CBSE Class 10 Mathematics Circles MCQs Set 17

Answer: (b) \( 2\sqrt{7} \) cm

Explanation:
1. The radius is perpendicular to the tangent at the point of contact, so the centre, the point of contact and the external point form a right triangle.
2. Tangent length = \( \sqrt{8^2 - 6^2} = \sqrt{28} \).
3. \( \sqrt{28} = 2\sqrt{7} \) cm.

Teacher's Note:
1. Tangent length from an external point at distance d from the centre is \( \sqrt{d^2 - r^2} \).
2. Always draw the radius to the point of contact first.

Answer: (d) \( \sqrt{119} \) cm

Explanation:
1. OP is the radius and OP is perpendicular to PQ, so triangle OPQ is right angled at P.
2. \( PQ^2 = OQ^2 - OP^2 = 144 - 25 = 119 \).
3. So \( PQ = \sqrt{119} \) cm.

Teacher's Note:
1. The line OQ is the hypotenuse because the right angle is at the point of contact P.
2. Do not confuse the hypotenuse with a leg: 13 cm is a tempting wrong answer.

Answer: (a) 50°

Explanation:
1. In quadrilateral OAPB, ∠OAP = ∠OBP = 90° and ∠APB = 80°.
2. ∠AOB = 360° − 90° − 90° − 80° = 100°.
3. OP bisects ∠AOB, so ∠POA = 100° ÷ 2 = 50°.

Teacher's Note:
1. The angle between two tangents and the angle between the radii at their points of contact are supplementary.
2. OP bisects both ∠AOB and ∠APB.

Answer: (b) 45°

Explanation:
1. Since PQ is a tangent, ∠OPQ = 90°.
2. An isosceles triangle cannot have two right angles, so the equal angles are the other two: ∠POQ = ∠OQP.
3. Each of them is (180° − 90°) ÷ 2 = 45°.

Teacher's Note:
1. The tangent-radius angle of 90° fixes the right angle of the triangle.
2. In an isosceles right triangle the two acute angles are 45° each.

Answer: (a) 60 cm²

Explanation:
1. Tangent length PQ = \( \sqrt{13^2 - 5^2} = 12 \) cm.
2. Triangle OQP is right angled at Q, so its area is \( \frac{1}{2} \times 5 \times 12 = 30 \) cm².
3. Quadrilateral PQOR is made of two such congruent triangles, so area = 2 × 30 = 60 cm².

Teacher's Note:
1. Tangents from an external point are equal, so triangles OQP and ORP are congruent.
2. Do not forget to double the area of one triangle.

Answer: (c) cyclic quadrilateral

Explanation:
1. ∠OAP = 90° and ∠OBP = 90° because a tangent is perpendicular to the radius at the point of contact.
2. So ∠OAP + ∠OBP = 180°, which means opposite angles of OAPB are supplementary.
3. A quadrilateral with supplementary opposite angles is cyclic.

Teacher's Note:
1. OAPB is a kite, not necessarily a rhombus or a square.
2. It becomes a square only if ∠APB is 90°.

Answer: (c) 40°

Explanation:
1. ∠AOB = 180° − 80° = 100° (angles of quadrilateral OAPB).
2. Triangle OAB is isosceles with OA = OB, so ∠OAB = (180° − 100°) ÷ 2 = 40°.

Teacher's Note:
1. Another way: ∠OAB = ∠OAP − ∠BAP = 90° − 50° = 40°.
2. Triangle PAB is isosceles with PA = PB, so ∠PAB = 50°.

Answer: (b) 14π

Explanation:
1. The sides of triangle ABC are AB = 2 + 1 = 3, BC = 1 + 3 = 4, CA = 3 + 2 = 5, so the radii satisfy x + y = 3, y + z = 4, x + z = 5.
2. Adding: x + y + z = 6, so x = 2, y = 1, z = 3.
3. Total area = π(4 + 1 + 9) = 14π.

Teacher's Note:
1. Circles touching externally have centre distance equal to the sum of their radii.
2. Three equations in three unknowns give each radius.

Answer: (d) 90°

Explanation:
1. Draw the common tangent at C meeting AB at M.
2. Tangents from M give MA = MC and MB = MC, so MA = MB = MC.
3. M is then the centre of the circle on AB as diameter, so ∠ACB = 90° (angle in a semicircle).

Teacher's Note:
1. The midpoint of AB is equidistant from A, B and C.
2. This standard result is worth remembering.

Answer: (c) 8 cm

Explanation:
1. Tangents from D to the first circle: DA = DC = 4 cm.
2. Tangents from D to the second circle: DB = DC = 4 cm.
3. AB = DA + DB = 8 cm.

Teacher's Note:
1. D is the midpoint of the common tangent AB.
2. Use the equal tangent lengths from one external point to each circle separately.

Answer: (a) 8.64

Explanation:
1. The triangle is right angled, so its hypotenuse is a diameter: 6 units.
2. The sides are 3k, 4k, 5k with 5k = 6, so k = 1.2 and the legs are 3.6 and 4.8.
3. Area = ½ × 3.6 × 4.8 = 8.64.

Teacher's Note:
1. A right triangle inscribed in a circle has its hypotenuse as the diameter.
2. Scale the 3, 4, 5 sides by the factor k.

Answer: (d) 8 cm

Explanation:
1. The chord CD is 8 cm from A, so it is 8 − 5 = 3 cm from the centre O.
2. Half the chord = \( \sqrt{5^2 - 3^2} = 4 \) cm.
3. CD = 2 × 4 = 8 cm.

Teacher's Note:
1. A chord parallel to the tangent at A is perpendicular to the diameter AB.
2. The perpendicular from the centre bisects the chord.

Answer: (c) 15 cm

Explanation:
1. Let the third tangent length from the opposite vertex be x, so the sides are 14, 6 + x and 8 + x and the semi-perimeter is s = 14 + x.
2. Area = rs = 4(14 + x). Heron: \( \sqrt{s(s-14)(s-6-x)(s-8-x)} = \sqrt{(14+x)(x)(8)(6)} \).
3. Equating: \( 16(14+x)^2 = 48x(14+x) \), so 14 + x = 3x and x = 7.
4. The sides are 13 and 15; the longer is 15 cm.

Teacher's Note:
1. The tangent lengths from each vertex are equal in pairs.
2. Compare the area computed as rs with the area from Heron's formula.

Answer: (c) 13 cm

Explanation:
1. Distances of the chords from the centre are \( \sqrt{r^2 - 25} \) and \( \sqrt{r^2 - 144} \).
2. They add up to 17: try r = 13, giving 12 + 5 = 17.
3. So the radius is 13 cm.

Teacher's Note:
1. The perpendicular from the centre bisects a chord.
2. Using the 5, 12, 13 triple avoids solving a square-root equation.

Answer: (b) \( r\sqrt{3} \)

Explanation:
1. The distance between the centres is r, and the common chord is the perpendicular bisector of the line joining the centres.
2. Half the chord = \( \sqrt{r^2 - (r/2)^2} = \frac{\sqrt{3}}{2} r \).
3. The full chord = \( r\sqrt{3} \).

Teacher's Note:
1. The common chord is perpendicular to the line of centres and cuts it at its midpoint.
2. Pythagoras in the half-triangle gives the half-chord.

Answer: (c) diameter

Explanation:
1. A chord through the centre has length equal to twice the radius.
2. Any other chord is shorter than this, so the longest chord is the diameter.

Teacher's Note:
1. The diameter is the largest chord of a circle.
2. The radius is only half of it and is not a chord.

Answer: (c) 40 cm

Explanation:
1. Tangents from A: AP = AR = 7 cm.
2. Tangents from B and C: BP = BQ and CQ = CR, so BQ + CQ = BC = 13 cm gives BP + CR = 13 cm.
3. Perimeter = 2(AP + BC) = 2(7 + 13) = 40 cm.

Teacher's Note:
1. The perimeter equals twice the sum of the three tangent lengths.
2. Here the tangent lengths are 7, and the other two add to 13.

Answer: (c) 14 cm

Explanation:
1. BQ = BP = 27 cm, so CQ = 38 − 27 = 11 cm and CR = CQ = 11 cm.
2. DR = CD − CR = 25 − 11 = 14 cm.
3. Since ∠D = 90° and the circle touches DA and DC, the quadrilateral formed by D, the two contact points and the centre is a square, so r = DR = 14 cm.

Teacher's Note:
1. Tangents from one external point are equal.
2. A circle touching the arms of a right angle has radius equal to the tangent length from the corner.

Answer: (b) 2 cm

Explanation:
1. Hypotenuse PR = \( \sqrt{144 + 25} = 13 \) cm.
2. For a right triangle, r = (sum of legs − hypotenuse) ÷ 2 = (12 + 5 − 13) ÷ 2 = 2 cm.
3. Check with area: ½ × 12 × 5 = 30 = r × s = r × 15, so r = 2.

Teacher's Note:
1. The incircle radius is also area divided by semi-perimeter.
2. Both methods must agree.

Answer: (c) 24 cm

Explanation:
1. AP = AQ = \( \sqrt{13^2 - 5^2} = 12 \) cm.
2. BP = BR and CQ = CR (tangents from B and C).
3. Perimeter = AB + BR + RC + CA = AB + BP + CQ + CA = AP + AQ = 24 cm.

Teacher's Note:
1. Replace each side's tangent piece by its equal on the other tangent.
2. The perimeter of the triangle cut off by a third tangent is constant.