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.
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Answer key: Geometry and measures worksheet — GCSE Foundation
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- (b) 113.04 cm² — Sector area is (angle ÷ 360) × π × radius², and radius means the RADIUS, not the diameter: here the diameter is 24 cm, so the radius is 12 cm. The fraction is 90 ÷ 360 = 1/4, and 12² = 144, so the area is 0.25 × 3.14 × 144 = 113.04 cm². Using the diameter itself as if it were the radius gives 0.25 × 3.14 × 576 = 452.16 cm². Using the arc-length formula, 2 × π × radius, instead of the area formula gives 0.25 × 2 × 3.14 × 12 = 18.84 cm². Using the radius instead of its square gives 0.25 × 3.14 × 12 = 9.42 cm².
- (d) positive — Moving from A to B, the x-coordinate increases from 1 to 4 and the y-coordinate also increases from 1 to 7, so the line rises as it goes from left to right, which means the gradient is positive. "negative" would need y to decrease as x increases, which is not the case here. "zero" would need the y-coordinate to stay the same, but it changes from 1 to 7. "cannot be determined" is wrong because the coordinates of both points are known, so the direction of the line can always be found.
- (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.
- (c) Reflex — Method: compare the measured angle to the key boundaries of 90 degrees, 180 degrees and 360 degrees. Working: 254 degrees is greater than 180 degrees but less than 360 degrees, so it lies in the reflex range. A student who answers obtuse knows the angle is bigger than 90 degrees but has not realised it is also bigger than 180 degrees; they may have worked with the smaller angle at the same point, 360 - 254 = 106 degrees, instead of the reading given. A student who answers acute has misread the scale and taken the reading as 54 degrees rather than 254 degrees. A student who answers right has guessed a commonly recognised angle type without comparing the size properly. Answer: reflex.
- (d) 60° — Method: the 4 cm side is next to the angle wanted and the 8 cm side is the hypotenuse, so the ratio built from them is cos θ = adjacent ÷ hypotenuse, and the angle comes from the inverse cosine. Working: cos θ = 4 ÷ 8 = 0.5, so θ = cos⁻¹(0.5). Answer: 60°. The distractors: 30° comes from using sin⁻¹(0.5), which treats the 4 cm side as the side opposite the angle when it is the side next to it; 45° comes from assuming the two acute angles of the triangle must be equal; 90° comes from writing down the right angle the question already gives instead of the angle it asks for.
- (b) 21 — Both legs of the journey are on the same bearing, 070°, so the ship travels in one straight line the whole way and the distances simply add: 12 + 9 = 21 km. A candidate who subtracts instead of adding gets 12 − 9 = 3 km. A candidate who multiplies the two distances gets 12 × 9 = 108. A candidate who assumes the ship changed direction and treats the two legs as the sides of a right-angled triangle works out √(12² + 9²) = √225 = 15 km — but the bearing does not change, so there is no triangle and no hypotenuse to find. Because the ship stays on one straight line, the total distance is 21 km.
- (b) £33.60 — The area of the parallelogram flower bed is base × height = 3.5 × 2 = 7 m². The cost is 7 × £4.80 = £33.60. £16.80 comes from using the triangle formula instead of the parallelogram formula: 3.5 × 2 = 7, and half of 7 is 3.5 m², then 3.5 × £4.80 = £16.80. £26.40 comes from adding the base and height, 3.5+2 = 5.5, instead of multiplying them, then multiplying by £4.80. £7.00 correctly finds the area, 7 m², but forgets to multiply it by the cost per m².
- (d) 46.2 cm — The perimeter of a sector is the arc length plus its two straight radii. The circumference is 2 × 3.14 × 10 = 62.8 cm, and the arc is 150 ÷ 360 of that: 62.8 × 150 ÷ 360 = 26.2 cm (1 d.p.). Adding the two radii, 26.2 + 10 + 10 = 46.2 cm. Giving just the arc length, without adding the straight edges, gives 26.2 cm. Adding only ONE radius instead of two gives 36.2 cm. Using 150 ÷ 180 instead of 150 ÷ 360 for the fraction gives an arc of 52.3 cm and a perimeter of 72.3 cm.
- (d) 7 cm — Area of a parallelogram = base × height, so height = area ÷ base = 84 ÷ 12 = 7 cm. A pupil who multiplies instead of dividing gets 84 × 12 = 1008 cm. A pupil who divides the base by the area instead of the area by the base gets 12 ÷ 84 ≈ 0.14 cm. A pupil who mistakenly halves the area first, as if this were a triangle, gets (84 ÷ 2) ÷ 12 = 3.5 cm. The correct height is 7 cm.
- (a) $\binom{1}{6.5}$ — After the first flight, the drone is at (3.5 − 6.2, −2 + 4.5) = (−2.7, 2.5). The second vector takes it from (−2.7, 2.5) to (−1.7, 9): subtract the coordinates, (−1.7 − (−2.7), 9 − 2.5) = (1, 6.5). The distractor $\binom{−4.4}{6.5}$ comes from treating the drone's position after the first flight as (2.7, 2.5) instead of (−2.7, 2.5), giving −1.7 − 2.7 = −4.4 for the top number. The distractor $\binom{−5.2}{11}$ comes from finding the vector straight from the start point (3.5, −2) to (−1.7, 9), ignoring the first flight altogether. The distractor $\binom{−1}{−6.5}$ comes from subtracting the wrong way round, (−2.7 − (−1.7), 2.5 − 9), which reverses both signs of the correct vector.
- (c) tan θ = opposite/adjacent — The tangent ratio is defined as tan θ = opposite/adjacent, so this is the ratio that connects the opposite and adjacent sides. "sin θ = opposite/adjacent" wrongly labels this ratio as sine, when sine actually connects the opposite side and the hypotenuse. "cos θ = opposite/adjacent" wrongly labels it as cosine, when cosine connects the adjacent side and the hypotenuse. "tan θ = adjacent/opposite" uses the correct ratio name but has the opposite and adjacent sides the wrong way round.
- (b) 4 m — The two triangles are similar, so height ÷ shadow is the same for both objects. For the post, height ÷ shadow = 1.2 ÷ 1.8. Flagpole height = 6 × (1.2 ÷ 1.8) = 4 m. A student who multiplies the flagpole's shadow by the post's height without applying the ratio gets 6 × 1.2 = 7.2 m. A student who assumes a constant DIFFERENCE between shadow and height instead of a constant ratio finds the post's shadow minus height, 1.8 − 1.2 = 0.6 m, then subtracts that from the flagpole's shadow to get 6 − 0.6 = 5.4 m.
- (b) 56.52 cm — Arc length = (216 ÷ 360) × 2 × 3.14 × 15 = 0.6 × 94.2 = 56.52 cm. (28.26 cm comes from leaving out the factor of 2, using πr instead of 2πr; 94.2 cm comes from finding the full circumference and forgetting the angle fraction; 113.04 cm comes from using the diameter, 30 cm, instead of the radius.)
- (a) 8.1 m — Multiply the length on the plan by the scale factor: 4.5 × 1.8 = 8.1, so the real length is 8.1 m. Choosing 6.3 m adds the two numbers instead of multiplying them (4.5 + 1.8 = 6.3). Choosing 2.5 m divides the plan length by the scale factor the wrong way round (4.5 ÷ 1.8 = 2.5) instead of multiplying. Choosing 9.0 m rounds the scale factor up to 2 before multiplying (4.5 × 2 = 9.0), losing the accuracy the 1.8 was giving.
- (a) 035° — A bearing is measured clockwise from north, so an angle of 35° clockwise from north is a bearing of 035° (written with three figures). Choosing 325° measures the angle anticlockwise instead of clockwise (360 − 35 = 325). Choosing 215° adds 180° to the angle, mixing this up with a back-bearing calculation (35 + 180 = 215). Choosing 350° reorders the digits of 035, writing the ones digit before the tens digit by mistake.
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