CBSE Class 9 Mathematics Circles MCQs Set 06

Answer: (a) falls inside the triangle

Explanation:
1. The circumcentre is the meeting point of the perpendicular bisectors of the sides.
2. When every angle is acute these bisectors meet inside the triangle.

Teacher's Note:
1. Right triangle: circumcentre on the hypotenuse. Obtuse: outside.
2. The circumradius is the distance from the circumcentre to any vertex.

Answer: (a) only if both are diameters

Explanation:
1. If chords AB and CD bisect each other at P, then the line from the centre to P is perpendicular to both chords (a line from the centre to a midpoint of a chord is perpendicular to it).
2. Two different chords cannot both be perpendicular to the same line through P unless that line is trivial, so P must be the centre; hence both are diameters.

Teacher's Note:
1. Only diameters pass through the centre, the only point that can be the midpoint of two chords.
2. Two diameters always bisect each other.

Answer: (c) less than a right angle

Explanation:
1. The angle in a semicircle is 90°. The angle in a major segment is half of the angle at the centre subtended by the minor arc, which is less than 180°.
2. So the angle is less than 90°, an acute angle.

Teacher's Note:
1. Major segment: acute. Semicircle: right. Minor segment: obtuse.
2. The angle in a segment equals half of the central angle on the opposite arc.

Answer: (a) It is symmetrical about every diameter

Explanation:
1. Reflecting a circle in any line through its centre maps every point to a point at the same distance from the centre.
2. So every diameter is a line of symmetry.

Teacher's Note:
1. A circle has infinitely many lines of symmetry.
2. Each diameter divides it into two congruent semicircles.

Answer: (a) bisects angle BAC

Explanation:
1. Equal chords are equidistant from the centre, so O is equidistant from AB and AC and lies on the bisector of angle BAC.
2. Alternatively triangles OAB and OAC are congruent by SSS, giving angle OAB = angle OAC.

Teacher's Note:
1. Use SSS: OA is common, OB = OC and AB = AC.
2. AO also bisects BC at right angles.

Answer: (a) \( \dfrac{27\pi}{4} \) cm²

Explanation:
1. Radius R = 3 cm and the centre is at distance R - 1.5 = 1.5 cm from the surface plane.
2. Radius of the section: \( r^2 = 3^2 - 1.5^2 = 6.75 \), so the area is \( \pi \times 6.75 = \dfrac{27\pi}{4} \) cm².

Teacher's Note:
1. Use \( r^2 = R^2 - d^2 \) for a plane section of a sphere.
2. The section is a circle.

Answer: (b) TA · TB

Explanation:
1. The tangent-secant theorem states that the square of the tangent length equals the product of the whole secant and its external part.
2. So \( TP^2 = TA \cdot TB \).

Teacher's Note:
1. Check using similar triangles TPA and TBP.
2. It is the limiting case of the intersecting secants result.

Answer: (c) supplementary

Explanation:
1. The two opposite angles stand on arcs that together make the whole circle, so they sum to 180°.

Teacher's Note:
1. The exterior angle therefore equals the interior opposite angle.
2. The converse also holds.

Answer: (c) AB / \( \sqrt{2} \)

Explanation:
1. Let the radius be r, so AB = 2r. Triangle AOC is right-angled at O with OA = OC = r, so \( AC = r\sqrt{2} \).
2. Hence \( AC = \dfrac{2r}{\sqrt{2}} = \dfrac{AB}{\sqrt{2}} \).

Teacher's Note:
1. AC is the side of the inscribed square.
2. Use Pythagoras in triangle AOC.

Answer: (a) 7 cm

Explanation:
1. Joining the centre to the vertices divides a regular hexagon into six equilateral triangles.
2. So the circumradius equals the side, 7 cm.

Teacher's Note:
1. Each central angle is 60° and the two radii are equal, giving equilateral triangles.
2. Its perimeter is 6 times the radius.

Answer: (a) outside the triangle

Explanation:
1. The altitudes of an obtuse-angled triangle meet only when produced, so their meeting point lies outside the triangle.

Teacher's Note:
1. Acute: inside. Right: at the right-angle vertex.
2. Draw the altitudes to the longest side.

Answer: (b) incentre

Explanation:
1. A point on an angle bisector is equidistant from the arms of that angle, so the meeting point of the angle bisectors is equidistant from all three sides.

Teacher's Note:
1. That point is the centre of the incircle.
2. The circumcentre is equidistant from the vertices instead.

Answer: (b) coincident

Explanation:
1. In an equilateral triangle the median, angle bisector, altitude and perpendicular bisector from each vertex coincide.
2. So all the centres are the same point.

Teacher's Note:
1. The orthocentre also coincides with them.
2. The common point is at 2/3 of the median from each vertex.

Answer: (b) the interior opposite angle

Explanation:
1. The interior angle adjacent to the exterior angle and the interior opposite angle are supplementary.
2. The exterior angle and its adjacent interior angle are also supplementary, so the exterior angle equals the interior opposite angle.

Teacher's Note:
1. Opposite angles of a cyclic quadrilateral are supplementary.
2. The exterior angle of a triangle, in contrast, equals the sum of two angles.

Answer: (b) \( \sqrt{d_1 d_2} \)

Explanation:
1. With radii r₁, r₂ the centres are r₁ + r₂ apart. The tangent length is \( \sqrt{(r_1 + r_2)^2 - (r_1 - r_2)^2} = 2\sqrt{r_1 r_2} \).
2. Since \( d_1 = 2r_1 \) and \( d_2 = 2r_2 \), this equals \( \sqrt{d_1 d_2} \).

Teacher's Note:
1. Drop a perpendicular from the smaller centre to the larger radius.
2. The common tangent is the geometric mean of the diameters.

Answer: (b) 18 cm

Explanation:
1. \( AT^2 = AB \cdot AC \Rightarrow 144 = 6 \cdot AC \Rightarrow AC = 24 \) cm.
2. So BC = AC - AB = 24 - 6 = 18 cm.

Teacher's Note:
1. Use the tangent-secant theorem and subtract the external part.
2. Do not stop at AC = 24.

Answer: (b) touch each other internally

Explanation:
1. The difference of the radii is 10 - 8 = 2 cm, which equals the distance between the centres.
2. Circles whose centre distance equals the difference of the radii touch internally.

Teacher's Note:
1. External touching needs the distance to equal the sum of the radii.
2. Intersection needs the distance between the difference and the sum.

Answer: (c) parallel

Explanation:
1. Each tangent is perpendicular to the diameter at its end point.
2. Two lines perpendicular to the same line are parallel.

Teacher's Note:
1. The tangent is perpendicular to the radius at the point of contact.
2. Parallel tangents are at a distance equal to the diameter.

Answer: (a) outside the segment

Explanation:
1. A minor segment is the smaller part cut off by a chord, on the side away from the centre.
2. So the centre lies in the major segment, outside the minor segment.

Teacher's Note:
1. A semicircle is the case where the centre is on the chord.
2. A major segment contains the centre.

Answer: (c) 40°

Explanation:
1. Opposite angles of a cyclic quadrilateral are supplementary: angle ABC = 180° - 130° = 50°.
2. Angle ACB = 90° (angle in a semicircle), so angle BAC = 180° - 90° - 50° = 40°.

Teacher's Note:
1. Angle in a semicircle is 90°.
2. Use the angle sum of triangle ABC.