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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- (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.
- (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.
- (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°.
- (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.
- (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°.
- (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°.
- (c) 12 cm — By Pythagoras' Theorem, QR² = PQ² + PR², so PR² = QR² − PQ² = 15² − 9² = 225 − 81 = 144. Square root: √144 = 12 cm.
- (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°.
- (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) 115° — Angle APE and angle BPQ are vertically opposite, so angle BPQ = 65°, equal to angle APE. Angle BPQ and angle PQD are co-interior (allied) angles between the parallel lines, and co-interior angles always add up to 180°, so angle PQD = 180 − 65 = 115°. "65°" comes from treating co-interior angles as equal to each other, the way alternate angles are, instead of adding to 180°. "295°" comes from applying the angles-round-a-point fact (360° − 65°) directly to the original 65°, skipping the correct co-interior step. "25°" comes from misremembering co-interior angles as adding to 90° instead of 180° (90 − 65 = 25).
- (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.
- (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°.
- (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) 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.
- (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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