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

Practicing Class 9 Maths MCQ and Ganita Manjari Class 9 Maths Chapter 5 I’m Up and Down and Round and Round MCQ Questions Online Test with Answers daily helps in time management.

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

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

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

Question 1.
A line segment that joins two points on a circle and passes through the centre is called a
(a) radius
(b) segment
(c) diameter
(d) circumference
Answer:
(c) diameter

Explanation:
Let us check with options.
Option (a) A radius is a straight line joining the centre and circumference.
Option (b) A segment of a circle is the region bounded by a chord and the arc.
Option (c) A diameter is specifically a chord that passes through the centre of the circle.
Option (d) The length of the complete circle is called its circumference.

Question 2.
Where is the circumcentre of an obtuse – angled triangle located?
(a) Inside the triangle
(b) On the hypotenuse
(c) Outside the triangle
(d) At one of the vertices
Answer:
(c) Outside the triangle

Explanation:
The circumcentre in an obtuse angled triangle is located outside the triangle.

Question 3.
In the given figure, O is the centre of the circle and CD = DE = EF = FG. ∠COD = 40° then ∠COG is equal to

(a) 160°
(b) 200°
(c) 240°
(d) 280°
Answer:
(b) 200°

Explanation:
Since, equal chords of a circle subtend equal angles at the centre.
∴ ∠COD = ∠DOE = ∠EOF
= ∠FOG = 40°
∴ ∠COG = 4 × 40° = 160°
∴ Reflex ∠COG = 360° – ∠COG
= 360° – 160° = 200°

Question 4.
In the given figure, if OA = 10 cm, AB = 12 cm and OD ⊥ AB, find CD.

(a) 2 cm
(b) 3 cm
(c) 4 cm
(d) 6 cm
Answer:
(a) 2 cm

Explanation:
Since, OD ⊥ AB,
AC = BC = 6 cm
[∵ perpendicular drawn from the centre of circle bisects the chords] and ∠OCA = 90° and OA = 10 cm
Thus, OA² = AC² + OC²
[by Pythagoras theorem]
⇒ 10² = 6² + OC²
⇒ OC² = 100 – 36 = 64
⇒ OC = 8cm
Since, OD = OA [∵ radii of same circle]
OD = 10 cm
Thus, CD = OD – OC
= 10 – 8
= 2 cm

Question 5.
In the given figure, OD is perpendicular to the chord AB of a circle, whose centre is O. If BC is a diameter and OD =4cm then A equals

(a) 6 cm
(b) 8 cm
(c) 10 cm
(d) 12 cm
Answer:
(b) 8 cm

Explanation:
Since, OD ⊥ AB and the perpendicular drawn from the centre to a chord bisects the chord.
∴ D is the mid-point of AB.
Also, O being the centre, is the mid-point of BC.
Thus, in ΔABC, D and O are mid-points of AB and BC, respectively.
∴ OD ∥ AC and OD = \(\frac{1}{2}\)CA
[by mid-point theorem]
⇒ CA = 2OD = 2 × 4 = 8 cm

Question 6.
In the given figure, a line intersect two concentric circles with centre O at A, B, C and D. If AB = 10 cm, find CD.

(a) 5 cm
(b) 10 cm
(c) 15 cm
(d) 20 cm
Answer:
(b) 10 cm

Explanation:
Let OP be perpendicular from O on line
∴ AP = DP and BP = CP

[∵ perpendicular from the centre of a circle to a chord bisects the chord]
⇒ AB = CD
CD = 10 cm

Question 7.
In the given figure, OA and OB are respectively perpendiculars to chords CD and EF of a circle, whose centre is O. If OA = OB then

(a) CE = DF
(b) CE > DF
(c) CE < DF
(d) None of these
Answer:
(a) CE = DF

Explanation:
We have, OA ⊥ CD and OB 1EF
OA = OB
⇒ Chords CD and EF are equidistant from O.
⇒ CD = EF
[Y chords equidistant from the centre of a circle are equal in length]
⇒ arc CD = arc EF
On subtracting arc ED (or ED) from both sides, we get

Question 8.
In the given figure, if ∠ABC = 20°, then ∠AOC is equal to

(a) 20°
(b) 40°
(c) 60°
(d) 10°
Answer:
(b) 40°

Explanation:
Given, ∠ABC = 20°
We know that angle subtended aI the centre by an arc is twice the angle subtended by it at any point on the remaining part of the circle.
∴ ∠AOC = 2∠ABC = 2 × 200 = 400

Question 9.
In the given figure. if ∠AOR = 120° then find the value of ∠ACB.

(a) 100°
(b) 110°
(c) 130°
(d) 120°
Answer:
(d) 120°

Explanation:
Reflex ∠AOB =360° -120° =240°
∴ ∠ACB = \(\frac{1}{2}\) reflex ∠AOB
[Y the angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle]
∴ ∠ACB = \(\frac{1}{2}\) × 240° = 120°

Question 10.
In the given figure, if ∠DAB = 60°, ∠ABD = 50° then ∠ACB is equal to

(a) 60°
(b) 50°
(c) 70°
(d) 80°
Answer:
(c) 70°

Explanation:
Given, ∠DAB = 60° and ∠ABD = 50°
Now, ∠ADB = ∠ACB …(i)
[∵ angles in same segment of a circle are equal]
In ΔABD, ∠ABD + ∠ADB + ∠DAB = 180°
[by angle sum property of a triangle]
50° + ∠ADB + 60° = 180°
⇒ ∠ADB = 180° – 110°
⇒ ∠ADB = 70°
From Eq. (i), ∠ACB = ∠ADB = 70°

Question 11.
In the given figure, BC is a diameter of the circle and ∠BAO = 60°. Then, ∠ADC is equal to

(a) 30°
(b) 45°
(c) 60°
(d) 120°
Answer:
(c) 60°

Explanation:
In ΔAOB, OA = OB
[∵ both are the radius of circle]
⇒ ∠OBA = ∠BAO
[Y angles opposite to equal sides are equal]
⇒ ∠OBA = 60° [∵ ∠BAO = 60°, given]
We know that ∠ABC = ∠ADC
angles in the same segment AC are equal]
∠ADC = 60°

Question 12.
In the given figure, O is the centre of the circle and chord AC and BD intersect at Q such that ∠AQB = 140° and ∠QBC = 25°, find the value of ∠ADB.

(a) 105°
(b) 140°
(c) 115°
(d) 135°
Answer:
(c) 115°

Explanation:
∠AQB = ∠QCB + ∠CBQ
[exterior angle property]
⇒ 140° = ∠QCB + 25°
⇒ ∠QCB = 140° – 25° = 115°
or ∠ACB = 115°
∴ ∠ADB = ∠ACB = 115°
[∵ angles in same segment of a circle are equal]

Question 13.
In the given figure, ∠ABC = 59°, ∠ACB = 40°, find ∠BDC.

(a) 99°
(b) 90°
(c) 81°
(d) 71°
Answer:
(c) 81°

Explanation:
In ΔABC,
∠ABC + ∠BAC + ∠ACB = 180°
[by angle sum property of triangle]
⇒ 59° + ∠BAC + 40° = 180°
⇒ ∠BAC = 180° – 59° – 40°
= 180° – 99° = 81°
∴ ∠BDC = ∠BAC = 81°
[∵ angles in same segment of a circle are equal]

Question 14.
In the given figure, if AOB is a diameter of the circle and AC = BC then ∠CAB is equal to

(a) 30°
(b) 60°
(c) 90°
(d) 45°
Answer:
(d) 45°

Explanation:
We know that diameter subtends a right angle to the circle.
∠BCA = 90° …(i)
Given, AC = BC
∠ABC = ∠CAB …(ii)
[∵ angles opposite to equal sides of a triangle are equal]

In ΔABC,
∠CAB + ∠ABC + ∠BCA = 180°
[by angle sum property of a triangle]
⇒ ∠CAB + ∠CAB + 90° = 180°
[from (i) and (ii)]
⇒ 2∠CAB = 180° – 90°
⇒ ∠CAB = 45°

Question 15.
The sum of opposite angles of a cyclic quadrilateral is
(a) 120°
(b) 90°
(c) 180°
(d) 150°
Answer:
(c) 180°

Explanation:
The sum of opposite angles of a cyclic quadrilateral is always supplementary (180°).

Question 16.
In the given figure, ABCD is a cyclic quadrilateral in which BD is diameter then find ∠BAD.

(a) 90°
(b) 100°
(c) 80°
(d) 110°
Answer:
(a) 90°

Explanation:
Since, ABCD is a cyclic quadrilateral and BD is diameter.
∴ ∠BAD = 90° [angle in semi-circle]

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

Direction (Q. Nos. 1-3) In the questions given below, there are two statements marked as Assertion (A) and Reason (R).
Read the statements and choose the correct option.

Question 1.
Assertion (A) The length of a chord which is at a distance of 5 cm from the centre of a circle of radius 10 cm is 17.32 cm.
Reason (R) The perpendicular from the centre of a circle to a chord bisects the chord.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Answer:
(a) Both A and R are true and R is the correct explanation of A.

Explanation:
Let PQ be a chord of a circle with centre O and radius 10 cm. Draw OR ⊥ PQ.

Now, OP = 10 cm and OR = 5 cm
In right angled ΔORP, we get
OP² = PR² + OR²
⇒ PR² = OP² – OR²
⇒ PR² =10² – 5² = 75
⇒ PR = \(\sqrt{75}\) = 8.66 cm
Since, the perpendicular from the centre to a chord bisects the chord.
Therefore, PQ = 2 × PR = 2 × 8.66 = 17.32 cm
So, Assertion is true.
Also, Reason is true and it is the correct explanation of Assertion.

Question 2.
Assertion (A) Two diameters of a circle intersect each other at right angles. Then, the quadrilateral formed by joining their end-point is a rectangle.
Reason (R) Angles in the same segment of a circle are equal.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Answer:
(b) Both A and R are true but R is not the correct explanation of A.

Explanation:
Let AB and CD be two perpendicular diameters of a circle with centre O.

In ΔAOC and ΔBOC, we have
OA = OB [radii of same circle]
∠AOC = ∠BOC [each equal to 90°)
and OC = OC (common]
∴ ΔAOC ≅ ΔBOC [by SAS congruence rule]
⇒ AC = BC [by CPCT]
Also, ∠ACB = 90° [angle in a semi-circle is a right angle]
Similarly, we get
BC = BDand ∠CBD = 90°
DB = DA and ∠BDA = 90°
Hence, ACBD is a square.
It implies that ACBD is a rectangle.
So, Assertion is true.
Also, Reason is true but it is not necessary that
Reason is the correct explanation.

Question 3.
Assertion (A) ABCD is a cyclic quadrilateral. AB and DC are produced to meet at E.
Then, ∠ECB = ∠EDA.

Reason (R) Exterior angle in a cyclic quadrilateral is equal to opposite interior angle.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Answer:
(d) A is false but R is true.

Explanation:
∠ECB = ∠EAD
[∵ exterior angle in a cyclic quadrilateral is equal to interior opposite angle]
Assertion is false but Reason is true.