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.How many faces does a cuboid have?
- 2.A jug holds 750 ml of water. Work out what fraction of a litre this is.
- 3.Every diameter of a circle is an example of a more general type of straight line joining two points on the circle. Which term names that more general type of line?
- 4.Work out the exact value of (sin 30°)² + (cos 30°)².
- 5.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.
- 6.A solid has 5 faces: one square face and four triangular faces, which all meet at a single point above the square. What is the name of this solid?
- 7.A shop sells a scale model of a van. The model and the real van are similar shapes, with lengths in the ratio 1 : 24. The model is 18.5 cm long. Work out the length of the real van, giving your answer in metres.
- 8.In a kite, one pair of opposite angles are equal in size. In kite WXYZ, angle X = 40° and angle Z = 100° are the two angles that are NOT equal to each other. The other two angles, W and Y, are equal to each other. Work out the size of angle W.
- 9.Triangle ABC has a right angle at B. Triangle DEF has a right angle at E. The hypotenuse of triangle ABC equals the hypotenuse of triangle DEF, and side AB equals side DE. Which condition proves that the two triangles are congruent?
- 10.A circular table mat has diameter 20 cm. Yasmin has a 70 cm length of ribbon and sews it around the edge of the mat with no overlap. Using π = 3.14, work out how much ribbon is left over after going around the mat once.
- 11.A written description of a diagram states: 'AB is marked with a single arrow, CD is marked with a single arrow, and EF is marked with a double arrow.' Using the standard convention for marking parallel lines, what can be concluded?
- 12.An isosceles triangle ABC has AB equal to AC. AM is a line from vertex A, perpendicular to BC, meeting BC at point M. Which condition proves that triangle ABM is congruent to triangle ACM?
- 13.A straight line touches the edge of a circle at exactly one point and does not cross into the circle at all. Write down the term for this line.
- 14.A transversal crosses a pair of parallel lines. At one line, the angle is (5x + 4)°. The corresponding angle at the other line is (3x + 24)°. Work out the value of x.
- 15.Quadrilateral PQRS has angle P = 92°, angle Q = 84° and angle R = 106°. Work out the size of angle S.
Answer key
- (c) 6 — A cuboid has six flat faces: a top, a bottom and four sides. Count each flat surface once: top, bottom, front, back, left, right — six faces in total, so the answer is 6. Choosing 8 counts the vertices (corners) instead of the faces. Choosing 12 counts the edges instead of the faces. Choosing 4 counts only the four side faces and forgets the top and the bottom.
- (a) 3/4 — 1 litre = 1000 ml, so 750 ml is 750/1000 of a litre. Dividing both the numerator and denominator by 250 simplifies this to 3/4. Writing the fraction upside down, as the litre out of the 750 ml, gives 4/3. Finding the fraction of the litre that is NOT filled, 250/1000, gives 1/4. Dividing the numerator by 250 but the denominator by only 100, an inconsistent simplification, gives 3/10.
- (c) chord — A diameter is a straight line joining two points on the circle that happens to pass through the centre. The general term for any straight line joining two points on a circle is a chord — every diameter is a chord, but not every chord is a diameter. Calling it a radius is wrong because a radius runs from the centre to just ONE point on the circle, not between two points on the circle. A tangent touches the circle at only one point, so it cannot be a diameter at all. An arc is a curved part of the circle's edge, not a straight line.
- (d) 1 — sin 30° = 1/2, so (sin 30°)² = 1/4. cos 30° = √3/2, so (cos 30°)² = 3/4. Adding these gives 1/4 + 3/4 = 1. '1/4' only calculates (sin 30°)² and forgets to add the cos 30° term. '3/4' only calculates (cos 30°)² and forgets to add the sin 30° term. '−1/2' comes from subtracting the two squared values instead of adding them: 1/4 − 3/4 = −1/2.
- (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) Square-based pyramid — A solid with one square base and four triangular faces meeting at a single apex above the base is a square-based pyramid. A triangular prism has two triangular faces and three rectangular faces, not a square base with four triangles, so that is a different solid. A cube has six square faces, and a cuboid has six rectangular faces — neither has any triangular faces at all. The solid described is a square-based pyramid.
- (d) 4.44 m — The real van's length is 18.5 × 24 = 444 cm, which converts to 4.44 m. '0.77 m' comes from dividing by 24 instead of multiplying, using the ratio the wrong way round (18.5 ÷ 24 = 0.77 cm), and then writing that figure down as metres. '44.4 m' comes from converting 444 cm to metres with the decimal point in the wrong place. '444 m' comes from working out 444 cm correctly but forgetting to convert it into metres at all.
- (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°.
- (d) RHS — Both triangles have a right angle, their hypotenuses are equal, and one further pair of sides (AB and DE) are equal, exactly the RHS condition, so RHS is correct. SAS needs two sides and the angle between them; here the right angle at B is not between the hypotenuse and AB, so this is not a valid SAS setup, making SAS wrong. SSS needs three pairs of equal sides, but only two sides are given here, so SSS is wrong. ASA needs two angles and the side between them, but only one angle, the right angle, is given, so ASA is wrong.
- (a) 7.2 cm — The distance around the mat is the circumference, π × diameter = 3.14 × 20 = 62.8 cm. Yasmin started with 70 cm, so the ribbon left over is 70 − 62.8 = 7.2 cm. 62.8 cm is the circumference itself, the amount of ribbon used, not what is left over. 50 cm comes from subtracting the diameter instead of the circumference: 70 − 20 = 50. 20 cm is simply the diameter of the mat and involves no calculation with the ribbon length at all.
- (c) AB ∥ CD only; EF not confirmed — By convention, lines marked with the same number of arrows are parallel to each other, but lines marked with a different number of arrows belong to a different, unrelated family of parallel lines. AB and CD both have a single arrow, so AB is parallel to CD. EF has a double arrow, showing it is not part of the same family as AB and CD; it may be parallel to some other line marked with a double arrow, but nothing here confirms it is parallel to AB or CD, so 'AB ∥ CD only; EF not confirmed' is correct. 'AB, CD and EF are all parallel' and 'EF is parallel to AB' both wrongly treat every arrow mark as showing the same relationship. 'None of the lines are parallel' wrongly assumes a different arrow count rules out any parallel relationship at all, when it actually just signals a different pairing.
- (d) RHS, using AM as common side — Triangle ABM and triangle ACM both have a right angle at M, since AM is perpendicular to BC. AB and AC are the hypotenuses of the two triangles and are equal, and AM is a side common to both triangles, giving a right angle, equal hypotenuses and one further equal side, exactly RHS, so 'RHS, using AM as common side' is correct. 'SAS, right angle as included angle' wrongly treats the right angle at M as included between AB and AM, but AB is the hypotenuse, not one of the two sides forming that right angle. 'SSS, using BM = CM as a fact' wrongly assumes BM equals CM as a given fact, when this is only true because of the RHS congruence, not before it, so it cannot be used to prove that congruence. 'ASA, AB as the included side' again wrongly labels a side as if it could sit between two angles when only one angle, the right angle, is actually known.
- (d) tangent — A line that touches a circle at exactly one point, without crossing into the circle, is called a tangent. A chord is a straight line joining two points ON the circle, so it touches at two points, not one. A radius runs from the centre to the circle's edge, not along the outside of it. A diameter is a chord that passes through the centre, also touching the circle at two points.
- (b) 10 — Corresponding angles are equal, so 5x + 4 = 3x + 24. Subtracting 3x from both sides gives 2x + 4 = 24, then subtracting 4 gives 2x = 20, so x = 10. 14 comes from adding the constants, 4 + 24, instead of subtracting them when rearranging. 19 comes from treating the angles as co-interior (summing to 180°): 5x + 4 + 3x + 24 = 180 gives 8x = 152, so x = 19. 20 correctly reaches 2x = 20 but stops without dividing by 2.
- (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°.
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