Circles Test

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Circles Test - 21

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Weekly Quiz Competition

Question 1
1 / -0

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

A
$$30^0$$

B
$$150^0$$

C
$$90^0$$

D
$$120^0$$

Solution

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.
Question 2
1 / -0

In the figure $$AD||BC;\ \angle CAB = 45^{\circ} ;\ \angle DBC = 55^{\circ}$$, then $$\angle DCB$$ equals:

A
$$55^{\circ}$$

B
$$80^{\circ}$$

C
$$100^{\circ}$$

D
$$120^{\circ}$$

Solution

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.
Question 3
1 / -0

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

A
Two chords of a circle equidistant from the centre are equal

B
Equal chords in a circle subtend equal angles at the centre

C
Angle in a semicircle is a right angle

D
All of the above

Solution

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.
Question 4
1 / -0

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

A
$$40^o$$

B
$$50^o$$

C
$$60^o$$

D
$$100^o$$

Solution

$$ABCD$$ 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.
Question 5
1 / -0

If $$O$$ is the centre of a circle, $$AB$$ its chord, $$C$$ is the mid point of $$AB$$ , then $$OAC$$ is

A
an acute angled

B
an obtuse angled

C
a right angled triangle

D
an isosceles triangle

Solution

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,
In $$\Delta OAC$$

$$\angle OAC + \angle AOC + \angle OCA = 180^{\circ}$$ (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
Question 6
1 / -0

In the figure, $$\angle B$$ is equal to:

A
$$80^{\circ}$$

B
$$95^{\circ}$$

C
$$70^{\circ}$$

D
$$115^{\circ}$$

Solution

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.
Question 7
1 / -0

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

A
$$ 90^0 $$

B
$$ 30 ^0$$

C
$$ 45^0 $$

D
None of these
Question 8
1 / -0

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:

A
$$115^o$$

B
$$120^o$$

C
$$125^o$$

D
$$130^o$$

Solution

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.
Question 9
1 / -0

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.

A
Statement - 1 is True, Statement - 2 is True, Statement - 2 is a correct explanation for Statement - 1

B
Statement - 1 is True, Statement - 2 is True : Statement 2 is NOT a correct explanation for Statement - 1

C
Statement - 1 is True, Statement - 2 is False

D
Statement - 1 is False, Statement - 2 is True

Solution

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.
Question 10
1 / -0

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

A
$$80^0$$

B
$$50^0$$

C
$$85^0$$

D
$$40^0$$

Solution

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 $$180^{\circ}$$.
So $$ \angle A+\angle C=180^{\circ}$$
$$\Rightarrow (2x-10+2y+30)^{\circ}=180^{\circ}$$
$$\Rightarrow (2x+2y)^{\circ}=160^{\circ}\quad ------\left( 1 \right) $$

$$ \& \quad \text{also,} \ \angle B+\angle D=180^{\circ}$$
$$\Rightarrow (2y-20+3x+10)^{\circ}=180^{\circ}$$
$$\Rightarrow (3x+2y)^{\circ}=190^{\circ}\quad ------\left( 2 \right) $$ .

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.