CBSE Class 10 Mathematics Circles MCQs Set 11

Answer: (d) 90°

Explanation:
1. 1. NS is a tangent and AS is the radius through the point of contact S.
2. 2. A tangent is perpendicular to the radius at the point of contact, so \( \angle NSA = 90^\circ \).

Teacher's Note:
1. Tangent and radius meet at 90 degrees at the point of contact.
2. Always join the centre to the point of contact first.

Answer: (b) 12 m

Explanation:
1. 1. In right \( \triangle NSA \) (right angle at S): \( NS^2 = NA^2 - AS^2 = 13^2 - 5^2 = 144 \).
2. 2. \( NS = 12 \) m.

Teacher's Note:
1. Use Pythagoras theorem in the right triangle formed by tangent, radius and the line to the centre.
2. 5, 12, 13 is a Pythagorean triplet.

Answer: (c) 12 m

Explanation:
1. 1. Tangents drawn from an external point to a circle are equal in length.
2. 2. \( NR = NS = 12 \) m.

Teacher's Note:
1. Two tangents from one external point are always equal.
2. No new calculation is needed once NS is known.

Answer: (d) \( \angle RAS = 180^\circ - \theta \)

Explanation:
1. 1. In quadrilateral NSAR, \( \angle NSA = \angle NRA = 90^\circ \).
2. 2. Angle sum is \( 360^\circ \), so \( \theta + \angle RAS + 90^\circ + 90^\circ = 360^\circ \).
3. 3. Hence \( \angle RAS = 180^\circ - \theta \).

Teacher's Note:
1. The angle between two tangents and the angle at the centre are supplementary.
2. Use the angle sum of the quadrilateral.

Answer: (a) \( 90^\circ - \dfrac{\theta}{2} \)

Explanation:
1. 1. AN bisects \( \angle SNR \) and \( \angle SAR \) (the two right triangles are congruent), so \( \angle ANS = \dfrac{\theta}{2} \).
2. 2. In right \( \triangle NSA \): \( \angle NAS = 90^\circ - \dfrac{\theta}{2} \).

Teacher's Note:
1. The line from the external point to the centre bisects the angle between the tangents.
2. Then use the angle sum in the right triangle.

Answer: (b) 8 cm

Explanation:
1. 1. Let \( AD = AF = x \), \( BD = BE = y \), \( CE = CF = z \) (equal tangents).
2. 2. \( x+y = 15 \), \( y+z = 13 \), \( z+x = 14 \), so \( x+y+z = 21 \).
3. 3. \( AD = x = 21 - 13 = 8 \) cm.

Teacher's Note:
1. Name the three tangent lengths x, y, z and add the three sides.
2. Semi-perimeter minus the opposite side gives the tangent length from a vertex.

Answer: (b) 7 cm

Explanation:
1. 1. \( BE = BD = y = 21 - 14 = 7 \) cm.

Teacher's Note:
1. Tangent length from B equals s - CA.
2. Check: AD + BD = 8 + 7 = 15 = AB.

Answer: (c) 6 cm

Explanation:
1. 1. \( CF = CE = z = 21 - 15 = 6 \) cm.

Teacher's Note:
1. Tangent length from C equals s - AB.
2. Check: BE + CE = 7 + 6 = 13 = BC.

Answer: (c) 30 sq cm

Explanation:
1. 1. OD is perpendicular to AB, so OD = 4 cm is the height of \( \triangle OAB \).
2. 2. Area \( = \dfrac{1}{2} \times 15 \times 4 = 30 \) sq cm.

Teacher's Note:
1. The radius to the point of contact is the height on that side.
2. Use half base times height.

Answer: (c) 84 sq cm

Explanation:
1. 1. Area of \( \triangle ABC \) = area \( \triangle OAB + \triangle OBC + \triangle OCA \).
2. 2. \( = \dfrac{1}{2} \times 4 \times (15+13+14) = 2 \times 42 = 84 \) sq cm.
3. 3. Heron's formula check: \( \sqrt{21 \times 8 \times 7 \times 6} = 84 \).

Teacher's Note:
1. Split the triangle into three triangles with height equal to the inradius.
2. Area = r x s, where s is the semi-perimeter.