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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Properties of triangles and quadrilaterals worksheet — GCSE Foundation
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- 1.A trapezium-shaped allotment plot has two parallel sides that face each other, and two sloping sides that are equal in length (an isosceles trapezium). One of the angles next to the shorter parallel side is 118°. Work out the angle next to the other end of the same shorter parallel side.
- 2.Two sides of a triangle are 9 cm and 15 cm long. Write down which one of these lengths is possible for the third side.
- 3.Quadrilateral PQRS has angle P = 92°, angle Q = 84° and angle R = 106°. Work out the size of angle S.
- 4.Write down the name given to a quadrilateral with exactly one pair of parallel sides.
- 5.Write down the mathematical name for a quadrilateral that has exactly one pair of parallel sides.
- 6.A picture frame is a rhombus with a diagonal of 16 cm and a diagonal of 12 cm. The two diagonals meet at right angles at their midpoints. Work out the length of one side of the rhombus.
- 7.A quadrilateral has exactly two lines of symmetry, its two pairs of opposite angles are equal, and its diagonals are not equal in length. Write down the name of this quadrilateral.
- 8.A rhombus has all four sides equal in length. Write down how many pairs of parallel sides a rhombus has.
- 9.A triangular bunting flag has two equal sides. The angle between those two equal sides is 80°, and the other two angles of the flag are equal to each other. Work out the size of each of the two equal angles.
- 10.A parallelogram has an area of 96 cm² and a base of 8 cm. Work out the perpendicular height of the parallelogram.
- 11.The cross-section of a tent is an isosceles triangle. The angle at the top, between the two equal sloping sides, is 62°. The other two angles are equal to each other. Work out the size of each of those two angles.
- 12.In kite WXYZ, WX = WZ and XY = ZY (so at each of X and Z, one side from each of the two unequal pairs meets). Angle X = 100°. Work out angle Z.
- 13.In an isosceles triangle each of the two base angles is 46°. Work out the size of the angle at the apex.
- 14.Two sides of a triangle are 20 cm and 8 cm long. Write down which one of these lengths is possible for the third side.
- 15.Write down how many right angles a right-angled triangle has.
Answer key
- (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°.
- (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.
- (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) Trapezium — A trapezium is defined as a quadrilateral with exactly one pair of parallel sides. A parallelogram has two pairs, and a kite and rhombus are defined by side lengths, not by having only one pair of parallel sides.
- (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) 10 cm — The diagonals of a rhombus bisect each other at right angles, splitting it into four congruent right-angled triangles with legs 8 cm (half of 16 cm) and 6 cm (half of 12 cm). By Pythagoras' Theorem, side² = 8² + 6² = 64 + 36 = 100. Square root: √100 = 10 cm.
- (b) Rhombus — A rhombus has exactly two lines of symmetry, formed by its two diagonals, and both pairs of opposite angles are equal, but its diagonals are unequal in length. A square also has opposite angles equal, but it has four lines of symmetry and its diagonals ARE equal, so it does not fit. A kite normally has only one line of symmetry and only one pair of opposite angles equal, so it does not fit. A general parallelogram has no lines of symmetry at all, so it does not fit. The correct answer is rhombus.
- (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.
- (b) 50° — Method: the three angles of a triangle add up to 180°, and the two angles opposite the equal sides are equal, so subtract the given angle from 180° and halve the remainder. Working: 180° − 80° = 100°, and 100° ÷ 2 = 50°. Answer: 50°. The distractors: 100° comes from subtracting from 180° and forgetting to halve, so it is the two equal angles together; 40° comes from halving the 80° that is given rather than halving what is left of the 180°; 80° comes from assuming that the two base angles must match the angle between the equal sides.
- (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) 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) 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) 88° — Method: the three angles add up to 180°, and here it is the two equal base angles that are known, so take both of them away from 180°. Working: the two base angles come to 2 × 46° = 92°, and 180° − 92° = 88°. Answer: 88°. The distractors: 134° comes from subtracting only one base angle, 180° − 46°, and forgetting that there are two of them; 92° comes from doubling the base angle and stopping there, which is the two base angles together rather than the apex; 46° comes from assuming that all three angles of the triangle are equal to the one that is given.
- (d) 15 cm — Method: any two sides of a triangle must together be longer than the third, so the third side must be longer than the difference of the two given sides and shorter than their sum. Working: the difference is 20 − 8 = 12 cm and the sum is 20 + 8 = 28 cm, so the third side must be between 12 cm and 28 cm, and 15 cm lies inside that range. Answer: 15 cm. The distractors: 12 cm is exactly the difference, so the three lengths would lie flat along a straight line and never close into a triangle; 5 cm is shorter than the difference — 5 + 8 = 13 cm cannot reach across the 20 cm side — and is chosen by candidates who check no lower limit at all; 30 cm is longer than the sum of the other two, so those two sides could never meet, and it is chosen by candidates who check no upper limit.
- (c) 1 — Method: a right angle measures 90°, the three angles of any triangle add up to 180°, and a right-angled triangle is defined as a triangle that contains a right angle. Working: taking one right angle out of the total leaves 180° − 90° = 90° to be shared between the other two angles, so both of those must be acute; a second right angle would use the whole of that remaining 90° and leave nothing at all for the third angle, which is impossible. The definition therefore fixes the count at exactly one. Answer: 1. The distractors: 2 comes from counting the two sides that form the right angle instead of counting the angles themselves; 3 comes from reading the name as a description of the whole triangle, so that all three of its angles are taken to be right angles, which would need an angle sum of 3 × 90° = 270°; 0 comes from over-applying the angle sum — a candidate who works out that 90° + 90° = 180° leaves nothing for a third angle can conclude from that alone that no triangle may contain a right angle at all.
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