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.
Answer key: Geometric reasoning and simple proofs worksheet — GCSE Foundation
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- (d) 71° — Alternate angles between parallel lines are equal. Since angle HMN and angle MNK are alternate angles, angle MNK = angle HMN = 71°.
- (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°.
- (c) 67° — Angles on a straight line add up to 180°. Add the two known angles: 48° + 65° = 113°. Subtract from 180°: 180° − 113° = 67°.
- (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.
- (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°.
- (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.
- (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) 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.
- (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.
- (d) 80° — The four angles of any quadrilateral add up to 360°. Three of the angles add up to 100 + 85 + 95 = 280°, so the fourth angle is 360 − 280 = 80°. "70°" comes from using 350° as the total instead of 360°, an easy slip on the quadrilateral angle sum. "280°" comes from stopping at the sum of the three given angles and writing that total down as the answer, instead of subtracting it from 360°. "260°" comes from subtracting only the one angle 100° from 360°, instead of subtracting all three given angles.
- (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°.
- (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.
- (a) 62° — Angle KPQ and angle MQP are co-interior (allied) angles between parallel lines, so they add up to 180°: angle MQP = 180° − 118° = 62°. 118° comes from treating the angles as equal, as if this were a corresponding or alternate angle pair, instead of using the co-interior rule. 90° comes from wrongly assuming the crossing line meets JK and LM at right angles. 59° comes from halving 118° instead of subtracting it from 180°.
- (c) RHS applies: the right angle, hypotenuse and one matching side are all equal — Both triangles are right-angled, and the hypotenuses (AC and DF) are equal, and one further pair of matching sides (AB and DE) are equal. This matches the Right angle-Hypotenuse-Side condition, which needs only these three facts, not a third side.
- (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.
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