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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Geometry and measures worksheet — GCSE Foundation
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- 1.Triangle ABC has a right angle at B and hypotenuse AC = 17 cm, with AB = 8 cm. Triangle DEF has a right angle at E and hypotenuse DF = 17 cm. Which extra fact about DEF would complete the RHS condition for proving that ABC ≅ DEF, with the vertices matching in that order?
- 2.Work out the exact value of sin 45° × cos 45°.
- 3.Four angles meet at a point. Three of them measure 82°, 105° and 96°. Work out the size of the fourth angle.
- 4.The diagram shows the plan, front elevation and side elevation of a solid. Which solid could this be?
- 5.Which of the following is NOT one of the four basic conditions used to prove two triangles are congruent?
- 6.The three angles of a triangle are 10°, 58° and 112°. Write down the name that describes this triangle.
- 7.The point R(−6, 9) is translated by the vector . Write down the coordinates of the image of R.
- 8.Point A and point B are 5 cm apart. How many points are exactly 4 cm from A and exactly 3 cm from B?
- 9.At a point on a straight line, two angles are formed. One of them is 63°. Work out the size of the other angle.
- 10.Write down the exact value of cos 30°.
- 11.A triangular badge has two sides of length 8 cm and a third side of length 5 cm. Work out the perimeter of the badge.
- 12.A solid is built from five centimetre cubes, standing side by side on a table in a single straight row. Looking down from directly above (the plan view), how many squares are visible?
- 13.A student says the reverse of the translation vector is . Write down the correct reverse translation vector.
- 14.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.
- 15.Two straight lines PQ and RS cross at point O. Angle POR = 52°. Work out angle QOS.
Answer key
- (d) DE = 8 cm — Method: use the stated correspondence ABC ≅ DEF to work out which side in DEF matches the known side AB in ABC. Working: the correspondence sends A to D, B to E and C to F, so AB corresponds to DE; the right angles at B and E and the equal hypotenuses AC = DF = 17 cm are already given, so DE = 8 cm supplies the third ingredient — Right angle, Hypotenuse, Side. Options: EF = 8 cm matches AB to the wrong side, since EF corresponds to BC, and BC = √(17² − 8²) = 15 cm, not 8 cm; BC = 8 cm states something about triangle ABC rather than the missing fact about DEF, and it is false as well, since BC = 15 cm; angle D = angle A does follow once the triangles are congruent, but RHS is completed by a matching side, not by a matching angle. Answer: DE = 8 cm.
- (c) 1/2 — sin 45° = √2/2 and cos 45° = √2/2, so sin 45° × cos 45° = √2/2 × √2/2 = 2/4 = 1/2. √2/2 comes from writing down only one of the two factors and forgetting to multiply by the other. √2 comes from adding the two exact values instead of multiplying them: √2/2 + √2/2 = √2. 1 comes from wrongly treating sin 45° × cos 45° as sin(45° + 45°) = sin 90° = 1 — multiplying two ratios is not the same as adding their angles.
- (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.
- (a) A cuboid — Method: work out which solid has flat faces only, no curved surfaces and no point where edges meet, since only that gives rectangles for all three views. Working: a solid whose plan, front elevation and side elevation are all rectangles has three pairs of flat rectangular faces meeting at right angles — that is a cuboid. Answer: a cuboid. The distractors: a cylinder is wrong because its plan view (from above) is a circle, not a rectangle. A cone is wrong because its plan view is a circle and its front and side elevations are triangles. A square-based pyramid is wrong because its front and side elevations come to a point at the apex, giving triangles rather than rectangles, even though its plan view could be a square.
- (c) SSA — Method: recall the four basic congruence criteria for triangles and compare them with each option. Working: the four basic conditions are SSS, SAS, ASA and RHS. SSA (two sides and a non-included angle) is not one of them, because knowing two sides and an angle that is not between them does not always fix a unique triangle. Options: SSS, ASA and RHS are each genuine congruence conditions, so none of them is the correct answer to 'which is NOT'. Answer: SSA.
- (c) Scalene — Method: equal angles in a triangle sit opposite equal sides, so compare the three angles with each other. Working: 10°, 58° and 112° are all different, so no two sides of the triangle are equal either, and a triangle with no equal sides is scalene. Answer: scalene. The distractors: isosceles is chosen by candidates who pair up the two acute angles, 10° and 58°, as base angles without checking that they are actually equal; equilateral is chosen by candidates who check that the three angles add to 180° and take that as meaning the triangle is regular; right-angled is chosen by candidates who see that 112° is larger than 90° and classify the triangle as containing a right angle, when in fact none of the three angles is 90°.
- (b) (−1, −2) — To translate R(−6, 9) by $\binom{5}{−11}$, add 5 to the x-coordinate and −11 to the y-coordinate: (−6 + 5, 9 + (−11)) = (−1, −2). (−1, 9) applies only the x-component and leaves the y-coordinate unchanged. (−6, −2) applies only the y-component and leaves the x-coordinate unchanged. (−11, 20) comes from subtracting the vector instead of adding it.
- (c) 2 — A point 4 cm from A lies on a circle of radius 4 cm centred at A; a point 3 cm from B lies on a circle of radius 3 cm centred at B. Since AB = 5 cm, and 4 + 3 = 7 is greater than 5 while 4 − 3 = 1 is less than 5, the two circles genuinely cross each other, at two separate points, one on each side of line AB. "1" comes from wrongly assuming the circles only touch rather than cross, which would need 4 + 3 to equal exactly 5. "0" comes from wrongly assuming the circles miss each other completely. "4" comes from counting where each circle crosses the line AB itself (two points each) instead of counting where the two circles cross each other.
- (b) 117 — Method: two angles meeting at a point on a straight line add up to 180°. Working: 180 − 63 = 117. Answer: 117°. A candidate who thinks the two angles on a straight line must be equal gives 63. A candidate who uses 90° instead of 180°, working out 90 − 63, gets 27. A candidate who uses 360° instead of 180°, working out 360 − 63, gets 297.
- (d) √3/2 — cos 30° is one of the exact values you must know: cos 30° = √3/2. The value 1/2 is the exact value of sin 30° (and of cos 60°), not cos 30°. The value √2/2 is the exact value of cos 45° (and of sin 45°). The value 1 is the exact value of cos 0°.
- (c) 21 cm — Method: the perimeter is the distance all the way round the edge, so every side is counted once and the three side lengths are added. Working: the two equal sides give 8 + 8 = 16 cm, and the third side adds 5 cm to that. Answer: the perimeter is 21 cm. The distractors: 16 cm comes from adding the two 8 cm sides and handing that total in before the third side has been included; 13 cm comes from adding one 8 cm side to the 5 cm side, as though the badge carried only the two different lengths printed on it rather than three sides; 24 cm comes from taking all three sides to be 8 cm and working out 3 × 8, which would be the perimeter only if the badge were equilateral.
- (a) 5 — Looking straight down on a row of 5 cubes standing side by side, each cube contributes exactly one square to the view from above, since the cubes do not overlap and none is hidden behind another — so the plan shows 5 squares in a row. "1" comes from treating the whole row as a single block instead of counting each cube. "10" comes from doubling the count, perhaps by also counting a front elevation's squares alongside the plan's. "25" comes from squaring the number of cubes (5 × 5) instead of counting them.
- (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) 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°.
- (c) 52° — When two straight lines cross, the angles opposite each other at the crossing point are vertically opposite and are equal. Ray OQ is opposite ray OP and ray OS is opposite ray OR, so angle QOS is vertically opposite angle POR. Therefore angle QOS = 52°.
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