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CBSE Questions for Class 9 Maths Circles Quiz 6 - MCQExams.com
CBSE
Class 9 Maths
Circles
Quiz 6
In the given figure, if $$O$$ is centre, then external $$\angle AOC=$$
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$$190^{\circ}$$
0%
$$160^{\circ}$$
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$$200^{\circ}$$
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$$185^{\circ}$$
Explanation
$$In\quad the\quad given\quad figure\\ \angle ADC+\angle CDE=180^° (Linear\quad pair\quad of\quad angles)\\ \angle CDE=80^° (Given)\\ \angle ADC=180^°-\angle CDE\\ \quad \quad \quad \quad \quad =180^°-80^°=100^°\\ Also,\angle AOC=2\angle ADC\\ [\because Angle\quad at\quad centre\quad is\quad double\quad the\quad angle\quad on\quad remaining\quad part\quad of\quad circle]\\ Hence\angle AOC=2\times 100^°=200^°\\ Hence\quad option(C)\quad is\quad right\quad answer$$
In the given figure, $$\angle DBC=22^{\circ}\;and\;\angle DCB=78^{\circ}$$ then $$\angle BAC$$ is equal to:
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$$90^{\circ}$$
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$$80^{\circ}$$
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$$78^{\circ}$$
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$$22^{\circ}$$
Explanation
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)
In the given figure determine the value of x.
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$$90^{\circ}$$
0%
$$115^{\circ}$$
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$$130^{\circ}$$
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$$65^{\circ}$$
Explanation
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 $$\angle FBA$$.
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$$30^{\circ}$$
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$$60^{\circ}$$
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$$80^{\circ}$$
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$$10^{\circ}$$
Explanation
$$CDEF\quad is\quad a\quad cyclic\quad quadrilateral.\\ \angle FED=80^o(Given)\\ \angle FED+\angle FCD=180^o ...(sum\quad of opposite\quad angles\quad of\quad cyclic\quad quadrilateral\quad is\quad 180^o)\\ \Rightarrow FCD=180^o-80^o=100^o.\\ Since\quad AB\parallel CD,\\ \Rightarrow ABCD\quad is\quad a\quad parallelogram.\\ \angle ABC+\angle BCD=180^o ...(Adjacent\quad angles\quad of\quad parallelogram.\quad are\quad supplementary)\\ \angle BCD=\angle FCD=100^o\\ \Rightarrow \angle ABC=180^o-100^o=80^o\\ \angle ABC=\angle FBA=80^o.\\ Hence,\quad option\quad C \quad is\quad correct.$$
In figure, $$O$$ is centre, then $$\angle BXD=$$
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$$65^{\circ}$$
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$$60^{\circ}$$
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$$70^{\circ}$$
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$$55^{\circ}$$
Explanation
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 $$BEXD$$
$$\angle BED+\angle EBX+\angle BXD+\angle XDE=360^o \\ 25^o+\cfrac{265^o}{2}+\angle BXD+\cfrac{265^o}{2}=360^o \\ \angle BXD=360^o-265^o-25^o \\ \angle BXD=70^o$$
In the figure, $$O$$ is the center. If $$\angle MON=80^o$$, then $$\angle MQN$$ equals
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$$40^o$$
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$$160^o$$
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$$100^o$$
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$$10^o$$
Explanation
Angle subtended by an arc at the center is double the angle subtended by the same are at any point on the circumference.
$$\angle MON\quad =80^o\\ \angle MQN=\frac { 80^o }{ 2 } =40^o$$
$$PQ$$ is a diameter and $$PQRS$$ is a cyclic quadrilateral. If $$\angle PSR=150^o$$, then measure of $$\angle RPQ$$ is:
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$$90^{\circ}$$
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$$60^{\circ}$$
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$$30^{\circ}$$
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None of these
Explanation
$$PQRS$$ 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.
$$ABCD$$ is a cyclic quadrilateral inscribed in a circle with the centre $$O$$. Then $$\angle OAD$$ is equal to:
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$$30^{\circ}$$
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$$40^{\circ}$$
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$$50^{\circ}$$
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$$60^{\circ}$$
Explanation
$$ Given-\quad \\ ABCD\quad is\quad a\quad cyclic\quad quadrilateral\quad inscribed\quad in\quad a\\ circle\quad with\quad centre\quad O.\\ OA,\quad OB,\quad OC\quad \& \quad OD\quad have\quad been\quad joined.\\ \angle OAB={ 40 }^{ o },\quad \angle OBC={ 30 }^{ o }\quad \& \quad \angle OCD={ 50 }^{ o }.\\ To\quad find\quad out-\\ \angle OAD=?\\ Solution-\\ OC=OB\quad (radii\quad of\quad the\quad same\quad circle)\\ \therefore \quad \Delta OBC\quad is\quad an\quad isosceles\quad triangle.\\ \therefore \quad \angle OCB=\angle OBC={ 30 }^{ o }.\\ \therefore \quad \angle BCD=\angle OCD+\angle OCB={ 50 }^{ o }+{ 30 }^{ o }={ 80 }^{ o }.\\ Now,\quad ABCD\quad is\quad a\quad cyclic\quad quadrilateral.\\ So,\quad by\quad angle\quad sum\quad property\quad of\quad a\quad cyclic\quad quadrilateral,\\ we\quad get,\\ \angle BAD={ 180 }^{ o }-\angle BCD={ 180 }^{ o }-{ 80 }^{ o }={ 100 }^{ o }.\\ So,\quad \angle OAD=\angle BAD-\angle OAB={ 100 }^{ o }-{ 40 }^{ o }={ 60 }^{ o }.\\ Hence, \quad option\quad D \quad is \quad correct.$$
$$ABCD$$ is a cyclic quadrilateral. Then, find $$\angle x^{\circ}$$ as given in the figure.
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$$50^{\circ}$$
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$$80^{\circ}$$
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$$90^{\circ}$$
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$$100^{\circ}$$
Explanation
$$\angle EDC+\angle ADC=180^o ...(Linear\quad pair\quad of\quad angles)\\ \angle EDC=80^o ...(Given)\\ Then, \angle EDC+\angle ADC=180^o\\ \angle ADC=180^o-\angle EDC=180^o-80^o=100^o.\\ ABCD\quad is\quad a\quad cyclic\quad quadrilateral.\\ Hence,\quad \angle ADC+\angle ABC=180^o ...(Sum\quad of\quad opposite\quad angles\quad is\quad 180^o)\\ \Rightarrow \angle ABC=180^o-\angle ADC\\ \quad \quad \quad \quad \quad \quad \quad =180^o-100^o\\ \quad \quad \quad \quad \quad \quad \quad =80^o.\\ Now, \angle ABF+\angle ABC=180^o\\ \angle ABF=180^o-\angle ABC\\ \angle ABF=\angle x ...(Given)\\ \Rightarrow \angle x=180^o-80^o\\ \Rightarrow \angle x=100^o.\\ Hence,\quad option \quad D \quad is\quad correct.$$
In the given figure, the value of $$'a'$$ is:
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$$30^{\circ}$$
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$$40^{\circ}$$
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$$60^{\circ}$$
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$$90^{\circ}$$
Explanation
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.
In the circle, reflex $$\angle AOC=x^o, \angle ABC=y^o$$. $$OABC$$ is a parallelogram, then find $$y$$.
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$$80^o$$
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$$120^o$$
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$$110^o$$
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Data insufficient
Explanation
In the circle, clearly, $$x=2y$$.
Then, $$\angle A OC=360-x=360-2y$$.
Here, $$ ABCD$$ is a parallelogram
Then, $${BCO}+ {AOC}=180^o$$
$$\Rightarrow {BCO}+360-2y=180^o$$
$$\Rightarrow 2y=BCO+180^o$$.
Similarly, $$ABC+BCO=180^o$$
$$\Rightarrow y+2y-180=180$$
$$\Rightarrow 3y=360^o$$
$$\Rightarrow y=120^o$$.
Hence, option $$B$$ is correct.
Find $$\angle ASR$$.
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$$52^{\circ}$$
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$$78^{\circ}$$
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$$102^{\circ}$$
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$$10^{\circ}$$
Explanation
$$PABQ\quad is\quad a\quad cyclic\quad quadilateral\\ \angle QPA=78^o...(Given). \\ Then, \angle QPA+\angle ABQ=180^o\\ \Rightarrow 78^o+\angle ABQ=180^o\\ \Rightarrow \angle ABQ=180^o-78^o=102^o....(1).\\ \angle ABR\quad and\quad \angle ABQ\quad are\quad linear\quad pair\quad of\quad angles\quad on\quad straight\quad line\quad QR\\ Then, \angle ABQ+\angle ABR=180^o\\ \Rightarrow \angle ABR=180^o-\angle ABQ=180^o-102^o=78^o.\\ Hence,\quad \angle ABR\quad =78^o.\\ In\quad cyclic\quad quadilateral\quad ABCD,\quad \\ \angle ASR+\angle ABR=180^o\\ \Rightarrow \angle ASR=180^o-78^o\\ \Rightarrow \angle ASR=102^o.\\ Hence,\quad option\quad C \quad is\quad correct.$$
In the given figure, $$PQRS$$ is a cyclic quadrilateral. Its diagonals $$PR$$ and $$QS$$ intersect each other at $$T$$. If $$\angle PRS=80^{\circ}\;and\;\angle RQS=50^{\circ}$$, calculate $$\angle PSR$$.
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$$30^{\circ}$$
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$$50^{\circ}$$
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$$70^{\circ}$$
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$$90^{\circ}$$
Explanation
$$Given:\\ PQRS\ is\ a\ cyclic\ quadilateral.\\ \angle PRS=80^o....(1)\\ \angle RQS=50^o\\ \angle RQS=\angle RPS\ (Angles\ in\ the\ same\ segment\ are\ equal)\\ \Rightarrow \angle RPS=50^o...(2)\\ In\quad \triangle PSR\\ \angle PRS+\angle RPS+\angle PSR=180^o\ (Angle\ sum\ property)\\ 80^o+50^o+\angle PSR=180^o\ (From\ (1)\ and\ (2))\\ \Rightarrow \angle PSR=180^o-130^o\\ \ \ \ \ \angle PSR=50^o\\ Hence\ option\ (B)\ is\ right\ $$
In figure, $$\angle BAC = 60^{\circ}$$ and $$\angle BCA = 20^{\circ}$$ , find $$\angle ADC$$.
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15$$^{\circ}$$
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50$$^{\circ}$$
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80$$^{\circ}$$
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40$$^{\circ}$$
Explanation
In $$\Delta ABC$$, we have
$$\angle BAC + \angle BCA + \angle ABC = 180^{\circ}$$....[Angle sum property]
$$\Rightarrow 60^{\circ} + 20^{\circ} + \angle ABC = 180^o$$
$$\therefore \angle ABC = 180^{\circ} - 80^{\circ} = 100^{\circ}$$.
$$ABCD$$ is a cyclic quadrilateral,
$$\therefore \angle ABC + \angle ADC = 180^{\circ}$$.....
[Since, opposite angles of a cyclic quadrilateral are supplementary]
$$\implies$$ $$100^{\circ} + \angle ADC = 180^{\circ}$$
$$\implies \angle ADC = 180^{\circ} - 100^{\circ} = 80^{\circ}$$.
Hence, option $$C$$ is correct.
In the figure, $$\angle$$ B is equal to:
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$$85^0$$
0%
$$95^0$$
0%
$$70^0$$
0%
$$115^0$$
Explanation
Since, $$APQD$$ is a cyclic quadrilateral,
therefore
$$\angle A + \angle DQP = 180^o$$ ...(opposite angles of cyclic quadrilateral are supplementary)
$$\implies$$ $$\angle DQP = 180^o -85^o =95^o$$.
Now,
$$\angle PQC = 180^o - \angle DQP=180^o-95^o=85^o$$ ...[linear pairs].
$$\therefore \angle B = 180^o -85^o =95^o$$
...(opposite angles of cyclic quadrilateral are supplementary).
Hence, option $$B$$ is correct.
In the figure above, if O is the centre of the circle, find the value of y.
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40$$^{\circ}$$
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70$$^{\circ}$$
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60$$^{\circ}$$
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30$$^{\circ}$$
Explanation
Given that, the centre of the circle is $$O$$ and $$\angle BOD=140^\circ$$
To find out: The value of $$y$$
We know that, in a circle, the angle subtended by an arc at the centre is twice the angle subtended by the same arc at any point on the circle.
$$\therefore \ \angle BOD = 2 \angle BAD$$
$$\Rightarrow 140^{\circ} = 2 \angle BAD$$
$$\displaystyle \Rightarrow \angle BAD = \frac{1}{2} \times 140^{\circ} = 70^{\circ}$$
Now, in cyclic quadrilateral ABCD,
$$\angle BAD + \angle BCD = 180^{\circ}$$ [Opposite angles of a cyclic quadrilateral are supplementary]
$$\therefore \ 70^{\circ} + \angle BCD = 180^{\circ}$$
$$\therefore \angle BCD = 180^{\circ} - 70^{\circ} = 110^{\circ}$$
Also, $$\angle BCD + \angle DCP = 180^{\circ}$$ [Linear pair]
$$\Rightarrow 110^{\circ} + y = 180^{\circ}$$
$$\therefore \ y = 70^{\circ}$$
Hence, option B is correct.
In a cyclic quadrilateral $$ABDC,\,\angle CAB=80^{\circ}$$ and $$\angle ABC=40^{\circ}$$. The measure of the $$\angle ADB$$ will be?
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$$120^{\circ}$$
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$$80^{\circ}$$
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$$60^{\circ}$$
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$$40^{\circ}$$
Explanation
In $$\Delta ACB$$,
$$\because\;\angle ACB=180^{\circ}-(80^{\circ}+40^{\circ})=60^{\circ}$$
$$\therefore\;\angle ADB=\angle ACB$$
$$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=60^{\circ}$$
In fig, a is the centre of the circle. If $$\angle$$ BOD 160$$^{\circ}$$ find the values of x and y.
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80$$^{\circ}$$, 100$$^{\circ}$$
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135$$^{\circ}$$, 45$$^{\circ}$$
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140$$^{\circ}$$, 40$$^{\circ}$$
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120$$^{\circ}$$, 60$$^{\circ}$$
Explanation
In the cyclic quadrilateral ABCD
$$\displaystyle \angle BCD = \frac{1}{2} \angle BOD$$
[Angle made by an arc at the centre is double the angle made by it, at any other point on the remaining part of the circle]
$$\displaystyle \therefore \angle x = \frac{1}{2} \times 160^{\circ} = 80^{\circ}$$
$$\therefore x = 80^{\circ}$$
$$\angle x + \angle y = 180^{\circ}$$
[opp. angles of a cyclic quadrilateral]
$$\therefore \angle y = 180^{\circ} - \angle x$$
$$\angle y = 180^{\circ} - 80^{\circ} = 100^{\circ}$$
$$y = 100^{\circ}$$
One of the base angle a cyclic trapezium is double the other. What is the measure of the larger angle?
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$$60^0$$
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$$80^0$$
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$$75^0$$
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$$120^0$$
Explanation
In cyclic trapezium,
Sum of opposite angle $$=180^o$$ ...[opposite angles of a cyclic quadrilateral are supplementary].
Let the smaller angle be $$x$$
and larger angle $$=2x$$.
Then, $$2x+x=180^o \implies 3x=180^o \implies x=60^o$$.
Therefore, Larger angle $$=2x=2\times 60^o=120^o$$.
Hence, option $$D$$ is correct.
Find $$x.$$
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$$40^{\circ}$$
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$$50^{\circ}$$
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$$30^{\circ}$$
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$$60^{\circ}$$
Explanation
ABCD is a quadrilateral inside the circle,
$$\angle D + \angle B = 180^o$$
$$130^o + \angle B = 180^o$$
$$\angle B = 50^o$$
AB is diameter $$\angle C = 90^o$$ [angle in a semicircle]
In $$\Delta ABC$$
$$\angle A + \angle B + \angle C = 180^o$$
$$\angle A + 50^o + 90^o = 180^o$$
$$\Rightarrow \angle A + 140^o = 180^o$$
$$\Rightarrow \angle A = 180^o - 140^o$$
$$=40^o$$
$$\therefore \angle CAB = 40^o \Rightarrow x = 40^o$$
option (A) is correct.
ABCD is cyclic quadrilateral. The tangents to a circle at A and C meet at P. If $$\angle$$ APC = 50$$^{\circ} $$, then the value of $$\angle$$ ADC is:
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$$ \displaystyle 65^{\circ} $$
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$$ \displaystyle 70^{\circ} $$
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$$ \displaystyle 80^{\circ} $$
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none of these
Explanation
Given $$\angle APC=50^0 $$
As the radius is always perpendicular to tangent at the point of contact,
$$\angle OCP=\angle OAP=90^0$$
As $$AOCP $$ is a quadrilateral, sum of all its interior angles are equal to $$360^0$$
$$\therefore\ \angle AOC=130^0\quad \quad [360^0-(90^0+90^0+50^0)]$$
We know that, angle subtended by an arc at the center of a circle is twice the angle subtended by the same arc on the circumference of the circle.
$$\therefore \ \angle ADC=\dfrac{130^0}{2}$$
$$=65^0$$
Hence, $$\angle ADC=65^0$$.
If $$O$$ is the centre of the circle and $$A, B$$ and $$C$$ are points on its circumference and $$\angle AOC = 130^{\circ}$$, find $$\angle ABC$$.
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$$105^{\circ}$$
0%
$$115^{\circ}$$
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$$110^{\circ}$$
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$$120^{\circ}$$
Explanation
Take any point $$P$$ on the circumference of the circle as shown.
Join $$AP$$ and $$CP$$.
$$\because ABC$$ subtends $$\angle AOC$$ at centre $$O$$ and $$\angle APC$$ at any point $$P$$ on the circumference of the circle.
$$\therefore \angle AOC = 2 \angle APC$$
[$$\because$$ Angle subtended by an arc at the centre is twice the angle subtended by the same arc at any point on the circumference]
$$\displaystyle \implies \angle APC = \frac{1}{2} \angle AOC ~~~[ \because AOC = 130^{\circ}]$$
$$\implies$$ $$\displaystyle \frac{1}{2} \times 130^{\circ} = 65^{\circ}$$.
$$\because$$ $$ABCP$$ is a cycle quadrilateral,
$$\implies \angle APC + \angle ABC = 180^{\circ}$$ ...
[$$\because$$ sum of opposite angles of a cyclic quadrilateral is $$180^{\circ}$$]
$$\implies 65^{\circ} + \angle ABC = 180^{\circ}$$
$$\therefore \angle ABC = 180^{\circ} - 65^{\circ} = 115^{\circ}$$.
Hence, option $$B$$ is correct.
$$ABCD$$ is a cyclic quadrilateral, in which $$BC$$ is parallel to $$AD$$, $$\displaystyle\angle ADC=\displaystyle 110^{\circ}$$ and $$\displaystyle \angle BAC=50^{\circ}$$. What is the value of $$\displaystyle \angle DAC$$?
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$$\displaystyle 40^{\circ}$$
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$$\displaystyle 50^{\circ}$$
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$$\displaystyle 60^{\circ}$$
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$$\displaystyle 64^{\circ}$$
Explanation
$$ABCD$$ is a cyclic quadrilateral.
Then, $$\angle ABC+\angle ADC=180^o$$ ...[Opposite angles of a cyclic quadrilateral are supplementary]
$$\Rightarrow$$ $$\angle ABC =180^{\circ}-110^{\circ} =70^{\circ}$$.
In $$\triangle ABC$$,
$$\angle ABC+\angle ACB+\angle BAC=180^o$$ ...[Angle sum property]
$$\Rightarrow$$
$$70^o+\angle ACB+50^o=180^o$$
$$\Rightarrow$$ $$ \angle ACB=180^{\circ}-(50^{\circ}+70^{\circ})=60^{\circ}$$.
Given, $$BC$$ and $$AD$$ are parallel,
$$\Rightarrow \angle ACB=\angle DAC =60^{\circ}$$ ...[Alternate interior angles].
Hence, option $$C$$ is correct.
In the given figure, AB=AC and $$\displaystyle \angle ABC=68^{\circ},$$ what is the value of $$\displaystyle \angle BPC$$?
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$$\displaystyle 40^{\circ}$$
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$$\displaystyle 42^{\circ}$$
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$$\displaystyle 44^{\circ}$$
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$$\displaystyle 48^{\circ}$$
Explanation
Theorem: Chord of a circle subtends equal angles at different points on the circumference of the circle on same side
$$\Rightarrow \angle BAP=\angle BPC$$
$$ AB=AC\Rightarrow \angle B=\angle C=68^{°}$$ (Opposite angles of equal sides are equal)
In $$\triangle ABC $$
$$\angle A+ \angle B + \angle C =180^o$$ (Sum of angles in a triangle )
$$68^o+68^o+ \angle A = 180^o$$
$$136^o+\angle A = 180^o$$
$$\Rightarrow \angle A=44^{°}$$
$$\Rightarrow \angle A=\angle BAC =\angle BPC =44^{°}$$ (Angles in same segment are equal )
In the given figure, $$BD=DC$$ and $$\displaystyle \angle DBC=30^{\circ}$$ what is the value of $$\displaystyle \angle BAC$$?
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$$\displaystyle 40^{\circ}$$
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$$\displaystyle 50^{\circ}$$
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$$\displaystyle 60^{\circ}$$
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$$\displaystyle 65^{\circ}$$
Explanation
As $$BD=DC$$ ......
(given)
$$\Rightarrow \angle DBC=\angle DCB=30^{\circ}$$
$$\Rightarrow \angle BDC =180^\circ -30^\circ-30^\circ=120^{\circ}$$
As $$ABCD$$ is a cyclic quadrilateral ; $$\angle BAC+\angle BDC =180^{\circ}$$
$$\Rightarrow \angle BAC =180^{\circ}-120^{\circ} =60^{\circ}$$
What is the value of $$ \displaystyle \angle ABC $$ ?
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$$ \displaystyle 105^{o} $$
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$$ \displaystyle 110^{o} $$
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$$ \displaystyle 115^{o} $$
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$$ \displaystyle 120^{o} $$
Explanation
Extending A and C onto the circumference to D as shown in the figure
Theorem: Angle subtended by an arc at the centre is double the angle subtended by the same arc at the circumference of a circle
$$\Rightarrow \angle AOC=2\times \angle ADC $$
$$\Rightarrow \angle ADC=\dfrac {130}{2}=65^o$$
As $$ABCD$$ is a cyclic quadrilateral, sum of its opposite interior angles are equal to $$180^o$$
$$\Rightarrow \angle ADC+\angle CBA =180^o$$
$$\Rightarrow \angle ABC =180^o - 65^o=115^o$$
ABCD is a cyclic trapezium and $$ \displaystyle AB\parallel CD $$. If $$AB$$ is the diameter of the circle and
$$ \displaystyle \angle CAB=30^{\circ} $$, then the value of $$ \displaystyle \angle ADC $$ is:
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$$ \displaystyle 110^{\circ}$$
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$$ \displaystyle 115^{\circ}$$
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$$ \displaystyle 120^{\circ}$$
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$$ \displaystyle 125^{\circ}$$
Explanation
Here, $$\angle CAB=30^o$$ ...[Given].
Also, $$\angle ACB=90^o$$ ...[Angle inscribed in a semi-circle].
Now, in $$\triangle ACB$$,
$$\angle ACB+\angle CAB+\angle ABC=180^o$$ ...[Angle sum property]
$$\Rightarrow$$ $$90^o+30^o+\angle ABC=180^o$$
$$\Rightarrow$$
$$\angle ABC=180^o-120^o$$
$$\Rightarrow$$
$$\angle ABC=60^o$$.
Here, $$ABCD$$ is a cyclic quadrilateral.
We know that, sum of opposite angles of a cyclic quadrilateral is $$180^o$$.
$$\therefore$$ $$\angle ABC+\angle ADC=180^o$$
$$\Rightarrow$$
$$60^o+\angle ADC=180^o$$
$$\Rightarrow$$
$$\angle ADC=180^o-60^o$$
$$\Rightarrow$$
$$\angle ADC=120^o.$$
Hence, option $$D$$ is correct.
Find the value of x from the given figure in which O is the center of the circle
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$$ \displaystyle x=50^{\circ} $$
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$$ \displaystyle x=60^{\circ} $$
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$$ \displaystyle x=80^{\circ} $$
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$$ \displaystyle x=90^{\circ} $$
Explanation
$$\textbf{Step - 1 : Find the angle at the center,}$$
$$\text{Given that, angle subtend by segment BC at point A on circle}=40^{\circ}$$
$$\text{Then, the angle subtend by segment BC at center O}=\angle{\text{BOC}}=40^{\circ}\times2=80^{\circ}$$
$$\textbf{Step - 2 : Find the value of x}$$
$$\text{In triangle OBC, OB = OC, as the radii of the same circle.}$$
$$\text{Hence, }\angle\text{OBC}=\angle\text{OCB}=x, \text{as the angles opposite to equal sides are also equal}$$
$$\text{Then by angle sum property of triangle,}$$
$$\angle\text{OBC}+\angle\text{OCB}+\angle\text{BOC}=180^{\circ}\Rightarrow80^{\circ}+x+x=180^{\circ}\Rightarrow2x=180^\circ-80^\circ=100^\circ$$
$$\therefore{x}=\dfrac{100^\circ}{2}=50^\circ$$
$$\textbf{Therefore, option A }\mathbf{x=50^\circ}\textbf{ is correct answer.}$$
Find the values of $$y$$ and $$z$$, where $$O$$ denotes the centre of the circle.
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$$y=70^o$$ and $$z=35^o$$
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$$y=290^o$$ and $$z=145^o$$
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$$y=145^o$$ and $$z=290^o$$
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$$y=35^o$$ and $$z=70^o$$
Explanation
Since the given quadrilateral is a cyclic quadrilateral,
$$\angle y + 35^o =180^o$$
$$\Rightarrow \angle y=180^o-35^o$$
$$\Rightarrow \angle y=145^o$$.
By the inscribed angle theorem, we have,
$$\angle z=2\times \angle y$$
$$\Rightarrow \angle z=2\times145^o$$
$$\Rightarrow \angle z = 290^o$$.
Hence, option $$C$$ is correct.
In given figure, BC is a diameter of the circle and $$\displaystyle \angle BAO=60^{\circ}\,$$. Then$$\,\angle ADC$$ is equal to
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$$\displaystyle 30^{\circ}$$
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$$\displaystyle 45^{\circ}$$
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$$\displaystyle 60^{\circ}$$
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$$\displaystyle 120^{\circ}$$
Explanation
In $$\triangle AOB$$,
$$OA=OB$$
$$=>\angle OAB=\angle OBA=60^{0}$$
We know that the exterior angle of a triangle is equal to sum of the two opposite angles of the triangle.
$$\angle AOC=\angle OAB+\angle OBA$$
$$\angle AOC=60^{0}+60^{0}$$
$$\angle AOC = 120^{0}$$
We know that the angle subtended at the centre of the circle by an arch is twice the angle subtended at the circumference by the same arc.
$$\therefore \angle ADC=\dfrac{1}{2}\angle AOC$$
$$=\dfrac{120}{2}$$
$$=60^{0}$$
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Practice Class 9 Maths Quiz Questions and Answers
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