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.A straight line passes through points A, B and C, with B between A and C. A fourth point D is not on the line. Angle ABD = 132°. Work out the size of angle DBC.
- 2.Triangle ABC is right-angled at B. Triangle DEF is right-angled at E. AB = DE = 6 cm and AC = DF = 10 cm (AC and DF are the hypotenuses of their triangles). A student says this is not enough information to prove the triangles are congruent, because only two sides are given. Which reason shows the student is wrong?
- 3.Triangle DEF is isosceles, with DE = DF. Angle E = 58°. Work out angle F.
- 4.Triangles LMN and XYZ have LM = XY, MN = YZ and LN = XZ. Give a reason why triangle LMN is congruent to triangle XYZ.
- 5.Triangle XYZ has angle X = 90°. Angle Y = 34°. Work out angle Z.
- 6.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.
- 7.A vertical flagpole casts a shadow 6 m long at the same time as a 1.2 m vertical post casts a shadow 1.8 m long. The flagpole and the post are both vertical, and the triangles formed by each object, its shadow, and the sun's ray are similar. Work out the height of the flagpole.
- 8.A children's slide is built as a right-angled triangle: the vertical ladder side is 2.4 m, the horizontal base along the ground is 3.2 m, and the sloping slide is the third side. Work out the exact length of the sloping slide.
- 9.Two triangular offcuts of wood, PQR and STU, are cut for a construction project. PQ = 8 cm, QR = 6 cm and angle PQR = 90°. ST = 8 cm, TU = 6 cm and angle STU = 90°. A carpenter wants to check the two pieces are identical in shape and size before using them as a matching pair. Using only the measurements given, and without working out any further lengths, which condition proves that triangle PQR is congruent to triangle STU?
- 10.Two angles lie on a straight line. One angle is 118°. Which reason correctly explains why the other angle is 62°?
- 11.A hexagon has interior angles of 130°, 142°, 125°, 150°, 135° and x°. Work out the value of x.
- 12.A quadrilateral has angles of 100°, 85° and 95° at three of its vertices. Work out the fourth angle.
- 13.Two straight lines PQ and RS cross at point O. Angle POR = 52°. Work out angle QOS.
- 14.Two angles lie on a straight line. One angle is 37°. Work out the size of the other angle.
- 15.In parallelogram PQRS, angle P = 65°. Which reason correctly explains why angle R = 65°?
Answer key
- (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) 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) 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) 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.
- (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°.
- (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.
- (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.
- (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.
- (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°.
- (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°.
- (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) 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.
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