Printable · GCSE Foundation · ages 14-16
Geometry and measures worksheet — GCSE Foundation
Fifteen questions across the geometry and measures statements at Foundation tier. Choose the non-calculator filter to rehearse Paper 1, which counts for a third of the marks.
Geometry and measures worksheet — GCSE Foundation
MathsUKwww.geekhero.co.uk
- 1.Triangle DEF has a right angle at E. Angle DFE = 60° and EF = 9 cm. Work out the exact length of DE.
- 2.a is the column vector with top number 3 and bottom number 2. b is the column vector with top number −1 and bottom number 4. Work out a + b, giving your answer as a column vector in the form (top, bottom).
- 3.A tent is shaped like a triangular prism: two triangular end faces and three rectangular faces. One of the rectangular faces lies flat on the ground as the groundsheet. How many of the tent's faces are above the ground, not touching the ground?
- 4.A bottle holds 2 litres of water. A glass holds 250 ml. Work out how many of these glasses can be filled from one full bottle.
- 5.A parallelogram has a base of 9 cm and a perpendicular height of 6 cm. Work out the area of the parallelogram.
- 6.Two of the angles in a triangle are 50° and 70°. Work out the size of the third angle.
- 7.A construction for the perpendicular bisector of AB begins by opening a pair of compasses to more than half the length of AB, then drawing an arc centred at A. What is the next step?
- 8.A sector of a circle has radius 9 cm and angle 40°. Work out the area of the sector, in terms of π.
- 9.A cake recipe takes 135 minutes from start to finish. Write this time as hours and minutes.
- 10.A circle has a radius of 3 cm. Work out the diameter of the circle.
- 11.Triangle JKL has a right angle at K, hypotenuse JL = 13 cm, and side JK = 5 cm. Triangle MNO has a right angle at N, hypotenuse MO = 13 cm, and side MN = 5 cm. Using only the facts given, and without working out any further lengths, write down the congruence condition that proves triangle JKL is congruent to triangle MNO.
- 12.In a construction, a point P is marked at the foot of the perpendicular drawn from point X to a straight line. Which statement correctly compares the distance XP with the distance from X to any other point on the line?
- 13.Four angles meet at a point. Three of them measure 82°, 105° and 96°. Work out the size of the fourth angle.
- 14.A triangle has a base of 8 cm and a perpendicular height of 5 cm. Work out its area.
- 15.Two similar hexagonal tiles have lengths in the ratio 4 : 7. A side of the smaller tile is 8.4 cm long. Work out the length of the corresponding side of the larger tile.
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.
- (b) (2, 6) — To add column vectors, add the top numbers together and add the bottom numbers together, separately: top = 3 + (−1) = 2, bottom = 2 + 4 = 6, giving (2, 6). A candidate who writes the top and bottom numbers of the correct answer in the wrong order gives (6, 2). A candidate who subtracts instead of adds, working out 3 − (−1) = 4 and 2 − 4 = −2, gives (4, −2). A candidate who adds the top numbers correctly but subtracts the bottom numbers by mistake, working out 2 − 4 = −2, gives (2, −2). The correct answer is (2, 6).
- (b) 4 — A triangular prism has 5 faces in total: 2 triangular ends and 3 rectangular faces. One rectangular face is the groundsheet, lying on the ground, so the other 5 − 1 = 4 faces are above the ground. A candidate who forgets one of the triangular ends when counting the remaining faces answers 3. A candidate who forgets to subtract the groundsheet at all answers 5. A candidate who only counts the two sloped rectangular faces, forgetting the two triangular ends, answers 2. The number of faces above the ground is 4.
- (a) 8 — Method: two capacities can only be divided once they are written in the same unit. Working: there are 1000 ml in 1 litre, so the bottle holds 2 × 1000 = 2000 ml. Then 2000 ÷ 250 = 8. Answer: 8 glasses. Using 1 litre = 10 000 ml gives 20 000 ÷ 250 = 80. Dividing the two numbers as they stand, 250 ÷ 2 = 125, ignores the units altogether. Converting the glass into litres with the wrong factor, as though 250 ml were 2.5 litres, gives 2 ÷ 2.5 = 0.8.
- (b) 54 cm² — Area of a parallelogram = base × perpendicular height = 9 × 6 = 54 cm². (27 cm² comes from halving the product as if it were a triangle, 1/2 × 9 × 6; 15 cm² comes from adding the base and height instead of multiplying, 9 + 6; 30 cm² comes from doubling the sum of the base and height, as if finding a perimeter, 2 × (9 + 6).)
- (a) 60° — Method: the three angles of a triangle add up to 180°, so add the two known angles and subtract the total from 180°. Working: 50 + 70 = 120, then 180 − 120 = 60. Answer: 60°. The distractors: 120° is the sum of the two known angles, given as the answer instead of being subtracted from 180°; 110° comes from subtracting only the 70° angle from 180° and forgetting the 50° one; 240° comes from subtracting the sum from 360°, using the angles at a point rather than the angle sum of a triangle.
- (b) an arc of the same radius, centred at B — The perpendicular bisector construction needs two arcs of equal radius, one centred at each end of the segment — after the arc from A, the next step is an arc of exactly the same radius from B, so the two arcs cross at two points; the line through those two crossing points is the perpendicular bisector. "an arc of the same radius, centred at the midpoint of AB" is not possible yet, since the midpoint is only found once both arcs are drawn — it is the RESULT of the construction, not a step in it. "a smaller arc, centred at A again" gives two different-sized arcs from the same point, which never cross to give the bisector. "a straight line joining the ends of the first arc" only connects points on one arc, and locates nothing.
- (b) 9π cm² — Sector area is the fraction θ/360 of the full circle's area, πr². Substitute θ = 40 and r = 9: 40 out of 360 is one ninth, and the full circle's area is π × 9² = 81π. One ninth of 81π is 9π, so the sector area is 9π cm². Choosing 2π cm² is this sector's ARC LENGTH, not its area — a different formula and a different quantity. Choosing 81π cm² is the area of the WHOLE circle, forgetting to scale down by the sector's angle. Choosing 18π cm² is the CIRCUMFERENCE of the whole circle, 2π × 9, mixed up with an area.
- (c) 2 hours 15 minutes — Divide the total minutes by 60: 135 ÷ 60 = 2 remainder 15, so that is 2 hours 15 minutes. Treating 100 minutes as one hour, a metric-style mistake, gives 135 − 100 = 35, so 1 hour 35 minutes. Subtracting 60 twice to reach the 15 minutes left over, but losing count and recording only one of the two hours removed, gives 1 hour 15 minutes. Writing 135 ÷ 60 = 2.25 and then reading the '25' as minutes, instead of converting the 0.25 of an hour into 15 minutes, gives 2 hours 25 minutes.
- (a) 6 cm — Method: a diameter runs right across a circle through its centre, so it is made of two radii laid end to end, which gives diameter = 2 × radius. Working: the radius is 3 cm, so the diameter is 2 × 3 = 6 cm. Answer: 6 cm. The distractors: 1.5 cm comes from dividing by 2 instead of multiplying by it, which is the relationship applied in the wrong direction; 3 cm comes from copying the radius straight down, treating the two words as names for the same measurement; 5 cm comes from adding 2 to the radius instead of multiplying the radius by 2.
- (c) RHS — Both triangles are right-angled, and have equal hypotenuses (13 cm) and one equal corresponding side (5 cm), so they are congruent by the RHS (right angle, hypotenuse, side) condition — the three facts given are exactly a right angle, a hypotenuse and one other side. The distractor SAS would apply if two sides and the angle between those two sides were matched, but the right angle here lies between the 5 cm side and the third side, not between the 5 cm side and the hypotenuse. The distractor SSS needs all three pairs of sides matched, and only two sides of each triangle are given — the third side would first have to be calculated by Pythagoras, so SSS is not the condition the given facts supply. The distractor AAS would need two angles and a non-included side matched, but only one angle in each triangle is given.
- (d) XP is the shortest distance from X to the line — The perpendicular from a point to a line always gives the shortest possible distance to that line — joining X to any other point on the line forms the hypotenuse of a right-angled triangle with XP as one of the shorter sides, and a hypotenuse is always longer than either of the other two sides. So XP is shorter than the distance to every other point on the line. "XP is the longest distance from X to the line" reverses this relationship. "XP equals every other distance from X to the line" would only be true if X were equidistant from every point on the line, which is impossible for a point and a straight line. "XP cannot be compared without knowing the line's length" is false — the shortest-distance fact holds whatever the line's length, since only the local right angle matters.
- (d) 77 — Method: angles that meet at a point add up to 360°. Working: 82 + 105 + 96 = 283; 360 − 283 = 77. Answer: 77°. A candidate who gives the sum of the three known angles and forgets to subtract it from 360° gets 283. A candidate who leaves out the 105° angle, working out 360 − 82 − 96, gets 182. A candidate who leaves out the 82° angle, working out 360 − 105 − 96, gets 159.
- (c) 20 cm² — The area of a triangle is half of base × height. First, base × height = 8 × 5 = 40. Half of 40 is 20 cm². 40 cm² forgets to halve and just gives base × height. 13 cm² adds the base and height together instead of multiplying them. 80 cm² doubles base × height instead of halving it.
- (c) 14.7 cm — The scale factor from the smaller tile to the larger tile is 7 ÷ 4 = 1.75, so the larger side is 8.4 × 1.75 = 14.7 cm. '4.8 cm' comes from scaling by 4 ÷ 7 instead, using the ratio the wrong way round. '11.4 cm' comes from adding the difference between the ratio numbers, 7 − 4 = 3, onto 8.4, instead of scaling. '58.8 cm' comes from multiplying by 7 on its own, using a ratio number as the scale factor instead of working out 7 ÷ 4 = 1.75 first.
Build your own mix at the worksheet builder.