CBSE Questions for Class 9 Maths Circles Quiz 4 - MCQExams.com

$$ Given-\\ A,\quad B,\quad C\quad \& \quad D\quad are\quad four\quad points\quad such\quad that\quad \angle BAC={ 30 }^{ o }\quad \& \quad \angle BDC={ 60 }^{ o }.\\ To\quad find\quad out-\\ if\quad D\quad is\quad the\quad centre\quad of\quad the\quad circle\quad which\quad passes\quad through\quad A,\quad B\quad \& \quad C.\\ Solution-\\ \angle BAC={ 30 }^{ o }\quad \& \quad \angle BDC={ 60 }^{ o }.\\ \therefore \quad \angle BDC=2\angle BAC.\\ We\quad know\quad that\quad the\quad angle\quad subtended\quad by\quad a\quad chord\quad at\quad the\quad \\ centre\quad of\quad a\quad circle\quad is\quad double\quad the\quad angle\quad at\quad the\quad circumference\quad \\ subtended\quad by\quad the\quad same\quad chord.\\ \therefore \quad \angle BDC\quad is\quad the\quad angle\quad at\quad the\quad centre\quad and\quad \angle BAC\quad is\quad the\\ angle\quad at\quad the\quad circumference\quad subtended\quad by\quad the\quad chord\quad BC.\\ So\quad D\quad is\quad the\quad centre\quad of\quad the\quad circle\quad which\quad passes\quad through\quad \\ A,\quad B\quad \& \quad C.\\ \therefore \quad The\quad statement\quad is\quad correct\quad by\quad fig\quad I\\ But\quad by\quad fig\quad II\quad this\quad is\quad not\quad true.\\ So\quad the\quad statement\quad is\quad not\quad always\quad true.\quad \quad \\ \\ Ans-\quad Option\quad A. $$

Given, $$ABCD$$ is a cyclic quadrilateral

$$\angle A+\angle C =180^o$$ .......(The sum of the opposite angles of a cyclic quadrilateral $$={ 180 }^{ o }$$)

$$ Given-\\ AOBEDC\quad is\quad a\quad polygon\quad inscribed\quad in\quad a\quad circle\quad with\quad centre\\ O\quad and\quad diameter\quad AOB.\\ To\quad find\quad out-\\ \angle ACD+\angle BED=?\\ Solution-\\ We\quad join\quad AE\quad \& \quad BC.\\ ACDE\quad is\quad a\quad cyclic\quad quadrilateral.\\ \therefore \quad \angle ACD+\angle AED={ 180 }^{ o }........(i)\\ Similarly\quad BCDE\quad is\quad a\quad cyclic\quad quadrilateral.\\ \therefore \quad \angle BED+\angle BCD={ 180 }^{ o }........(ii).\\ (\because \quad The\quad sum\quad of\quad the\quad opposite\quad pair\quad of\quad angles\quad of\quad \\ a\quad cyclic\quad quadrilateral={ 180 }^{ o }\quad )\\ Again\quad \angle ACB={ 90 }^{ o }=\angle AEB.........(iii\quad \& \quad iv)\quad \\ (\because \quad The\quad angle\quad subtended\quad by\quad the\quad diameter\quad of\quad a\quad circle\\ at\quad the\quad cicumference={ 90 }^{ o }).\\ Adding\quad (i),\quad (ii),\quad (iii)\quad \& \quad (iv)\quad we\quad get\\ \angle ACD+\angle AED+\angle BED+\angle BCD+\angle ACB+\angle AEB={ 180 }^{ o }+{ 180 }^{ o }+{ 90 }^{ o }+{ 90 }^{ o }={ 540 }^{ o }\\ \Longrightarrow (\angle ACD+\angle BED)+(\angle BCD+\angle BCD)+(\angle AED+\angle AEB)={ 540 }^{ o }\\ \Longrightarrow (\angle ACD+\angle BED)+\angle ACD+\angle BED={ 540 }^{ o }\\ \Longrightarrow 2(\angle ACD+\angle BED)={ 540 }^{ o }\\ \Longrightarrow \angle ACD+\angle BED={ 270 }^{ O }.\\ Ans-\quad Option\quad B.\\ \\ \\ \\  $$ 

The  circumference of circle with radius $${ R }_{ 1 }=2\pi { R }_{ 1 }$$.

We know, Opposite angles in a cyclic quadrilateral are supplementary, i.e. they sum up to $$180^0$$.

The area of a triangle is equal to the base times the height.
In a semi circle, the diameter is the base of the semi-circle.
This is equal to $$2\times r$$ (r = the radius)
If the triangle is an isosceles triangle with an angle of $$45^\circ$$ at each end, then the height of the triangle is also a radius of the circle.
A = $$\frac{1}{2} \times b \times h$$ formula for the area of a triangle becomes
A = $$\frac{1}{2}\times 2 \times r \times r$$ because:
The base of the triangle is equal to $$2\times r$$
The height of the triangle is equal to r
A = $$\frac{1}{2} \times 2 \times r \times r$$ becomes:
A = $$r^2$$

Given, $$ABCD$$ is a quadrilateral inscribed in a circle.

Given, $$\angle AOC=110^o$$.
By property of circles, we have $$\angle AOC=2\angle ADC$$
$$\Rightarrow \angle ADC=110^o\div 2=55^o$$

Quadrilateral $$APBC$$ is a cyclic quadrilateral.
We know, the opposite angles of a cyclic quadrilateral are supplementary.

Here, $$\angle DAC = \angle DBC = 55^o$$ ...(angles in the same segment).

Radii of one circle are the same and two chords are equidistant from the center so by Pythagoras theorem, chords have to be equal.
Since the chords are equal, so their distance from the center will also be equal, and hence congruent triangles will form always and so the chords will always subtend equal angles at the center.
Also, the angle in a semicircle is a right angle.

$$ABCD$$ is a cyclic quadrilateral.

Also, $$\angle ADC = 130^o$$.

We know, opposite angles of a cyclic quadrilateral are supplementary.

Then, $$\angle ADC  + \angle CBA = 180^o$$

$$\angle CBO = \angle CBA = 180^o-30^o =50^\circ   $$.

Therefore, option $$B$$ is correct.

Since, $$APQD$$ is a cyclic quadrilateral,

Line joining the midpoint of chord to the centre of circle is perpendicular to chord.

In cyclic quadrilateral, the sum of opposite angles $$= 180^o$$
$$\Longrightarrow \angle A + \angle C = 180^o$$
$$\Longrightarrow   2x -1+2y+15=180^o$$
$$\Longrightarrow  x + y = 83$$                   ...(i).

In $$\triangle PQC,$$

We know $$PC = QC$$

So, $$ \angle CQP = \angle CPQ$$

$$\Rightarrow \angle CPQ = 25^\circ$$

In $$\triangle PRC,$$

We know $$PC = RC$$

Hence $$\angle CRP = \angle CPR$$

$$\therefore \angle CPR = 15^\circ$$

Now,

$$\angle QPR = \angle QPC + \angle CPR$$

              $$= 25^\circ + 15^\circ$$

              $$= 40^\circ$$

We know "Angle subtended by an arc at the centre is double the angle subtended by the same arc at the circumference of a circle."

Thus,

$$∠QCR = 2 \times ∠QPR$$

              $$= 2 \times  40^\circ$$

              $$= 80^\circ$$

Hence Statement 1 is false and Statement 2 is true.

Given $$ \angle A=(2x-10)^{\circ},\quad \angle B=(2y-20)^{\circ},\quad \angle C=(2y+30)^{\circ}$$ and $$\angle D=(3x+10)^{\circ}$$.

Given: $$O$$ is the center of the circle. $$K$$ and $$L$$ are mid points of chords $$AB$$ and $$CD$$, respectively.
$$\therefore OK \perp AB$$ and $$OL \perp CD$$.             {The line joining the centre of a circle to the mid-point of a chord is perpendicular to the chord}
As $$AB = CD$$.
$$\therefore$$ $$OK = OL$$          ...(Equal chords are equidistant from centre)

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

$$ Given-\\ PQ=30cm\quad \& \quad RS=16cm\quad are\quad the\quad chords\quad of\quad a\quad circle\quad of\quad \\ radius=17cm\quad with\quad centre\quad O.\\ Also\quad they\quad are\quad on\quad the\quad opposite\quad sides\quad of\quad O.\quad \quad \quad \\ PQ\parallel RS.\\ To\quad find\quad out\quad -\quad The\quad distance\quad between\quad PQ\quad \& \quad RS=?\\ the\quad Solution-\\ We\quad join\quad OP\quad \& \quad OR\quad and\quad drop\quad a\quad perpendicular\quad ON\quad from\quad O\\ to\quad RS\quad at\quad N.\\ Since\quad PQ\parallel RS,\quad ON\quad will\quad be\quad perpendicular\quad to\quad PQ\quad at\quad M.\\ \left( \angle ONR={ 90 }^{ o }=\angle OMP\quad as\quad they\quad are\quad corresponding\quad angles. \right) \\ So\quad the\quad distance\quad between\quad PQ\quad \& \quad RS=ON+OM=MN.\\ Here\quad PM=\frac { 1 }{ 2 } PQ=\frac { 1 }{ 2 } \times 30cm=15cm\quad since\quad \\ the\quad perpendicular\quad from\quad the\quad centre\quad of\quad a\quad circle\quad to\quad a\quad chord\quad \\ bisects\quad the\quad latter.\\ \quad Now\quad \Delta OPM\quad is\quad a\quad right\quad one\quad with\quad OP\quad as\quad hypotenuse.\\ \therefore \quad By\quad Pythagoras\quad theorem,\quad we\quad have\quad OM=\sqrt { { OP }^{ 2 }-{ PM }^{ 2 } } \\ =\sqrt { { 17 }^{ 2 }-{ 15 }^{ 2 } } cm=8cm.\\ Again\quad \quad RN=\frac { 1 }{ 2 } RS=\frac { 1 }{ 2 } \times 16cm=8cm\quad since\quad the\quad perpendicular\quad from\quad the\quad \\ centre\quad of\quad a\quad circle\quad to\quad a\quad chord\quad bisects\quad the\quad latter.\\ We\quad have,\quad \Delta ORN\quad is\quad a\quad right\quad one\quad with\quad OR\quad as\quad hypotenuse.\\ \therefore \quad By\quad Pythagoras\quad theorem,\quad we\quad have\quad ON=\sqrt { { OR }^{ 2 }-{ RN }^{ 2 } } \\ =\sqrt { { 17 }^{ 2 }-{ 8 }^{ 2 } } cm=15cm.\\ Now\quad the\quad distance\quad beteen\quad PQ\quad \& \quad RS=MN=ON-OM\\ =(15+8)cm=23cm.\\ Ans-\quad Option\quad C.\\ \\  $$ 

Given- $$\overline { POQ }$$  is a diameter of a given circle. $$PQRS$$ is a cyclic quadrilateral. $$SQ$$ is joined. Also, $$\angle SRQ={ 138 }^{ \circ }$$.

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

Again,

$$\angle QPS+\angle QRS=180^\circ$$ ...[ sum of opposite angles  of a cyclic quadrilateral is $${ 180 }^{ \circ }$$]

$$\Rightarrow \angle QPS={ 180 }^{ \circ }-\angle QRS$$

$$\Rightarrow \angle QPS={ 180 }^{ \circ }-138^{ \circ }$$

$$\Rightarrow\angle QPS={ 42 }^{ \circ}$$.

 In $$\triangle PSQ,$$

$$  \angle PQS +\angle PSQ+\angle QPS=180^\circ$$

                                 $$\angle PQS={ 180 }^{ \circ }-(\angle PSQ+\angle QPS)$$

                                               $$={ 180 }^{ \circ }-({ 90 }^{ \circ }+{ 42 }^{ \circ })$$

                                               $$={ 48 }^{ \circ }$$

Hence, option $$C$$ is correct.

Given, $$ \angle A =\left ( 2x-1 \right )^{\circ}$$, $$ \angle B =\left ( y+5 \right )^{\circ}$$, $$ \angle C=\left ( 2y+15 \right )^{\circ}$$ and $$\angle D =\left ( 4x-7 \right )^{\circ}$$.

$$ABCD $$ is a cyclic quadrilateral. 

Given,
$$\angle BAD=70^{\circ}$$, $$\angle ABD=56^{\circ}$$ and $$\angle ADC=72^{\circ}$$.

O$$Y = OZ = radius = r$$

Given $$YZ = r$$

$$\implies \triangle OYZ $$ is equilateral

$$\implies \angle YOZ = 60^\circ$$

We know that angle made by a chord on any point on the circle is half the angle made by the chord at the center

$$\implies \angle YXZ = \dfrac{\angle YOZ}{2}$$

$$\implies \angle YXZ = \dfrac{60}{2} = 30^\circ$$

Given $$ABCD$$ is a cyclic quadrilateral.

Given, $$AB$$ is the common chord of two intersecting circles.