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
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- 1.p is the column vector with top number 5 and bottom number 2. q is the column vector with top number 1 and bottom number 6. Work out p + q, giving your answer as a column vector in the form (top, bottom).
- 2.Triangle ABC has a right angle at B, hypotenuse AC = 13 cm and AB = 5 cm. Triangle DEF has a right angle at E, hypotenuse DF = 13 cm and DE = 5 cm. Which condition proves the two triangles are congruent?
- 3.A perpendicular bisector is drawn through the midpoint of a line segment AB. Which set of points does this construction show?
- 4.A circle has diameter 10 cm. Using π = 3.14, work out the circumference of the circle.
- 5.Two similar triangular tiles have lengths in the ratio 3 : 5. The longest side of the smaller tile is 9 cm. Work out the length of the longest side of the larger tile.
- 6.Triangle ABC has AB = 8 cm, BC = 10 cm and CA = 6 cm. Triangle LMN has LM = 10 cm, MN = 6 cm and NL = 8 cm. The two triangles are congruent by SSS. Which of the following correctly matches the corresponding vertices?
- 7.Lines AB and CD cross at a point, forming four right angles where they meet. Which notation correctly describes the relationship between AB and CD?
- 8.A ramp is built from two straight metal supports that rest on the flat ground and meet each other at the top, forming a triangle with the ground. One support meets the ground at 34°. The angle between the two supports where they meet is 71°. Work out the angle between the second support and the ground.
- 9.The curved part of a circle's circumference, between two points on the circle, has its own name. Write down this name.
- 10.A student says the reverse of the translation vector is . Write down the correct reverse translation vector.
- 11.A line segment joins A(1, 1) and B(4, 7). Write down the sign of the gradient of this line.
- 12.Two triangles are proved congruent using the ASA condition. What can be concluded about the two remaining pairs of corresponding sides that were not part of the original ASA facts?
- 13.Two angles are supplementary. One of them is twice the size of the other. Work out the size of the smaller angle.
- 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.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.
Answer key
- (d) (6, 8) — Method: add column vectors by adding the top numbers together and the bottom numbers together. Working: top numbers 5 + 1 = 6; bottom numbers 2 + 6 = 8. Answer: p + q = (6, 8). A candidate who subtracts instead of adds gets (4, −4). A candidate who swaps the top and bottom of the correct answer gets (8, 6). A candidate who only adds the top numbers and leaves the bottom number of p unchanged gets (6, 2).
- (b) RHS - right angle, hypotenuse and one side equal — A right angle, the hypotenuse (13 cm) and one other side (5 cm) are equal in both triangles, so this is RHS. SAS would need the equal angle to be the one INCLUDED between the two equal sides, but the right angle at B is not between AB and the hypotenuse AC — it is opposite the hypotenuse instead, so SAS does not apply directly here. SSS needs all three sides, but only two sides are stated. ASA needs two angles, but only one angle (the right angle) is given.
- (d) points the same distance from A as from B — The perpendicular bisector of AB is, by definition, the locus of every point equidistant from A and B — any point on it forms two congruent right-angled triangles with A and B, which is exactly the property the construction guarantees. "points as far from A as the length AB" describes a circle centred at A with radius AB, not a bisector. "the single point exactly halfway along AB" names only the midpoint, one point, not the whole locus the construction produces. "points twice as far from A as from B" describes a different curve entirely, not a straight line construction.
- (b) 31.4 cm — Circumference = π × diameter, so 3.14 × 10 = 31.4 cm. 15.7 cm comes from using the radius, 5 cm, in place of the diameter: 3.14 × 5 = 15.7, which is only half the circumference. 78.5 cm comes from using the area formula π × radius² instead of the circumference formula: 3.14 × 5² = 3.14 × 25 = 78.5. 62.8 cm comes from keeping the 2 from the radius form of the formula, C = 2 × π × radius, but putting the full diameter into it: 2 × 3.14 × 10 = 62.8.
- (c) 15 cm — Method: going from the smaller tile to the larger one, every length is multiplied by the scale factor 5 ÷ 3. Working: 9 ÷ 3 = 3, and 3 × 5 = 15. Answer: 15 cm. The distractors: 11 cm comes from adding the difference between the parts of the ratio, 5 − 3 = 2, to the 9 cm side, treating a ratio as a gap rather than a multiplier; 45 cm comes from multiplying by 5 and forgetting to divide by 3; 27 cm comes from multiplying by 3, which is the part of the ratio belonging to the smaller tile.
- (c) A↔N, B↔L, C↔M (ABC≅NLM) — Matching equal side lengths: AB (8 cm) equals NL (8 cm), BC (10 cm) equals LM (10 cm), and CA (6 cm) equals MN (6 cm). This gives the correspondence A with N, B with L, and C with M, so triangle ABC is congruent to triangle NLM, making 'A↔N, B↔L, C↔M (ABC≅NLM)' correct. 'A↔L, B↔M, C↔N (ABC≅LMN)' simply matches the vertices in the order they are written without checking the side lengths: AB (8 cm) would need to equal LM (10 cm), which is false. 'A↔M, B↔N, C↔L (ABC≅MNL)' also fails this check, since AB (8 cm) would need to equal MN (6 cm), which is false. 'A↔N, B↔M, C↔L (ABC≅NML)' gets A correct but swaps B and C, so AB (8 cm) would need to equal NM (6 cm), which is also false.
- (c) AB ⊥ CD — Method: match the description of the lines to the correct symbol. Working: the symbol ⊥ means 'is perpendicular to', used when two lines meet at right angles, which is exactly what is described. Options: ∥ means 'is parallel to', for lines that never meet, so it does not apply here; = compares two lengths as equal, but no lengths are mentioned; ≅ means 'is congruent to', used for whole shapes, not for describing how two lines cross. Answer: AB ⊥ CD.
- (a) 75° — The three angles in the triangle formed by the two supports and the ground add up to 180°. Two of the angles are 34° and 71°, so the third angle is 180 − 34 − 71 = 75, which is 75°. 105° comes from adding the two given angles together instead of subtracting them from 180° (34 + 71 = 105). 146° comes from computing 180 − 34 = 146 and forgetting to also subtract 71°. 37° comes from subtracting the two given angles from each other instead of using the triangle's angle sum (71 − 34 = 37).
- (a) arc — A curved part of a circle's circumference, between two points, is called an arc. A chord is the straight line joining those two points, not the curved part. A segment is the region enclosed between a chord and an arc, and a sector is the region enclosed between two radii and an arc — both segment and sector are regions, not curved lengths.
- (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.
- (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) They must also be equal — Once two triangles are proved congruent by any condition, including ASA, they are identical in every respect: every pair of corresponding sides and every pair of corresponding angles must be equal, not just the ones originally used to prove the congruence. So the two remaining pairs of corresponding sides must also be equal, making 'they must also be equal' correct. 'They might be equal or not' and 'not enough information to say' both wrongly suggest that congruence only guarantees the specific facts used to prove it, when congruence actually guarantees the triangles are identical overall. 'They must be different' is backwards: the triangles being identical is the entire point of proving congruence, not a reason for a side to differ.
- (d) 60 — Method: write the two supplementary angles as x and 2x, since one is twice the other, and solve x + 2x = 180. Working: 3x = 180, so x = 60. Answer: the smaller angle is 60°. A candidate who gives the larger angle, 2x, instead of the smaller angle gets 120. A candidate who uses a complementary sum of 90° instead of a supplementary sum of 180° gets 30. A candidate who divides 180 by 2 instead of by 3 gets 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.
- (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.
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