General Instructions
- The question paper has 19 questions divided into four sections, carrying a total of 40 marks.
- Section A: Questions 1–8, one mark each (8 marks).
- Section B: Questions 9–12, two marks each (8 marks).
- Section C: Questions 13–16, three marks each (12 marks).
- Section D: Questions 17–19, four marks each (12 marks).
- All questions are compulsory. Write the full working wherever marks are awarded for method, not just the final answer.
- Use of a calculator is not permitted. For the constructions question, use only a ruler and a pair of compasses and show all construction arcs.
Section A – Multiple Choice Questions
8 questions × 1 mark = 8 marks
Each question below has four options (A, B, C, D). Choose the one correct answer.
In the given diagram, \(PS\) and \(PT\) are tangents to the circle. \(SQ \parallel PT\) and \(\angle SPT = 80^\circ\). The value of \(\angle QST\) is:
The areas of two similar triangles are in the ratio \(4 : 9\). The ratio of their corresponding sides is:
In \(\triangle ABC\), \(DE \parallel BC\) with \(AD = 2.5\text{ cm}\), \(DB = 5\text{ cm}\) and \(AE = 3\text{ cm}\). Then \(EC\) equals:
The locus of a point equidistant from two given points \(A\) and \(B\) is:
In a circle with centre \(O\), the angle subtended by chord \(AB\) at the centre is \(110^\circ\). The angle subtended by \(AB\) at a point on the major arc is:
\(ABCD\) is a cyclic quadrilateral with \(\angle A = 68^\circ\). Then \(\angle C\) equals:
In the given diagram, \(RT\) is a tangent touching the circle at \(S\). If \(\angle PST = 30^\circ\) and \(\angle SPQ = 60^\circ\), then \(\angle PSQ\) equals:
Two chords \(AB\) and \(CD\) of a circle intersect at \(P\) inside the circle. If \(PA = 6\text{ cm}\), \(PB = 8\text{ cm}\) and \(PC = 4\text{ cm}\), then \(PD\) equals:
Section B – Short Answer Questions
4 questions × 2 marks = 8 marks
In \(\triangle ABC\), \(DE \parallel BC\). If \(AD = 4\text{ cm}\), \(DB = x\text{ cm}\), \(AE = 6\text{ cm}\) and \(EC = 9\text{ cm}\), find the value of \(x\).
\(\triangle ABC \sim \triangle DEF\) with \(AB = 5\text{ cm}\) and \(DE = 3\text{ cm}\). If the area of \(\triangle DEF\) is \(18\text{ cm}^2\), find the area of \(\triangle ABC\).
From an external point \(P\), a tangent \(PT\) of length \(12\text{ cm}\) is drawn to a circle, and a secant \(PAB\) cuts the circle at \(A\) and \(B\) with \(PA = 8\text{ cm}\). Find the length of \(PB\).
A chord of length \(24\text{ cm}\) is drawn in a circle of radius \(13\text{ cm}\). Find the perpendicular distance of the chord from the centre of the circle.
Section C – Application Questions
4 questions × 3 marks = 12 marks
\(AB\) is a diameter of a circle and \(C\) is a point on the circle such that \(\angle ABC = 35^\circ\). Find \(\angle ACB\) and \(\angle BAC\).
\(ABCD\) is a cyclic quadrilateral in which \(\angle B = 3x\) and \(\angle D = 2x\). Find the value of \(x\) and hence the measures of \(\angle B\) and \(\angle D\).
From an external point \(P\), tangents \(PA\) and \(PB\) are drawn to a circle with centre \(O\). If \(\angle APB = 70^\circ\), find \(\angle AOB\) and \(\angle OAB\).
Two chords \(AB\) and \(CD\) of a circle intersect at a point \(P\) inside the circle. If \(PA = 5\text{ cm}\), \(PB = 12\text{ cm}\) and \(PC = 3\text{ cm}\), find \(PD\) and the length of chord \(CD\).
Section D – Structured / Long Answer Questions
3 questions × 4 marks = 12 marks
Using a ruler and a pair of compasses, construct a triangle \(ABC\) in which \(AB = 4\text{ cm}\), \(BC = 5\text{ cm}\) and \(CA = 6\text{ cm}\). Construct the circle passing through \(A\), \(B\) and \(C\) (the circumcircle) and measure its radius.
In \(\triangle ABC\), \(DE \parallel BC\) with \(AD : DB = 2 : 3\). If the area of \(\triangle ADE\) is \(16\text{ cm}^2\), find the area of the trapezium \(DBCE\).
From an external point \(P\), \(PA\) and \(PB\) are tangents to a circle with centre \(O\), touching the circle at \(A\) and \(B\). Prove that \(\angle APB + \angle AOB = 180^\circ\). If \(\angle APB = 65^\circ\), find \(\angle OAB\).
Check your work: Section A (8) + Section B (8) + Section C (12) + Section D (12) = 40 marks. Attempt the whole paper before revealing the answers below.
Answer Key
Section A – Multiple Choice Questions
| Question | Answer |
|---|---|
| 1 | D (\(50^\circ\)) |
| 2 | A (\(2 : 3\)) |
| 3 | A (\(6\text{ cm}\)) |
| 4 | A (perpendicular bisector of \(AB\)) |
| 5 | A (\(55^\circ\)) |
| 6 | A (\(112^\circ\)) |
| 7 | D (\(90^\circ\)) |
| 8 | A (\(12\text{ cm}\)) |
Sections B to D – Final Answers
| Question | Answer |
|---|---|
| 9 | \(x = 6\) |
| 10 | Area \(= 50\text{ cm}^2\) |
| 11 | \(PB = 18\text{ cm}\) |
| 12 | \(5\text{ cm}\) |
| 13 | \(\angle ACB = 90^\circ\); \(\angle BAC = 55^\circ\) |
| 14 | \(x = 36\); \(\angle B = 108^\circ\); \(\angle D = 72^\circ\) |
| 15 | \(\angle AOB = 110^\circ\); \(\angle OAB = 35^\circ\) |
| 16 | \(PD = 20\text{ cm}\); \(CD = 23\text{ cm}\) |
| 17 | Construction; circumradius \(\approx 3.0\text{ cm}\) |
| 18 | Area of \(DBCE = 84\text{ cm}^2\) |
| 19 | Proof; \(\angle OAB = 32.5^\circ\) |
Detailed Solutions
Section A – Multiple Choice Questions
Tangents from an external point are equal, so \(PS = PT\) and \(\angle PST = \angle PTS\). In \(\triangle PST\), \(2\angle PTS = 180^\circ - 80^\circ = 100^\circ\), so \(\angle PTS = 50^\circ\). Since \(SQ \parallel PT\) with \(ST\) as transversal, alternate angles are equal: \(\angle QST = \angle PTS = 50^\circ\).
For similar figures, the ratio of areas equals the square of the ratio of corresponding sides. \(\left(\dfrac{s_1}{s_2}\right)^2 = \dfrac{4}{9}\), so \(\dfrac{s_1}{s_2} = \dfrac{2}{3}\).
By the basic proportionality theorem, \(\dfrac{AD}{DB} = \dfrac{AE}{EC}\), i.e. \(\dfrac{2.5}{5} = \dfrac{3}{EC}\). Thus \(EC = \dfrac{5 \times 3}{2.5} = 6\text{ cm}\).
A point equidistant from \(A\) and \(B\) lies on the perpendicular bisector of segment \(AB\).
The angle at the centre is twice the angle at the circumference on the same arc. A point on the major arc sees the chord via the minor arc, so the angle is \(\dfrac{110^\circ}{2} = 55^\circ\).
In a cyclic quadrilateral, opposite angles are supplementary: \(\angle A + \angle C = 180^\circ\), so \(\angle C = 180^\circ - 68^\circ = 112^\circ\).
By the alternate segment theorem, the angle between the tangent \(RT\) and the chord \(PS\) equals the angle in the alternate segment: \(\angle PQS = \angle PST = 30^\circ\).
In \(\triangle PQS\), \(\angle PSQ = 180^\circ - 60^\circ - 30^\circ = 90^\circ\).
For two chords intersecting inside a circle, \(PA \times PB = PC \times PD\). So \(6 \times 8 = 4 \times PD\), giving \(PD = 12\text{ cm}\).
Section B – Short Answer Questions
By the basic proportionality theorem, \(\dfrac{AD}{DB} = \dfrac{AE}{EC}\): \(\dfrac{4}{x} = \dfrac{6}{9}\). So \(6x = 36\).
Ratio of areas \(= \left(\dfrac{AB}{DE}\right)^2 = \left(\dfrac{5}{3}\right)^2 = \dfrac{25}{9}\). So \(\dfrac{\text{area}(ABC)}{18} = \dfrac{25}{9}\), giving area \((ABC) = 18 \times \dfrac{25}{9} = 50\text{ cm}^2\).
By the tangent–secant theorem, \(PT^2 = PA \times PB\). So \(12^2 = 8 \times PB\), giving \(PB = \dfrac{144}{8} = 18\text{ cm}\).
The perpendicular from the centre bisects the chord, so half-chord \(= 12\text{ cm}\). By Pythagoras, \(d = \sqrt{13^2 - 12^2} = \sqrt{169 - 144} = \sqrt{25}\).
Section C – Application Questions
Use the angle in a semicircle, then the angle sum of a triangle.
Since \(AB\) is a diameter, the angle in the semicircle is a right angle: \(\angle ACB = 90^\circ\).
In \(\triangle ABC\), \(\angle BAC = 180^\circ - 90^\circ - 35^\circ = 55^\circ\).
Opposite angles of a cyclic quadrilateral are supplementary.
\(\angle B + \angle D = 180^\circ\), so \(3x + 2x = 180^\circ\), giving \(5x = 180^\circ\) and \(x = 36\).
Thus \(\angle B = 3 \times 36 = 108^\circ\) and \(\angle D = 2 \times 36 = 72^\circ\).
A tangent is perpendicular to the radius at the point of contact.
\(\angle OAP = \angle OBP = 90^\circ\). In quadrilateral \(OAPB\), \(\angle AOB = 360^\circ - 90^\circ - 90^\circ - 70^\circ = 110^\circ\).
In \(\triangle OAB\), \(OA = OB\) (radii), so \(\angle OAB = \angle OBA = \dfrac{180^\circ - 110^\circ}{2} = 35^\circ\).
Use the intersecting chords theorem for two chords meeting inside a circle.
\(PA \times PB = PC \times PD\), so \(5 \times 12 = 3 \times PD\), giving \(PD = \dfrac{60}{3} = 20\text{ cm}\).
\(CD = PC + PD = 3 + 20 = 23\text{ cm}\).
Section D – Structured / Long Answer Questions
Draw \(BC = 5\text{ cm}\). With centre \(B\) and radius \(4\text{ cm}\), and centre \(C\) and radius \(6\text{ cm}\), draw arcs meeting at \(A\). Join \(AB\) and \(AC\).
Construct the perpendicular bisectors of \(AB\) and \(BC\); let them meet at \(O\). With centre \(O\) and radius \(OA = OB = OC\), draw the circumcircle.
Measuring the radius \(OA\) from the drawing gives approximately \(3.0\text{ cm}\). (Check: \(R = \dfrac{abc}{4\Delta} = \dfrac{4 \times 5 \times 6}{4 \times 9.92} \approx 3.0\text{ cm}\).)
Find the ratio of the areas of \(\triangle ADE\) and \(\triangle ABC\), then subtract.
\(AD : DB = 2 : 3\) means \(AD : AB = 2 : 5\). Since \(DE \parallel BC\), \(\triangle ADE \sim \triangle ABC\), so \(\dfrac{\text{area}(ADE)}{\text{area}(ABC)} = \left(\dfrac{2}{5}\right)^2 = \dfrac{4}{25}\).
\(\text{area}(ABC) = 16 \times \dfrac{25}{4} = 100\text{ cm}^2\).
\(\text{area}(DBCE) = 100 - 16 = 84\text{ cm}^2\).
A tangent is perpendicular to the radius at the point of contact, so \(\angle OAP = \angle OBP = 90^\circ\). In quadrilateral \(OAPB\), the angles sum to \(360^\circ\): \(\angle APB + \angle AOB = 360^\circ - 90^\circ - 90^\circ = 180^\circ\).
Given \(\angle APB = 65^\circ\), then \(\angle AOB = 180^\circ - 65^\circ = 115^\circ\).
In \(\triangle OAB\), \(OA = OB\) (radii), so \(\angle OAB = \dfrac{180^\circ - 115^\circ}{2} = 32.5^\circ\).