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) 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.
- (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.
- (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).
- (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.
- (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°.
- (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°.
- (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.
- (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) 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°.
- (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) 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) 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°.
- (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.
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