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.
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Answer key: Geometric reasoning and simple proofs worksheet — GCSE Foundation
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- (d) 4 m — The sloping slide is the hypotenuse of a right-angled triangle with the other two sides 2.4 m and 3.2 m. By Pythagoras' Theorem, hypotenuse² = 2.4² + 3.2² = 5.76 + 10.24 = 16. Square root: √16 = 4 m. Adding the two sides instead of squaring them gives 2.4 + 3.2 = 5.6 m. Truncating each squared side to a whole number before adding (5.76 to 5 and 10.24 to 10) gives √15 ≈ 3.87 m. Forgetting to take the square root and leaving the sum of the squares as the answer gives 5.76 + 10.24 = 16, written as 16 m.
- (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) 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).
- (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) 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.
- (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.
- (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°.
- (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°.
- (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°.
- (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).
- (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) 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) 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) 12 cm — By Pythagoras' Theorem, QR² = PQ² + PR², so PR² = QR² − PQ² = 15² − 9² = 225 − 81 = 144. Square root: √144 = 12 cm.
- (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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