CBSE Class 10 Mathematics Circles MCQs Set 07

Answer: (a) \(62\frac{1}{2}^\circ\)

Explanation:
1. \(\angle ACB = 180^\circ - 55^\circ = 125^\circ\), since the radii and the tangents form a quadrilateral with two right angles.
2. \(\angle AQB = \frac{1}{2}\angle ACB = \frac{125^\circ}{2} = 62\frac{1}{2}^\circ\).

Answer: (b) 5

Explanation:
1. The radius is perpendicular to the tangent, so \(OP^2 + PQ^2 = OQ^2\).
2. \(OP^2 = 13^2 - 12^2 = 169 - 144 = 25\), so \(OP = 5\) cm.

Answer: (b) \(45^\circ\)

Explanation:
1. The radius is perpendicular to the tangent, so \(\angle OPQ = 90^\circ\).
2. In the isosceles triangle, the two equal angles are \(\angle OQP = \angle POQ\). So \(2\angle OQP + 90^\circ = 180^\circ\), giving \(\angle OQP = 45^\circ\).

Teacher's Note:
1. Figure and the labelling of \(R\) are reconstructed from the source. Please check against the printed book.

Answer: (b) \(6r\)

Explanation:
1. The side of a regular hexagon inscribed in a circle equals the radius \(r\).
2. Perimeter \(= 6 \times r = 6r\).

Answer: (c) 32 cm

Explanation:
1. \(OT \perp PT\), so \(OP^2 = OT^2 + PT^2 = 49 + 576 = 625\), giving \(OP = 25\) cm.
2. \(R\) is on the circle, so \(OR = 7\) cm and \(PR = OP + OR = 25 + 7 = 32\) cm.

Teacher's Note:
1. The figure is not fully readable in the extraction. The stem is taken from the source's working.

Answer: (b) 51 cm

Explanation:
1. The common chord \(AB\) is bisected at right angles by \(O_1O_2\). So \(AC = 12\) cm.
2. \(CO_1 = \sqrt{37^2 - 12^2} = \sqrt{1225} = 35\) cm and \(CO_2 = \sqrt{20^2 - 12^2} = \sqrt{256} = 16\) cm.
3. \(O_1O_2 = 35 + 16 = 51\) cm.

Answer: (b) \(70^\circ\)

Explanation:
1. \(OP \perp TP\) and \(OQ \perp TQ\), so the two right angles are at \(P\) and \(Q\).
2. In the quadrilateral \(OPTQ\), \(\angle PTQ = 360^\circ - 90^\circ - 90^\circ - 110^\circ = 70^\circ\).

Answer: (a) 3 cm

Explanation:
1. For a tangential quadrilateral, \(AB + CD = BC + AD\).
2. \(6 + 4 = 7 + AD\), so \(AD = 3\) cm.

Answer: (b) 12 cm

Explanation:
1. The chord is bisected by the smaller circle at the point of contact \(P\), and \(OP \perp AB\).
2. \(AP = \sqrt{10^2 - 8^2} = 6\) cm, so the chord \(AB = 2AP = 12\) cm.

Answer: (a) \(25^\circ\)

Explanation:
1. \(\angle OAP = 90^\circ\) since the tangent is perpendicular to the radius. \(\angle AOP = 180^\circ - 115^\circ = 65^\circ\).
2. \(\angle APO = 180^\circ - 90^\circ - 65^\circ = 25^\circ\).

Answer: (c) 28 cm

Explanation:
1. \(PA = PB = 14\) cm.
2. Tangents from \(C\) and \(D\) are equal, so the perimeter of \(\triangle PCD = PC + CD + DP = PA + PB = 28\) cm.

Answer: (d) \(55^\circ\)

Explanation:
1. In a quadrilateral circumscribing a circle, opposite angles at the centre are supplementary.
2. \(\angle COD = 180^\circ - 125^\circ = 55^\circ\).

Teacher's Note:
1. The figure is reconstructed from the source's working.

Answer: (c) \(2\sqrt{3}\) cm

Explanation:
1. \(\angle OAT = 90^\circ\), so in \(\triangle OAT\), \(\cos 30^\circ = \frac{AT}{OT}\).
2. \(AT = 4 \times \frac{\sqrt{3}}{2} = 2\sqrt{3}\) cm.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. Both statements are true. The tangents at the ends of a diameter are both perpendicular to the same diameter, so they are parallel.
2. The reason is true but does not explain the assertion.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. \(\sqrt{5^2 - 3^2} = 4\) cm, so the assertion is true.
2. The Pythagoras relation used for the tangent length is the reason, so it explains the assertion.

Teacher's Note:
1. The source's key says option (b) for this item. Pythagoras is the method used for the tangent length, so option (a) is shown here. Please check against the printed book.