Circles Test

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Circles Test - 26

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Weekly Quiz Competition

Question 1
1 / -0

In the figure, $$\angle BDC=$$

A
$$95^{\circ}$$

B
$$105^{\circ}$$

C
$$100^{\circ}$$

D
$$110^{\circ}$$

Solution

$$\quad In\angle BOC\quad is\quad a\quad central\quad angle\quad =150(Given)\\ \angle BAC=\frac { 1 }{ 2 } \angle BOC[\because Angle\quad at\quad centre\quad is\quad double\quad the\quad angle\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad on\quad remaining\quad part\quad of\quad circle]\\ \Rightarrow \angle BAC=75...(1)\\ In\quad a\quad cyclic\quad quadilateral\quad ACDB\\ \angle BAC+\angle BDC=180\\ \Rightarrow \angle BDC=180-\angle BAC\\ \Rightarrow \angle BDC=180-75.....(From\quad 1)\\ \Rightarrow \angle BDC=105$$
Question 2
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$$ABCD$$ is a cyclic quadrilateral $$AE$$ is drawn parallel to $$CD$$ and $$BA$$ is produced. If $$\displaystyle \angle ABC=92^{\circ}$$ and $$\displaystyle \angle FAE=20^{\circ},$$ then $$\displaystyle \angle BCD=$$

A
$$\displaystyle 88^{\circ}$$

B
$$\displaystyle 108^{\circ}$$

C
$$\displaystyle 115^{\circ}$$

D
$$\displaystyle 72^{\circ}$$

Solution

We know that exterior angle of a cyclic quadrilateral is equal to the interior opposite angle
Also, in a cyclic quadrilateral $$ABCD,$$ sum of opposite angles is $$180^{\circ}$$

$$\angle ADC+ \angle ABC= 180^{\circ}$$

$$\angle ADC+92^{\circ}$$ = $$180^{\circ}$$

$$\angle ADC =88^{\circ}$$

$$\angle EAD=\angle ADC = 88^{\circ}$$ (Alternate interior angles of parallel lines $$CD$$ and $$AE$$)

$$\angle FAD=20^{\circ}$$+$$88^{\circ}=108^{\circ}$$

$$\angle BCD= \angle FAD =108^{\circ}$$ (Exterior angle of a cyclic quadrilateral)
Question 3
1 / -0

In the given figure, if $$O$$ is centre, then external $$\angle AOC=$$

A
$$190^{\circ}$$

B
$$160^{\circ}$$

C
$$200^{\circ}$$

D
$$185^{\circ}$$

Solution

$$In\quad the\quad given\quad figure\\ \angle ADC+\angle CDE=180^° (Linear\quad pair\quad of\quad angles)\\ \angle CDE=80^° (Given)\\ \angle ADC=180^°-\angle CDE\\ \quad \quad \quad \quad \quad =180^°-80^°=100^°\\ Also,\angle AOC=2\angle ADC\\ [\because Angle\quad at\quad centre\quad is\quad double\quad the\quad angle\quad on\quad remaining\quad part\quad of\quad circle]\\ Hence\angle AOC=2\times 100^°=200^°\\ Hence\quad option(C)\quad is\quad right\quad answer$$
Question 4
1 / -0

In the given figure, $$\angle DBC=22^{\circ}\;and\;\angle DCB=78^{\circ}$$ then $$\angle BAC$$ is equal to:

A
$$90^{\circ}$$

B
$$80^{\circ}$$

C
$$78^{\circ}$$

D
$$22^{\circ}$$

Solution

Let $$ \angle BAC = x $$
$$ \therefore \angle BDC = x $$ (angle subtended by chord on same segment are equal)
In $$ \Delta BDC $$,
$$ x+22^{\circ}+78^{\circ} = 180^{\circ} $$
$$ x+100 = 180^{\circ} $$
$$ \boxed {x = 80^{\circ}} = \angle BAC$$
Ans (B)
Question 5
1 / -0

In the given figure, $$CDEF$$ is a cyclic quadrilateral, $$DE\;$$and$$\;CF$$ are produced to $$A\;$$and$$\;B$$ respectively such that $$AB\;\parallel\;CD$$. If $$\angle FED=80^{\circ}$$, find $$\angle FBA$$.

A
$$30^{\circ}$$

B
$$60^{\circ}$$

C
$$80^{\circ}$$

D
$$10^{\circ}$$

Solution

$$CDEF\quad is\quad a\quad cyclic\quad quadrilateral.\\ \angle FED=80^o(Given)\\ \angle FED+\angle FCD=180^o ...(sum\quad of opposite\quad angles\quad of\quad cyclic\quad quadrilateral\quad is\quad 180^o)\\ \Rightarrow FCD=180^o-80^o=100^o.\\ Since\quad AB\parallel CD,\\ \Rightarrow ABCD\quad is\quad a\quad parallelogram.\\ \angle ABC+\angle BCD=180^o ...(Adjacent\quad angles\quad of\quad parallelogram.\quad are\quad supplementary)\\ \angle BCD=\angle FCD=100^o\\ \Rightarrow \angle ABC=180^o-100^o=80^o\\ \angle ABC=\angle FBA=80^o.\\ Hence,\quad option\quad C \quad is\quad correct.$$
Question 6
1 / -0

$$PQ$$ is a diameter and $$PQRS$$ is a cyclic quadrilateral. If $$\angle PSR=150^o$$, then measure of $$\angle RPQ$$ is:

A
$$90^{\circ}$$

B
$$60^{\circ}$$

C
$$30^{\circ}$$

D
None of these

Solution

$$PQRS$$ is a cyclic quadrilateral.
Then, $$\angle PQR +\angle PSR= 180^{\circ}$$ ...[Opposite angles of cyclic quadrilateral are supplementary]
$$\implies$$ $$\angle PQR +150^o= 180^{\circ}$$
$$\implies$$ $$\angle PQR = 180^{\circ} - 150^{\circ} = 30^{\circ}$$.

In $$\triangle PQR$$,
$$\angle PRQ = 90^{\circ}$$ (Angle of a semicircle)
Then, $$\angle RPQ + 90^{\circ} + 30^{\circ} = 180^{\circ}$$ ...[Angle sum property]
$$\Rightarrow \angle RPQ + 120^{\circ} = 180^{\circ}$$
$$\Rightarrow \angle RPQ = 60^{\circ}$$.

Hence, option $$B$$ is correct.
Question 7
1 / -0

In figure, $$O$$ is centre, then $$\angle BXD=$$

A
$$65^{\circ}$$

B
$$60^{\circ}$$

C
$$70^{\circ}$$

D
$$55^{\circ}$$

Solution

Given $$\angle AOC={ 95 }^{ o }$$
$$\angle ABC=\angle ADC=\cfrac { 95^o }{ 2 } \; \; (Half\; Angle)\\ \angle EBX=\angle EOX={ 180 }^{ o }-\cfrac { 95^o }{ 2 } =\cfrac { 265^o }{ 2 } $$
In quadrilateral $$BEXD$$
$$\angle BED+\angle EBX+\angle BXD+\angle XDE=360^o \\ 25^o+\cfrac{265^o}{2}+\angle BXD+\cfrac{265^o}{2}=360^o \\ \angle BXD=360^o-265^o-25^o \\ \angle BXD=70^o$$
Question 8
1 / -0

$$ABCD$$ is a cyclic quadrilateral inscribed in a circle with the centre $$O$$. Then $$\angle OAD$$ is equal to:

A
$$30^{\circ}$$

B
$$40^{\circ}$$

C
$$50^{\circ}$$

D
$$60^{\circ}$$

Solution

$$ Given-\quad \\ ABCD\quad is\quad a\quad cyclic\quad quadrilateral\quad inscribed\quad in\quad a\\ circle\quad with\quad centre\quad O.\\ OA,\quad OB,\quad OC\quad \& \quad OD\quad have\quad been\quad joined.\\ \angle OAB={ 40 }^{ o },\quad \angle OBC={ 30 }^{ o }\quad \& \quad \angle OCD={ 50 }^{ o }.\\ To\quad find\quad out-\\ \angle OAD=?\\ Solution-\\ OC=OB\quad (radii\quad of\quad the\quad same\quad circle)\\ \therefore \quad \Delta OBC\quad is\quad an\quad isosceles\quad triangle.\\ \therefore \quad \angle OCB=\angle OBC={ 30 }^{ o }.\\ \therefore \quad \angle BCD=\angle OCD+\angle OCB={ 50 }^{ o }+{ 30 }^{ o }={ 80 }^{ o }.\\ Now,\quad ABCD\quad is\quad a\quad cyclic\quad quadrilateral.\\ So,\quad by\quad angle\quad sum\quad property\quad of\quad a\quad cyclic\quad quadrilateral,\\ we\quad get,\\ \angle BAD={ 180 }^{ o }-\angle BCD={ 180 }^{ o }-{ 80 }^{ o }={ 100 }^{ o }.\\ So,\quad \angle OAD=\angle BAD-\angle OAB={ 100 }^{ o }-{ 40 }^{ o }={ 60 }^{ o }.\\ Hence, \quad option\quad D \quad is \quad correct.$$
Question 9
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$$ABCD$$ is a cyclic quadrilateral. Then, find $$\angle x^{\circ}$$ as given in the figure.

A
$$50^{\circ}$$

B
$$80^{\circ}$$

C
$$90^{\circ}$$

D
$$100^{\circ}$$

Solution

$$\angle EDC+\angle ADC=180^o ...(Linear\quad pair\quad of\quad angles)\\ \angle EDC=80^o ...(Given)\\ Then, \angle EDC+\angle ADC=180^o\\ \angle ADC=180^o-\angle EDC=180^o-80^o=100^o.\\ ABCD\quad is\quad a\quad cyclic\quad quadrilateral.\\ Hence,\quad \angle ADC+\angle ABC=180^o ...(Sum\quad of\quad opposite\quad angles\quad is\quad 180^o)\\ \Rightarrow \angle ABC=180^o-\angle ADC\\ \quad \quad \quad \quad \quad \quad \quad =180^o-100^o\\ \quad \quad \quad \quad \quad \quad \quad =80^o.\\ Now, \angle ABF+\angle ABC=180^o\\ \angle ABF=180^o-\angle ABC\\ \angle ABF=\angle x ...(Given)\\ \Rightarrow \angle x=180^o-80^o\\ \Rightarrow \angle x=100^o.\\ Hence,\quad option \quad D \quad is\quad correct.$$
Question 10
1 / -0

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

A
$$30^{\circ}$$

B
$$40^{\circ}$$

C
$$60^{\circ}$$

D
$$90^{\circ}$$

Solution

Given- $$\overline { AOB }$$ is a diameter of a given circle. $$ABCD$$ is a cyclic quadrilateral. $$AC$$ is joined. Also, $$\angle ADC={ 130 }^{ \circ }$$.

Since, $$ \overline { AOB }$$ is the diameter of the given circle, it subtends $$\angle ACB$$ to the circumference at $$C$$, i.e. $$ \angle ACB={ 90 }^{ \circ }$$ ...[ since it is an angle in a semicircle].

Again,
$$\angle ADC+\angle ABC=180^\circ$$ ...[ sum of opposite angles of a cyclic quadrilateral is $${ 180 }^{ \circ }$$]
$$\Rightarrow \angle ABC={ 180 }^{ \circ }-\angle ADC$$
$$\Rightarrow \angle ABC={ 180 }^{ \circ }-130^{ \circ }$$
$$\Rightarrow\angle ABC={ 42 }^{ \circ}$$.

In $$\triangle ABC,$$
$$ \angle ABC +\angle ACB+\angle CAB=180^\circ$$
$$\angle CAB={ 180 }^{ \circ }-(\angle ACB+\angle ABC)$$
$$={ 180 }^{ \circ }-({ 90 }^{ \circ }+{ 50 }^{ \circ })$$
$$={ 40 }^{ \circ }$$

Hence, option $$B$$ is correct.