Circles Test

Result

Circles Test - 33

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, O is the centre of the circle .What is the value of $$x?$$

A
$$115^o$$

B
$$125^o$$

C
$$155^o$$

D
$$230^o$$

Solution

Given: $$O$$ is the centre of the circle.
So, $$OP,OR,OQ$$ are the radius of the circle.
So, $$OP=OR=OQ$$

Now,
Since $$OP=OQ$$
$$\therefore \triangle OPQ$$ is isosceles.

Hence $$\angle OPQ=\angle OQP={ 25 }^{ \circ }$$

From angle sum property,
$$ \angle POQ={ 180 }^{ \circ }-(\angle OPQ+\angle OQP)$$
$$\\ ={ 180 }^{ \circ }-({ 25 }^{ \circ }+{ 25 }^{ \circ })\\ ={ 180 }^{ \circ }-50^{ \circ }\\ =130^{ \circ }$$

Again, $$OQ=OR$$
$$\therefore \triangle OQR$$ is isosceles.

Hence $$\angle ORQ=\angle OQR={ 30 }^{ \circ }$$

By angle sum property

$$ \angle QOR={ 180 }^{ \circ }-(\angle ORQ+\angle OQR)$$
$$={ 180 }^{ \circ }-(30^{ \circ }+30^{ \circ })\\ ={ 180 }^{ \circ }-60^{ \circ }\\ =120^{ \circ }$$

We know that, in a circle the central angle is twice of any inscribed angle subtended by the same arc.

Here, $$ \angle POR$$ is the central angle and $$\angle PSR$$ is the inscribed angle subtended by the same arc.

$$\therefore \angle POR=2\angle PSR$$

$$\Rightarrow \angle POQ+\angle QOR=2\angle PSR$$

$$ \Rightarrow { 130 }^{ \circ }+{ 120 }^{ \circ }=2x$$

$$\Rightarrow 2x={ 250 }^{ \circ }$$

$$ \Rightarrow x=125^{ \circ }$$
Question 2
1 / -0

If the perimeter and the area of a circle are numerically equal, then the radius of the circle is

A
$$2\ units$$

B
$$p\ units$$

C
$$4\ units$$

D
$$7\ units$$

Solution

Given,
Numerically perimete of the circle is equal to area of circle

$$\Rightarrow 2\pi r=\pi r^2$$

$$\Rightarrow 2=r$$

$$\therefore r=2$$
Question 3
1 / -0

In the figure ,O is the center of the circle. The value of $$(2a + b + C)$$ in

A
$${192^ \circ }$$

B
$${180^ \circ }$$

C
$${278^ \circ }$$

D
$${266^ \circ }$$

Solution
Question 4
1 / -0

In the figure given, $$O$$ is the center of the circle, if $$\angle PAB=108^{o}$$, then find $$\angle BCQ$$.

A
$$152^{\circ}$$

B
$$120^{\circ}$$

C
$$150^{\circ}$$

D
$$162^{\circ}$$

E
None of these
Question 5
1 / -0

In the given quadrilateral $$PQRS $$, if $$ \angle RSP = {80}^{0} $$, then $$ \angle RQT= $$ ____.

A
$$ {100}^{0} $$

B
$$ {80}^{0} $$

C
$$ {70}^{0} $$

D
$$ {110}^{0} $$

Solution

Given is a cyclic quadrilateral $$PQRS$$.
We know that,
In cyclic quadrilaterals sum of opposite angles is $$180^0$$.

So,
$$\angle RSP + \angle PQR = 180^0$$
$$\Rightarrow 80^0 + \angle PQR = 180^0$$
$$\Rightarrow \angle PQR = 100^0$$.

Also, we know that sum of angles on a straight line is $$180^0$$,

So,
$$\angle PQR + \angle RQT = 180^0$$
$$\Rightarrow 100^0 + \angle RQT = 180^0$$
$$\Rightarrow \angle RQT = 80^0$$.

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

An arc subtends an angle of $$45^{o}$$ in its alternate segment then the angle made by its corresponding major arc is at the centre

A
$$270^{o}$$

B
$$90^{o}$$

C
$$45^{o}$$

D
$$315^{o}$$

Solution

Here, major arc $$AB$$ is subtending $$ 45^{\circ}$$ in the alternate (minor) segment
To find : angle $$x$$
Soluton: As angle around a point is $$ 360^{\circ}$$
Around $$O$$,
$$x + 45^o =360^o$$
$$ \Rightarrow x = 315^{\circ} $$
Question 7
1 / -0

In the given circle, $$AD=BC,\angle BAC={30}^{o}$$ and $$\angle CBD={70}^{o}$$
Find
$$(i)\,\angle BCD$$
$$(ii)\,\angle BCA$$
$$(iii)\,\angle ABC$$
$$(iv)\,\angle ADB$$

A
$$(i)\,50^\circ\,(ii)\,80^\circ\,(iii)\,50^\circ\,(iv)\,100^\circ$$

B
$$(i)\,80^\circ\,(ii)\,50^\circ\,(iii)\,100^\circ\,(iv)\,50^\circ$$

C
$$(i)\,80^\circ\,(ii)\,130^\circ\,(iii)\,100^\circ\,(iv)\,50^\circ$$

D
None of these

Solution
Question 8
1 / -0

In figure, $$O$$ is the centre of the circle, then

A
$$\angle x= \angle y+ \angle z$$

B
$$\angle y= \angle x+ \angle z$$

C
$$\angle z= \angle x+ \angle y$$

D
$$none\ of\ these$$

Solution

On the basis of given figure, we can conclude
$$\angle 3=\angle 4$$ and $$\angle x=2\angle 3$$

$$\Rightarrow \angle x=\angle 3+\angle 4$$.....(i)

$$\angle y=\angle 3+\angle 1$$........(ii)

$$(i)-(ii)$$ gives,

$$\angle x-\angle y=\angle 4-\angle 1$$.......(iii)

$$\angle 4=\angle z+\angle 1$$

$$\Rightarrow \angle 4-\angle 1=\angle z$$.........(iv)

From (iii) and (iv) we get,

$$\angle x-\angle y=\angle z$$

$$\Rightarrow \angle z+\angle y=\angle x$$

$$\therefore \angle x=\angle y+\angle z$$
Question 9
1 / -0

BC is the diameter of a circle. Points A and D are situated on the circumference of the semi circle $$\angle$$ABD$$=35^o$$ and $$\angle$$BCD$$=60^o$$, $$\angle$$ADB equal to?

A
$$20^o$$

B
$$25^o$$

C
$$30^o$$

D
$$115^o$$

Solution
Question 10
1 / -0

$$ABCD$$ is a cyclic quadrilateral such that $$\angle ADB=30^{o}$$ and $$\angle DCA=80^{o}$$, then $$\angle DAB=$$?

A
$$70^{o}$$

B
$$100^{o}$$

C
$$150^{o}$$

D
$$125^{o}$$

Solution

Given,
$$\angle ADB = 30^0$$ and $$ \angle DCA = 80^0$$.

Now,
$$\angle DCA = \angle DBA = 80^0$$ (Angle subtended by same chord are equal).

In $$\triangle ADB$$,
$$\angle ADB + \angle DBA + \angle DAB = 180^0$$ (Sum of angles in triangle is $$180^0$$)
$$\implies$$ $$\angle DAB = 180^0 - 80^0 - 30^0 = 70^0$$.

Hence, the measure of $$\angle DAB$$ is $$70^0.$$
Therefore, option $$A$$ is correct.