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.Triangle ABC is similar to triangle PQR. AB corresponds to PQ, and BC corresponds to QR. AB = 5 cm, PQ = 15 cm and BC = 7 cm. Work out the length of QR.
- 2.A ramp is built from two straight metal supports that rest on the flat ground and meet each other at the top, forming a triangle with the ground. One support meets the ground at 34°. The angle between the two supports where they meet is 71°. Work out the angle between the second support and the ground.
- 3.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.
- 4.Triangle ABC is extended so that side BC is extended beyond C to a point D, forming exterior angle ACD. Angle BAC = 48° and angle ABC = 67°. Which single angle fact gives angle ACD = 115° in one step, without first working out the interior angle at C?
- 5.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.
- 6.Two straight lines PQ and RS cross at point O. Angle POR = 52°. Work out angle QOS.
- 7.Triangle DEF is isosceles, with DE = DF. Angle E = 58°. Work out angle F.
- 8.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.
- 9.A hexagon has interior angles of 130°, 142°, 125°, 150°, 135° and x°. Work out the value of x.
- 10.Triangles LMN and XYZ have LM = XY, MN = YZ and LN = XZ. Give a reason why triangle LMN is congruent to triangle XYZ.
- 11.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.
- 12.In parallelogram PQRS, angle P = 65°. Which reason correctly explains why angle R = 65°?
- 13.Two angles lie on a straight line. One angle is 37°. Work out the size of the other angle.
- 14.Triangle DEF is isosceles, with DE = DF. Angle E = 58°. Give a reason why angle F = 58°.
- 15.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.
Answer key
- (d) 21 cm — The scale factor from triangle ABC to triangle PQR is PQ ÷ AB = 15 ÷ 5 = 3. QR = BC × scale factor = 7 × 3 = 21 cm. A student who divides BC by the scale factor instead of multiplying gets 7 ÷ 3 = 2.33 cm. A student who multiplies BC by AB instead of by the scale factor gets 7 × 5 = 35 cm.
- (a) 75° — The three angles in the triangle formed by the two supports and the ground add up to 180°. Two of the angles are 34° and 71°, so the third angle is 180 − 34 − 71 = 75, which is 75°. 105° comes from adding the two given angles together instead of subtracting them from 180° (34 + 71 = 105). 146° comes from computing 180 − 34 = 146 and forgetting to also subtract 71°. 37° comes from subtracting the two given angles from each other instead of using the triangle's angle sum (71 − 34 = 37).
- (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.
- (c) exterior angle = sum of the two interior opposite angles — Angle ACD is the exterior angle at C, and the exterior angle of a triangle is always equal to the sum of the two interior angles at the other two vertices — here, angle BAC and angle ABC — so angle ACD = 48° + 67° = 115°. "angles in a triangle add up to 180°" is a true fact, but on its own it only gives the INTERIOR angle at C (180 − 48 − 67 = 65°), not the exterior angle 115° — reaching 115° that way still needs a further step. "angles on a straight line add up to 180°" is the fact that links the interior and exterior angles at C to each other, not the fact that gives 115° directly from the two OTHER angles. "exterior angle is always double the smallest angle" is not a genuine angle fact: double the smallest angle here is 48° × 2 = 96°, not 115°.
- (d) 71° — Alternate angles between parallel lines are equal. Since angle HMN and angle MNK are alternate angles, angle MNK = angle HMN = 71°.
- (c) 52° — When two straight lines cross, the angles opposite each other at the crossing point are vertically opposite and are equal. Ray OQ is opposite ray OP and ray OS is opposite ray OR, so angle QOS is vertically opposite angle POR. Therefore angle QOS = 52°.
- (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°.
- (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°.
- (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°.
- (a) SSS – all three corresponding sides are equal — All three pairs of corresponding sides are stated as equal — LM = XY, MN = YZ and LN = XZ — with no angle mentioned. This matches the SSS condition, so triangle LMN is congruent to triangle XYZ.
- (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) opposite angles of a parallelogram are equal — P and R are opposite vertices of the parallelogram, and opposite angles of a parallelogram are always equal, which is why angle R equals angle P, 65°. Co-interior angles adding up to 180° is the correct fact for angle Q or angle S, the angles adjacent to P along a side, not for the opposite angle R. Alternate angles are equal is a fact about a transversal crossing two parallel lines, which explains other angle relationships in the parallelogram, not the one between opposite angles P and R directly. Angles on a straight line adding up to 180° applies to two angles that sit together on one straight line, which P and R do not.
- (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.
- (d) Base angles of an isosceles triangle are equal — DE = DF, so triangle DEF is isosceles with DE and DF as the two equal sides. The base angles opposite those equal sides, angle E and angle F, are therefore equal to each other. Angle F = 58°. A student who instead quotes 'Angles in a triangle sum to 180°' has picked a true fact about triangles, but that fact finds a missing angle from the other two — it does not explain why two angles are equal to each other. A student who quotes 'Angles on a straight line sum to 180°' has confused this with a straight-line angle fact, but no straight line of angles is described in this triangle.
- (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.
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