MCQ Questions for Class 9 Maths Circles Quiz 4

If A, B, C, D are four points such that

$\angle BAC={ 30 }^{ \circ }$ &

$\angle BDC={ 60 }^{ \circ },$ then D is the center of the circle through A, B and C. The statement is

0%
sometimes true
0%
false
0%
true
0%
none of the above.

Explanation

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

In the given figure,

$ABCD$ is a cyclic quadrilateral in which

$\angle BAD = 120^o$ . Then

$m\angle BCD$ is:

0%
$240^o$
0%
$60^o$
0%
$120^o$
0%
$180^o$

Explanation

Given,

$ABCD$ is a cyclic quadrilateral

and

$\angle BAD={ 120 }^{ o }$ .

Since,

$ABCD$ is a cyclic quadrilateral,

$\therefore \angle BCD+\angle BAD={ 180 }^{ o }$ ...[Opposite angles of cyclic quadrilateral are supplementary].

$\implies \angle BCD={ 180 }^{ o }-\angle BAD\\={ 180 }^{ o }-{ 120 }^{ o }\\={ 60 }^{ o }$ .

Hence, option

$B$ is correct.

The angles of a cyclic quadrilateral

$ABCD$ are

$A = (6x + 10)$ ,

$B = (5x)$ ,

$C = (x + y)$ ,

$D = (3y -10)$ .
Find

$x$ and

$y$ , and hence find the values of the four angles.

0%
$x = 10, y = 30, A =120^o, B=100^o, C= 506O, D=806o$
0%
$x = 20, y = 30, A =130^o, B=100^o, C= 50^o, D=80^o$
0%
$x = 18, y = 30, A =140^o, B=100^o, C= 50^o, D=84^o$
0%
$x = 30, y = 30, A =180^o, B=100^o, C= 50^o, D=80^o$

Explanation

$ABCD$ is a cyclic quadrilateral, we have

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

$={ 180 }^{ o }$ )

$\implies$

$6x+ 10^o +x+y =180^o$

$\implies$

$7x+y=170^o$ .....(1)

and

$\angle B+\angle D=180^o$ ............(The sum of the opposite angles of a cyclic quadrilateral

$={ 180 }^{ o }$ )

$\implies$

$(3y-10) +5x =180^o$

$\implies$

$3y+5x =190^o$ ....(2).

Solving (1) and (2), we have,

$x=20^o$

$y=30^o$

$A=6x+10 =6 \times 20^o +10^o =130^o$

$B=5x=5 \times 20^o =100^o$

$C =x+y =30^o+20^o=50^o$

$D=3y-10 =3 \times30-10=80^o$ .

Therefore, option

$B$ is correct.

In given figure, AOB is a diameter of the circle and C, D, E are any three points on the semi-circle. Find the value of

$\angle ACD+\angle BED.$

0%
${ 540 }^{ o }$
0%
${ 270 }^{ o }$
0%
${ 180 }^{ o }$
0%
${ 120 }^{ o }$

Explanation

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

If the sum of the circumferences of two circles with radii

$R_1$ and

$R_2$ is equal to the circumference of a circle of radius

$R$ , then

0%
$R_{1}+R_{2}=R$
0%
$R_{1}+R_{2}>R$
0%
$R_{1}+R_{2} < R$
0%
Nothing definite can be said about the relation among $R_1, R_2$ and R.

Explanation

The circumference of circle with radius

${ R }_{ 1 }=2\pi { R }_{ 1 }$ .

and the circumference of circle with radius

${ R }_{ 2 }=2\pi { R }_{ 2 }.$

$\therefore$ The Sum of Circumferences,

$Sum=2\pi \left( { R }_{ 1 }+{ R }_{ 2 } \right)$ .

Again the circumference of circle with radius

${ R }=2\pi { R }.$

$\therefore$ By given condition,

$2\pi \left( { R }_{ 1 }+{ R }_{ 2 } \right) =2\pi { R }\Longrightarrow { R }_{ 1 }+{ R }_{ 2 }=R$ .

0%
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
0%
Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
0%
Assertion is correct but Reason is incorrect
0%
Assertion is incorrect but Reason is correct

Explanation

Given, one angle

$40^o$ and opposite angle

$140^o$ .

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

$180^0$ .

Then,

$40^o+140^o=180^o$ .

Therefore, the given assertion is correct but reason is incorrect.

Hence, option

$C$ is correct.

Area of the largest triangle that can be inscribed in a semi-circle of radius

$r$ units is

0%
$r^2$ sq. units
0%
$\dfrac{1}{2} r^2$ sq. units
0%
$2r^2$ sq. units
0%
$\sqrt{2} r^2$ sq. units

Explanation

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 =

$r^2$

A quadrilateral

$ABCD$ is inscribed in a circle such that

$AB$ is a diameter and

$\angle ADC={ 130 }^{ o }$ , then

$m\angle BAC=$ ?

0%
${ 90 }^{ o }$
0%
${ 50 }^{ o }$
0%
${ 40 }^{ o }$
0%
${ 30 }^{ o }$

Explanation

Given,

$ABCD$ is a quadrilateral inscribed in a circle.

Also,

$\angle ADC={ 130 }^{ o }$ .

We have to find:

$\angle BAC$ .

Now join

$AD$ .

Then,

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

$={ 180 }^{ o }$ )

i.e.

$\angle ABC={ 180 }^{ o }-\angle ADC= { 180 }^{ o }-{ 130 }^{ o }={ 50 }^{ o }$ .

So, in

$\Delta ABC$ , we have

$\angle BAC={ 180 }^{ o }$

$-(\angle ACB+\angle ABC)$ ...(Since

$AB$ is the diameter, the angle subtended by the diameter of a circle at the circumference

$=90^o$ i.e.

$\angle BAC=90^o$ )

$={ 180 }^{ o }-({ 50 }^{ o }+{ 90 }^{ o })={ 40 }^{ o }$ .

Therefore, option

$C$ is correct.

In the given figure, if

$\angle AOC=110^o$ , then the values of

$\angle D$ and

$\angle B$ , respectively are:

0%
$55^o, 125^o$
0%
$55^o,110^o$
0%
$110^o,25^o$
0%
$125^o,55^o$

Explanation

Given,

$\angle AOC=110^o$ .
By property of circles, we have

$\angle AOC=2\angle ADC$

$\Rightarrow \angle ADC=110^o\div 2=55^o$

$\Rightarrow$

$\angle D=55^o$ .

Also,

$\angle B + \angle D=180^o$ ...[Opposite angles of cyclic quadrilateral are supplementary]

$\implies$

$\angle B=180^o-\angle D=180^o-55^o=125^o$ .

Hence, option '

$A$ is correct.

In the below diagram,

$O$ is the centre of the circle,

$AC$ is the diameter and if

$\angle APB=120^0$ , then

$\angle BQC$ is

0%
$30^0$
0%
$150^0$
0%
$90^0$
0%
$120^0$

Explanation

Quadrilateral

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

Hence,

$\angle APB + \angle ACB = 180^0$ .

Since,

$\angle APB = 120^0$

$\Rightarrow \angle ACB =180^o-120^o= 60^0$ .

Here,

$AC$ is a diameter and hence

$\angle ABC = 90^0$ .
Thus, in

$\Delta ABC,$

$\angle CAB+\angle ABC+\angle ACB=180^o$ ...[Angle sum property]

$\implies$

$\angle CAB+90^o+60^o=180^o$

$\implies$

$\angle CAB = 180^0 - 90^0 - 60^0 = 30^0$ .

Now, quadrilateral

$ABQC$ is another cyclic quadrilateral.

So,

$\angle CAB + \angle CQB = 180^0$ ...[Opposite angles of a cyclic quadrilateral]

$\therefore \angle CQB=180^o-30^o = 150^0$ .

Hence, option

$B$ is correct.

In the figure

$AD||BC;\ \angle CAB = 45^{\circ} ;\ \angle DBC = 55^{\circ}$ , then

$\angle DCB$ equals:

0%
$55^{\circ}$
0%
$80^{\circ}$
0%
$100^{\circ}$
0%
$120^{\circ}$

Explanation

Here,

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

Then,

$\angle DAB = \angle DAC + \angle CAB = 55^o+45^o=100^o$ .

Now,

$\angle DAB + \angle DCB = 180^o$ ...(opposite angles of cyclic quadrilateral

$ABCD$ )

$\Rightarrow \angle DCB =180^o-100^o= 80^o$ .

Therefore, option

$B$ is correct.

Which of the following statement (s) is/ are true?

0%
Two chords of a circle equidistant from the centre are equal
0%
Equal chords in a circle subtend equal angles at the centre
0%
Angle in a semicircle is a right angle
0%
All of the above

In the figure

$\angle ADC={ 130 }^{ o }$ and chord BC = chord BE. Find

$\angle CBO$ .

0%
$40^o$
0%
$50^o$
0%
$60^o$
0%
$100^o$

Explanation

$ABCD$ 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.

In the figure,

$\angle B$ is equal to:

0%
$80^{\circ}$
0%
$95^{\circ}$
0%
$70^{\circ}$
0%
$115^{\circ}$

Explanation

Since,

$APQD$ is a cyclic quadrilateral,

therefore,

$\angle A + \angle DQP = 180^{\circ}$ ...[Opposite angles of cyclic quadrilateral are supplementary]

$\Rightarrow \angle DQP = 180^{\circ} - 85^{\circ} = 95^{\circ}$ .

Now,

$\angle PQC = 180^{\circ} - \angle DQP = 85^{\circ}$ ...[Straight line property].

Also,

$PBCQ$ is a cyclic quadrilateral,

therefore,

$\angle PQC +\angle PBC =180^0$ ...[Opposite angles of cyclic quadrilateral are supplementary]

$\Rightarrow \angle PBC = 180^{\circ} - 85^o = 95^o$ .

Hence, option

$B$ is correct.

$O$ is the centre of a circle,

$AB$ its chord,

$C$ is the mid point of

$AB$ , then

$OAC$ is

0%
an acute angled
0%
an obtuse angled
0%
a right angled triangle
0%
an isosceles triangle

Explanation

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

$\therefore OC \bot AB$

$\therefore \angle DCA = 90^{\circ}$

Now,

$\Delta OAC$

$\angle OAC + \angle AOC + \angle OCA = 180^{\circ}$ (Sum of interior angle of triangle is

$180^{\circ}$ )

$\Rightarrow \angle OAC + \angle AOC + 90^{\circ} = 180^{\circ}$

$\angle OAC + \angle AOC = 180^{\circ} - 90^{\circ}$

$\angle OAC + \angle AOC = 90^{\circ}$

$\therefore \angle OAC < 90^{\circ}$

$\therefore \angle OAC$ is acute angle

The angle made by the line from the centre with the chord which it bisects is:

0%
$90^0$
0%
$30 ^0$
0%
$45^0$
0%
None of these

The greatest angle of a cyclic quadrilateral

$ABCD$ in which

$\angle A = (2x-1)^o, \angle B = (y+5)^o, \angle C = (2y+15)^o$ and

$\angle D = (4x-7)^o$ is:

0%
$115^o$
0%
$120^o$
0%
$125^o$
0%
$130^o$

Explanation

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).

Also,

$\angle B + \angle D = 180^o$

$\Longrightarrow 4x+y=182$ ....(ii).

Then, by solving (i) and (ii),

$x = 33$ and

$y = 50$ .

Thus,

$\angle A = 65^o, \angle B = 55^o, \angle C = 115^o, \angle D = 125^o$ .
So, greatest angle

$\angle D = 125^0$ .
Hence, option

$C$ is correct.

Statement 1: In given figure, if C is centre of the circle and

$\angle PQC = 25^{\circ}$ and

$\angle PRC = 15^{\circ}$ , then

$\angle QCR$ is

$40^{\circ}.$

Statement 2: Angle subtended by arc at the centre is twice angled subtended by it at circumference of the circle.

0%
Statement - 1 is True, Statement - 2 is True, Statement - 2 is a correct explanation for Statement - 1
0%
Statement - 1 is True, Statement - 2 is True : Statement 2 is NOT a correct explanation for Statement - 1
0%
Statement - 1 is True, Statement - 2 is False
0%
Statement - 1 is False, Statement - 2 is True

Explanation

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.

Find the smallest angle of a cyclic quadrilateral

$ABCD$ in which

$\angle A= (2x-10)^o, \, \, \angle B=(2y-20)^o, \, \, \angle C= (2y+30)^o$ and

$\angle D=(3x+10)^o.$

0%
$80^0$
0%
$50^0$
0%
$85^0$
0%
$40^0$

Explanation

Given

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

$\angle D=(3x+10)^{\circ}$ .

We know, the sum of the opposite angles of a cyclic quadrilateral is

$180^{\circ}$ .

$\angle A+\angle C=180^{\circ}$

$\Rightarrow (2x-10+2y+30)^{\circ}=180^{\circ}$

$\Rightarrow (2x+2y)^{\circ}=160^{\circ}\quad ------\left( 1 \right)$

$\& \quad \text{also,} \ \angle B+\angle D=180^{\circ}$

$\Rightarrow (2y-20+3x+10)^{\circ}=180^{\circ}$

$\Rightarrow (3x+2y)^{\circ}=190^{\circ}\quad ------\left( 2 \right)$ .

Subtracting

$\left( 1 \right)$ from

$\left( 2 \right)$ , we get,

$(3x-2x)^{\circ}=190^{\circ}-160^{\circ}$

$\Rightarrow x^{\circ}=30^{\circ}$ .

Using

$x^{\circ}=30^{\circ}$ in

$\left( 1 \right)$ , we get,

$(2y+2\times 30)^{\circ}=160^{\circ}$

$\Rightarrow (2y+60)^{\circ}=160^{\circ}$

$\Rightarrow (2y)^{\circ}=100^{\circ}$

$\Rightarrow y^{\circ}=50^{\circ}$ .

So,

$\angle A=(2x-10)^{\circ}=(2\times30-10)^{\circ}=(60-10)^{\circ}={ 50 }^{ o }$

$\angle B=(2y-20)^{\circ}=(2\times 50-20)^{\circ}=(100-20)^{\circ}={ 80 }^{ o }$

$\angle C=(2y+30)^{\circ}=(2\times 50+30)^{\circ}=(100+30)^{\circ}={ 130 }^{ o }$

and

$\angle D=(3x+10)^{\circ}=(3\times 30+10)^{\circ}=(90+10)^{\circ}={ 100 }^{ o }$ .

Hence, the smallest angle is

$\angle A=50^o$ .

Therefore, option

$B$ is correct.

In the given circle with centre

$'O'$ , the mid points of two equal chords

$AB$ &

$CD$ are

$K$ &

$L$ , respectively. If

$\angle$

$OLK =$

$25^{\circ}$ , Then

$\angle$

$LKB$ is equal to:

0%
$125^{\circ}$
0%
$115^{\circ}$
0%
$105^{\circ}$
0%
$90^{\circ}$

Explanation

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)

So,

$\bigtriangleup$

$OKL$ is an isosceles triangle.

$\therefore \angle OKL = \angle OLK = 25^{\circ}$

Therefore,

$\angle LKB = \angle OKL + \angle OKB$

$= 25^{\circ} + 90^{\circ} = 115^{\circ}$

$ABCD$ is a cyclic quadrilateral. If

$\angle A-\angle C=30^{\circ}$ , then

$\angle C =$ ?

0%
$90^{\circ}$
0%
$115^{\circ}$
0%
$60^{\circ}$
0%
$75^{\circ}$

Explanation

Given,

$ABCD$ is a cyclic quadrilateral.

We know, opposite angles of cyclic quadrilateral are supplementary.

Then,

$\angle A+\angle C$

$=180^{\circ}$ .......(i).

But given,

$\angle A-\angle C$

$=30^{\circ}$ .......(ii).

(i)

$-$ (ii), we get,

$2 \angle C=150^o$

$\implies$

$\angle C=75^o$ .

Hence, option

$C$ is correct.

In a circle of radius 17cm two parallel chords of the length 30cm and 16cm respectively are drawn on the opposite sides of the centre. Then the distance between them is

0%
7cm
0%
12cm
0%
23cm
0%
32cm

Explanation

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

In a circle of radius

$13$ cm,

$PQ$ and

$RS$ are two parallel chords of length

$24$ cm and l0cm respectively. The chords are on the same side of the centre then the distance between the chords is?

0%
$7$ cm
0%
$12$ cm
0%
$5$ cm
0%
$10$ cm

Explanation

$OP=OR=13cm,PQ=24cm$ and

$RS=10cm$

$PM=12cm$ and

$RN = 5cm$ [perpendicular from the centre bisects the chord]

In right angled triangle OMP and ONR ,by pythogoras theorem

${OP}^{2}={OM}^{2}+{MP}^{2}$

$OM =\sqrt{{OP}^{2}-{MP}^{2}}$

$OM =\sqrt{{13}^{2}-{12}^{2}}$

$OM=5cm$

${OR}^{2}={ON}^{2}+{RN}^{2}$

$ON =\sqrt{{OR}^{2}-{RN}^{2}}$

$ON =\sqrt{{13}^{2}-{5}^{2}}$

$OM=12cm$

$MN=ON-OM$

$MN=12-5$

$MN=7cm$

Therefore distance between the chords is

$7 cm$

option A is the answer.

In the given figure,

$POQ$ is the diameter of the circle with center

$O$ . Quadrilateral

$PQRS$ is a cyclic quadrilateral and

$SQ$ is joined. If

$\angle R = 138^\circ$ , then

$m\angle PQS$ is:

0%
$90^\circ$
0%
$42^\circ$
0%
$48^\circ$
0%
$38^\circ$

Explanation

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.

Find the four angles of a cyclic quadrilateral ABCD in which

$\angle A=(2x-1)^o, \angle B=(y+5)^o, \angle C=(2y+15)^o$ and

$\angle D(4x-7)^o$

0%
$\angle A=25^o, \angle B=45^o, \angle C=105^o, \angle D=135^o$
0%
$\angle A=35^o, \angle B=75^o, \angle C=95^o, \angle D=135^o$
0%
$\angle A=45^o, \angle B=75^o, \angle C=95^o, \angle D=125^o$
0%
$\angle A=65^o, \angle B=55^o, \angle C=115^o, \angle D=125^o$

Explanation

Given,

$\angle A =\left ( 2x-1 \right )^{\circ}$ ,

$\angle B =\left ( y+5 \right )^{\circ}$ ,

$\angle C=\left ( 2y+15 \right )^{\circ}$ and

$\angle D =\left ( 4x-7 \right )^{\circ}$ .

In cyclic quadrilateral, the sum of opposite angles

$= 180^o$

$\Rightarrow \angle A + \angle C = 180^o$

$\Rightarrow \left ( 2x-1 \right )+\left (2y+15 \right )=180^{\circ}$

$\Rightarrow 2x+2y=166^{\circ}\Rightarrow x+y=83$ .....(1)

Also,

$\angle B + \angle D = 180^o$

$\Rightarrow \left ( y+5 \right )+\left (4x-7 \right )=180^{\circ}$

$\rightarrow 4x+y=182^{\circ}$ ....(2)

Substract (2) from (1), we get,

$\Rightarrow -3x=-99^o \Rightarrow x=33^o$ .

Now, substitute

$x=33$ in (1), we get,

$\Rightarrow 33^o+y=83^o \Rightarrow y=50^o$ .

Hence,

$\angle A =\left ( 2x-1 \right )=2\times33-1=65^{\circ}$

$\angle B =\left ( y+5 \right )=50+5=55^{\circ}$

$\angle C=\left ( 2y+15 \right )=2\times50+15=115^{\circ}$

$\angle D =\left ( 4x-7 \right )=4\times33-7=125^{\circ}$ .

Therefore, option

$D$ is correct.

$ABCD$ is a cyclic quadrilateral, then the angles of the quadrilateral in the same order are:

0%
$70^{\circ} , 120^{\circ} , 110^{\circ}, 60^{\circ}$
0%
$120^{\circ} , 110^{\circ} , 70^{\circ}, 60^{\circ}$
0%
$110^{\circ} , 700^{\circ} , 60^{\circ}, 120^{\circ}$
0%
$60^{\circ} , 120^{\circ} , 70^{\circ}, 110^{\circ}$

Explanation

$ABCD$ is a cyclic quadrilateral.

In a cyclic quadrilateral, the sum of opposite angles is

$180^{\circ}$ and sum of all the angles

$=$

$360^{\circ}$

Option 1:

$70^{\circ}, 120^{\circ}, 110^{\circ}, 60^{\circ}$
Sum of Opposite angles:

$\angle A + \angle C = 70 + 110 = 180^{\circ}$
and

$\angle B + \angle D = 120 + 60 = 180^{\circ}$ .
The opposite angles are

$180^{\circ}$ and sum of all the angles is

$360^{\circ}$ .

Option 2:

$120^{\circ}, 110^{\circ}, 70^{\circ}, 60^{\circ}$
Sum of Opposite angles:

$\angle A + \angle C = 120 + 70 = 190^{\circ}\ne 180^o$
and

$\angle B + \angle D = 110 + 60 = 170^{\circ} \ne 180^o$ .
The opposite angles are not

$180^{\circ}$ .

Option 3:

$110^{\circ}, 700^{\circ}, 60^{\circ}, 100^{\circ}$
Sum of Opposite angles:

$\angle A + \angle C = 110 + 60 = 170^{\circ}\ne 180^o$
and

$\angle B + \angle D = 700 + 100 = 800^{\circ} \ne 180^o$ .
The opposite angles are not

$180^{\circ}$ .

Option 4:

$60^{\circ}, 120^{\circ}, 70^{\circ}, 110^{\circ}$
Sum of Opposite angles:

$\angle A + \angle C = 60 + 70 = 130^{\circ}\ne 180^o$
and

$\angle B + \angle D = 120 + 110 = 2300^{\circ} \ne 180^o$ .
The opposite angles are not $$180^{\circ}$.

Thus, the angles of the cyclic quadrilateral are:

$70^{\circ}, 120^{\circ}, 110^{\circ}, 60^{\circ}$ .

Therefore, option

$A$ is correct.

In the figure,

$\angle BAD = 70^{\circ}$ ,

$\angle ABD = 56^{\circ}$ and

$\angle ADC = 72^{\circ}$ .

Calculate (i)

$\angle BDC$ (ii)

$\angle BCD$ (iii)

$\angle CBD$ .

0%
(i) $\angle BDC = 18^{\circ}$

(ii) $\angle BCD = 110^{\circ}$

(iii) $\angle CBD = 52^{\circ}$
0%
(i) $\angle BDC = 18^{\circ}$

(ii) $\angle BCD = 110^{\circ}$

(iii) $\angle CBD = 54^{\circ}$
0%
(i) $\angle BDC = 18^{\circ}$

(ii) $\angle BCD = 120^{\circ}$

(iii) $\angle CBD = 52^{\circ}$
0%
(i) $\angle BDC = 10^{\circ}$

(ii) $\angle BCD = 110^{\circ}$

(iii) $\angle CBD = 52^{\circ}$

Explanation

Given,

$\angle BAD=70^{\circ}$ ,

$\angle ABD=56^{\circ}$ and

$\angle ADC=72^{\circ}$ .

ABCD is a cyclic quadrilateral.

We know, the sum of opposite angles of a cyclic quadrilateral is

$180^o$ .
Therefore,

$\angle BAD+\angle BCD=180^{\circ}$

$\angle BCD =180^{\circ} - \angle BAD$

$\angle BCD=180^{\circ}-70^{\circ}$

$=110^{\circ}$ .....(i).

Also,

$\angle ADC+\angle ABC=180^o$

$72^{\circ}+\left( 56^{\circ}+\angle CBD \right)=180^{\circ}$

$\angle CBD=180^{\circ}-128^{\circ}$

$=52^{\circ}$ .....(ii).

$\triangle BCD$ ,

$\angle BCD+\angle DBC+\angle BDC=180^{\circ}$ ...[Angle sum property]

$110^{\circ}+52^{\circ}+\angle BDC=180^{\circ}$

$\angle BDC=18^{\circ}$ ...(i).

Therefore, option

$A$ is correct.

In the given figure,

$\triangle XYZ$ is inscribed in a circle with centre

$O$ . If the length of chord

$YZ$ is equal to the radius of the circle

$OY$ , then

$\angle YXZ$ is equal to

0%
$60^o$
0%
$30^o$
0%
$80^o$
0%
$100^o$

Explanation

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

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

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

In the given figure,

$ABCD$ is a cyclic quadrilateral in which

$\angle CAD = 25^{o}, \angle ABC = 50^{o}$ and

$\angle ACB = 35^{o}$ .

Then: (i)

$\angle CBD$ (ii)

$\angle DAB$ (iii)

$\angle ADB$ are respectively?

0%
(i) $\angle CBD = 35^{o}$

(ii) $\angle DAB = 80^{o}$

(iii) $\angle ADB = 85^{o}$
0%
(i) $\angle CBD = 25^{o}$

(ii) $\angle DAB = 35^{o}$

(iii) $\angle ADB = 85^{o}$
0%
(i) $\angle CBD = 25^{o}$

(ii) $\angle DAB = 70^{o}$

(iii) $\angle ADB = 35^{o}$
0%
(i) $\angle CBD = 85^{o}$

(ii) $\angle DAB = 70^{o}$

(iii) $\angle ADB = 95^{o}$

Explanation

Given

$ABCD$ is a cyclic quadrilateral.

$AD$ &

$BC$ have been joined.

$\angle CAD={ 25 }^{ o }, \angle ACB=35^{ o }, \angle ABC={ 5 }0^{ o }.$

$\angle CBD$ &

$\angle CAD$ are angles subtended by the chord

$CD$ to the circumference.

$\therefore \angle CBD=\angle CAD={ 25 }^{ o }$ ...[since the angles, subtended by a chord of a circle to the circumference of the same circle, a are equal].

Similarly,

$\angle ACB$ &

$\angle ADB$ are angles subtended by the chord

$AB$ to the circumference.

$\therefore \quad \angle ACB=\angle ADB={ 35 }^{ o }$ ...[since the angles, subtended by a chord of a circle to the circumference of the same circle, are equal].

Also,

$\angle ADC$ &

$\angle ABC$ are angles subtended by the chord

$AB$ to the circumference.

$\therefore \quad \angle ADC=\angle ABC={ 50 }^{ o }$ ...[since the angles, subtended by a chord of a circle to the circumference of the same circle, are equal].

$\therefore \angle BDC=\angle ADC+\angle ADB={ 50 }^{ o }+35^{ o }={ 85 }^{ o }.$

Now

$ABCD$ is a cyclic quadrilateral.

$\therefore$ The sum of its opposite angles

$={ 180 }^{ o }.$

$\angle BAC+\angle BDC={ 180 }^{ o }$

$\Longrightarrow \angle BAC={ 180 }^{ o }-\angle BDC={ 180 }^{ o }-{ 85 }^{ o }=95^{ o }.$

$\therefore \angle DAB=\angle BAC-\angle CAD=95^{ o }-25^{ o }=70^{ o }.$

So, (i)

$\angle CBD=25^{ o }$ , (ii)

$\angle DAB=70^{ o }$ , (iii)

$\angle ADB=35^{ o }$ .

Hence, Option

$C$ is correct.

Two circle intersect at

$A$ and

$B$ . Quadrilaterals

$PCBA$ and

$ABDE$ are inscribed in these circles such that

$PAE$ and

$CBD$ are line segments. Also,

$\angle$ P = 95

$^o$ and

$\angle$ C = 40

$^o$ . The value of

$Z$ is:

0%
$65^o$
0%
$105^o$
0%
$95^o$
0%
$85^o$

Explanation

Given,

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

$ABCP$ &

$ABDE$ are two cyclic quadrilaterals inscribed in the circles in such a way that

$\overline { PAE }$ &

$\overline { CBD }$ are line segments,

i.e.

$\overline { PAE}$ as well as

$\overline { CBD }$ are straight lines.

Also,

$\angle APC={ 95 }^{ o }$ &

$\angle BCP={ 40 }^{ o }.$

Then,

$\angle APC+\angle ABC=180^{ o }$ .......(i) [since the sum of the opposite angles of a cyclic quadrilateral is

${ 180 }^{ o }$ ]

and

$\angle ABC+\angle AED=180^{ o }$ .......(ii) (linear pairs).

So, from (i) & (ii), we get,

$$\angle APC=\angle ABD={ 95 }^{ o }.

Again

$\angle ABD+\angle AED(=Z)=180^{ o }$ ...[since the sum of the opposite angles of a cyclic quadrilateral is

${ 180 }^{ o }$

$\therefore Z=180^{ o }-\angle ABD=180^{ o }-95^{ o }=85^{ o }.$

Hence, option

$D$ is correct.

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

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Practice Class 9 Maths Quiz Questions and Answers

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

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Practice Class 9 Maths Quiz Questions and Answers

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