# Free Circles 03 Practice Test - 9th Grade

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

A.

rectangle

B.

square

C.

rhombus

D.

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.

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

A.

trapezium

B.

C.

kite

D.

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.

The longest chord of a circle is called a ___ of the circle.

#### 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.

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

A.

50

B.

40

C.

30

D.

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.

GivenABC=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=180AOC=180100=80

x=40

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

A.

30

B.

40

C.

50

D.

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

AB is a chord of a circle with centre O.

x=90

A.

True

B.

False

#### SOLUTION

Solution : A

The line drawn through the centre of a circle perpendicularly bisects a chord.
Hence, x=90.

Which of the following cannot be a cyclic quadrilateral?

A.

Square

B.

Rectangle

C.

Parallelogram

D.

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.

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.

A.

(902)

B.

45

C.

(602)

D.

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.

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.

A.

150,30

B.

120,60

C.

30,150

D.

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=12AOB=12×60=30.

Now in the cyclic quadrilateral ACBD,

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

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

A.

60

B.

80

C.

50

D.

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=180120=60