Circles Test

Result

Circles Test - 25

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 figure, $$AB$$ is parallel to $$DC$$, $$\angle BCD=80^o$$ and $$\angle BAC = 25^o$$. Then m$$\angle ADC$$ is:

A
$$100^o$$

B
$$80^o$$

C
$$65^o$$

D
$$55^o$$

Solution

Given, $$ ABCD$$ is a cyclic quadrilateral and $$ AC$$ is its diagonal.

Also, $$ \angle BAD={ 25 }^{ o }, \angle BCD={ 80 }^{ o }$$

Since, $$ ABCD$$ is a cyclic quadrilateral.
$$ \therefore \angle BCD+\angle BAD=1{ 80 }^{ o }$$ ....(since the sum of the opposite angles of a cyclic quadrilateral is $${ 180 }^{ o }$$)

$$ \Rightarrow \angle BAD=1{ 80 }^{ o }-\angle BCD=1{ 80 }^{ o }-{ 80 }^{ o }=1{ 00 }^{ o }$$.

Again, $$AB\parallel CD$$...[Alternate interior angles]
$$\Rightarrow \angle BAC=\angle ACD=25^{ o }$$.

$$ \therefore \angle ACB=\angle BCD-\angle ACD=80^{ o }-25^{ o }=55^{ o }$$.

Now in $$\Delta ABC$$,
$$\angle ABC=180^{ o }-(\angle BAC+\angle BCA)$$ ....(angle sum property of triangles)

$$ \Rightarrow \angle ABC=180^{ o }-(25^{ o }+55^{ o })=100^{ o }$$

Again $$ ABCD$$ is a cyclic quadrilateral.
$$\therefore \angle ABC+\angle ADC={ 180 }^{ o }$$ ....(since the sum of the opposite angles of a cyclic quadrilateral is $$ { 180 }^{ o }$$)

$$ \therefore \angle ADC={ 180 }^{ o }-\angle ABC={ 180 }^{ o }-100^{ o }=80^{ o }$$.

Hence, option $$B$$ is correct.
Question 2
1 / -0

Then m$$\angle EBA$$ is:

A
$$25^o$$

B
$$20^o$$

C
$$35^o$$

D
$$70^o$$

Solution

Given- $$ABDE$$ is a cyclic quadrilateral in a circle with centre $$O$$ and diameter $$AB$$.
$$BE$$ is a diagonal ofthe given cyclic quadrilateral.
$$ \angle EAB={ 65 }^{ o }$$

To find out- $$\angle EBA=?$$

Solution-
$$ AB$$ is the diameter.
Then, $$\angle AEB$$ is an angle in the semicircle.
$$\therefore \angle AEB={ 90 }^{ o }$$

And $$\angle EBA ={ 180 }^{ o }-(\angle AEB+\angle EAB)$$ .....(angle sum property of triangles).
$$\Rightarrow \angle EBA ={ 180 }^{ o }-({ 90 }^{ o }+{ 65 }^{ o })={ 25 }^{ o }$$.

Hence, option $$A$$ is correct.
Question 3
1 / -0

$$ABC$$ is an isosceles triangle in the given circle with centre $$O,$$ if $$\angle ABC=42^{\circ},$$ then find the measure of $$ \angle CDE.$$

A
$$84^{\circ}$$

B
$$138^{\circ}$$

C
$$96^{\circ}$$

D
$$148^{\circ}$$

Solution

In $$\triangle ABC,$$
$$AB=AC$$ [Given]

We know that angles opposite to equal sides are equal in a triangle. So,
$$\angle ACB=\angle ABC =42^{\circ}$$

Now, by using angle sum property,
$$\begin{aligned}{}\angle ACB + \angle ABC + \angle A& = 180^\circ\\42^\circ + 42^\circ + \angle A& = 180^\circ\\84^\circ + \angle A &= 180^\circ\\\angle A &= 96^\circ\end{aligned}$$

Quadrilateral $$ABCD$$ is a cyclic quadrilateral so, by using the property that the exterior angle of a cyclic quadrilateral is equal to the opposite interior angle we get,
$$\angle CDE=\angle A$$
$$x=96^\circ$$
Question 4
1 / -0

In the given diagram, $$AB$$ is the diameter of the given circle with centre $$O.$$ $$C$$ and $$D$$ are points on the circumference of the circle. If $$\angle ABD=35^{\circ}$$ and $$\angle CDB=15^{\circ},$$ then $$\angle CBD$$ equals:

A
$$55^{\circ}$$

B
$$75^{\circ}$$

C
$$40^{\circ}$$

D
$$25^{\circ}$$

Solution

Given, $$ABCD$$ is cyclic quadrilateral.

Now $$\angle ADB = 90^{\circ}$$ ....(Angle in a semi circle).
Given, $$\angle BDC= 15^{\circ}$$
Thus, $$\angle D = \angle ADB + \angle BDC$$
$$\implies$$ $$\angle D = 90^o + 15^o = 105^{\circ}$$.

Now, $$\angle B + \angle D = 180^o$$ ....(Sum of opposite angles of a cyclic quadrilateral is $$180^o$$)
$$\implies$$ $$\angle B + 105^o = 180^o$$
$$\implies$$ $$\angle B = 75^{\circ}$$.

Then, $$\angle DBC + \angle DBA = 75^o$$
$$\implies$$ $$\angle DBC + 35^o = 75^o$$
$$\implies$$ $$\angle DBC = 40^{\circ}$$.

Hence, option $$C$$ is correct.
Question 5
1 / -0

Find the value of $$(a + b)$$.

A
$$\displaystyle 40^{\circ}$$

B
$$\displaystyle 80^{\circ}$$

C
$$\displaystyle 120^{\circ}$$

D
$$\displaystyle 160^{\circ}$$

Solution

Here, $$\angle DCF = \angle BCE = 30^{\circ}$$ ...(vertically opposite angles)

Now, $$\angle ADC = \angle DFC + \angle DCF$$ ...(Exterior angle property)
$$\implies$$ $$\angle ADC = 30^o + b$$.

Also, $$\angle ABC = \angle BCE + \angle BEC$$ ...(Exterior angle property)
$$\implies$$ $$\angle ABC = 30^o + a$$.

Now, $$\angle ADC + \angle ABC = 180^o$$ (Opposite angles of cyclic quadrilateral)
$$\implies$$ $$30^o + a + 30^o + b = 180^o$$
$$\implies$$ $$a + b = 120^{\circ}$$.

Hence, option $$C$$ is correct.
Question 6
1 / -0

In figure, AB is a chord of a circle with centre O and AP is the tangent at A such that $$\angle BAP=75^{\circ}$$. Then $$\angle ACB$$ is equal to:

A
$$135^{\circ}$$

B
$$120^{\circ}$$

C
$$105^{\circ}$$

D
$$90^{\circ}$$

Solution

Given,
AB is a chord of a circle with centre O and AP is the tangent at A and $$\angle BAP=75^{\circ}$$
$$\therefore \angle OAP=90^{\circ}$$
$$\therefore \angle OAB=90^{\circ}-75^{\circ}$$
$$=15^{\circ}$$
So, $$\angle AOB=180^{0}-2(15^{0})$$
$$=150^{0}$$
$$\therefore$$ Exterior $$\angle AOB=360^{0}-150^{0}$$
$$=210^{0}$$
So, $$\angle ACB=\dfrac{1}{2}\times 210^{0}$$
$$=105^{0}$$
[Since angle subtended by an arc at the centre is double than the angle subtended by it at the remaining part of the circle]
Question 7
1 / -0

In the given figure, $$AB = BC = CD$$. If $$\displaystyle \angle BAC=25^{\circ}$$, then value of $$\displaystyle \angle AED$$ is:

A
$$\displaystyle 50^{\circ}$$

B
$$\displaystyle 60^{\circ}$$

C
$$\displaystyle 65^{\circ}$$

D
$$\displaystyle 75^{\circ}$$

Solution

In $$\triangle ABC$$, $$AB = AC$$
Thus, $$\angle BAC = \angle BCA$$ ....(Isosceles triangle property).

Also, $$\angle BDC = \angle BAC = 25^{\circ}$$ ....(Angles in same segment).

In $$\triangle BDC$$,
by angle sum property, the sum of angles $$= 180^o$$
$$\implies$$ $$\angle CBD + \angle BCA + \angle ACD + \angle BDC = 180$$
$$\implies$$ $$\angle ACD = 180 - 75 = 105^{\circ}$$.

Now, in cyclic quadrilateral $$ ACDE$$,
$$\angle ACD + \angle AED = 180$$ ....(Opposite angles of cyclic quadrilateral)
$$\implies$$ $$\angle AED = 180 - 105$$
$$\implies$$ $$\angle AED = 75^{\circ}$$

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

At one end $$A$$ of a diameter $$AB$$ of a circle of radius $$5$$ cm, tangent $$XAY$$ is drawn to the circle. The length of the chord $$CD$$ parallel to $$XY$$ at a distance $$8$$ cm from $$A$$ is:

A
$$4$$ cm

B
$$5$$ cm

C
$$6$$ cm

D
$$8$$ cm

Solution

We have , $$AN=8$$ cm
So, $$ON=AN-OM$$
$$\Rightarrow ON=8-5$$
$$\Rightarrow ON=3$$ cm
Since $$ON$$ is perpendicular to the chord $$CD$$,
So, In $$\triangle OCM$$
By Pythagoras theorem, we get
$$CN=\sqrt {(OC)^2-(OM)^2}$$
$$\Rightarrow CN=\sqrt {(5)^2-(3)^2}$$
$$\Rightarrow CN=\sqrt{(25-9)}$$
$$\Rightarrow CN=\sqrt{16}$$
$$\Rightarrow CN=4$$
Hence, $$CD=2 \times CN$$
$$=2\times 4$$
$$=8$$ cm
Question 9
1 / -0

If two chords of a circle are equidistant from the center of the circle then they are

A
Equal to each other

B
Not equal to each other

C
Intersect each other

D
None of these

Solution

Let $$AB, PQ$$ be two chords, $$OC$$ and $$OR$$ be their distance from center.

Given

$$OC = OR$$

We know that $$BC = AC $$ and $$PR = SR$$ because perpendicular line from center bisects the chord.

In $$\triangle OBC$$

$$BC^2 = r^2 – OC^2 \qquad –(1)$$

In $$\triangle OPR$$

$$OP^2 = RP^2 + OR^2$$

$$PR^2 = r^2 – OC^2 \qquad –(2)$$

$$(1) = (2)$$

$$\implies BC = PR \implies 2BC = 2PR$$

$$AB = PQ$$
Question 10
1 / -0

In the given figure, the value of $$a$$ is:

A
$$\displaystyle 30^{\circ}$$

B
$$\displaystyle 40^{\circ}$$

C
$$\displaystyle 60^{\circ}$$

D
$$\displaystyle 90^{\circ}$$

Solution

$$ABCD$$ is a cyclic quadrilateral.
Thus, $$\angle D + \angle B = 180^o$$ ...[Opposite angles of cyclic quadrilateral are supplementary]
$$\implies$$ $$130 + \angle B = 180$$
$$\implies$$ $$\angle B = 50^{\circ}$$.

And, $$AB$$ is a diameter, hence, $$\angle ACB = 90^o$$ (Angle in a semi circle)
Thus, In $$\triangle ABC$$,
$$\angle ABC + \angle ACB + \angle CAB = 180^o$$ ...[Angle sum property]
$$\implies$$ $$50^o + 90^o + \angle CAB = 180^o$$
$$\implies$$ $$\angle CAB = 40^{\circ}$$.

Hence, option $$B$$ is correct.