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.A straight line is drawn from the centre of a circle to a point on the circle. Write down the name of this line.
- 2.A parallelogram has a base of 20 cm and a sloping side of 13 cm. The perpendicular drawn from the top of that sloping side down to the base meets the base 5 cm from the foot of the sloping side, so the perpendicular, the 5 cm and the 13 cm sloping side form a right-angled triangle. Work out the area of the parallelogram.
- 3.A solid has one curved surface and two flat circular faces of equal size, one at each end. What is the name of this solid?
- 4.A line segment joins two points on the circumference of a circle, without necessarily passing through the centre. Write down the general term for this line segment.
- 5.Triangle DEF is isosceles, with DE = DF. Angle E = 58°. Work out angle F.
- 6.In triangle ABC the angle at A and the angle at C are equal. The side AB is extended beyond B, and the exterior angle formed at B measures 98°. Work out the size of the angle at A.
- 7.In an isosceles triangle each of the two base angles is 46°. Work out the size of the angle at the apex.
- 8.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.
- 9.A triangle has vertices A(1, 1), B(5, 1) and C(5, 4). Work out the lengths of AB, BC and CA, and write down whether triangle ABC is right-angled.
- 10.Work out the y-coordinate of the midpoint of the line segment joining (2, 4) and (8, 10).
- 11.A rectangle has vertices A(1, 1), B(4, 1), C(4, 5) and D(1, 5). Work out the perimeter of the rectangle.
- 12.A right-angled triangle has angles of 90°, 40° and x°. Work out the value of x.
- 13.In this question, a curved surface counts as a face. How many faces does a sphere have?
- 14.A postcard measuring 10 cm by 15 cm is enlarged by a scale factor of 2 to make a poster. Work out the area of the poster, in cm².
- 15.An angle measures 108°. Write down the name given to an angle of this size.
Answer key
- (b) The radius — Method: each of the four names describes a line in a fixed position relative to the circle, so match the description in the question against those positions. Working: the line described has one end at the centre and its other end on the circle. A chord joins two points that both lie on the circle, so it is ruled out. A diameter joins two points on the circle by passing through the centre, so it is twice as long as the line described. A tangent touches the circle at exactly one point and never reaches the centre. The line with one end at the centre and one end on the circle is a radius. Answer: The radius. The distractors: The diameter comes from remembering only that the line involves the centre, and not noticing that it stops there instead of carrying on to the far side; The chord comes from seeing one end on the circle and assuming both ends are; The tangent comes from matching the phrase 'a point on the circle' to the line that touches at exactly one point, ignoring that a tangent never reaches the centre.
- (d) 240 cm² — Method: the area of a parallelogram is base × perpendicular height, and the perpendicular height is not the sloping side, so it must be found first from the right-angled triangle. Working: the sloping side is the hypotenuse, so the height squared is 13² − 5² = 169 − 25 = 144, giving a height of √144 = 12 cm; then 20 × 12 = 240. Answer: 240 cm². The distractors: 260 cm² comes from using the 13 cm sloping side as the height, 20 × 13, without going through the right-angled triangle at all; 120 cm² comes from finding the height of 12 cm correctly and then halving the product, (20 × 12) ÷ 2, which is the rule for a triangle and not for a parallelogram; 100 cm² comes from using the 5 cm along the base as the height, 20 × 5.
- (d) Cylinder — Method: check how many faces are flat and how many are curved, and how many flat faces there are. Working: this solid has a curved surface joining two flat circular faces of equal size, which is exactly a cylinder. A student who answers cone has forgotten a cone has only one flat circular face, not two. A student who answers sphere has forgotten a sphere has no flat faces at all. A student who answers triangular prism has spotted that the solid has two identical end faces but has taken those ends to be triangles rather than circles, and a prism's side faces are flat rectangles, not one curved surface. Answer: cylinder.
- (a) Chord — Method: identify what the two endpoints of the line segment are, and whether it must pass through the centre. Working: a line joining any two points on the circumference, whether or not it passes through the centre, is a chord, and the diameter is just a special case of it. A student who answers diameter has wrongly assumed the line must pass through the centre. A student who answers radius has confused joining two circumference points with joining the centre to the circumference. A student who answers tangent has confused a line crossing through the circle with one that only touches its outside. Answer: chord.
- (c) 58° — In an isosceles triangle, the base angles opposite the equal sides are equal. Since DE = DF, angle F is the base angle equal to angle E, so angle F = 58°.
- (a) 49° — Method: an exterior angle of a triangle equals the sum of the two interior angles that are not next to it, which here are the angles at A and at C; since those two are equal, the exterior angle is twice the angle at A. Working: 2 × angle A = 98, so angle A = 98 ÷ 2 = 49. Answer: 49°. The distractors: 82° is the interior angle at B, 180 − 98, given in place of the angle at A; 41° comes from finding that interior angle of 82° and halving it, 82 ÷ 2, instead of halving the exterior angle; 98° comes from taking the exterior angle to be equal to the angle at A on its own, with no halving at all.
- (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) 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.
- (d) Yes, since 3² + 4² = 5² — AB is horizontal with length 5 − 1 = 4, BC is vertical with length 4 − 1 = 3, and CA = √(4² + 3²) = √25 = 5. Since the two shorter sides satisfy 3² + 4² = 5², the triangle is right-angled, with the right angle at B. "No, since 3 + 4 ≠ 5" wrongly tests Pythagoras' theorem by adding the sides instead of squaring them first. "No, since AB, BC and CA are not all equal" confuses a right-angled triangle with an equilateral one — a triangle does not need equal sides to have a right angle. "Yes, since 4² + 5² = 3²" reaches the correct conclusion but puts the longest side, 5, on the wrong side of the equation, as if it were one of the two shorter sides instead of the hypotenuse.
- (b) 7 — Method: each coordinate of a midpoint is the mean of the matching pair of coordinates, so the y-coordinate of the midpoint depends on the two y-coordinates alone. Working: the y-coordinates are 4 and 10, so the mean is (4 + 10) ÷ 2 = 14 ÷ 2 = 7. Answer: 7. The distractors: 5 comes from working out the x-coordinate of the midpoint, (2 + 8) ÷ 2 = 5, and writing that down in place of the y-coordinate the question asked for; 3 comes from halving the difference of the y-coordinates, (10 − 4) ÷ 2 = 3, which is half the vertical gap rather than a position; 14 comes from adding the two y-coordinates and forgetting to halve the total.
- (c) 14 units — Method: the perimeter of a rectangle is the distance all the way round its outside, 2 × (length + width), so the two side lengths must be found first; on a coordinate grid a side's length is the difference between the coordinates that change along it. Working: along AB, from (1, 1) to (4, 1), only x changes, so AB = 4 − 1 = 3. Along BC, from (4, 1) to (4, 5), only y changes, so BC = 5 − 1 = 4. Perimeter = 2 × (3 + 4) = 2 × 7 = 14. Answer: 14 units. The distractors: 18 units comes from reading the vertex numbers 4 and 5 as the side lengths instead of subtracting, giving 2 × (4 + 5); 12 units comes from working out the area, 3 × 4, in place of the perimeter; 7 units comes from adding one length to one width and stopping there, without doubling for the opposite pair of sides.
- (d) 50 — The angles in any triangle add up to 180°, so x = 180 − 90 − 40 = 50. "90" comes from subtracting only the 90° angle from 180° and forgetting to also subtract the 40°. "130" comes from adding 90° and 40° together instead of subtracting them from 180°. "45" comes from halving the 90° angle instead of using the angle sum of the triangle at all.
- (c) 1 — Method: the instruction states that a curved surface counts as a face, so count the surfaces of the sphere on that basis. Working: a sphere has exactly one continuous curved surface and no flat surfaces at all. A student who answers 0 has ignored the instruction and refused to count the curved surface. A student who answers 2 has confused the sphere with a cylinder, which has two flat circular faces. A student who answers 3 has imagined extra hidden surfaces that do not exist. Answer: 1 face.
- (d) 600 cm² — Enlarging by scale factor 2 makes the new dimensions 10 × 2 = 20 cm and 15 × 2 = 30 cm, so the poster's area = 20 × 30 = 600 cm². A pupil who scales the original area, 150 cm², by the scale factor itself instead of by its square gets 150 × 2 = 300 cm². A pupil who adds the scale factor to each dimension instead of multiplying gets (10 + 2) × (15 + 2) = 204 cm². A pupil who forgets to enlarge the postcard at all just uses the original area, 150 cm². The correct area of the poster is 600 cm².
- (d) An obtuse angle — Method: compare the angle with the two markers that separate the angle names, a right angle at 90° and a straight line at 180°. Working: 108° is greater than 90° and smaller than 180°, so it lies between the right angle and the straight line. Answer: an obtuse angle. The distractors: an acute angle is one below 90°, and is chosen by a candidate who checks only that 108° is less than 180°; a reflex angle is one above 180°, and is chosen by a candidate who checks only that 108° is more than 90° and then takes the largest category; a right angle is exactly 90°, and is chosen by reading 108° as near enough to 90° instead of comparing it properly.
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