Circles Test

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Circles Test - 30

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Weekly Quiz Competition

Question 1
1 / -0

Given a circle with the center indicated, find $$x.$$

A
$$100^°$$

B
$$80^°$$

C
$$50^°$$

D
$$40^°$$

Solution

We know that the angle at the centre is twice is the angle at the circumference subtended by the same arc.
Here angle at the circumference is $$x$$
Let the triangle be $$ABC$$.
Thus the angle at the center $$B$$ is $$2x$$.
Since $$AB$$ and $$BC$$ are the radius of the circle, they are of the same length.
So, $$\angle \; A = \angle \; C = 50^{\circ}$$
Now,
$$2x+50^°+50^°=180^°$$
$$\Rightarrow 2x=180^°-100^°$$
$$\Rightarrow 2x=80^°$$
$$\Rightarrow x=40^°$$
Question 2
1 / -0

$$ABCD$$ is a cyclic quadrilateral with $$AD$$ parallel to $$BC$$. $$\angle DCB = 65^o$$. The magnitude of $$\angle CBE$$ is:

A
$$100^o$$

B
$$110^o$$

C
$$115^o$$

D
$$120^o$$

Solution

Given, $$AD \parallel BC$$.

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

Then, $$\angle OCB + \angle DAB = 180^\circ$$ ...[Opposite angles of cyclic quadrilateral are supplementary]
$$ \implies \angle DAB = 115^\circ$$.

Since, $$AD \parallel BC$$

$$\implies$$ $$\angle DAB = \angle CBE = 115^\circ$$ ...[Corresponding angles].
Hence, option $$C$$ is correct.
Question 3
1 / -0

The portion of a circle between two radii and an arc is called

A
Sector

B
Segment

C
Chord

D
Secant

Solution

Sector of a circle is the portion of a disk enclosed by two radii and an arc. The smaller area is known as the minor sector and the larger as the major sector.
So option A is the correct answer.
Question 4
1 / -0

Find $$PM$$ if $$PQ=8 \ cm$$

A
$$3 cm$$

B
$$4 cm$$

C
$$5 cm$$

D
$$8 cm$$

Solution

The perpendicular from centre to the chord bisects the chord.
$$PM=\dfrac{1}{2}\times8=4$$
So, option B is correct.
Question 5
1 / -0

Find the measure of $$\widehat{AOB}$$.

A
$$100^\circ$$

B
$$104^\circ$$

C
$$102^\circ$$

D
$$103^\circ$$

Solution

Given, $$\angle ABC=38^\circ,$$ $$ \angle BCD=12^\circ$$ and $$BC$$ is diameter

Now, $$\angle BAC=\angle BDC=90^\circ$$ (triangle with diameter as a base)

In $$\triangle ABC$$

$$\angle ACB+\angle BAC+\angle ABC=180^\circ$$

$$ \angle ACB=180^\circ-90^\circ-38^\circ=52^\circ$$

Also, $$\angle AOB=2\angle ACB$$ (half angle theorem)

$$\Rightarrow \angle AOB=2\times 52^\circ=104^\circ$$
Question 6
1 / -0

Find the value of $$x$$ and $$y$$.

A
$$x=150^o, y=120^o$$

B
$$x=120^o, y=150^o$$

C
$$x=100^o, y=130^o$$

D
$$x=130^o, y=100^o$$

Solution

The given quadrilateral is a cyclic quadrilateral.
We know, Opposite angles of cyclic quadrilateral are supplementary.

Therefore, $$\angle x +60^o = 180^o$$
$$\implies$$ $$\angle x = 180 - 60 = 120^o$$.

Similarly, $$\angle y +30^o = 180^o$$
$$\implies$$ $$\angle y = 180 - 30 = 150^o$$.

Hence, $$x = 120^o$$ and $$y=150^o$$.
Therefore, option $$B$$ is correct.
Question 7
1 / -0

In the figure given above, what is $$\angle BYX$$ equal to?

A
$$85^{\circ}$$

B
$$50^{\circ}$$

C
$$45^{\circ}$$

D
$$90^{\circ}$$

Solution

We know that angles subtended by an arc on the circumference of the circle on the same segment are equal.
$$\angle XBY = \angle XPY = 45^{\circ}$$
$$\Rightarrow BYX = 180^{\circ} - (50^{\circ} + 45^{\circ}) = 85^{\circ}$$.
Question 8
1 / -0

In the given figure, $$O$$ is the centre of the circle and $$\angle QPR = x^{\circ}; \angle ORQ = y^{\circ}$$. Which statement is true about $$x^{\circ}$$ and $$y^{\circ}$$?

A
$$x^{\circ} + y^{\circ} = 120^{\circ}$$

B
$$x^{\circ} + y^{\circ} = 180^{\circ}$$

C
$$x^{\circ} + y^{\circ} = 90^{\circ}$$

D
$$x^{\circ} + y^{\circ} = 150^{\circ}$$

Solution

In $$\triangle OQR \quad OR=OQ=$$ radius
$$\Rightarrow \angle OQR=\angle ORQ=y^o$$
Thus by angle sum property, we have
$$\angle OQR+\angle ORQ+\angle QOR=180^o \\ 2\angle ORQ+\angle QOR=180^o \\ \angle QOR=180-2\angle ORQ=180-2x^o$$
Also, $$\angle QOR=2\angle QPR$$ (half angle property)
$$180-2x=2y \\ 180=2x+2y \\ x^o+y^o=90^o$$
Question 9
1 / -0

$$A,B,C,D$$ are four distinct points on a circle whose centre is at $$O$$.
If $$\angle OBD-\angle CDB=\angle CBD-\angle ODB$$, then what is $$\angle A$$ equal to ?

A
$${45}^{o}$$

B
$${60}^{o}$$

C
$${120}^{o}$$

D
$${135}^{o}$$

Solution
Question 10
1 / -0

What is the relation between '$$p$$' and '$$q$$'?

A
$$q = 2p$$

B
$$p = 2q$$

C
$$p = q$$

D
$$q > p$$

Solution

$$ABCD$$ is a cyclic quadrilateral. Hence,
$$\angle AOC=180^{\circ}-p$$

At point $$O$$ there is a complete angle hence,
$$\begin{aligned}{}\angle AOC + q& = {360^\circ }\\q& = {360^\circ } - \angle AOC\\q& = {360^\circ } - ({180^\circ } - p)\\q& = {180^\circ } + p\\q& > p\end{aligned}$$