Circles Test

Result

Circles Test - 18

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

In the given figure, $$AB$$ is the diameter of a circle with $$O$$ and $$AT$$ is a tangent. If $$\angle AOQ = 58^{\circ}$$, then the value of $$\angle ATQ$$ is _________.

A
$$52^{\circ}$$

B
$$61^{\circ}$$

C
$$46^{\circ}$$

D
$$75^{\circ}$$

Solution

Angle subtended by an arc(here, $$AQ$$) at the centre is twice the angle subtended by it on the circumference on same side of the chord(here, $$AQ$$)

Given, $$\angle AOQ = 58^{\circ}$$

So, $$\angle ABT = \angle ABQ= 29^{\circ}$$
Since, radius of circle is perpendicular to the tangent at the point of contact, $$\angle BAT = 90^{\circ}$$
In $$\triangle ABT$$, $$\angle BAT + \angle ABT + \angle BTA = 180^{\circ}$$
$$\Rightarrow 90^{\circ} + 29^{\circ} + \angle BTA = 180^{\circ}$$
$$\Rightarrow \angle BTA = 61^{\circ}$$

$$\angle ATQ = \angle BTA = 61^{\circ}$$

So, Option (B)
Question 2
1 / -0

Identify the center of the circle.

A
$$A$$

B
$$O$$

C
$$D$$

D
$$C$$

Solution

Center of a circle is a point inside the circle and is at an equal distance from all of the points on its circumference.
In the given figure, $$O$$ is such point.
So, $$O$$ is the center of the given circle.
Hence, option $$B$$ is correct.
Question 3
1 / -0

Given circle with center $$C$$ shows an inscribed quadrilateral. Find $$y$$.

A
$$100^o$$

B
$$80^o$$

C
$$60^o$$

D
$$40^o$$

Solution

Here, the given quadrilateral is cyclic.
We know, the sum of opposite angles of a cyclic quadrilateral is $$180^o$$.
$$\therefore y^o+100^o=180^o$$
Thus, $$y^o=180^o-100^o=80^o$$.

Hence, option $$B$$ is correct.
Question 4
1 / -0

Two concentric circles intersect at _____ number of points.

A
$$2$$

B
$$0$$

C
$$1$$

D
None of these

Solution

Two concentric circles do not intersect. They only share a common centre as shown in the figure.
Question 5
1 / -0

In the given figure, $$O$$ is the centre of the circle. Find the value of $$\angle BDC$$.

A
$$30^{o}$$

B
$$45^{o}$$

C
$$60^{o}$$

D
$$75^{o}$$

Solution

In the given figure, $$\angle BDC=\angle BAC$$
$$\Rightarrow \angle BDC=45^{\circ}$$ (Angle subtended by an arc in the same segment are equal.)
Question 6
1 / -0

In the given circle ABCD, O is the centre and $$\angle BDC$$ = $$42^0$$, The $$\angle ACB$$ is equal to

A
$$42^0$$

B
$$45^0$$

C
$$48^0$$

D
$$60^0$$

Solution

In the given figure,
$$\angle BDC={ 42 }^{ o }$$
$$\angle BAC={ 42 }^{ o }$$ (Angles subtended by the same arc in same segment of a circle are equal )
Now, since $$\angle ABC={ 90 }^{ o }$$ (angle in semicircle is a right angle)

By angle sum property,
$$\angle BAC+ \angle ABC + \angle ACB={ 180}^{o}$$
$$ 42^{o}+90^{o}+\angle ACB=180^{o}$$
$$\Rightarrow \angle ACB={ 48 }^{ o }$$
Question 7
1 / -0

In the given figure, chord $$BE=BD$$, $$\angle CBD=33^{o}$$ and $$OB \bot AC$$, then $$x+y-z$$ is equal to:

A
$$230^{o}$$

B
$$237^{o}$$

C
$$337^{o}$$

D
None of these

Solution

In $$\Delta BED$$,
$$BE=BD$$ (given)
$$\angle BED=\angle BDE$$ (opposite angles are equals).

Given, $$x=\angle BDE$$.
Also, $$\angle OBC=90^{\circ}$$ (given)
$$\angle OBD+\angle CBD=90^{\circ}$$
$$\implies$$ $$\angle OBD+33^{\circ}=90^{\circ}$$ ....[Since $$\angle CBD=33^o$$]
$$\implies$$ $$\angle OBD=90^o-33^o=57 ^{\circ}$$.

Now, $$x+z=180^{\circ}$$ (opposite angles of a cyclic quadrilateral)
$$\implies$$ $$z=180^{\circ}-x$$ ----- (i).

In $$\Delta BED$$,
$$\angle BED + \angle BDE + \angle EBD=180^{\circ}$$ (By angle sum property of triangle)
$$\implies$$ $$x+y+57^{\circ}=180^{\circ}$$
$$\implies$$ $$x+y+57^{\circ}=180^{\circ}-x$$
$$\implies$$ $$x+y+57^{\circ}=z$$ .....(from (i))
$$\implies$$ $$x+y-z=-57^{\circ}$$.

Hence, option $$D$$ is correct.
Question 8
1 / -0

In the above figure, $$\overline {AB}$$ and $$\overline {AC}$$ are equal chords and $$O$$ is the centre. If $$\angle BOC=100^{o}$$, then find $$\angle ACO$$.

A
$$10^{o}$$

B
$$20^{o}$$

C
$$30^{o}$$

D
$$25^{o}$$

Solution

Consider the diagram shown in the question.

Given:

$$AB=AC$$

$$\angle BOC=100{}^\circ $$

$$\Rightarrow \angle BAC=50{}^\circ $$ (angle at the circle is half of angle at the centre by same chord $$\left( BC \right)$$)

Consider the quadrilateral $$OBAC$$.

$$\angle BOC$$ (of the quadrilateral) $$=360{}^\circ -100{}^\circ =260{}^\circ $$

Consider triangle $$OBC$$.

$$\angle OBC=\angle OCB$$ ($$OB$$ and $$OC$$ are radius)

Similarly, in $$\Delta ABC$$,

$$\angle ABC=\angle ACB$$ (Since, $$AB=AC$$)

$$ \angle ABO+\angle OBC=\angle ACO+\angle OCB $$

$$ \Rightarrow \angle ABO=\angle ACO $$

We know that interior angles of a quadrilateral are $$360{}^\circ $$. Therefore,

$$ 50{}^\circ +260{}^\circ +\angle ABO+\angle ACO=360{}^\circ $$

$$ 2\angle ACO=50{}^\circ $$

$$ \angle ACO=25{}^\circ $$

Hence, this is the required result.
Question 9
1 / -0

In figure, the value of $$x$$ is

A
$$60^{\circ}$$

B
$$15^{\circ}$$

C
$$30^{\circ}$$

D
$$None\ of\ these$$

Solution

$$\text{Theorem: Angles in the same segments are equal}$$
$$\therefore x^{\circ}=30^{\circ}$$
Question 10
1 / -0

In figure $$A,B$$ and $$C$$ are three points on a circle with centre $$O$$ such that $$\angle{BOC}={30}^{o}$$ and $$\angle {AOB}={60}^{o}$$. If $$D$$ is a point on the circle other than the arc $$ABC$$, find $$\angle {ADC}$$.

A
$$30^\circ$$

B
$$90^\circ$$

C
$$45^\circ$$

D
$$60^\circ$$

Solution

According to theorem of circles $$\angle AOC = 2\angle ADC$$ .....(i)

$$\angle AOC=\angle AOB+\angle BOC = 60^{\circ}+30^{\circ}=90^{\circ}$$

from equation(i)

$$90^{\circ}=2\angle ADC$$

$$\angle ADC=45^{\circ}$$