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.A sector of a circle has radius 6 cm and an area of 4.71 cm². Using π = 3.14, work out the angle of the sector.
- 2.A student is proving that opposite angles of a cyclic quadrilateral ABCD sum to 180°, letting a = angle DAB and c = angle BCD, and using the fact that the angle at the centre is twice the angle at the circumference. She has already written: 'By the angle at the centre theorem, the reflex angle BOD equals 2a and the non-reflex angle BOD equals 2c.' Which statement must come immediately after this one in a correct proof?
- 3.AT is a diameter of a circle with centre O. PT is a tangent to the circle at the point T, and P is a point on this tangent such that A, T and P form a triangle. Angle PAT = 28°. Work out the size of angle APT.
- 4.Shape S is enlarged by a scale factor of −3 about a fixed centre. Which statement correctly describes the image compared to the original shape?
- 5.A party hat is in the shape of a cone with a base radius of 6 cm and a slant height of 10 cm. Work out the curved surface area of the party hat. Use π = 3.14 and the formula curved surface area = πrl.
- 6.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?
- 7.A rectangular name card has a perimeter of 30 cm. The card is 9 cm long. Work out the width of the card.
- 8.Quadrilateral WXYZ has WX parallel to ZY, and WZ = XY (the two non-parallel sides are equal in length). Give a reason why angle W = angle X.
- 9.A chord divides a circle into two regions, one larger than the other. Write down the term used for the smaller of these two regions.
- 10.The diagram shows the plan view and the front elevation of a cuboid-shaped box, each drawn as a rectangle. Work out the area of the box's side elevation, in square centimetres.
- 11.A footpath is 3750 m long. Work out this distance in kilometres.
- 12.A right-angled triangle has a hypotenuse of 10 cm. One of its other angles is 45°. Work out the exact length of one of the two shorter sides.
- 13.A scale drawing of a park uses a scale of 1 : 2000. A path is drawn 8.5 cm long on the drawing. A cyclist rides the length of the path and then rides straight back again along the same path. How far does the cyclist travel in total, in metres?
- 14.Points P and Q have coordinates P(0, 0, 0) and Q(5, 7, 9) in a three-dimensional coordinate system, with all lengths in centimetres. Work out the distance PQ. Give your answer correct to 1 decimal place.
- 15.A triangular prism has a cross-section that is a triangle with a base of 3.5 cm and a perpendicular height of 4 cm. The prism is 10 cm long. Work out the volume of the prism.
Answer key
- (d) 15° — Sector area = (angle ÷ 360) × π × r², so 4.71 = (angle ÷ 360) × 3.14 × 36 = (angle ÷ 360) × 113.04. Dividing gives angle ÷ 360 = 4.71 ÷ 113.04 = 1/24, so angle = 360 ÷ 24 = 15°. (90° comes from forgetting to square the radius, using (angle ÷ 360) × 3.14 × 6 = 18.84 in place of 113.04; 3.75° comes from using the diameter, 12 cm, in place of the radius, giving (angle ÷ 360) × 3.14 × 144 = 452.16; 45° comes from using the arc length formula, (angle ÷ 360) × 2 × 3.14 × 6 = 37.68, instead of the sector area formula.)
- (a) Reflex + non-reflex angle BOD = 360°, so 2a + 2c = 360°. — A correct proof must build in a strictly logical order: the reflex and non-reflex angles at O, which together make a complete turn about the centre, must be added to 360° before 2a and 2c can be combined and divided. The statement 'Reflex + non-reflex angle BOD = 360°, so 2a + 2c = 360°' is the only one that follows directly from having 2a and 2c already established, and it is exactly what is needed before the final division step. Jumping straight to 'Divide by 2 throughout to get a + c = 180°' skips the step that justifies why 2a + 2c equals 360° in the first place — there is nothing yet to divide. Repeating 'Angle BOD is 2a and 2c, by the centre theorem' does not move the proof forward at all, since that fact has already been established in the sentence given. Introducing 'OB = OD are radii, so triangle OBD is isosceles' brings in an unrelated triangle and an unrelated method that plays no part in this particular proof.
- (b) 62° — Because AT is a diameter, the radius OT lies along AT, and the tangent PT meets every radius at 90°, by the tangent–radius theorem. So PT meets AT at T at a right angle: angle ATP = 90°. The angles of triangle APT sum to 180°, so angle APT = 180° − 90° − 28° = 62°. Copying the given angle PAT straight across, as though the triangle were isosceles, gives 28°. Adding the two known angles instead of subtracting them from 180° gives 90° + 28° = 118°. Naming the right angle itself, angle ATP, instead of the angle that was actually asked for gives 90°. Subtract both known angles from 180°, and 62° is angle APT.
- (a) 3 times the size, opposite side, rotated 180° — Method: for any enlargement, the MAGNITUDE of the scale factor gives the size ratio between image and object, while the SIGN decides which side of the centre the image falls on; a negative scale factor puts the image on the opposite side, which is equivalent to a 180° rotation about the centre. Working: the scale factor is −3, so the size ratio is the magnitude, which is 3, and the negative sign puts the image on the opposite side of the centre, rotated 180° relative to the original. Answer: 3 times the size, opposite side, rotated 180°. The magnitude of the scale factor controls the SIZE only: do not let the sign leak into it and turn 3 into 1/3. The sign controls the SIDE and orientation, which a positive-only view of enlargement, just 'further away' with the same orientation, misses entirely.
- (a) 188.4 cm² — Curved surface area of a cone = πrl. With r = 6 cm, l = 10 cm and π = 3.14, curved surface area = 3.14 × 6 × 10 = 188.4 cm². A student who uses the cylinder's curved surface area formula, 2πrl, instead of the cone's gets 2 × 3.14 × 6 × 10 = 376.8 cm². A student who uses the circle-area formula πr² instead of πrl gets 3.14 × 36 = 113.04 cm². A student who multiplies r × l but leaves out π entirely gets 6 × 10 = 60 cm².
- (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) 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) Isosceles trapezium: base angles are equal — WX is parallel to ZY and the two non-parallel sides WZ and XY are equal in length, so WXYZ is an isosceles trapezium. In an isosceles trapezium the two angles at each of the parallel sides are equal, so angle W = angle X (and angle Z = angle Y). A student who answers with the parallelogram property has quoted a fact that is true of a parallelogram, but WZ and XY are given as non-parallel so WXYZ is not a parallelogram — and in a parallelogram "opposite angles" pairs W with Y, not W with X. A student who answers with the kite property has again taken a true fact about the wrong shape: a kite's equal sides are two pairs of adjacent sides, not the two non-parallel sides of a trapezium. A student who answers with the rhombus property has used something no rhombus has — a rhombus has two pairs of parallel sides and only two pairs of equal angles; all four angles are equal only in a square.
- (c) minor segment — A chord splits a circle into two segments; the smaller of the two is called the minor segment and the larger one the major segment. "major segment" names the larger region, the opposite of what is asked for. "sector" is a different region altogether, enclosed by two radii and an arc, not by a chord. "arc" is a curved length along the circumference, not an enclosed region at all.
- (d) 15 cm² — Method: identify the length that is common to both views, since it is the box's length; then read off the width from the plan view and the height from the front elevation, and multiply those two measurements to find the area of the side elevation. Working: both rectangles share a side of 8 cm, which is the box's length; the plan view's other side gives a width of 5 cm, and the front elevation's other side gives a height of 3 cm. The side elevation is bounded by the width and the height: 5 cm × 3 cm = 15 cm². Answer: 15 cm². The distractors: 24 cm² comes from giving the area of the front elevation shown (8 cm × 3 cm) instead of working out a new rectangle for the side elevation. 40 cm² comes from giving the area of the plan view shown (8 cm × 5 cm) instead of working out the side elevation. 120 cm² comes from multiplying all three measurements together (8 cm × 5 cm × 3 cm), finding the volume of the box instead of the area of one face.
- (a) 3.75 km — To convert metres to kilometres, divide by 1000: 3750 ÷ 1000 = 3.75 km. Dividing by 100 instead of 1000 gives 37.5 km. Dividing by 10 instead of 1000 gives 375 km. Dividing by 10 000 instead of 1000 gives 0.375 km.
- (a) 5√2 cm — Method: the angles of a triangle add to 180°, so the third angle is 45° as well and the two shorter sides are equal. Take one of them as the side opposite a 45° angle and use sin 45° = opposite ÷ hypotenuse. Working: the exact value of sin 45° is √2/2, so the shorter side = 10 × √2 ÷ 2, and half of 10 is 5. Answer: 5√2 cm, which is about 7.07 cm. Remembering sin 45° as √2 rather than as √2 halved gives 10√2 cm, which is longer than the hypotenuse. Halving the hypotenuse because 45° is half of 90° gives 5 cm. Taking the value from the other special triangle, sin 60° = √3/2, gives 5√3 cm.
- (b) 340 — The real length of the path is 8.5 × 2000 = 17000 cm, which converts to 170 m by dividing by 100. The cyclist rides the path there and back, so the total distance is 170 × 2 = 340 m. A candidate who forgets the return journey gives only the one-way distance, 170 m. A candidate who halves the one-way distance instead of doubling it, reading “there and back” as splitting the journey, gets 85 m. A candidate who doubles the scale factor to 4000 by mistake, getting a one-way length of 340 m, and then doubles that for the return journey, gets 680 m. The total distance the cyclist travels is 340 m.
- (c) 12.4 cm — In three dimensions, the distance between two points extends Pythagoras' theorem to three squared terms: PQ² = 5² + 7² + 9² = 25 + 49 + 81 = 155. Taking the square root, PQ = √155 = 12.4 cm (1 d.p.). Using only the x- and y-coordinates, and ignoring the third dimension entirely, gives PQ = √(5² + 7²) = √74 = 8.6 cm (1 d.p.), which is not the distance in 3D space. Adding the three coordinates directly, 5 + 7 + 9 = 21 cm, treats the coordinates as if they were lengths along a single straight path rather than the sides of a right-angled arrangement. Squaring and adding all three coordinates but forgetting to take the square root leaves 155 cm, the squared distance rather than the distance itself.
- (b) 70 cm³ — Area of the triangular cross-section = base × height ÷ 2. Base × height = 3.5 × 4 = 14 cm², and half of that is 14 ÷ 2 = 7 cm². Volume of the prism = cross-sectional area × length = 7 × 10 = 70 cm³. A pupil who forgets to halve when finding the triangle's area gets 3.5 × 4 × 10 = 140 cm³. A pupil who correctly finds the cross-sectional area but forgets to multiply by the length stops at 7 cm³. A pupil who ignores the height altogether, multiplying base by length, gets 3.5 × 10 = 35 cm³. The correct volume is 70 cm³.
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