# Free Circles 03 Practice Test - 9th Grade

### Question 1

The collection of all the points in a plane, which are at a fixed distance from a fixed point in the plane is called ______.

rectangle

square

rhombus

circle

#### SOLUTION

Solution :D

Circle is the set of all the points in a plane which are at a given distance from a fixed point in the plane. That fixed point is known as the centre of the circle.

### Question 2

If sum of opposite angles of a quadrilateral is 180∘, then the quadrilateral is a _____.

trapezium

cyclic quadrilateral

kite

parallelogram

#### SOLUTION

Solution :B

A quadrilateral is called cyclic if all the four vertices of it lie on a circle and the sum of its opposite angles is 180∘.

### Question 3

The longest chord of a circle is called a

#### SOLUTION

Solution :Chord is a line segment that joins two points on the circumference of a circle. Diameter is the longest chord of a circle which passes through centre joining the two points on the circumference of a circle.

### Question 4

In the given figure, O is the centre of the circle. Find x.

50∘

40∘

30∘

20∘

#### SOLUTION

Solution :B

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.

Given∠ABC=50∘,

∴∠AOC=2 ∠ABC=2×50∘=100∘

Also, since OA and OC are radii of the circle, OA=OC.

⟹∠OAC=∠OCA

In △AOC,

x+x+∠AOC=180∘.

⟹2x=180∘−∠AOC=180∘−100∘=80∘

⟹x=40∘

### Question 5

In the given figure, O is the centre of the circle. Find the value of X.

30∘

40∘

50∘

100∘

#### SOLUTION

Solution :C

In the figure, ∠ABD=50∘

Since the angles subtended by an arc in the same segment are equal, we have ∠ABD=∠ACD.

⇒X=∠ACD=50∘

### Question 6

AB is a chord of a circle with centre O.

∠x=90∘

True

False

#### SOLUTION

Solution :A

The line drawn through the centre of a circle perpendicularly bisects a chord.

Hence, ∠x=90∘.

### Question 7

Which of the following cannot be a cyclic quadrilateral?

Square

Rectangle

Parallelogram

Isosceles Trapezium

#### SOLUTION

Solution :C

A quadrilateral is cyclic if its opposite angles are supplementary.

In square, rectangle and isosceles trapezium, the opposite angles are supplementary, that is, the sum of the opposite angles is 180∘.

In a parallelogram, the opposite angles are equal and may not add to 180∘. Hence, a parallelogram cannot be a cyclic quadrilateral.

### Question 8

In the given figure, A, B and C are three points on the circle with centre O such that ∠BOC=30∘ and ∠AOB=60∘. If D is a point on the circle that is not on the arc ABC, then find ∠ADC.

(902)∘

45∘

(602)∘

30∘

#### SOLUTION

Solution :A and B

It can be observed that ∠AOC=∠AOB+∠BOC=60∘+30∘=90∘.

We know that 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.

So,∠ADC=12∠AOC=12×90∘=45∘

∴∠ADC=45∘

### Question 9

A chord AB of a circle is equal to the radius of the circle. Find the angles subtended by the chord at points on the major arc and the minor arc.

150∘,30∘

120∘,60∘

30∘,150∘

60∘,120∘

#### SOLUTION

Solution :C

In ΔOAB,

AB = OA = OB = radius of the circle.

∴ΔOAB is an equilateral triangle.

Therefore, each interior angle of this triangle will be equal to 60∘.

∴∠AOB=60∘.

Since the angle subtended by an arc of the circle at its centre is double the angle subtended by it at any point on the remaining part of the circle, we have

∠ACB=12∠AOB=12×60∘=30∘.

Now in the cyclic quadrilateral ACBD,

∠ACB+∠ADB=180∘. [Opposite angles in a cyclic quadrilateral are supplementary]

∴∠ADB=180∘−∠ACB=180∘−30∘=150∘

Therefore, the angles subtended by the chord AB at a point on the major arc and the minor arc are 30∘ and 150∘ respectively.

### Question 10

ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If ∠DBC=80∘ and ∠BAC=40∘, then find ∠BCD.

60∘

80∘

50∘

40∘

#### SOLUTION

Solution :A

Given that ∠BAC=40∘ and ∠DBC=80∘.

Since the angles formed by the same segment are equal,

∠BDC=∠BAC=40∘.

In ΔBDC,

∠BDC+∠DBC+∠BCD=180∘. [Angle sum property]

i.e., 40∘+80∘+∠BCD=180∘

⟹∠BCD=180∘−120∘=60∘