Printable · GCSE Foundation · ages 14-16
Properties of triangles and quadrilaterals worksheet — GCSE Foundation
Fifteen questions on "properties of triangles and quadrilaterals" — DfE statement G4. Print it, or print three versions so neighbours cannot copy by letter; the key gives the letter for each version.
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Answer key: Properties of triangles and quadrilaterals worksheet — GCSE Foundation
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- (c) 13 — Area of the trapezium = 1/2 × (8 + 12) × 5 = 1/2 × 100 = 50 m². Number of bags = 50 ÷ 4 = 12.5, which rounds up to 13 bags since seed is sold only in whole bags. A student who mistakenly uses 2 m² of coverage per bag instead of 4 m² finds 50 ÷ 2 = 25 bags.
- (b) Kite — Method: name a quadrilateral by matching what is given — which sides are equal, whether those equal sides lie next to each other or opposite each other, and whether any sides are parallel — against the definitions of the special quadrilaterals. Working: the two 6 cm sides meet at B and the two 9 cm sides meet at D, so each pair of equal sides is a pair of neighbours rather than a pair of opposites, and the stem rules out any parallel sides. The quadrilateral with two pairs of equal adjacent sides and no parallel sides is a kite. Answer: kite. The distractors: a rhombus is chosen by candidates who see two pairs of equal sides and read that as all four sides being equal, which the two different lengths of 6 cm and 9 cm rule out; a parallelogram is chosen by candidates who remember that a parallelogram has two pairs of equal sides but not that in a parallelogram the equal sides are the opposite ones, and who pass over the statement that nothing is parallel; an isosceles trapezium is chosen by candidates who notice that the shape is symmetrical about the line BD and treat symmetry on its own as the mark of a trapezium, when a trapezium needs a pair of parallel sides.
- (a) 100° — In a kite, the pair of angles between the unequal sides are equal to each other. Angle X and angle Z are both between one side from the WX/WZ pair and one side from the XY/ZY pair, so angle Z = angle X = 100°.
- (d) 10 cm — For a triangle to exist, any two sides must add up to more than the third side. 9 + 10 = 19 > 15, and 15 − 9 = 6 < 10, so 10 cm satisfies the triangle inequality. The other lengths fail: 6 cm gives 9 + 6 = 15, which is not more than 15; 24 cm and 26 cm are each at least as large as 9 + 15 = 24.
- (b) Trapezium — A trapezium is defined as a quadrilateral with exactly one pair of parallel sides. A parallelogram has two pairs of parallel sides, not one, so it does not fit the definition. A kite has no parallel sides at all. A rhombus, like a parallelogram, has two pairs of parallel sides. The correct name is trapezium.
- (c) 78° — The angles in any quadrilateral add up to 360°. Add the three given angles: 92° + 84° + 106° = 282°. Angle S = 360° − 282° = 78°. A pupil who only adds angle P and angle Q, forgetting angle R, gets 360° − (92° + 84°) = 184°. A pupil who only adds angle Q and angle R, forgetting angle P, gets 360° − (84° + 106°) = 170°. A pupil who makes a carrying slip adding the three angles, getting 292° instead of 282°, gets 360° − 292° = 68°. The correct answer is 78°.
- (a) 59° — Method: the three angles of a triangle add up to 180°, and in an isosceles triangle the two angles opposite the equal sides are equal, so take the known angle away from 180° and share what is left equally between the other two. Working: 180° − 62° = 118°, and 118° ÷ 2 = 59°. Answer: 59°. The distractors: 118° comes from taking 62° from 180° and stopping there, which gives the two angles together rather than one of them; 31° comes from halving the 62° that is given instead of halving what is left; 62° comes from assuming that all three angles of the triangle are equal to the one that is given.
- (a) 110° — The angles in a quadrilateral add up to 360°. So 40° + 100° + W + W = 360°, giving 2W = 360° − 140° = 220°, so W = 110°. A pupil who works out 2W = 220° but forgets to divide by 2, since there are two equal angles W, gives 220°. A pupil who mistakenly uses the angle sum of a triangle, 180°, instead of 360°, gets 180° − 140° = 40°. A pupil who simply adds the two given angles together instead of subtracting from 360° gets 40° + 100° = 140°. The correct answer is 110°.
- (b) 2 — A rhombus is a special parallelogram, so opposite sides are parallel — that is two pairs of parallel sides. A quadrilateral with only one pair of parallel sides is a trapezium, not a rhombus, so 1 is wrong. A kite has 0 pairs of parallel sides, not a rhombus. There are only two pairs of opposite sides in a quadrilateral altogether, so 4 is not possible. The correct answer is 2.
- (c) Trapezium — A trapezium is defined as a quadrilateral with exactly one pair of parallel sides.
- (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) 68° — Method: two properties are needed. Angle A and angle D are co-interior angles between the parallel sides AB and DC, so they add up to 180°; and because the trapezium is isosceles, the two angles on the side AB are equal, so angle B = angle A. Working: angle A = 180° − 112° = 68°, and angle B = angle A = 68°. Answer: 68°. The distractors: 112° comes from assuming that angles B and D are equal, which is the property of a parallelogram, not of a trapezium; 90° comes from assuming that the angles on the other parallel side must be right angles; 248° comes from using the 360° angle sum of a quadrilateral and taking away only the one angle that is given.
- (d) 118° — In an isosceles trapezium, the two angles next to the same parallel side are equal, because the sloping sides are equal in length. So the angle at the other end of the shorter parallel side also equals 118°.
- (a) 12 cm — Area of a parallelogram = base × perpendicular height, so height = area ÷ base = 96 ÷ 8 = 12 cm. A student who adds the area and base instead of dividing gets 96 + 8 = 104 cm. A student who multiplies the area and base instead of dividing gets 96 × 8 = 768 cm. A student who uses the triangle area formula, area = 1/2 × base × height, instead of the parallelogram formula solves 96 = 1/2 × 8 × h and gets h = 24 cm.
- (a) Square — Method: two conditions are being asked for at once, so test each shape against both — the two diagonals must always be the same length as each other, and they must always meet at 90°. Working: in a rectangle the diagonals are equal but they meet at 90° only in the special case where the rectangle is also a rhombus; in a rhombus the diagonals do meet at 90° but they are of different lengths unless the rhombus is also a rectangle; the shape that satisfies both conditions for every example of it is the one that is both, and its diagonals are equal and perpendicular. Answer: the square. The distractors: the rectangle is where a candidate stops who tests only the equal-length condition and never checks the angle at the crossing; the rhombus is where a candidate stops who tests only the right-angle condition and never checks the two lengths; the parallelogram is chosen by a candidate who remembers that the diagonals of a parallelogram bisect each other and treats bisecting each other as being equal to each other, which is a different property.
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