Printable · GCSE Higher · ages 14-16
Geometry and measures worksheet — GCSE Higher
Fifteen questions across the geometry and measures statements at Higher 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 Higher
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- 1.A carpenter cuts two triangular metal brackets for a shelf. Bracket P has sides of 12 cm, 16 cm and 20 cm. Bracket Q has sides of 16 cm, 20 cm and 12 cm, listed in a different order. Before fitting them as a matching pair, the carpenter wants to check they are exactly the same shape and size. Which of the following correctly justifies this?
- 2.A circle has a radius of 1 cm. Work out the circumference of the circle. Give your answer in terms of π.
- 3.A designer plots a logo point at (3, 4) on a grid. She reflects it in the line y = 1 and then rotates the image 90° clockwise about the point (1, 1) to create a repeating tile pattern. Work out the coordinates of the point after both transformations.
- 4.In pentagon PQRST, which of the following correctly names the interior angle at vertex R, using standard three-letter angle notation?
- 5.PT is a tangent to a circle with centre O, touching the circle at T. C is a point on the circle such that C lies inside angle OTP (the right angle between the radius OT and the tangent PT), and OC is a radius. Angle TOC = 130°. Work out the size of angle PTC, the angle between the tangent PT and the chord TC.
- 6.Triangle JKL has a right angle at K, hypotenuse JL = 13 cm, and side JK = 5 cm. Triangle MNO has a right angle at N, hypotenuse MO = 13 cm, and side MN = 5 cm. Using only the facts given, and without working out any further lengths, write down the congruence condition that proves triangle JKL is congruent to triangle MNO.
- 7.A rectangular gate is braced with a diagonal strut. The gate is 1.2 m wide and 0.9 m tall. Work out the exact length of the diagonal strut.
- 8.A sector of a circle has radius 8 cm and takes up three-quarters of the full circle. Work out the perimeter of the sector, in terms of π.
- 9.Work out the exact value of sin 30° + cos 60°.
- 10.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?
- 11.A is the point (1, 3) and B is the point (7, 3). Work out the coordinates of the midpoint of AB.
- 12.A crane's hook starts at the point (3.4, 9.5) on a construction site plan measured in metres. It moves along the vector to pick up a beam, then along the vector to place it. Work out the single column vector that describes the hook's overall movement from its start position to where it places the beam.
- 13.A cylinder has a volume of 942 cm³ and a height of 12 cm. Using π = 3.14, work out the radius of the cylinder.
- 14.A robot on a grid moves by the vector , then by the vector , then by the vector . Write down the single column vector that has the same overall effect as these three moves.
- 15.A pinhole camera is set up at a pitch-side stand, with its aperture at the point O = (2, 1) on a grid. Light from the corner F of a triangular flag, at the point (4, 2), passes through the aperture and forms an inverted image on the film behind it. The projection is an enlargement of scale factor −1.5 centred at O. Work out the coordinates of the image of corner F.
Answer key
- (a) Yes - SSS, the three side lengths all match — Bracket P's sides (12 cm, 16 cm, 20 cm) can each be matched to one of Bracket Q's sides (16 cm, 20 cm, 12 cm) — the same three lengths, just listed differently — so the brackets are congruent by SSS. SAS is wrong here because no angle is stated for either bracket, only three sides. The order the sides are listed in does not matter for SSS — only whether the SET of three lengths matches, and it does, so 'listed in a different order' is not a reason to say no. Nothing extra is needed: SSS proves congruence from side lengths alone, without any angles, so it can be determined.
- (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) (−2, −1) — Method: apply the reflection to the point first, then rotate the image about the given centre, in the order the question states them. Working: reflecting (3, 4) in the line y = 1 keeps x = 3 and puts the image as far below the line as the point is above it: 4 is 3 units above y = 1, so the image is 3 units below, at 1 − 3 = −2 (the same as 2 × 1 − 4 = −2). The reflected point is (3, −2). Rotating (3, −2) by 90° clockwise about (1, 1): subtracting the centre gives 3 − 1 = 2 and −2 − 1 = −3, the clockwise rule swaps and negates these to give −3 and −2, and adding the centre back gives 1 + (−3) = −2 and 1 + (−2) = −1. The final image is (−2, −1). Answer: (−2, −1). Reflect before you rotate, exactly as the design process is described, and rotate about the CENTRE (1, 1) given in the question rather than the origin: either mistake, or reversing the two steps, sends the tile to a different point.
- (a) ∠QRS — Method: the middle letter in three-letter angle notation is always the vertex of the angle, and the outer two letters are the neighbouring vertices along the shape's sides. Working: at vertex R, the two adjacent vertices along the pentagon are Q and S, so the interior angle is written ∠QRS, with R in the middle. Options: ∠PQR names the angle at Q, not R, since Q is the middle letter there; ∠RST puts R first rather than in the middle, so it actually names the angle at S; ∠TRP does have R in the middle, but T and P are not the vertices adjacent to R along the pentagon's sides, so it does not describe R's interior angle. Answer: ∠QRS.
- (a) 65° — OT and OC are both radii, so triangle OTC is isosceles with OT = OC, and its base angles are equal: angle OTC = angle OCT = (180° − 130°) ÷ 2 = 25°. A tangent is perpendicular to the radius at the point of contact, so angle OTP = 90°. Since C lies inside angle OTP, the chord TC splits this right angle into angle OTC and angle PTC, so angle PTC = angle OTP − angle OTC = 90° − 25° = 65°. Stopping after finding the base angle of the isosceles triangle, without subtracting it from the right angle at T, leaves 25° instead of the angle actually asked for. Adding the base angle to the right angle instead of subtracting it, 90° + 25° = 115°, reverses the direction the two angles combine in. Finding the sum of the two base angles of the isosceles triangle, 180° − 130° = 50°, and stopping there, gives another wrong value entirely.
- (c) RHS — Both triangles are right-angled, and have equal hypotenuses (13 cm) and one equal corresponding side (5 cm), so they are congruent by the RHS (right angle, hypotenuse, side) condition — the three facts given are exactly a right angle, a hypotenuse and one other side. The distractor SAS would apply if two sides and the angle between those two sides were matched, but the right angle here lies between the 5 cm side and the third side, not between the 5 cm side and the hypotenuse. The distractor SSS needs all three pairs of sides matched, and only two sides of each triangle are given — the third side would first have to be calculated by Pythagoras, so SSS is not the condition the given facts supply. The distractor AAS would need two angles and a non-included side matched, but only one angle in each triangle is given.
- (d) 1.5 m — By Pythagoras' theorem, diagonal² = 1.2² + 0.9² = 1.44 + 0.81 = 2.25, so diagonal = √2.25 = 1.5 m. 2.1 m comes from simply adding the two sides (1.2 + 0.9) instead of using Pythagoras' theorem. 0.3 m comes from subtracting the two sides (1.2 − 0.9) instead. 2.25 m comes from correctly finding 1.2² + 0.9² = 2.25 but forgetting to take the square root at the end.
- (a) 12π + 16 cm — A three-quarter sector's perimeter is the curved arc plus the two straight radii that close the shape. The full circumference is 2 × π × 8 = 16π cm, and three-quarters of that is 12π cm. Adding the two straight radii, 8 cm each, gives 12π + 16 cm. Leaving out the straight edges gives just 12π cm. Using one-quarter of the circumference, the piece left over rather than the piece asked for, gives 4π + 16 cm. Adding only one radius instead of two gives 12π + 8 cm.
- (c) 1 — sin 30° = 1/2 and cos 60° = 1/2, so sin 30° + cos 60° = 1/2 + 1/2 = 1. 0 comes from subtracting the two values instead of adding them (1/2 − 1/2). 1/2 comes from writing down only one of the two exact values and forgetting to add the other. √3 comes from swapping the two angles and working out sin 60° + cos 30° = √3/2 + √3/2 = √3.
- (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.
- (c) (4, 3) — Method: the midpoint of a segment is the mean of its two end points, so its x-coordinate is the mean of the two x-coordinates and its y-coordinate is the mean of the two y-coordinates. Working: for x, (1 + 7) ÷ 2 = 8 ÷ 2 = 4. For y, (3 + 3) ÷ 2 = 6 ÷ 2 = 3. The midpoint is therefore (4, 3). Answer: (4, 3). The distractors: (3, 3) comes from halving the difference of the x-coordinates, (7 − 1) ÷ 2 = 3, which measures half the distance instead of locating the point; (3.5, 3) comes from halving only the larger x-coordinate and leaving the smaller one out of the working; (4, 0) comes from averaging the x-coordinates correctly but then subtracting the y-coordinates, 3 − 3, rather than averaging them.
- (b) $\binom{-0.4}{-7}$ — The overall movement is the sum of the two vectors: (−1.2 + 0.8, −4.5 + (−2.5)) = (−0.4, −7). '$\binom{-2}{-2}$' comes from subtracting the second vector from the first instead of adding them. '$\binom{-0.4}{7}$' gets the top number right but drops the negative sign on the bottom number. '$\binom{2}{-7}$' comes from treating −1.2 + 0.8 as if the signs did not matter, giving +2 instead of −0.4.
- (c) 5 cm — Volume of a cylinder = πr²h, so r² = V ÷ (πh) = 942 ÷ (3.14 × 12) = 942 ÷ 37.68 = 25, and r = √25 = 5 cm. A pupil who finds r² = 25 but forgets to take the square root gives 25 cm. A pupil who forgets to divide by π, using r² = 942 ÷ 12 = 78.5, gets r = √78.5 ≈ 8.9 cm. A pupil who forgets to divide by the height, using r² = 942 ÷ 3.14 = 300, gets r = √300 ≈ 17.3 cm. The correct radius is 5 cm.
- (d) $\binom{0}{4}$ — Add the three vectors component by component: x: 3 + (−7) + 4 = 0; y: −2 + 5 + 1 = 4, giving $\binom{0}{4}$. $\binom{−4}{3}$ comes from adding only the first two vectors and forgetting the third. $\binom{14}{−6}$ comes from reading the second vector as $\binom{7}{−5}$ instead of $\binom{−7}{5}$, flipping its signs. $\binom{4}{0}$ comes from swapping the final x-total and y-total.
- (a) (−1, −0.5) — Method: for an enlargement centred at a point O that is not the origin, first find the point's position relative to O, multiply that by the scale factor, then add O's coordinates back on — image = O + k × (F − O). Working: F relative to O is (4, 2) − (2, 1) = (2, 1). Multiplying by the scale factor, −1.5 × (2, 1) = (−3, −1.5). Adding O back on: (2, 1) + (−3, −1.5) = (−1, −0.5). Answer: (−1, −0.5). Measure F's position from the aperture O, not from the origin, apply the full signed scale factor −1.5 to that displacement, and add O's own coordinates back on afterwards: measuring from the origin instead of O, treating the factor as +1.5, or adding O onto the wrong displacement all place the image mark somewhere else.
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