CBSE Class 10 Mathematics Circles Assignment Set 13

Exercise-1: Subjective Questions

Teacher's Note:
1. Use AB + CD = BC + DA.
2. Solve for the unknown side.

Teacher's Note:
1. The centre lies on the bisector of the angle between the tangents.
2. Use the angle sum of the quadrilateral OATB.

Question A-3. Find the length of the tangent drawn from a point 8 cm away from the centre of a circle of radius 6 cm.
Answer:
The radius to the point of contact is perpendicular to the tangent: \( \text{tangent}^2 = 8^2 - 6^2 = 28 \).
Length of the tangent = \( \sqrt{28} = 2\sqrt{7} \) cm.
In simple words: The tangent is a leg of a right triangle whose hypotenuse is the distance to the centre.

Teacher's Note:
1. Draw the radius to the point of contact.
2. Use Pythagoras theorem.

Question A-5. ABCD is a quadrilateral such that ∠D = 90°. A circle C(O, r) touches the sides AB, BC, CD and DA at P, Q, R and S respectively. If BC = 38 cm, CD = 25 cm and BP = 27 cm, find r.
Answer:
BQ = BP = 27 cm (tangents from B), so CQ = 38 − 27 = 11 cm and CR = CQ = 11 cm. Then DR = 25 − 11 = 14 cm.
∠D = 90° and OR ⊥ CD, OS ⊥ DA, OR = OS, so ORDS is a square. Hence r = DR = 14 cm.
In simple words: In the right-angled corner D, the radius equals the tangent length DR.

Teacher's Note:
1. Find the tangent lengths from C and D.
2. ORDS is a square.

Teacher's Note:
1. Find the hypotenuse first.
2. Use equal tangent lengths or area = r × s.

Question A-7. From an external point P, two tangents PA and PB are drawn to a circle with centre O. Prove that OP is the perpendicular bisector of AB.
Answer:
PA = PB (tangents from an external point) and OA = OB (radii).
So both P and O are equidistant from A and B, hence both lie on the perpendicular bisector of AB. The line OP is therefore the perpendicular bisector of AB.
(Alternatively, △OPA ≅ △OPB gives ∠APM = ∠BPM and then △APM ≅ △BPM gives AM = BM and ∠AMP = 90°.)
In simple words: Two points each equidistant from A and B determine the perpendicular bisector.

Teacher's Note:
1. Use equal tangents and equal radii.
2. A point equidistant from two points lies on their perpendicular bisector.

Question A-9. A circle touches the sides of a quadrilateral ABCD at P, Q, R and S respectively. Show that the angles subtended at the centre by a pair of opposite sides are supplementary.
Answer:
Let O be the centre. The tangents from each vertex subtend equal angles at O: ∠AOP = ∠AOS = 1, ∠BOP = ∠BOQ = 2, ∠COQ = ∠COR = 3, ∠DOR = ∠DOS = 4 (marking each by a number).
The angles at O add up to 360°: 2(1 + 2 + 3 + 4) = 360°, so 1 + 2 + 3 + 4 = 180°.
∠AOB = 1 + 2 and ∠COD = 3 + 4, so ∠AOB + ∠COD = 180°. Similarly ∠BOC + ∠DOA = 180°. Hence opposite sides subtend supplementary angles at the centre.
In simple words: The eight half-angles around the centre fall into four equal pairs.

Teacher's Note:
1. Use the result that the two tangents from a point subtend equal angles at the centre.
2. The full angle at O is 360°.

Teacher's Note:
1. Find the angle OPA using sine.
2. An isosceles triangle with a 60° angle is equilateral.

Question A-11. The radius of the incircle of a triangle is 4 cm and the segments into which one side is divided by the point of contact are 6 cm and 8 cm. Determine the other two sides of the triangle.
Answer:
Let the incircle touch BC at D with BD = 6 cm and DC = 8 cm, and let the tangent length from A be x cm. Then AB = 6 + x, AC = 8 + x, BC = 14, and s = 14 + x.
Area \( = r \times s = 4(14 + x) \) and, by Heron's formula, \( \text{Area}^2 = s(s - 14)(s - 6 - x)(s - 8 - x) = (14 + x) \cdot x \cdot 8 \cdot 6 \).
\( 16(14 + x)^2 = 48x(14 + x) \), so \( 16(14 + x) = 48x \), that is 14 + x = 3x and x = 7.
The other two sides are 6 + 7 = 13 cm and 8 + 7 = 15 cm. (Check: s = 21, area = \( \sqrt{21 \times 7 \times 8 \times 6} = 84 = 4 \times 21 \).)
In simple words: Equal tangent lengths give the sides in terms of x, and area = r × s gives the equation.

Teacher's Note:
1. Use area = inradius × semi-perimeter.
2. Compare it with Heron's formula.

Teacher's Note:
1. The tangent is perpendicular to the radius.
2. A perpendicular from the centre bisects a chord.

Teacher's Note:
1. Use equal tangents from C and from B.
2. B lies between C and Q.

Exercise-1: Objective Questions

Objective A-1. The length of the tangent drawn from a point 8 cm away from the centre of a circle of radius 6 cm is
(A) \( \sqrt{7} \) cm   (B) \( 2\sqrt{7} \) cm   (C) 10 cm   (D) 5 cm
Answer:
Ans. (B) \( 2\sqrt{7} \) cm
Tangent length = \( \sqrt{8^2 - 6^2} = \sqrt{28} = 2\sqrt{7} \) cm.
In simple words: The right angle is at the point of contact. | Use Pythagoras theorem.

Teacher's Note:
1. The tangent, the radius and the line to the centre form a right triangle.

Objective A-2. A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at a point Q, so that OQ = 12 cm. The length of PQ is:
(A) 12 cm   (B) 13 cm   (C) 8.5 cm   (D) \( \sqrt{119} \) cm
Answer:
Ans. (D) \( \sqrt{119} \) cm
OP ⊥ PQ, so \( PQ = \sqrt{12^2 - 5^2} = \sqrt{119} \) cm.
In simple words: Use the right angle at P. | Square root of 144 − 25.

Teacher's Note:
1. OQ is the hypotenuse of the right triangle OPQ.

Objective A-3. If tangents PA and PB from a point P to a circle with centre O are inclined to each other at an angle of 80°, then ∠POA is equal to:
(A) 50°   (B) 60°   (C) 70°   (D) 80°
Answer:
Ans. (A) 50°
∠AOB = 180° − 80° = 100°, and OP bisects it, so ∠POA = 50°.
In simple words: The angle at the centre is the supplement of the angle between the tangents. | OP bisects it.

Teacher's Note:
1. The centre, the contact points and P form a kite with angle sum 360°.

Objective A-4. PQ is a tangent to a circle with centre O at the point P. If △OPQ is an isosceles triangle, then ∠OQP is equal to:
(A) 30°   (B) 45°   (C) 60°   (D) 90°
Answer:
Ans. (B) 45°
∠OPQ = 90°, so the equal sides are the two legs OP = PQ.
So ∠OQP = ∠POQ = 45°.
In simple words: The tangent meets the radius at 90°. | The two equal sides must be the legs.

Teacher's Note:
1. A right triangle that is isosceles has two angles of 45°.

Objective A-5. From a point P which is at a distance of 13 cm from the centre O of a circle of radius 5 cm, the pair of tangents PQ and PR to the circle are drawn. Then the area of the quadrilateral PQOR is:
(A) 60 cm²   (B) 65 cm²   (C) 30 cm²   (D) 32.5 cm²
Answer:
Ans. (A) 60 cm²
PQ = PR = \( \sqrt{13^2 - 5^2} \) = 12 cm.
Area of PQOR = 2 × area of △OQP = \( 2 \times \frac{1}{2} \times 5 \times 12 = 60 \) cm².
In simple words: Find the tangent length first. | Double the area of one right triangle.

Teacher's Note:
1. The quadrilateral is made of two congruent right triangles.

Teacher's Note:
1. Two opposite right angles make a quadrilateral cyclic.

Teacher's Note:
1. The tangent lengths from A, B and C are used to rebuild AB.

Teacher's Note:
1. The tangent-chord angle equals the angle in the segment on the opposite side of the chord.

Teacher's Note:
1. The triangle OAB is isosceles with an apex angle of 100°.