CBSE Questions for Class 9 Maths Circles Quiz 5 - MCQExams.com

Given, $$ PQRS$$ is a cyclic trapezium with $$PQ\parallel SR, \angle QPS={ 82 }^{ o }.$$

Given, $$ABCD$$ is a cyclic quadrilateral. $$AC$$ is its diagonal. $$\angle BAD={ 25 }^{ o }, \angle BCD={ 80 }^{ o }$$.

Here, $$ABCD$$ is a cyclic quadrilateral. 

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

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

But $$ \angle BAD=\angle BAC+\angle CAD$$

$$\therefore  \angle BAC+\angle CAD=1{ 00 }^{ o } \Rightarrow \angle CAD=1{ 00 }^{ o }-\angle BAC=1{ 00 }^{ o }-25^{ o }=75^{ o }$$.

Hence, option $$C$$ is correct.

$$ \\ Given-\\ The\quad \Delta ABC,\quad whose\quad incentre\quad is\quad I,\quad has\quad been\quad inscribed\quad in\quad a\quad cicle.\\ AI\quad is\quad produced\quad to\quad meet\quad the\quad circumference\quad at\quad D.\\ BD\quad \& \quad DC\quad have\quad been\quad joined.\\ \angle ABC={ 55 }^{ o }\quad \& \quad \angle ACB={ 65 }^{ o }.\\ To\quad find\quad out-\\ (i)\quad \angle BCD=?\quad \quad (ii)\quad \angle CBD=?\quad \quad (iii)\quad \angle DCI=?\quad \quad (iv)\quad \angle BIC=?\\ Solution-\\ \angle ABC=\angle ADC={ 55 }^{ o }\quad [since\quad both\quad the\quad angles\quad have\quad been\quad subtended\quad \\ by\quad the\quad chord\quad AC\quad to\quad the\quad circumference\quad at\quad B\quad \& \quad D].\\ Also, \quad  \angle ACB=\angle ADB={ 65 }^{ o }\quad [since\quad both\quad the\quad angles\quad have\quad been\quad subtended\quad \\ by\quad the\quad chord\quad AB\quad to\quad the\quad circumference\quad at\quad C\quad \& \quad D].\\ So,\quad \angle BDC=\angle ADC+\angle ADB={ 55 }^{ o }+{ 65 }^{ o }=120^{ o }.\\ Now,\quad reflex\angle BIC=2\angle BDC\quad (since\quad reflex\angle BIC\quad \& \quad \angle BDC\quad are\quad angles\quad \\ at\quad the\quad centre\quad \& \quad angle\quad at\quad the\quad circumference\quad at\quad D\quad subtended\quad by\\ the\quad chord\quad BC).\\ \therefore \quad reflex\angle BIC=2\times 120^{ o }=240^{ o }\Longrightarrow \angle BIC=360^{ o }-240^{ o }=120^{ o }.\\ By\quad the\quad same\quad reasoning\quad \angle BAC=\quad \frac { 1 }{ 2 } \times \angle BIC=\frac { 1 }{ 2 } \times 120^{ o }=60^{ o }.\\ Again,\quad I\quad is\quad the\quad incentre\quad of\quad \Delta ABC.\\ \therefore \quad AI,\quad BI\quad \& \quad CI\quad are\quad angular\quad bisector\quad of\quad \\ \angle BAC,\quad \angle ABC\quad \& \quad \angle ACB\quad respectively.\\ \therefore \quad \angle BAI=\frac { 1 }{ 2 } \times \angle BAC=\frac { 1 }{ 2 } \times 60^{ o }=30^{ o }=\angle CAI.\\ So,\quad \angle BCD=\angle BAI=30^{ o }\quad [since\quad \angle BCD\quad \& \quad \angle BAI\quad are\quad angles\quad \\ at\quad the\quad centre\quad \& \quad angle\quad at\quad the\quad circumference\quad at\quad A\quad subtended\quad by\\ the\quad chord\quad BD].\\ By\quad the\quad same\quad reasoning\quad \angle CBD=\angle CAI=30^{ o }.\\ Again,\quad CI\quad is\quad the\quad angular\quad bisector\quad of\quad \angle ACD.\\ \therefore \quad \angle BCI=\angle ACI=\frac {1 }{ 2 } \times 65^{ o }=32.5^{ o }.\\ \therefore \quad \angle DCI=\angle BCD+\angle BCI=32.5^{ o }+30^{ o }=62.5^{ o }.\\ So\quad \angle BCD,\quad \angle CBD,\quad \angle DCI\quad \& \quad \angle BIC\quad are \quad { 30 }^{ o },\quad { 30 }^{ o },\quad { 62.5 }^{ o }\quad \& \quad { 120 }^{ o }\quad respectively.\\ Hence,\quad option\quad B \quad is \quad correct. $$

Since $$AB$$ is diameter,

$$\angle AEB = 90^\circ$$ ...[Angle formed in a semi circle].

In $$\triangle AEB$$

$$\angle ABE = 180 - \angle EAB - \angle AEB$$ ...[Angle sum property]

$$\angle ABE = 180 – 65 -90 = 25^\circ$$.

Given, $$ED \parallel AB$$.

$$\implies \angle DEB = \angle EBA = 25^\circ$$ ...[Alternate interior angles].

Since $$EDCB$$ is a cyclic quadrilateral opposite angles are supplementary

Then, $$\angle DEB + \angle DCB  = 180^\circ $$

$$\implies \angle DCB = 180 – 25 = 155^\circ $$.

Hence, option $$D$$ is correct.

Given that $$O$$ is the center of the smaller circle and $$\angle APB=75^{\circ}$$.

$$ Given-\\ Two\quad circles\quad intersect\quad at\quad P\quad \& \quad Q.\\ The\quad lines\quad APB\quad \& \quad \quad DQC\quad intersect\quad the\quad circles\quad at\\ A,\quad P,\quad B\quad and\quad D,\quad Q,\quad C\quad respectively.\\ AD\quad \& \quad BC\quad have\quad been\quad joined.\\ \angle ADQ={ 80 }^{ o }\quad and\quad \angle DAP={ 84 }^{ o }.\\ To\quad find\quad out-\\ \angle QBC=?\\ \angle BCP=?\\ Solution-\\ We\quad join\quad PQ.\\ ADPQ\quad is\quad a\quad cyclic\quad quadrilateral.\\ \therefore \quad \angle PQA={ 180 }^{ o }-\angle ADP={ 180 }^{ o }-{ 84 }^{ o }=96^{ o }\quad [since \quad \\ the\quad sum\quad of\quad the\quad opposite\quad angles\quad of\quad a\quad cyclic\quad \\ quadrilateral={ 180 }^{ o }].\\ Also\quad \angle PQB={ 180 }^{ o }-\angle PQA={ 180 }^{ o }-{ 96 }^{ o }={ 84 }^{ o }(linear\quad pair).\\ Again\quad QBCP \quad is\quad a\quad cyclic\quad quadrilateral.\\ \therefore \quad \angle BCP={ 180 }^{ o }-\angle BQP={ 180 }^{ o }-{ 84 }^{ o }=96^{ o }.....(i)\quad [since\\ the\quad sum\quad of\quad the\quad opposite\quad angles\quad of\quad a\quad cyclic\quad \\ quadrilateral={ 180 }^{ o }].\\ Similarly, \quad ADPQ\quad is\quad a\quad cyclic\quad quadrilateral.\\ \therefore \quad \angle QPD={ 180 }^{ o }-\angle QAD={ 180 }^{ o }-{ 80 }^{ o }=100^{ o }\quad [since\\ the\quad sum\quad of\quad the\quad opposite\quad angles\quad of\quad a\quad cyclic\quad \\ quadrilateral={ 180 }^{ o }].\\ Also,\quad \angle QPC={ 180 }^{ o }-\angle QPD={ 180 }^{ o }-{ 100 }^{ o }={ 80 }^{ o }(linear\quad pair).\\ Again\quad QBCP\quad \quad is\quad a\quad cyclic\quad quadrilateral.\\ \therefore \quad \angle QBC={ 180 }^{ o }-\angle QPC={ 180 }^{ o }-{ 80 }^{ o }=100^{ o }......(ii)\quad [since\\ the\quad sum\quad of\quad the\quad opposite\quad angles\quad of\quad a\quad cyclic\quad \\ quadrilateral={ 180 }^{ o }].\\ \therefore \quad \quad \angle QBC=100^{ o }\\ and\quad \angle BCP=96^{ o }.\\ Hence, \quad option\quad A \quad is \quad correct.\\ \\ \\  $$

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

$$ Given-\\ ABCD\quad is\quad a\quad cyclic\quad quadrilateral.\\ Its\quad diagonals\quad AB\quad \& \quad CD\quad intersect\quad at\quad E.\\ \angle DBC={ 70 }^{ o },\quad \angle BAC={ 30 }^{ o }.\\ To\quad find\quad out-\\ \angle BCD=?\\ Also,\quad if\quad AB=BC\quad then\quad \angle ECD=?\\ Solution-\\ BC\quad subtends\quad \angle BDC\quad \& \quad \angle BAC\quad to\quad the\quad circumference\quad \\ of\quad the\quad given\quad circle\quad at\quad D\quad \& \quad A\quad respectively.\\ \therefore \quad \angle BDC=\angle BAC={ 30 }^{ o }\quad [since\quad the\quad angles,\quad subtended\quad by\quad \\ a\quad chord\quad of\quad a\quad circle\quad to\quad different\quad points\quad of\quad the\quad circumferece\quad \\ of\quad the\quad same\quad circle,\quad are\quad equal].\\ \therefore \quad \angle BCD=180^{ o }-(\angle DBC+\angle BDC)\quad \left( angle\quad sum\quad property\quad of\quad triangles \right) \\ \Longrightarrow \angle BCD=180^{ o }-({ 70 }^{ o }+{ 30 }^{ o })={ 80 }^{ o }.\\ Again,\quad in\quad \Delta ABC\quad we\quad have\quad AB=BC.\\ i.e\quad \Delta ABC\quad is\quad isosceles\quad with\quad AC\quad as\quad base.\\ \therefore \quad \angle BAC=\angle BCA={ 30 }^{ o }.\\ So\quad \angle ECD=\angle BCD-\angle BCA=\quad { 80 }^{ o }-{ 30 }^{ o }={ 50 }^{ o }.\\ \therefore \quad \angle BCD\quad \& \quad \angle ECD\quad are\quad respectively \quad { 80 }^{ o }\quad \& \quad { 50 }^{ o }.\\ Hence,\quad option\quad C \quad is \quad correct.  $$ 

$$ Given-\\ Two\quad circles\quad intersect\quad at\quad A\quad \& \quad B.\\ The\quad quadrilateral\quad PCBA\quad is\quad inscribed\quad in\quad one\quad circle\quad and\\ the\quad quadrilateral\quad EDBA\quad is\quad inscribed\quad in\quad another\quad circle\\ such\quad that\quad the\quad points\quad P,\quad A\quad \& \quad B\quad are\quad collinear\quad and\\ C,\quad B\quad \& \quad D\quad are\quad collinear.\\ \angle AED=z,\quad \angle APC=95^{ o }\quad \& \quad \angle PCB=40^{ o }.\\ To\quad find\quad out-\\ z=?\\ solution-\\ \angle APC+\angle ABC={ 180 }^{ o }\quad (since\quad the\quad sum\quad of\quad the\quad opposite\quad angles\quad \\ of\quad a\quad cyclic\quad quadrilateral\quad is\quad { 180 }^{ o }).\\ \Longrightarrow \angle ABC={ 180 }^{ o }-\angle APC={ 180 }^{ o }-95^{ o }=85^{ o }.\\ Again\quad \angle ABC+\angle ABD={ 180 }^{ o }(linear\quad pair).\\ \therefore \quad \angle ABD={ 180 }^{ o }-\angle ABC={ 180 }^{ o }-85^{ o }=95^{ o }.\\ Now\quad \angle ABD+\angle AED={ 180 }^{ o }\quad (since\quad the\quad sum\quad of\quad the\quad opposite\quad angles\quad \\ of\quad a\quad cyclic\quad quadrilateral\quad is\quad { 180 }^{ o }).\\ \therefore \quad \angle AED=z={ 180 }^{ o }-\angle ABD={ 180 }^{ o }-95^{ o }=85^{ o }.\\ Hence, \quad option\quad D \quad is \quad correct.$$

Given, $$O$$ is the centre of a circle in which a quadrilateral $$ ABCD$$ has been inscribed.

$$ Given-\\ O\quad is\quad the\quad centre\quad of\quad a\quad circle\quad with\quad diameter\quad AB.\\ AD\quad \& \quad BC\quad are\quad chords\quad which\quad have\quad been\quad extended\quad  to\quad intersect\quad at\quad E.\\ AB\quad \& \quad CD\quad are\quad joined.\\ \angle BAD={ 35 }^{ o }\quad \& \quad \angle BED={ 25 }^{ o }.\\ \\ To\quad find\quad out-\\ \angle DCB,\quad \angle DBC\quad \& \quad \angle BDC\\ \\ Solution-\\ \angle BAD=\angle DCB={ 35 }^{ o }......(i)\quad [since\quad both\quad of\quad them\quad have\quad been\quad \\ subtended\quad by\quad the\quad chord\quad BD\quad to\quad the\quad circumference\quad at\quad A\quad \& \quad C].\\ Now\quad \angle ACB={ 90 }^{ o }\quad since\quad \angle ACB\quad is\quad angle\quad in\quad a\quad semicircle,\quad AB\\ being\quad the\quad diameter.\\ \therefore \quad \angle ACD=\angle ACB-\angle DCB={ 90 }^{ o }-{ 35 }^{ o }={ 55 }^{ o }.\\ Again\quad \angle ABD=\angle ACD={ 55 }^{ o }\quad [since\quad both\quad of\quad them\quad have\quad been\quad \\ subtended\quad by\quad the\quad chord\quad AD\quad to\quad the\quad circumference\quad at\quad B\quad \& \quad C].\\ Now\quad in\quad \Delta ACE,\quad we\quad have\quad \angle CAD={ 180 }^{ o }-(\angle ACB+\angle BEA)={ 180 }^{ o }-({ 90 }^{ o }+{ 25 }^{ o })=65^{ o }.\\ \therefore \quad \angle BAC=\angle CAD-\angle BAD=65^{ o }-35^{ o }=30^{ o }.\\ \therefore \quad \angle BDC=\angle BAC=30^{ o }.......(ii)\quad [since \quad both\quad of\quad them\quad have\quad been\quad \\ subtended\quad by\quad the\quad chord\quad BC\quad to\quad the\quad circumference\quad at\quad A\quad \& \quad D].\\ Again\quad ACBD\quad is\quad a\quad cyclic\quad quadrilateral.\quad \\ \therefore \quad \angle CAD+\angle DBC={ 180 }^{ o }\quad (The\quad sum\quad of\quad the\quad opposite\quad angles\quad of\quad a\quad cyclic\quad \\ quadrilateral\quad is\quad { 180 }^{ o }).\\ So\quad \angle DBC={ 180 }^{ o }-\angle CAD={ 180 }^{ o }-65^{ o }=115^{ o }.......(iii).\\ \therefore \quad \quad \angle DCB,\quad \angle DBC\quad \& \quad \angle BDC\quad are \quad 35^{ o },\quad 115^{ o }\quad \& \quad 30^{ o }\quad respectively.\\ Hence, \quad option\quad C \quad is \quad correct. $$

$$\implies \angle \text{ADC}=\dfrac{1}{2}\angle \text{AOC}$$

Since $$AB$$ is diameter, $$\angle AEB = 90^\circ$$.

In $$\triangle AEB$$,

$$\angle ABE + \angle EAB +\angle AEB=180^o$$ ...[Angle sum property]

$$\implies$$ $$\angle ABE = 180^o - \angle EAB - \angle AEB$$

$$\implies$$ $$\angle ABE = 180^o – 65^o -90^o = 25^\circ$$.

Given, $$ED \parallel AB$$,

$$\implies$$ $$ \angle DEB = \angle EBA = 25^\circ$$ ....[Alternate interior angles].

Since $$EDCB$$ is a cyclic quadrilateral,

$$\angle EAB + \angle EDB  = 180^\circ$$ ...[Opposite angles of cyclic quadrilateral are supplementary]

$$\implies$$ $$\angle EDB = 180^o – 65^o = 115^\circ $$.

In $$\triangle EDB$$,

$$\angle EBD + \angle DEB+ \angle BDE=180^o$$

$$\implies$$ $$\angle EBD = 180 - \angle DEB - \angle BDE$$

$$\implies$$ $$\angle ABE = 180 – 25 -115 = 40^\circ$$.

Hence, option $$D$$ is correct.

$$\textbf{Step - 1: Verify, If we join any points on a circle we get a diameter of the circle}$$

                   $$\text{A Chord is a line segment that joins any two points of the circle.}$$

                   $$\text{The endpoints of this line segments lie on the circumference of the circle.}$$

                   $$\therefore \text{Option (A) is false}$$

$$\textbf{Step - 2: Verify, A diameter of a circle contains the center of the circle}$$

                   $$\text{Any interval joining two points on the circle and passing through the center is called a}$$

                   $$\text{diameter of the circle.}$$

                   $$\therefore \text{Option (B) is True}$$

$$\textbf{Step - 3: Verify, A semicircle is an arc}$$

                   $$\text{The arc of a circle consists of two points on the circle and all of the points on the circle that lie}$$

                   $$\text{between those two points.}$$

                   $$\text{It's like a segment that was wrapped partway around a circle.}$$

                   $$\text{An arc whose measure equals 180 degrees is called a semicircle since it divides the circle in two}$$

                   $$\therefore \text{Option (C) is True}$$

$$\textbf{Step - 4: Verify, the length of a circle is called its circumference}$$

                   $$\text{A Circle is a round closed figure where all its boundary points are equidistant from a fixed point}$$

                   $$\text{called the center.}$$ 

                   $$\text{The two important metrics of a circle is the area of a circle and the circumference of a circle.}$$

                   $$\therefore \text{Option (D) is True}$$

$$\textbf{Hence, option A is correct as it is false}$$        

We know, opposite angles of a cyclic quadrailateral are supplementary.
$$\therefore 3x+x=180^{\circ}$$ and $$2y+y=180^{\circ}$$
$$\Rightarrow 4x=180^{\circ}$$ and $$3y=180^{\circ}$$
$$\Rightarrow x=45^{\circ}$$ and $$y=60^{\circ}$$.

The perpendicular drawn from the center of the circle to the chord bisects the chord. So, it divides the chord in $$1:1$$ 

Construction: Join A and C. This forms a right triangle ACB   [ angle in a semicircle is $$90^{\circ}$$]

Given, $$ABCD$$ is cyclic quadrilateral.

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 $$\triangle OPR$$

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

$$PR^2 = r^2 – OC^2 \qquad –(2)$$

$$(1) = (2)$$

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

$$AB = PQ$$

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

Given- $$ABDE$$ is a cyclic quadrilateral in a circle with centre $$O$$ and diameter $$AB$$.

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)
$$\implies$$ $$30^o + a + 30^o + b = 180^o$$
$$\implies$$ $$a + b = 120^{\circ}$$.

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

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

Given,

We have , $$AN=8$$ cm

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

$$\quad In\angle BOC\quad is\quad a\quad central\quad angle\quad =150(Given)\\ \angle BAC=\frac { 1 }{ 2 } \angle BOC[\because Angle\quad at\quad centre\quad is\quad double\quad the\quad angle\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad on\quad remaining\quad part\quad of\quad circle]\\ \Rightarrow \angle BAC=75...(1)\\ In\quad a\quad cyclic\quad quadilateral\quad ACDB\\ \angle BAC+\angle BDC=180\\ \Rightarrow \angle BDC=180-\angle BAC\\ \Rightarrow \angle BDC=180-75.....(From\quad 1)\\ \Rightarrow \angle BDC=105$$