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 square-based pyramid has a square base and four identical triangular sloping faces, with its apex directly above the centre of the base. How many planes of symmetry does it have?
- 2.A sector of a circle has an angle of 90° and an area of 78.5 cm². Using π = 3.14, work out the radius of the circle.
- 3.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.
- 4.The diagram shows the plan of the base layer of a solid built from identical cubes. A second layer of cubes is added on top, filling the entire base layer to a height of 2 cubes everywhere. Work out how many squares are visible in the plan view of this solid.
- 5.A designer creates a repeating tile pattern. Each tile is translated from the one before it by the column vector with top number 4.5 and bottom number −2.5 (in centimetres). The first tile has its bottom-left corner at (1.5, 3). Work out the coordinates of the bottom-left corner of the third tile.
- 6.Triangle ABC and triangle DEF both contain a right angle, at B and E respectively, and the hypotenuses AC and DF are equal in length. Which extra fact would prove the triangles are congruent by RHS?
- 7.A cuboid has a length, width and height that are all different from each other. How many planes of symmetry does it have?
- 8.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.
- 9.An isosceles trapezium has exactly one pair of parallel sides, and its two non-parallel sides are equal in length. How many lines of symmetry does it have?
- 10.A solid has 5 faces, 9 edges and 6 vertices. Two of its faces are triangles and the other three are rectangles. Which solid is it?
- 11.A circle has centre O and radius 10 cm. Points A and B lie on the circle such that angle AOB = 130°. Work out the area of the minor segment cut off by the chord AB. Give your answer to 1 decimal place.
- 12.In a phone game, a character starts at the point (−5, 2) on a grid. It moves by the vector to collect a coin. The player now wants the character's next single move to finish at the point (3, 0). Write down the column vector of that second move.
- 13.In triangle ABC and triangle DEF, AB = DE, AC = DF, and the angle at A equals the angle at D. Write down the congruence condition that proves the two triangles are congruent.
- 14.Triangles PQS and RQS share the common side QS. PQ equals RQ, and PS equals RS. Which condition proves that triangle PQS is congruent to triangle RQS?
- 15.Two similar signs have lengths in the ratio 9 : 4. The larger sign has a side of length 14.4 cm. Work out the length of the corresponding side of the smaller sign.
Answer key
- (d) 4 — Method: a plane of symmetry must pass through the apex and cut the base along one of the base's own lines of symmetry. Working: a square has 4 lines of symmetry (2 through opposite edge midpoints, 2 through opposite corners), and each of these, combined with the apex, gives one plane of symmetry of the pyramid. A student who answers 2 has only found the planes through the edge midpoints, or only the ones through the corners, and missed the other pair. A student who answers 8 has doubled the correct count, perhaps confusing it with a different solid. A student who answers 1 has only spotted the one obvious front-to-back plane. Answer: 4.
- (a) 10 cm — Sector area is (angle ÷ 360) × π × radius². Here 90 ÷ 360 = 1/4, and 1/4 × 3.14 = 0.785, so radius² is 78.5 ÷ 0.785 = 100, and the radius is √100 = 10 cm. Stopping after finding radius² and not taking the square root gives 100 cm. Treating 78.5 as the area of the WHOLE circle, ignoring the 90° fraction, gives radius² = 78.5 ÷ 3.14 = 25, so a radius of 5 cm. Correctly finding a radius of 10 cm but then doubling it, mistaking the question for asking the diameter, gives 20 cm.
- (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) 6 squares — Method: the plan view shows the footprint of the solid; a second layer stacked on top of floor positions that are already covered does not create any new squares in the plan. Working: the row of 4 cubes and the row of 2 cubes attached at the end do not overlap, so the footprint has 4 + 2 = 6 distinct squares. Answer: 6 squares. The distractors: 12 squares comes from counting the total number of cubes used, including the second layer (6 floor positions × 2 layers = 12), instead of the footprint. 4 squares comes from counting only the row of four and forgetting the attached row of two. 5 squares comes from wrongly treating the corner square as shared between the two rows (4 + 1 instead of 4 + 2).
- (c) (10.5, −2) — Method: the vector from the first tile to the third tile is the pattern's vector doubled, since two translations happen between them. Working: doubling (4.5, −2.5) gives (9, −5); adding this to the starting corner (1.5, 3) gives x-coordinate 1.5 + 9 = 10.5 and y-coordinate 3 − 5 = −2. Answer: (10.5, −2). A candidate who only applies the vector once, translating to the second tile instead of the third, gets (6, 0.5). A candidate who adds 2.5 instead of subtracting it in the y-coordinate gets (10.5, 8). A candidate who doubles the x-part of the vector correctly but forgets to change the y-coordinate at all gets (10.5, 3).
- (b) AB = DE — RHS needs a right angle, the hypotenuse, and one OTHER side to be equal; the right angles and hypotenuses are already equal, so a matching pair of the remaining sides, AB = DE, completes RHS. Angle A = angle D is an extra ANGLE fact, not the extra SIDE fact that RHS specifically requires. AC being parallel to DF says nothing about either triangle's side lengths, so it cannot complete a congruence condition. Being drawn the same way up is about orientation on the page, not about any measurement, so it proves nothing about congruence.
- (b) 3 — A cuboid with all different edge lengths has three planes of symmetry: one parallel to each pair of opposite faces, cutting the solid exactly in half. Choosing 9 is the number of planes of symmetry a CUBE has (where all edges are equal) — this cuboid's edges are all different, so it has fewer. Choosing 1 counts only one of the three planes and forgets the other two, each parallel to a different pair of faces. Choosing 6 double-counts each of the three planes, as if counting each one from both sides.
- (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.
- (d) 1 — An isosceles trapezium has one line of symmetry, running through the midpoints of the two parallel sides. 0 would be true for a scalene trapezium, whose non-parallel sides are unequal, but this trapezium's non-parallel sides are equal, so it is symmetrical. 2 is the number of lines of symmetry of a rectangle, not a trapezium. 4 comes from wrongly applying the 'number of sides equals number of lines of symmetry' rule, which only holds for REGULAR polygons — a trapezium is not regular.
- (a) Triangular prism — Count the given properties against each solid. A triangular prism has 5 faces (2 triangular ends + 3 rectangular sides), 9 edges and 6 vertices — this matches exactly, so the solid is a triangular prism. A square-based pyramid also has 5 faces, but 8 edges and 5 vertices, and only one face is a square rather than three rectangles — the edge and vertex counts don't match. A cuboid has 6 faces, 12 edges and 8 vertices, none of which match the numbers given. A tetrahedron has 4 faces, all triangles, with 6 edges and 4 vertices — too few faces, and no rectangular faces at all.
- (b) 75.1 cm² — Method: a segment is a sector with its triangle cut away, so find the sector area and the triangle area (using Area = (1/2)r² sin C on the two radii) and subtract. Working: sector area = (130 ÷ 360) × π × 10² = 113.4 cm² (1 d.p.); triangle area = 1/2 × 10 × 10 × sin 130° = 38.3 cm² (1 d.p.); segment area = 113.4 − 38.3 = 75.1 cm². Answer: 75.1 cm². Reporting the sector area on its own, without subtracting the triangle, gives 113.4 cm²; reporting the triangle area on its own gives 38.3 cm²; and using the reflex angle, 360° − 130° = 230°, in the sector but still subtracting the triangle gives 162.4 cm², which is neither segment — the major segment would be the 230° sector PLUS the triangle, 239.0 cm². The minor segment is the smaller piece, cut off by the shorter arc, so use the angle actually given, 130°, and subtract the triangle from that sector.
- (c) $\binom{2}{7}$ — First find the character's position after the first move: (−5 + 6, 2 + (−9)) = (1, −7). The second move takes it from (1, −7) to (3, 0), so subtract the current position from the target: (3 − 1, 0 − (−7)) = $\binom{2}{7}$. $\binom{8}{−2}$ works from the starting point (−5, 2) and ignores the first move — (3 − (−5), 0 − 2) = $\binom{8}{−2}$. $\binom{−2}{−7}$ subtracts the wrong way round, current position minus target, instead of target minus current position — (1 − 3, −7 − 0) = $\binom{−2}{−7}$. $\binom{4}{−7}$ adds the target's coordinates to the current position instead of subtracting — (1 + 3, −7 + 0) = $\binom{4}{−7}$.
- (d) SAS — Method: a congruence condition is named by the parts that are given equal and the order in which they sit round the triangle, so count the sides and the angles first. Working: AB = DE and AC = DF are two pairs of equal sides, and the equal angle at A and D lies between AB and AC, so the given parts read side, included angle, side. Answer: SAS. The distractors: SSS needs three pairs of equal sides, and the third pair, BC and EF, is not given — it follows from the proof rather than being part of it; ASA reads the two equal sides as two equal angles, swapping which facts are which; RHS applies only when the triangles contain a right angle and the equal pair includes the hypotenuse, and nothing here says the angle at A is 90°.
- (d) SSS, using shared side QS — PQ equals RQ and PS equals RS are two given pairs of equal sides, and QS is common to both triangles, so QS equals itself and gives a third pair of equal sides. Three pairs of equal sides is exactly the SSS condition, so 'SSS, using shared side QS' is correct. 'SAS, using the angle at Q' is wrong because no angle is given anywhere in this question; angle PQS and angle RQS are not stated to be equal, and assuming they are would be assuming the very thing being proved. 'Only two pairs of sides — not enough' is wrong because it forgets that the shared side QS is itself a third pair of equal sides. 'Cannot prove — no angle given' is wrong because SSS is one of the four basic congruence conditions and specifically requires no angle at all.
- (d) 6.4 cm — The scale factor from the larger sign to the smaller sign is 4 ÷ 9, so the smaller side is 14.4 × 4 ÷ 9 = 6.4 cm. The distractor 32.4 cm comes from using the ratio the wrong way round, 14.4 × 9 ÷ 4 = 32.4. The distractor 9.4 cm comes from subtracting the difference between the ratio numbers (9 − 4 = 5) from the given length, 14.4 − 5 = 9.4. The distractor 3.6 cm comes from dividing the given length by 4 only, 14.4 ÷ 4 = 3.6, without also using the other ratio number.
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