CBSE Class 9 Mathematics Circles MCQs Set 05

Answer: (d) \( 8\sqrt3 \) cm

Explanation:
1. Radius 8 cm; the chord is 4 cm from the centre, so half the chord is \( \sqrt{8^2-4^2}=\sqrt{48}=4\sqrt3 \).
2. The chord is \( 2\times4\sqrt3=8\sqrt3 \) cm.

Teacher's Note:
1. The perpendicular from the centre bisects the chord.
2. 8 sqrt3 is about 13.86 cm.

Answer: (d) 4 cm

Explanation:
1. Radius 10 cm and half-chord 8 cm: the centre is \( \sqrt{10^2-8^2}=6 \) cm from the chord.
2. The height of the minor segment is \( 10-6=4 \) cm.

Teacher's Note:
1. The perpendicular from the centre bisects the chord.
2. The major segment would have height 16 cm.

Answer: (c) 130 m

Explanation:
1. Let the centre be \( x \) m from the longer path: \( x^2+56^2=(93-x)^2+25^2 \), giving \( x=33 \).
2. Radius \( =\sqrt{33^2+56^2}=65 \) m, so the diameter is 130 m.

Teacher's Note:
1. Half-chords are 56 m and 25 m.
2. Check the other path: 60^2 + 25^2 = 4225 = 65^2.

Answer: (a) 44.5 m

Explanation:
1. \( (r-25)^2+40^2=r^2 \) gives \( 50r=2225 \).
2. So \( r=44.5 \) m.

Teacher's Note:
1. Half the span is 40 m.
2. 44.5^2 = 19.5^2 + 40^2.

Answer: (d) 0.4 m

Explanation:
1. Distances from the centre: \( \sqrt{2.5^2-1.5^2}=2 \) m and \( \sqrt{2.5^2-0.7^2}=2.4 \) m.
2. On the same side, the distance between the chords is \( 2.4-2=0.4 \) m.

Teacher's Note:
1. Use half-chords 1.5 m and 0.7 m.
2. On opposite sides the distance would be 4.4 m.

Answer: (a) a circle concentric with the given circle, of radius 3 cm

Explanation:
1. Each midpoint is at distance \( \sqrt{5^2-4^2}=3 \) cm from the centre.
2. So the midpoints lie on the concentric circle of radius 3 cm.

Teacher's Note:
1. Equal chords are equidistant from the centre.
2. The perpendicular from the centre bisects the chord.

Answer: (b) 4 cm

Explanation:
1. The radius to the point of contact is perpendicular to the tangent.
2. Tangent length \( =\sqrt{5^2-3^2}=4 \) cm.

Teacher's Note:
1. Use the right triangle with sides radius, tangent and distance.
2. This is the 3-4-5 triangle.

Answer: (b) 19 cm

Explanation:
1. \( BD=\sqrt{13^2-8^2}=\sqrt{105} \). With the centre at the origin, B(-13, 0), A(13, 0), the point D is \( \left(-\dfrac{64}{13},\dfrac{8\sqrt{105}}{13}\right) \).
2. \( AD^2=\left(13+\dfrac{64}{13}\right)^2+\dfrac{6720}{169}=361 \), so \( AD=19 \) cm.

Teacher's Note:
1. Use coordinates for the tangent point.
2. Check with the cosine rule in triangle ABD.

Answer: (c) 7.74 cm

Explanation:
1. Distances from the centre: \( \sqrt{r^2-3.5^2} \) and \( \sqrt{r^2-7.5^2} \), differing by 5.
2. Solving \( \sqrt{r^2-12.25}-\sqrt{r^2-56.25}=5 \) gives \( r^2=59.86 \), so \( r\approx7.74 \) cm.

Teacher's Note:
1. Use half-chords 3.5 cm and 7.5 cm.
2. Check: sqrt(47.61) - sqrt(3.61) = 6.9 - 1.9 = 5.

Answer: (d) 9.6 cm

Explanation:
1. AO is the perpendicular bisector of BC; if it meets BC at M, then \( AB^2=AM\cdot AD \) where AD is the diameter, so \( AM=\dfrac{36}{10}=3.6 \).
2. \( BM=\sqrt{36-3.6^2}=4.8 \), so \( BC=9.6 \) cm.

Teacher's Note:
1. Triangle ABD is right-angled at B (angle in a semicircle).
2. BM is the altitude of that right triangle.

Answer: (a) \( \dfrac{27}{4}\pi \) sq. cm

Explanation:
1. The surface circle has radius \( \sqrt{3^2-(3-1.5)^2}=\sqrt{6.75} \) cm.
2. Area \( =\pi\times6.75=\dfrac{27}4\pi \) sq. cm.

Teacher's Note:
1. The cross-section is a chord circle of the sphere at distance 1.5 cm from the centre.
2. 6.75 = 27/4.

Answer: (a) \( \dfrac{AB}{\sqrt2} \)

Explanation:
1. With radius \( r \), triangle AOC is right-angled at O: \( AC=\sqrt{r^2+r^2}=r\sqrt2 \).
2. Since \( AB=2r \), \( AC=\dfrac{AB}{\sqrt2} \).

Teacher's Note:
1. The diameters meet at the centre at right angles.
2. AC is the side of the inscribed square.

Answer: (c) 7 cm

Explanation:
1. The centre and two adjacent vertices form an equilateral triangle (central angle \( 60^\circ \) with equal radii).
2. So the radius equals the side, 7 cm.

Teacher's Note:
1. Six equal triangles meet at the centre.
2. Each central angle is 360 / 6 = 60 degrees.

Answer: (c) the interior opposite angle

Explanation:
1. The interior angle at that vertex and its opposite angle are supplementary.
2. The exterior angle is also supplementary to the interior angle, so it equals the opposite interior angle.

Teacher's Note:
1. Linear pair at the vertex adds to 180 degrees.
2. Opposite angles add to 180 degrees.

Answer: (a) \( \sqrt{d_1d_2} \)

Explanation:
1. The centres are \( r_1+r_2 \) apart and the radii to the contact points are parallel, so the tangent segment is \( \sqrt{(r_1+r_2)^2-(r_1-r_2)^2}=2\sqrt{r_1r_2} \).
2. In diameters, \( 2\sqrt{r_1r_2}=\sqrt{4r_1r_2}=\sqrt{d_1d_2} \).

Teacher's Note:
1. Use the right triangle formed with the line of centres.
2. Two circles of diameters 4 and 9 give length 6.

Answer: (c) 18 cm

Explanation:
1. \( AT^2=AB\cdot AC \), so \( 144=6\times AC \) and \( AC=24 \) cm.
2. \( BC=AC-AB=18 \) cm.

Teacher's Note:
1. Tangent-secant theorem.
2. AC is the whole secant.

Answer: (c) parallel to each other

Explanation:
1. Each tangent is perpendicular to the radius at its point of contact.
2. Both are perpendicular to the same diameter, so they are parallel.

Teacher's Note:
1. Tangent is perpendicular to the radius.
2. Two lines perpendicular to the same line are parallel.

Answer: (a) outside the segment

Explanation:
1. The centre lies on the perpendicular bisector of the chord, on the side of the major segment.
2. So it is not in the minor segment.

Teacher's Note:
1. A semicircle has the centre on the chord.
2. In a major segment the centre lies inside.

Answer: (c) The angle in a segment greater than a semicircle is greater than a right angle

Explanation:
1. The angle in a major segment (greater than a semicircle) is half the central angle of the minor arc, so it is less than \( 90^\circ \).
2. The other three statements are true.

Teacher's Note:
1. The angle in a semicircle is exactly 90 degrees; in a major segment it is smaller.
2. In a minor segment the angle is obtuse.