Printable · GCSE Foundation · ages 14-16
Geometric reasoning and simple proofs worksheet — GCSE Foundation
Fifteen questions on "geometric reasoning and simple proofs" — DfE statement G6. Print it, or print three versions so neighbours cannot copy by letter; the key gives the letter for each version.
Geometric reasoning and simple proofs worksheet — GCSE Foundation
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- 1.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.
- 2.Lines PQ and RS are parallel. A straight line crosses PQ at point X and crosses RS at point Y. Angle PXY = 63°. Angle PXY and angle XYS are co-interior (allied) angles, which add up to 180°. Work out the size of angle XYS.
- 3.Lines GH and JK are parallel. A straight line crosses GH at point M and crosses JK at point N. Angle HMN = 71°. Angle HMN and angle MNK are alternate angles. Work out angle MNK.
- 4.A roof truss is shaped like an isosceles triangle resting on a horizontal ceiling joist. The truss's sloping edges are equal in length, and the angle at its apex is 40°. The ceiling joist continues in a straight line beyond the foot of the truss. Work out the angle between the truss's sloping edge and the extended joist, on the outside of the truss.
- 5.Two straight lines AB and CD cross at point O. Angle AOC = 74°. Work out angle AOD.
- 6.A straight line passes through points A, B and C, with B between A and C. A fourth point D is not on the line. Angle ABD = 132°. Work out the size of angle DBC.
- 7.In triangle ABC, point D lies on side AB and point E lies on side AC, so that DE is parallel to BC. AD = 3 cm, DB = 6 cm and DE = 4 cm. Work out the length of BC.
- 8.Triangle PQR has angle P = 90°. The side PQ = 9 cm and the hypotenuse QR = 15 cm. Work out the length of PR.
- 9.A vertical flagpole casts a shadow 6 m long at the same time as a 1.2 m vertical post casts a shadow 1.8 m long. The flagpole and the post are both vertical, and the triangles formed by each object, its shadow, and the sun's ray are similar. Work out the height of the flagpole.
- 10.Triangle XYZ has angle X = 90°. Angle Y = 34°. Work out angle Z.
- 11.Two angles lie on a straight line. One angle is 118°. Which reason correctly explains why the other angle is 62°?
- 12.Two triangular offcuts of wood, PQR and STU, are cut for a construction project. PQ = 8 cm, QR = 6 cm and angle PQR = 90°. ST = 8 cm, TU = 6 cm and angle STU = 90°. A carpenter wants to check the two pieces are identical in shape and size before using them as a matching pair. Using only the measurements given, and without working out any further lengths, which condition proves that triangle PQR is congruent to triangle STU?
- 13.Two angles lie on a straight line. One angle is 37°. Work out the size of the other angle.
- 14.A hexagon has interior angles of 130°, 142°, 125°, 150°, 135° and x°. Work out the value of x.
- 15.A builder props a straight plank against a vertical wall to reach a window ledge. The foot of the plank is 2.1 m from the base of the wall, and the plank is 3.5 m long. Work out how high up the wall the plank reaches.
Answer key
- (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.
- (c) 117° — Co-interior (also called allied) angles between parallel lines add up to 180°. Angle PXY and angle XYS are co-interior angles, so angle PXY + angle XYS = 180°. 180° − 63° = 117°, so angle XYS = 117°. A student who mistakes co-interior angles for alternate angles, which are equal rather than supplementary, answers 63° instead. A student who treats the two angles as complementary, subtracting from 90° instead of 180°, gets 90° − 63° = 27°. A student who assumes the transversal meets the parallel lines at right angles answers 90°.
- (d) 71° — Alternate angles between parallel lines are equal. Since angle HMN and angle MNK are alternate angles, angle MNK = angle HMN = 71°.
- (a) 110° — The truss is isosceles, so its two base angles are equal; since the three angles of the triangle add up to 180° and the apex is 40°, each base angle is (180 − 40) ÷ 2 = 70°. The angle between the sloping edge and the joist, extended beyond the base of the truss, sits on a straight line with that 70° base angle, and angles on a straight line add up to 180°, so the required angle is 180 − 70 = 110°. "70°" comes from stopping after finding the base angle and giving it directly, without also using the straight-line fact to find the angle on the OUTSIDE of the truss. "140°" comes from doubling the base angle instead of using the straight-line fact. "20°" comes from taking half of the apex angle (40° ÷ 2), mistaking it for the required angle instead of properly using the triangle's angle sum.
- (a) 106° — Angle AOC and angle AOD share the arm OA and together they make up the straight line CD, so they add up to 180°: angle AOD = 180° − 74° = 106°. 74° comes from treating AOC and AOD as vertically opposite angles, which are equal — but they are adjacent angles on a straight line, not vertically opposite. 16° comes from subtracting 74° from 90°, treating the angles as complementary. 148° comes from doubling 74° instead of subtracting it from 180°.
- (b) 48° — Angles on a straight line add up to 180°, so angle ABD + angle DBC = 180°. 180° − 132° = 48°, so angle DBC = 48°. A student who uses angles around a point (360°) instead of a straight line (180°) finds 360° − 132° = 228°. A student who halves angle ABD by mistake, thinking DBC must be half of ABD, gets 132° ÷ 2 = 66°.
- (d) 12 cm — Since DE is parallel to BC, triangle ADE is similar to triangle ABC (the two triangles share angle A, and the parallel lines make the angles at D and E equal to the angles at B and C). The whole side AB = AD + DB = 3 + 6 = 9 cm. The scale factor from the small triangle to the large triangle is AB ÷ AD = 9 ÷ 3 = 3, so BC = DE × 3 = 4 × 3 = 12 cm. Scaling by DB ÷ AD instead of AB ÷ AD gives 4 × (6 ÷ 3) = 8 cm. Adding DB directly onto DE, instead of scaling, gives 4 + 6 = 10 cm. Treating the triangles as congruent rather than similar, so assuming corresponding sides are simply equal, gives BC = DE = 4 cm.
- (c) 12 cm — By Pythagoras' Theorem, QR² = PQ² + PR², so PR² = QR² − PQ² = 15² − 9² = 225 − 81 = 144. Square root: √144 = 12 cm.
- (b) 4 m — The two triangles are similar, so height ÷ shadow is the same for both objects. For the post, height ÷ shadow = 1.2 ÷ 1.8. Flagpole height = 6 × (1.2 ÷ 1.8) = 4 m. A student who multiplies the flagpole's shadow by the post's height without applying the ratio gets 6 × 1.2 = 7.2 m. A student who assumes a constant DIFFERENCE between shadow and height instead of a constant ratio finds the post's shadow minus height, 1.8 − 1.2 = 0.6 m, then subtracts that from the flagpole's shadow to get 6 − 0.6 = 5.4 m.
- (a) 56° — The angles of a triangle add up to 180°. Add the two known angles: 90° + 34° = 124°. Subtract from 180°: 180° − 124° = 56°.
- (b) angles on a straight line add up to 180° — 118° and 62° add up to 180°, matching the fact that angles on a straight line always add up to 180° — that is the reason the second angle is 62°. Angles round a point add up to 360° is a different fact, used for angles surrounding a single point, not two angles on a line. Vertically opposite angles are equal would make the second angle 118° too, not 62°. Angles in a triangle add up to 180° is a true fact, but it applies to the three angles inside a triangle, not two angles on a straight line.
- (b) SAS (two sides and the included angle) — PQ = ST and QR = TU are two pairs of matching sides, and the angle between them (angle Q and angle T) is 90° in both triangles — two sides and the angle between them are equal, so the triangles are congruent by SAS, using only the measurements given. RHS needs the hypotenuses to be known equal; here only the two legs are given, so you would first have to work out each hypotenuse by Pythagoras' theorem — extra working the question rules out. SSS has the same problem: the third side of each triangle is not given, only calculable. ASA needs two angles and the side between them, but here it is two sides and the angle between them that are given, not two angles.
- (a) 143° — Angles on a straight line add up to 180°. Set up 37° + x = 180°. Subtract: x = 180° − 37° = 143°. 53° comes from using 90° as the total, as if the two angles made a right angle, instead of the 180° of a straight line.
- (c) 38° — The interior angles of a hexagon sum to (6 − 2) × 180° = 720°. 130° + 142° + 125° + 150° + 135° = 682°. x = 720° − 682° = 38°. A student who uses 6 × 180° instead of (6 − 2) × 180°, forgetting to subtract the 2, gets 1080° − 682° = 398°. A student who mis-adds the five given angles as 680° instead of 682° gets x = 720° − 680° = 40°. A student who answers with one of the given angles instead of solving for x might write 142°.
- (b) 2.8 m — Use Pythagoras' Theorem: the plank is the hypotenuse (3.5 m) of a right-angled triangle formed with the wall and the ground (2.1 m). height² = 3.5² − 2.1² = 12.25 − 4.41 = 7.84. height = √7.84 = 2.8 m. A student who subtracts the two given lengths directly instead of using Pythagoras gets 3.5 − 2.1 = 1.4 m. A student who doubles the distance from the wall by mistake gets 2.1 × 2 = 4.2 m.
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