I’m Up and Down and Round and Round Class 9 MCQ Maths Chapter 5

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MCQ on I’m Up and Down and Round and Round Class 9

Class 9 Maths I’m Up and Down and Round and Round MCQ

Choose the correct option from the given options:

Question 1.
The least possible radius of a circle through two points A and B which are 4 cm apart is :
(a) 3 cm
(b) 1 cm
(c) 2 cm
(d) 0 cm
Solution:
(c) 2 cm

Explanation:
Minimum possible length of the radius will be for the circle which will have the diameter equal to AB.
Since, AB = 4 cm
∴ diameter = 4 cm
⇒ radius = \(\frac{4}{2}\) cm
(radius of any circle is half of the diameter)
⇒ radius = 2 cm
So, least possible radius for the given situation will be 2 cm.

Question 2.
Through three collinear points how many different circles can be drawn?
(a) 2
(b) 4
(c) 1
(d) 0
Solution:
(d) 0

Explanation:
No circle can pass through three collinear points.

Question 3.
Through three non-collinear points how many different circles can be drawn?
(a) 2
(b) 4
(c) 1
(d) 0
Solution:
(c) 1

Explanation:
A unique circle can be drawn through three non-collinear points.

Question 4.
In the following figure 50, if AB = DE, then ∠ACB is :

(a) 35°
(b) 70°
(c) 45°
(d) 140°
Solution:
(b) 70°

Explanation:
Since, equal chords subtend equal angles at
the centre of the circle.
So, ABDE
∠ACB = ∠DCE
∠ACB = 70°.

Question 5.
In the following figure 51, if ∠ACB = ∠DCE, then DE =

(a) AB
(b) CE
(c) BC
(d) DC
Solution:
(a) AB

Explanation:
Since, chords of a circle that subtend equal angles at the centre are equal.
So, ∠ACB = ∠DCE
⇒ DE = AB.

I’m Up and Down and Round and Round MCQ Class 9

Question 6.
In figure, AB = 4.2 cm, CE ⊥ AB, BE =

(a) 2 cm
(b) 1.9 cm
(c) 2.1cm
(d) 4.2 cm
Solution:
(c) 2.1cm

Explanation:
In a circle, perpendicular drawn from the centre on to the chord of the circle bisects the chord.
So. CE ⊥ AB ⇒ BE = EA = \(\frac{1}{2}\)AB
⇒ BE = \(\frac{1}{2}\) × 4.2 = 2.1 cm.

Question 7.
In figure, AB < GF, CE > ____________.

(a) CG
(b) CA
(c) CF
(d) CH
Solution:
(d) CH

Explanation:
In a circle, larger chord is nearer to the centre of the circle.
Here, AB < GF
Therefore, CE < CH.

Question 8.
The length of the chord of a circle with radius 10 cm and whose perpendicular distance from the centre is 8 cm, is :
(a) 12 cm
(b) 6 cm
(c) 4 cm
(d) 8 cm
Solution:
(a) 12 cm

Explanation:

Using Baudhayana Pythagoras theorem in ∆ACM, we have
AM² = AC² – CM² = 10² – 8² = 100 – 64 = 36
⇒ AM = \(\sqrt{36}\) = 6 cm
We know that the perpendicular on the chord, drawn from the centre, bisects the chord.
So, AM = BM
⇒ BM = 6 cm
Now, AB = AM + BM = 6 + 6 = 12 cm.

Question 9.
The length of the chord of a circle is 16 cm and its radius is 10 cm, its perpendicular distance from the centre is :
(a) 12 cm
(b) 6 cm
(c) 4 cm
(d) 8 cm
Solution:
(b) 6 cm

Explanation:
Using Baudhayana Pythagoras theorem in ∆ACM, we have
CM² = AC² – AM² = 10² – 8² = 100 – 64 = 36 (since, AM =\(\frac{1}{2}\) × AB, the perpendicular on the chord, drawn from the centre, bisects the chord.)
⇒ CM = \(\sqrt{36}\) = 6 cm.

Question 10.
Angle in a semicircle is :
(a) 60°
(b) 90°
(c) 120°
(d) 45°
Solution:
(b) 90°

Explanation:
Angle in a semicircle is of 90°. This fact is established on the basis of the theorem which states that the angle subtended by an arc at the centre is double of the angle subtended by it at any point on the alternate segment.

So, ∠BCA = 180° (straight angle)
⇒ ∠ADB = \(\frac{1}{2}\) × ∠BCA = \(\frac{1}{2}\) × 180° = 90°.

Question 11.
In figure, x =

(a) 50°
(b) 100°
(c) 80°
(d) 60°
Solution:
(c) 80°

Explanation:
Since, in a cyclic quadrilateral the sum of opposite angles is 180°
So, ∠B + ∠D = 180°
⇒ ∠B + 100° = 180°
⇒ ∠B = 180° – 100° = 80°.

Question 12.
An arc of a circle subtends an angle of 80° at the centre. The measure of the angle subtended by the arc at a point on the circle is :
(a) 80°
(b) 40°
(c) 20°
(d) 160°
Solution:
(b) 40°

Explanation:
Since, the angle subtended by an arc at the centre is twice the angle subtended by it at any point in the alternate segment.
Since, angle subtended by the arc at the centre is 80°.
Therefore, angle subtended by the arc at a point on the circle will be \(\frac{80^{\circ}}{2}\) = 40°

Question 13.
The least possible radius of a circle through two points A and B which are 13 cm apart is :
(a) 6.5 cm
(b) 6 cm
(c) 5.5 cm
(d) 3.25 cm
Solution:
(a) 6.5 cm

Question 14.
Through two points how many different circles can be drawn?
(a) 2
(b) 4
(c) infinitely many
(d) 0
Solution:
(c) infinitely many

Question 15.
Name of the quadrilateral whose vertices lie on a circle is:
(a) Para-quadrilateral
(b) Cyclic Quadrilateral
(c) Meta-quadrilateral
(d) Tera-quadrilateral
Solution:
(b) Cyclic Quadrilateral

Question 16.
In the following figure 63, if AB = DE, then ∠ACB is :

(a) 85°
(b) 70°
(c) 42.5°
(d) 170°
Solution:
(a) 85°

Question 17.
In the following figure 64, if AB = DE, then ∠DCE = :

(a) ∠BAC
(b) ∠DEC
(c) ∠ABC
(d) ∠ACB
Solution:
(d) ∠ACB

Question 18.
In figure AB = 1.2 cm, CE ⊥ A, BE =

(a) 2 cm
(b) 0.6 cm
(c) 2.4 cm
(d) 1.2 cm
Solution:
(b) 0.6 cm

Question 19.
In figure, AB = GF, CE = ____________.

(a) CG
(b) CA
(c) CF
(d) CH
Solution:
(d) CH

Question 20.
The length of the chord of a circle with radius 13 cm and whose perpendicular distance from the centre is 5 cm, is :
(a) 12 cm
(b) 6 cm
(c) 24 cm
(d) 18 cm
Solution:
(c) 24 cm

Question 21.
The length of the chord of a circle is 8 cm and its radius is 5 cm, its perpendicular distance from the centre is :
(a) 3 cm
(b) 6 cm
(c) 4 cm
(d) 1.5 cm
Solution:
(a) 3 cm

Question 22.
Angle subtended by a diameter of a circle at any point on the circle is :
(a) 60°
(b) 90°
(c) 120°
(d) 45°
Solution:
(b) 90°

I’m Up and Down and Round and Round Class 9 Assertion and Reason Questions

Direction: A statement of Assertion (A) is followed by a statement of Reason (R). Choose the correct option from the following options.
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).
(c) Assertion (A) is true but Reason (R) is false.
(d) Assertion (A) is false but Reason (R) is true.

Question 1.
Assertion (A): In a circle AB is a diameter. CD is a chord AB > CD.
Reason (R): Diameter is the largest chord of any circle.
Solution:
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

Explanation:
Since, CD is a chord of the circle, Hence, it will be at some distance from the centre.
Since, AB is a diameter of the circle,
So, it will pass through the centre of the circle. Therefore, its distance form the centre = 0. Since, chord whose distance from the centre is more is smaller then the chord whose distance is less.
So, AB > CD.
Hence, Assertion (A) is true.
Also, in any circle, diameter is the largest chord.
Thus, Reason (R) is true and it explains the existence of the Assertion (A).

Question 2.
Assertion (A): In figure, the value of 2 is 92.

Reason (R): Sum of opposite angles of a cyclic quadrilateral is 180°.
Solution:
(d) Assertion (A) is false but Reason (R) is true.

Explanation:
Here, ∠M + ∠E = 180° (sum of opposite angles of a cyclic quadrilateral is 180°).
⇒ 92° + 0° = 180°
⇒ z° = 180° – 92 = 88°
So, Assertion (A) is false.
Sum of opposite angles of a cyclic quadrilateral is 180°, is a fact.
So, Reason (R) is true.

Question 3.
Assertion (A): Angle in a semicircle is 90°.
Reason (R): Angles in the same segment are equal.
Solution:
(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).

Explanation:
Since, angle subtended by an arc at the centre is double the angle subtended by it at any point in the alternate segment. Angle subtended by a semi-circle at the centre is 180°. Therefore, angle in a semicircle is 90°. So, Assertion (A) is true.
Also, we know that the angles in same segment are equal.
So, Reason (R) is true.
But Reason (R) is not the correct explanation for the Assertion (A).

Question 4.
Assertion (A): If ABCD is a cyclic quadrilateral (vertices are taken in the order of ABCD), then
∠A + ∠B = ∠C + ∠D.
Reason (R): Opposite angles of a cyclic quadrilateral are supplementary.
Solution:
(d) Assertion (A) is false but Reason (R) is true.

Explanation:
In a cyclic quadrilateral
ABCD, ∠A and ∠C are the pair of opposite angles.
Therefore, ∠A + ∠C = 180°, and ∠B and ∠D are the pair of opposite angles.
Therefore, ∠B + ∠D = 180°. So, ∠A + ∠C = ∠B + ∠D, but ∠A + ∠B may not be equal to ∠C + ∠D, for all cyclic quadrilaterals.
So, Assertion (A) is false.
Opposite angles are supplementary in a cyclic quadrilateral, is a fact.
So, Reason (R) is true.

Question 5.
Assertion (A): A unique circle can pass through the points P(2, 3), Q( -1, 3) and R(-4,-4).
Reason (R): Through three non-collinear points a unique circle can pass.
Solution:
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

Explanation:
Given points are non-collinear points. Thus, a unique circle will pass through these points. Therefore, Assertion (A) is true. We know that through two points infinitely many circles can pass while through three non-collinear points a unique circle can pass. So, Reason (R) is true and Reason (R) explains the existence of Assertion (A).

Question 6.
Assertion (A): In a circle AB is a diameter and O is its centre. CD is a chord such that CD = \(\frac{1}{2}\) AB. ∠COD = 90°.
Reason (R): Angle in a semicircle is a right angle.
Solution:
(d) Assertion (A) is false but Reason (R) is true.

Question 7.
Assertion (A): In figure, the value of z is 88.

Reason (R): Sum of opposite angles of a cyclic quadrilateral is 180°.
Solution:
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

Question 8.
Assertion (A): Angle in a semicircle is 90°.
Reason (R): Diameter passes trough the centre, so it makes 180° at the centre.
Solution:
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

Question 9.
Assertion (A): If ABCD is a cyclic quadrilateral (vertices are taken in the order of ABCD), then
∠A + ∠C = ∠B + ∠D.
Reason (R): Opposite angles of a cyclic quadrilateral are supplementary.
Solution:
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

Question 10.
Assertion (A): A unique circle can pass through the points A(0, 0), B( -1, 0) and C(0, – 4).
Reason (R): The centre of the circle will be at the point A(0, 0).
Solution:
(c) Assertion (A) is true but Reason (R) is false.