting the ratio, tan(angle) = 10 ÷ 4 = 2.5, gives 68.2°, the complement of the angle rather than the angle itself. Using the height and the other base edge, tan(angle) = 4 ÷ 6, gives 33.7°.
- (d) 118.0 cm² — Method: a rhombus is made of two congruent triangles either side of a diagonal, each with area 1/2 × 12 × 12 × sin 55°, so the whole rhombus has area 12 × 12 × sin 55° (side² × sin of the interior angle). Working: area = 12² × sin 55° = 118.0 cm². Stopping at one triangle's area, 1/2 × 12² × sin 55°, and forgetting to double it gives 59.0 cm²; using cos 55° instead of sin 55° gives 82.6 cm²; and multiplying the two sides together with no trig term at all gives 144.0 cm². Splitting the rhombus into its two triangles is the safest way to see why the 1/2 disappears from the whole-shape formula.
- (a) 4700 cm² — Method: split the T-shaped outline into the two rectangles it is made from, find the area of each, then add them together. Working: the wide base gives a rectangle 90 cm × 30 cm = 2700 cm²; the narrower block on top gives a rectangle 40 cm × 50 cm = 2000 cm²; adding these, 2700 + 2000 = 4700 cm². Answer: 4700 cm². The distractors: 2700 cm² comes from finding only the area of the base rectangle and forgetting to add the block on top. 2000 cm² comes from finding only the area of the top block and forgetting the base. 7200 cm² comes from treating the whole outline as one large rectangle, 90 cm wide by (30 + 50) = 80 cm tall, instead of splitting it into the two separate rectangles that actually make up the shape.
- (d) (11, −2) — First undo the original translation to find the vertex on the original shape: (1 − (−8), 6 − 3) = (9, 3). Then apply the second vector to that original vertex: (9 + 2, 3 + (−5)) = (11, −2). (3, 1) comes from applying the second vector to the image point (1, 6) instead of to the original vertex — (1 + 2, 6 + (−5)) = (3, 1). (7, 8) comes from subtracting the second vector from the original vertex (9, 3) instead of adding it — (9 − 2, 3 − (−5)) = (7, 8). (11, 3) comes from applying only the x-component of the second vector to the original vertex and leaving the y-coordinate unchanged.
- (d) 4.44 m — The real van's length is 18.5 × 24 = 444 cm, which converts to 4.44 m. '0.77 m' comes from dividing by 24 instead of multiplying, using the ratio the wrong way round (18.5 ÷ 24 = 0.77 cm), and then writing that figure down as metres. '44.4 m' comes from converting 444 cm to metres with the decimal point in the wrong place. '444 m' comes from working out 444 cm correctly but forgetting to convert it into metres at all.
- (c) 3028 cm² — Split the window into a rectangle and a semicircle. Rectangle area = 40 × 60 = 2400 cm². The semicircle has radius 40 ÷ 2 = 20 cm, so its area = 0.5 × 3.14 × 20² = 628 cm². Total area = 2400 + 628 = 3028 cm². A student who uses a full circle instead of a semicircle on top of the rectangle gets 3.14 × 20² = 1256 cm² for the circle, plus 2400 cm² for the rectangle, totalling 3656 cm². A student who uses the rectangle's full width (40 cm) as the radius instead of halving it gets 0.5 × 3.14 × 40² = 2512 cm² for the semicircle, plus 2400 cm² for the rectangle, totalling 4912 cm².
- (d) (0, −2) — To rotate (4, 2) by 90° clockwise about (1, 1), first find its position relative to the centre: (4 − 1, 2 − 1) = (3, 1). A 90° clockwise rotation sends (a, b) to (b, −a), so (3, 1) becomes (1, −3); adding the centre back gives (1 + 1, 1 − 3) = (2, −2). Reflecting (2, −2) in the line x = 1 gives (2 × 1 − 2, −2) = (0, −2). Doing the two transformations in the opposite order, reflecting first and then rotating, gives a different result, (2, 4), which shows the order matters. Stopping after the rotation and forgetting the reflection gives (2, −2). Stopping after only reflecting P in x = 1 and forgetting the rotation entirely gives (−2, 2). Rotate first, then reflect, in that order, and the final image is (0, −2).
Build your own mix at the worksheet builder.