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.In an isosceles triangle each of the two base angles is 46°. Work out the size of the angle at the apex.
- 2.Triangle PQR has vertices P(0, 0), Q(6, 0) and R(0, 8). Work out the perimeter of the triangle.
- 3.p is the column vector with top number 5 and bottom number 1. q is the column vector with top number −2 and bottom number 3. Work out p − 2q, giving your answer as a column vector in the form (top, bottom).
- 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.A stained-glass window panel is designed as a kite. Two adjacent sides are each 25 cm, and the other two adjacent sides are each 40 cm. A frame is fitted around the whole panel. The frame costs £2.50 per metre, sold only in whole metres. Work out the total cost of the frame.
- 6.A sign-maker is told a metal bracket outline must have two pairs of adjacent sides equal in length (not opposite pairs), exactly one pair of opposite angles equal, and exactly one line of symmetry. Which quadrilateral matches this description?
- 7.Point B is at (1, 6). It is reflected in the x-axis. Work out the coordinates of the image of point B.
- 8.An angle is measured with a protractor as 47°. What type of angle is this?
- 9.A sector of a circle has an angle of 90° at the centre. Write down what fraction of the whole circle this sector represents.
- 10.A builder lays a low wall made from identical bricks. The diagram shows the plan of the wall's footprint on a grid, where each square is one brick. The wall is built up to a height of 4 bricks everywhere. Work out the total number of bricks used.
- 11.A circle has a radius of 1 cm. Work out the circumference of the circle. Give your answer in terms of π.
- 12.To construct a perpendicular to a line at a given point P on the line, compasses are first opened to draw two arcs crossing the line, one on each side of P. What is the correct next step?
- 13.A rectangular name card has a perimeter of 30 cm. The card is 9 cm long. Work out the width of the card.
- 14.In a construction, a point P is marked at the foot of the perpendicular drawn from point X to a straight line. Which statement correctly compares the distance XP with the distance from X to any other point on the line?
- 15.Triangle ABC is isosceles, with AB = AC. Angle B = 70°. Which reason correctly explains why angle C = 70°?
Answer key
- (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.
- (c) 24 — PQ lies along the x-axis with length 6, and PR lies along the y-axis with length 8, meeting at a right angle at P, so QR = √(6² + 8²) = √100 = 10. The perimeter is 6 + 8 + 10 = 24. "14" adds only the two shorter sides, PQ and PR, and leaves out the hypotenuse QR completely. "28" comes from finding QR incorrectly as 6 + 8 = 14 instead of using Pythagoras' theorem, then adding 6 + 8 + 14. "48" comes from multiplying the two shorter sides, 6 × 8, instead of finding and adding all three sides of the triangle.
- (a) (9, −5) — Method: multiply every part of q by 2, then subtract the matching part from p. Working: 2q = (−4, 6); p − 2q gives top 5 − (−4) = 9 and bottom 1 − 6 = −5. Answer: p − 2q = (9, −5). A candidate who forgets to double q first, working out p − q instead, gets (7, −2). A candidate who doubles p instead of q, working out 2p − q, gets (12, −1). A candidate who adds 2q instead of subtracting it gets (1, 7).
- (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.
- (b) £5.00 — Perimeter of the kite = 25 + 25 + 40 + 40 = 130 cm = 1.3 m. The frame is sold only in whole metres, so 2 m must be bought. Cost = 2 × £2.50 = £5.00. A student who buys the exact 1.3 m instead of rounding up to whole metres gets 1.3 × £2.50 = £3.25. A student who never converts the perimeter from centimetres to metres and costs 130 × £2.50 gets £325.00.9 m, which rounds up to 3 whole metres, costing 3 × £2.50 = £7.50.
- (a) Kite — Method: check each named quadrilateral's properties against the three facts given, one at a time. Working: a kite has two pairs of adjacent sides equal (not opposite pairs), one pair of opposite angles equal (the two angles where an unequal pair of sides meet), and exactly one line of symmetry — matching all three facts. Options: a rhombus does have equal adjacent sides, but all four of its sides are equal, both pairs of its opposite angles are equal, and it has two lines of symmetry rather than exactly one; a parallelogram has its opposite sides equal rather than adjacent pairs, both pairs of opposite angles equal, and no line of symmetry at all; a trapezium does not generally have any pair of equal adjacent sides or a line of symmetry. Answer: kite.
- (d) (1, −6) — Reflecting in the x-axis keeps the x-coordinate the same and changes the sign of the y-coordinate: (1, 6) → (1, −6). A pupil who reflects in the y-axis instead gets (−1, 6). A pupil who changes the sign of both coordinates gets (−1, −6). A pupil who forgets to change any sign leaves the point unmoved at (1, 6). The correct image is (1, −6).
- (b) Acute — An acute angle is any angle less than 90°. Since 47° is less than 90°, it is acute. An obtuse angle is between 90° and 180°, which 47° is not. A reflex angle is greater than 180°, which is far too big for 47°. A right angle is exactly 90°, not 47°. The angle is acute.
- (c) 1/4 — A full turn at the centre of a circle is 360°, so a sector's fraction of the circle is its angle divided by 360°: 90 ÷ 360 = 1/4. 1/2 would be the fraction for a sector with an angle of 180°, not 90°. 3/4 is the fraction of the rest of the circle, the major sector left over from the 270° that is not part of this sector. 1/8 would be the fraction for a sector with an angle of 45°, half of 90°.
- (b) 28 — Method: total bricks = (number of squares in the footprint) × (the height of the wall in bricks). Working: the footprint has the row of 5 squares plus the 2 squares in the arm that stands out from the middle of that row, and they do not overlap, giving 5 + 2 = 7 squares; multiplying by the height of 4 bricks gives 7 × 4 = 28. Answer: 28. The distractors: 20 comes from using only the row of 5 and ignoring the arm (5 × 4). 24 comes from treating the bottom square of the arm as if it were one of the row's own squares, so the arm is counted as adding only 1 new square instead of 2 (5 + 1 = 6, then 6 × 4). 32 comes from counting the square of the row directly below the arm a second time as part of the arm (5 + 3 = 8, then 8 × 4).
- (c) 2π cm — Method: the circumference of a circle is 2πr, where r is the radius, or equivalently πd, where d is the diameter. Working: r = 1, so the circumference is 2 × π × 1 = 2π cm. Answer: 2π cm. The distractors: π cm comes from using the formula πd but substituting the radius in place of the diameter; 4π cm comes from doubling twice — changing the radius into the diameter of 2 cm and then putting that diameter into 2πr as though it were a radius; π cm² is the area of this circle, π × 1², and comes from reaching for the area formula when a distance round the outside was asked for, which is why it carries a squared unit.
- (b) Draw wider arcs from each crossing point, meeting above. — Once the two arcs cross the line at points either side of P, compasses are opened to a radius greater than before and arcs are drawn from each of those two points so that they meet above (or below) the line; joining that meeting point to P gives the perpendicular. (Joining the two crossing points with a straight line only retraces part of the original line, since both points already lie on it; drawing a circle centred at P through both crossing points does not locate any new point needed for the perpendicular; drawing an arc centred at P through only one crossing point repeats the first step instead of moving on to the second pair of arcs.)
- (a) 6 cm — Method: a rectangle is made of two lengths and two widths, so the perimeter is 2 × (length + width). Half the perimeter is therefore one length plus one width, and the width is what is left of that half once the length is taken away. Working: 30 ÷ 2 = 15 cm is one length added to one width, and 15 − 9 = 6. Answer: 6 cm. The distractors: 21 cm comes from taking the length off the whole perimeter, 30 − 9, so three sides are still counted in the figure that is written down; 12 cm comes from removing both lengths correctly, 30 − 2 × 9 = 12, and then recording that remainder as the width instead of halving it, because it is in fact the two widths together; 7.5 cm comes from dividing the perimeter by four, 30 ÷ 4, which is the calculation for the side of a square and throws away the length that was given.
- (d) XP is the shortest distance from X to the line — The perpendicular from a point to a line always gives the shortest possible distance to that line — joining X to any other point on the line forms the hypotenuse of a right-angled triangle with XP as one of the shorter sides, and a hypotenuse is always longer than either of the other two sides. So XP is shorter than the distance to every other point on the line. "XP is the longest distance from X to the line" reverses this relationship. "XP equals every other distance from X to the line" would only be true if X were equidistant from every point on the line, which is impossible for a point and a straight line. "XP cannot be compared without knowing the line's length" is false — the shortest-distance fact holds whatever the line's length, since only the local right angle matters.
- (c) base angles of an isosceles triangle are equal — Because AB = AC, triangle ABC is isosceles, and the base angles of an isosceles triangle — the two angles opposite the equal sides — are always equal, which is why angle C equals angle B, 70°. Angles in a triangle adding up to 180° is a true fact about the triangle as a whole, but it is not the reason two specific angles are equal to each other. Corresponding angles are equal is a fact about parallel lines cut by a transversal, which does not apply inside a single triangle like this. Vertically opposite angles are equal is a fact about two lines crossing, not about a triangle's base angles.
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