Circles Test

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Circles Test - 16

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Weekly Quiz Competition

Question 1
1 / -0

In the given figure, $$O$$ is the centre of the circle, $$\angle ACB=54^{\circ}$$ and $$BCE$$ is a straight line. Find $$x.$$

A
$$126^{\circ}$$

B
$$54^{\circ}$$

C
$$108^{\circ}$$

D
$$90^{\circ}$$

Solution

$$\angle ACE=180^{\circ}-54^{\circ}=126^{\circ}$$ ($$BCE$$ is a straight line)
$$\Rightarrow $$ $$\angle ADE=\angle ACE=126^{\circ}$$ (Angles in the same segement are equal)
Theorem: Angle subtended by an arc at the centre is double the angle subtended by the same arc at the circumference of a circle
$$\therefore $$ Reflex $$\angle AOE=2\times \angle ADE=2\times 126^{\circ}=252^{\circ}$$
$$\therefore $$ $$x=360^{\circ}-252^{\circ}=108^{\circ}$$
Question 2
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In the given figure, $$\angle M = 75^{\circ}$$, then $$ \angle O $$ is equal to:

A
$$105^{\circ}$$

B
$$100^{\circ}$$

C
$$85^{\circ}$$

D
$$110^{\circ}$$

Solution

A cyclic quadrilateral is a quadrilateral whose vertices all lie on the circumference of a circle.
We know, opposite angles of a cyclic quadrilateral add up to $$180^o$$.

In the given cyclic quadrilateral,
$$\angle{O }+\angle{ M}=180^o$$ and $$\angle{L }+\angle{ N}=180^o$$ ...[Opposite angles of cyclic quadrilateral].

Then, $$\angle{O }+\angle{ M}=180^o$$
$$\implies$$ $$\angle{O }+75^o=180^o$$
$$\implies$$ $$\angle{O}$$ $$=180^o-75^o={105}^{\circ}$$.

Therefore, option $$A$$ is correct.
Question 3
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A set of points equidistant from a fixed point in a plane figure is called

A
triangle

B
circle

C
square

D
none of these

Solution

A set of points equidistant from a fixed point in a plane figure is called a circle where the distance between each of the set of the points and the fixed point forms the radius of the circle.
Question 4
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If the length of the largest chord of a circle is $$17$$ cm, find the radius of a circle.

A
$$34$$ cm

B
$$8.5$$ cm

C
$$ \sqrt { 17 } $$ cm

D
$$ \sqrt { 34 } $$ cm

Solution

Given, the largest chord of a circle $$=17cm$$.
We know that the largest chord of a circle is its diameter.
$$\therefore$$ The diameter of the given circle $$=17 cm$$.
Hence, radius of the same circle$$=\dfrac { \text {diameter} }{ 2 } =\dfrac { 17 }{ 2 } cm=8.5cm$$.
Therefore, option $$B$$ is correct.
Question 5
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In fig. O is the centre of the circle. Determine $$\angle$$ AEC

A
50$$^{\circ}$$

B
68$$^{\circ}$$

C
22$$^{\circ}$$

D
55$$^{\circ}$$

Solution

ABCD is a cyclic quadrilateral in which side AB is produced to D.
Also we know that exterior $$\angle $$ of a cyclic quadrilateral is equal to the interior opposite angle
$$\therefore \angle CBD = \angle AEC$$
But $$\angle CBD = 68^{\circ}$$
$$\therefore \angle AEC = 68^{\circ}$$
Question 6
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In the above figure, $$\angle POR$$ is (there $$O$$ is centre of Circle)

A
$$60^\circ$$

B
$$80^\circ$$

C
$$100^\circ$$

D
$$160^\circ$$

Solution

$$\angle PQR=50^o+30^o=80^o$$
Theorem: Angle subtended by an arc at the centre is double the angle subtended by the same arc at the circumference of a circle
$$\therefore \angle POR = 2 \times 80 = 160^°$$
Question 7
1 / -0

In the given figure, the value of $$x$$ and $$y$$ is:

A
$$20^{\circ}$$, $$30^{\circ}$$

B
$$36^{\circ}$$, $$60^{\circ}$$

C
$$15^{\circ}$$, $$30^{\circ}$$

D
$$25^{\circ}$$, $$30^{\circ}$$

Solution

In a cyclic quadrilateral, the opposite angles are supplementary.

That is, $$2x+3x = 180^o$$

$$\Rightarrow 5x=180^o$$

$$ \Rightarrow x = 36^o$$

Similarly, $$y+2y = 180^o$$
$$\Rightarrow 3y=180^o$$
$$\Rightarrow y = 60^o$$.

Hence, option $$(B)$$ is correct.
Question 8
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What is the sum of $$\angle{BAD}+\angle{BPR}+\angle{BCD}+\angle{BQR}$$ in the above figure?

A
$$540^\circ$$

B
$$360^\circ$$

C
$$240^\circ$$

D
None of the above

Solution

In cyclic quadrilateral $$BADC$$,
$$\angle BAD +\angle BCD = 180^{ 0 }$$.....[opposite angles of a cyclic quadrilateral are supplementary]

In cyclic quadrilateral $$BPRQ$$,
$$\angle BPR+\angle BQR=180^0$$...[opposite angles of a cyclic quadrilateral are supplementary]

Hence the sum of
$$\angle{BAD}+\angle{BPR}+\angle{BCD}+\angle{BQR}$$ $$=180^o + 180^o = 360^o$$.

Therefore, option $$B$$ is correct.
Question 9
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Which of the following pair of angles are opposite angles of a cyclic quadrilateral?

A
$$131^{\circ}, 28^{\circ}$$

B
$$95^{\circ}, 55^{\circ}$$

C
$$123^{\circ}, 57^{\circ}$$

D
$$64^{\circ}, 52^{\circ}$$

Solution

We know that, the sum of opposite angles of every cyclic quadrilateral is $$180^{\circ}$$.
If $$ABCD$$ is cyclic quadrilateral,
then $$\angle A +\angle C=180^o$$.

The sum of the angles $${123}^{\circ} $$ and $${57}^{\circ}$$ is equal to $$180^{\circ}$$.
The sum of two angles given in every other option, other than option $$C$$ is not $$180^{\circ}$$.

Therefore, option $$C$$ is correct.
Question 10
1 / -0

In Fig 10.9 $$\displaystyle \angle AOB= 90^{\circ}\,and\,\angle ABC=30^{\circ}\,then\,\angle CAO$$ is equal to

A
$$\displaystyle 30^{\circ}$$

B
$$\displaystyle 45^{\circ}$$

C
$$\displaystyle 90^{\circ}$$

D
$$\displaystyle 60^{\circ}$$

Solution

In $$\triangle AOB$$
$$OA=OB$$
$$\therefore \angle OAB+\angle OBA+\angle AOB=180^{0}$$
$$=>2\angle OAB+90^{0}=180^{0}$$
$$=>2\angle OAB=180^{0}-90^{0}$$
$$=90^{0}$$
$$=>\angle OAB=\frac{90}{2}$$
$$=45^{0}$$
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.
$$\therefore \angle ACB=\frac{1}{2}\angle AOB$$
$$=\frac{90}{2}$$
$$=45^{0}$$
Again in $$\triangle ABC$$
$$\angle ACB+\angle CBA+\angle BAC=180^{0}$$
$$=>45^{0}+30^{0}+\angle BAC=180^{0}$$
$$=>\angle BAC=180^{0}-75^{0}$$
$$=105^{0}$$
$$\therefore \angle CAO=\angle BAC-\angle OAB$$
$$=>\angle CAO=105^{0}-45^{0}$$
$$=60^{0}$$