Properties and Constructions

1. 1

   Two parallel lines are cut by a transversal. One co-interior angle is 67°. What is the other co-interior angle?

   • A

     $67°$

   • B

     $113°$

   • C

     $90°$

   • D

     $23°$

   Check answer

   Answer & explanation

   Correct: option B

   Co-interior angles sum to 180°. $180° - 67° = 113°$.

2. 2

   What is the sum of the interior angles of a heptagon (7 sides)?

   • A

     $720°$

   • B

     $900°$

   • C

     $1080°$

   • D

     $540°$

   Check answer

   Answer & explanation

   Correct: option B

   $(n - 2) \times 180° = (7 - 2) \times 180° = 5 \times 180° = 900°$.

3. 3

   A regular polygon has an exterior angle of 24°. How many sides does it have?

   • A

     $12$

   • B

     $15$

   • C

     $18$

   • D

     $20$

   Check answer

   Answer & explanation

   Correct: option B

   $n = 360° \div 24° = 15$ sides.

4. 4

   Which of the following is a valid congruence criterion for triangles?

   • A

     AAA

   • B

     SSA

   • C

     SAS

   • D

     ASS

   Check answer

   Answer & explanation

   Correct: option C

   SAS (two sides and the included angle) is a valid congruence criterion. AAA proves similarity only; SSA is not valid because two triangles can share two sides and a non-included angle without being congruent.

5. 5

   $AB$ is a diameter of a circle. $C$ is a point on the circle. What is angle $ACB$?

   • A

     $45°$

   • B

     $60°$

   • C

     $90°$

   • D

     $180°$

   Check answer

   Answer & explanation

   Correct: option C

   By the circle theorem: the angle in a semicircle (subtended by a diameter) is 90°.

6. 6

   $ABCD$ is a cyclic quadrilateral. Angle $A = 78°$. What is angle $C$?

   • A

     $78°$

   • B

     $102°$

   • C

     $90°$

   • D

     $282°$

   Check answer

   Answer & explanation

   Correct: option B

   Opposite angles of a cyclic quadrilateral sum to 180°. $\angle C = 180° - 78° = 102°$.

7. 7

   Two similar triangles have a linear scale factor of 3. If the area of the smaller triangle is 8 cm², what is the area of the larger triangle?

   • A

     $24 \text{ cm}^2$

   • B

     $48 \text{ cm}^2$

   • C

     $72 \text{ cm}^2$

   • D

     $27 \text{ cm}^2$

   Check answer

   Answer & explanation

   Correct: option C

   Area scale factor $= 3^2 = 9$. Area of larger triangle $= 8 \times 9 = 72 \text{ cm}^2$.

8. 8

   A tangent touches a circle at point $T$. $O$ is the centre. What is the angle between the radius $OT$ and the tangent?

   • A

     $45°$

   • B

     $60°$

   • C

     $90°$

   • D

     $180°$

   Check answer

   Answer & explanation

   Correct: option C

   By the circle theorem: a tangent to a circle is perpendicular to the radius at the point of tangency, so the angle is 90°.

Exam-style multi-part questions. Try each part on paper, then reveal the model answer to compare your working.

1. Q17 marks total

   Show all answers

   Scenario

   $O$ is the centre of a circle. $A$, $B$ and $C$ are points on the circle. Angle $AOB = 130°$.

   1. a

      Find angle $ACB$. State the circle theorem you use.

      [2]
      Show answer

      Model answer · 2 marks

      $\angle ACB = 130° \div 2 = 65°$. Circle theorem used: the angle subtended at the centre is twice the angle subtended at the circumference by the same arc.

   2. b

      $ABCD$ is a cyclic quadrilateral. Find angle $ADB$. Give a reason for your answer.

      [2]
      Show answer

      Model answer · 2 marks

      $\angle ADB = 65°$. Reason: angles in the same segment (both $\angle ACB$ and $\angle ADB$ are subtended by arc $AB$ on the same side) are equal.

   3. c

      Angle $BOC = 80°$. Find angle $BAC$. Show your working.

      [3]
      Show answer

      Model answer · 3 marks

      Reflex angle $\angle BOC = 360° - 80° = 280°$ (angles at a point). Wait — the question likely means the minor arc. $\angle BAC = \angle BOC \div 2 = 80° \div 2 = 40°$ (angle at centre twice angle at circumference, same arc $BC$).

   Examiner note

   Theorem names must be stated for full marks — numerical steps without named reasons score method marks only.

2. Q26 marks total

   Show all answers

   Scenario

   Triangle $ABC$ has $AB = 8$ cm, $BC = 6$ cm and $\angle ABC = 90°$. Triangle $PQR$ is similar to triangle $ABC$ with $PQ = 20$ cm.

   1. a

      Find the scale factor from triangle $ABC$ to triangle $PQR$.

      [1]
      Show answer

      Model answer · 1 mark

      Scale factor $= PQ \div AB = 20 \div 8 = 2.5$.

   2. b

      Find $QR$.

      [2]
      Show answer

      Model answer · 2 marks

      $QR = BC \times 2.5 = 6 \times 2.5 = 15$ cm.

   3. c

      Calculate the area of triangle $PQR$.

      [3]
      Show answer

      Model answer · 3 marks

      Area of $\triangle ABC = \frac{1}{2} \times 8 \times 6 = 24 \text{ cm}^2$ (right-angled, legs $AB$ and $BC$). Area scale factor $= 2.5^2 = 6.25$. Area of $\triangle PQR = 24 \times 6.25 = 150 \text{ cm}^2$.

   Examiner note

   Part (c) can also be solved directly: $\frac{1}{2} \times PQ \times QR \times \sin 90° = \frac{1}{2} \times 20 \times 15 = 150$ cm². Both approaches are acceptable.

3. Q37 marks total

   Show all answers

   Scenario

   Using ruler and compasses only, construct the following. Show all construction arcs.

   1. a

      Construct the perpendicular bisector of line segment $AB$, where $AB = 8$ cm. Hence mark the midpoint $M$ of $AB$.

      [2]
      Show answer

      Model answer · 2 marks

      Open compasses to more than 4 cm. From $A$, draw arcs above and below $AB$. From $B$, draw arcs (same radius) to intersect both arcs. Join the two intersection points — this is the perpendicular bisector. Mark $M$ where it crosses $AB$.

   2. b

      Construct the locus of all points that are equidistant from the two sides of angle $XYZ = 60°$ (drawn in the question).

      [2]
      Show answer

      Model answer · 2 marks

      Construct the angle bisector of $\angle XYZ$: draw an arc from $Y$ to get equal points on $YX$ and $YZ$; from those points, draw equal arcs to intersect; join the intersection to $Y$. The angle bisector is the required locus.

   3. c

      A point $P$ must be within 5 cm of point $A$ and closer to $B$ than to $A$. Shade the region satisfying both conditions (with $AB = 8$ cm).

      [3]
      Show answer

      Model answer · 3 marks

      Draw a circle of radius 5 cm centred on $A$ (locus of points exactly 5 cm from $A$ — the region within is 'within 5 cm'). Draw the perpendicular bisector of $AB$ (locus of points equidistant from $A$ and $B$ — the region closer to $B$ is on $B$'s side). Shade the region that is inside the circle and on $B$'s side of the perpendicular bisector.

   Examiner note

   All construction arcs must be visible. Freehand lines or measured midpoints score zero for the construction mark.