Circles

Correct option: 3 — QT = 3 units

Correct option: 2 — The sum of the length of BP and BQ is same as the perimeter of ΔABC.

Correct option: 3 — Rectangle

Correct option: 1 — 40°

Correct option: 2 — Only 2

Correct option: 2 — Both (A) and (R) are true and (R) is not the correct explanation for (A).

Correct option: 4 — (A) is true but (R) is false.

• Writes that the measure of ∠ABC is 90°. [0.5]

• States that AC is the diameter of the circle and the angle subtended by diameter on the circumference of the circle is 90°. [0.5]

• States that since diameter subtends right angle on the circumference of circle, ΔABC is a right angled triangle with right angle at point C. [0.5]

• States that ∠ABC + ∠CAB = 90°. Therefore, ∠CAB = 40°. [0.5]

• Finds the length of PQ as √(OP² − OQ²) = √((12 + 3)² − 9²) = √(15² − 9²) = √(225 − 81) = √144 = 12 units. [1]

• States that the angle subtended by a diameter of a circle on it's circumference is 90°. Hence, measure of ∠OCB is 45°. [0.5]

• Uses angle sum property of triangle in ΔBCO to find the measure of ∠BOC = 180° - 45° - 33° = 102°. [0.5]

• States that tangent of a circle is perpendicular to its radius. [1]

• Uses pythagoras theorem to find the length of QR as, QR = √(OR² − OQ²) = √(5² − 3²) = 4 cm. [0.5]

• Finds PR = 2 × QR = 2 × 4 = 8 cm [0.5]

• i) Draws a circle with centre O and two tangents AB and AC. Joins OA, OB and OC. [0.5]

• ii) Writes that, in ΔOAB and ΔOAC, OA is common, OB = OC (radii of the same circle), ∠OBA = ∠OCA = 90°. Concludes that, ΔOAB is congruent to ΔOAC by RHS congruence criteria. Hence, writes that line segment AO joining the external point and the center of the circle, divides the quadrilateral made by both the tangents and the radius of the circle into two equal parts. (Award full marks if any other correct method is used. For example, this can also be proved by showing that the heights of two triangles are the same.) [1.5]

• States that in a circle angle made between radius and tangent at the point of contact of tangent is 90°. [0.5]

• Joins line segment OB as shown in image below. [0.5]

• Finds measure of ∠AOB = 180° - 90° - ∠ABO = 90° - ∠ABO. Similarly, ∠COB = 180° - 90° - ∠CBO = 90° - ∠CBO. [0.5]

• Finds ∠AOC = ∠AOB + ∠COB = 180° - ∠ABO - ∠CBO = 180° - ∠ABC = 180° - 43° = 137° (Award full marks if any other correct method is used.) [1] [sic — the official PDF's own rubric marks sum to 2.5 against the printed [2] question header]

• Writes that in ΔRPS: ∠PRS = 30° (alternate interior ∠'s since RS||PQ); ∠SPR = 90° (Angle subtended in semi-circle is of 90°); ∠PSR = (180° - 90° - 30°) = 60° (Angle sum property of triangle.) [1.5]

• Writes that ∠PSR = ∠PTR (angles in the same segment on chord PR). Hence, measure of ∠PTR = 60°. [0.5]

• Writes that tangents from external points are equal in length. Hence, PQ = PR = 4 units. [1]

• States that since PQ and PR are tangents to the circle, ∠PRO and ∠PQO are right angles. Hence, ΔPQO and ΔPRO are right angled triangle. [0.5]

• Finds area of ΔPQO = ½ × OQ × PQ = ½ × 3 × 4 = 6 sq units. Similarly, area of ΔPRO = 6 sq units. [1]

• Finds the area of quadrilateral PQOR = area of ΔPQO + area of ΔPRO = 12 sq units. [0.5]

• Draws OA. [0.5]

• Writes that the tangent to a circle is perpendicular to the radius of the circle at the point of contact and hence ∠OAP = 90°. [0.5]

• Uses the exterior angle property of triangles in ΔOAP as: ∠AOB = ∠OAP + ∠APO => ∠AOB = 90° + 43° = 133°. [0.5]

• Writes that ΔOAB is an isoceles triangle as OA = OB and hence ∠OAB = ∠OBA. [0.5]

• Uses the angle sum property of a triangle in ΔOAB and gets: ∠OAB = ∠OBA = (180-133)/2 = 23.5°. [0.5]

• Finds the measure of ∠PAB as 90° + 23.5° = 113.5°. (Award full marks if any other correct method is used.) [0.5]

• Joins OA and OC. Mentions that in ΔOBA and ΔOBC: i) OA = OC (radii of the circle) ii) AB = CB (tangents from an external point) iii) ∠OAB = ∠OCB = 90° (AB and BC are tangents) [1]

• Uses the above step to conclude that ΔOBA ≅ ΔOBC by RHS congruency. [0.5]

• Writes that ∠OBA = ∠OBC since the corresponding angles of congruent triangles are equal. [0.5]

• In ABCO, finds ∠AOC as 360° - (90° + 90° + 70°) = 110°. [0.5]

• Writes that angle made by the chord AC at the circumference is half of the angle made at the centre and hence finds the measure of ∠CDA as 110°/2 = 55°. [0.5]

• States that in ΔQRS, ∠RQS = ∠RSQ = 37° giving reason that angles on the equal sides of a triangle are equal in measure. [1]

• States that angles in the same segment of a circle are equal. Hence, ∠RQS = ∠RPS = 37°. [1]

• Finds that in ΔPRS, ∠PRS = (180 − 37)/2 = 71.5° since ∠PRS = ∠RSP as they are angles on the equal sides of the triangle. [1]

• Joins OP so that it intersects MN at point Q. Then, ΔMPN is isosceles and PO is the angle bisector of ∠MPN. So, OP ⊥ MN and therefore, OP bisects MN which gives NQ = QM = 12 cm. Also, OQ = √(ON² − NQ²) = √(13² − 12²) cm = 5 cm. [2]

• States that since OP ⊥ MN, ΔPQN is right angled triangle with ∠PQN = 90°. Now, ∠PNQ + ∠QNO = 90° = ∠PNQ + ∠NPQ. So, ∠QNO = ∠NPQ [1]

• Therefore, By AA similarity, ΔPQN ∼ ΔNQO. Hence, PN/NO = QN/QO. Substitutes the value of NO, QN and QO, and finds PN/13 = 12/5 => PN = 156/5 cm [1]

• States that, since, PM and PN are tangents to the circle from an external point, PM = PN. Hence, PM + PN = 2 × 156/5 cm = 312/5 cm (Provide full marks for any other correct methods used.) [1]

• Writes that tangents from an external point to a circle are equal in length. Hence finds GF = GH = 1.7 cm and CD = BC = 2.3 cm. [1]

• Finds AH as 4 - 1.7 = 2.3 cm and AB as 6 - 2.3 = 3.7 cm. [1]

• Writes that the perpendicular from the centre to a chord bisects the chord. Thus finds BD = HF = 2 × 1.3 = 2.6 cm. [1]

• Writes that distinct pair of adjacent sides in a kite are equal and states with reference from question that AG = EG and AC = EC. Finds AG = EG = 4 cm and AC = EC = 6 cm. Thus finds FE = 4 - 1.7 = 2.3 cm and ED = 6 - 2.3 = 3.7 cm. [1]

• Finds the perimeter of the hexagon ABDEFH as: 3.7 + 2.6 + 3.7 + 2.3 + 2.6 + 2.3 = 17.2 cm. Concludes that Sahid will need 17.2 cm of red coloured thread. [1]

• States that the tangent to a circle is perpendicular to the radius of the circle, therefore, ΔAOF is right angled triangle. [0.5]

• By using Pythagoras theorem, AO² = OF² + AF², Finds AO as √13 m. Finds AD as: AD = AO + OD = (√13 + 2) m [1]

• Finds the total cost of laying wire correct to nearest hundreds as (√13 + 2) × 500 = 500√13 + 1000 = Rs 2800. [0.5]

• States that the length of tangents drawn from an external point to the circle are same and concludes BF = BD = 7 m, AE = AF = 3 m and CD = CE = 4 m. [1]

• Finds perimeter of ΔABC as 28 m. Calculates area to be painted as 2 × 28 m² = 56 m². Concludes that 56 × ¼ = 14 liters of paint is required to cover the entire internal surface area of the wall and gates. [1]

• States the reason that angle between tangent and radius to the circle at the point of contact of tangent to the circle is 90°. Hence, concludes that the person must turn by a measure of 90°. [1]