Circles Test

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Circles Test - 27

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Weekly Quiz Competition

Question 1
1 / -0

In the circle, reflex $$\angle AOC=x^o, \angle ABC=y^o$$. $$OABC$$ is a parallelogram, then find $$y$$.

A
$$80^o$$

B
$$120^o$$

C
$$110^o$$

D
Data insufficient

Solution

In the circle, clearly, $$x=2y$$.

Then, $$\angle A OC=360-x=360-2y$$.

Here, $$ ABCD$$ is a parallelogram
Then, $${BCO}+ {AOC}=180^o$$
$$\Rightarrow {BCO}+360-2y=180^o$$
$$\Rightarrow 2y=BCO+180^o$$.

Similarly, $$ABC+BCO=180^o$$
$$\Rightarrow y+2y-180=180$$
$$\Rightarrow 3y=360^o$$
$$\Rightarrow y=120^o$$.

Hence, option $$B$$ is correct.
Question 2
1 / -0

Find $$\angle ASR$$.

A
$$52^{\circ}$$

B
$$78^{\circ}$$

C
$$102^{\circ}$$

D
$$10^{\circ}$$

Solution

$$PABQ\quad is\quad a\quad cyclic\quad quadilateral\\ \angle QPA=78^o...(Given). \\ Then, \angle QPA+\angle ABQ=180^o\\ \Rightarrow 78^o+\angle ABQ=180^o\\ \Rightarrow \angle ABQ=180^o-78^o=102^o....(1).\\ \angle ABR\quad and\quad \angle ABQ\quad are\quad linear\quad pair\quad of\quad angles\quad on\quad straight\quad line\quad QR\\ Then, \angle ABQ+\angle ABR=180^o\\ \Rightarrow \angle ABR=180^o-\angle ABQ=180^o-102^o=78^o.\\ Hence,\quad \angle ABR\quad =78^o.\\ In\quad cyclic\quad quadilateral\quad ABCD,\quad \\ \angle ASR+\angle ABR=180^o\\ \Rightarrow \angle ASR=180^o-78^o\\ \Rightarrow \angle ASR=102^o.\\ Hence,\quad option\quad C \quad is\quad correct.$$
Question 3
1 / -0

In the given figure, $$PQRS$$ is a cyclic quadrilateral. Its diagonals $$PR$$ and $$QS$$ intersect each other at $$T$$. If $$\angle PRS=80^{\circ}\;and\;\angle RQS=50^{\circ}$$, calculate $$\angle PSR$$.

A
$$30^{\circ}$$

B
$$50^{\circ}$$

C
$$70^{\circ}$$

D
$$90^{\circ}$$

Solution

$$Given:\\ PQRS\ is\ a\ cyclic\ quadilateral.\\ \angle PRS=80^o....(1)\\ \angle RQS=50^o\\ \angle RQS=\angle RPS\ (Angles\ in\ the\ same\ segment\ are\ equal)\\ \Rightarrow \angle RPS=50^o...(2)\\ In\quad \triangle PSR\\ \angle PRS+\angle RPS+\angle PSR=180^o\ (Angle\ sum\ property)\\ 80^o+50^o+\angle PSR=180^o\ (From\ (1)\ and\ (2))\\ \Rightarrow \angle PSR=180^o-130^o\\ \ \ \ \ \angle PSR=50^o\\ Hence\ option\ (B)\ is\ right\ $$
Question 4
1 / -0

In the figure, $$O$$ is the center. If $$\angle MON=80^o$$, then $$\angle MQN$$ equals

A
$$40^o$$

B
$$160^o$$

C
$$100^o$$

D
$$10^o$$

Solution

Angle subtended by an arc at the center is double the angle subtended by the same are at any point on the circumference.
$$\angle MON\quad =80^o\\ \angle MQN=\frac { 80^o }{ 2 } =40^o$$
Question 5
1 / -0

In the given figure $$\displaystyle \angle AOB=80^{\circ}$$ The value of x is

A
$$\displaystyle 10^{\circ}$$

B
$$\displaystyle 25^{\circ}$$

C
$$\displaystyle 40^{\circ}$$

D
$$\displaystyle 160^{\circ}$$

Solution

We have,
$$\displaystyle \angle AOB=2\angle ACB$$
[ Angle of the centre is twice at th angle of circumference \right ]
$$\displaystyle 80^{\circ}=2x$$
$$\displaystyle \therefore x=\frac{80}{2}=40^{\circ}.$$
Question 6
1 / -0

In figure, $$\angle BAC = 60^{\circ}$$ and $$\angle BCA = 20^{\circ}$$ , find $$\angle ADC$$.

A
15$$^{\circ}$$

B
50$$^{\circ}$$

C
80$$^{\circ}$$

D
40$$^{\circ}$$

Solution

In $$\Delta ABC$$, we have
$$\angle BAC + \angle BCA + \angle ABC = 180^{\circ}$$....[Angle sum property]
$$\Rightarrow 60^{\circ} + 20^{\circ} + \angle ABC = 180^o$$
$$\therefore \angle ABC = 180^{\circ} - 80^{\circ} = 100^{\circ}$$.

$$ABCD$$ is a cyclic quadrilateral,
$$\therefore \angle ABC + \angle ADC = 180^{\circ}$$.....[Since, opposite angles of a cyclic quadrilateral are supplementary]
$$\implies$$ $$100^{\circ} + \angle ADC = 180^{\circ}$$
$$\implies \angle ADC = 180^{\circ} - 100^{\circ} = 80^{\circ}$$.

Hence, option $$C$$ is correct.
Question 7
1 / -0

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

A
$$85^0$$

B
$$95^0$$

C
$$70^0$$

D
$$115^0$$

Solution

Since, $$APQD$$ is a cyclic quadrilateral,
therefore $$\angle A + \angle DQP = 180^o$$ ...(opposite angles of cyclic quadrilateral are supplementary)
$$\implies$$ $$\angle DQP = 180^o -85^o =95^o$$.

Now, $$\angle PQC = 180^o - \angle DQP=180^o-95^o=85^o$$ ...[linear pairs].

$$\therefore \angle B = 180^o -85^o =95^o$$ ...(opposite angles of cyclic quadrilateral are supplementary).

Hence, option $$B$$ is correct.
Question 8
1 / -0

In the figure above, if O is the centre of the circle, find the value of y.

A
40$$^{\circ}$$

B
70$$^{\circ}$$

C
60$$^{\circ}$$

D
30$$^{\circ}$$

Solution

Given that, the centre of the circle is $$O$$ and $$\angle BOD=140^\circ$$
To find out: The value of $$y$$

We know that, in a circle, the angle subtended by an arc at the centre is twice the angle subtended by the same arc at any point on the circle.
$$\therefore \ \angle BOD = 2 \angle BAD$$
$$\Rightarrow 140^{\circ} = 2 \angle BAD$$
$$\displaystyle \Rightarrow \angle BAD = \frac{1}{2} \times 140^{\circ} = 70^{\circ}$$

Now, in cyclic quadrilateral ABCD,
$$\angle BAD + \angle BCD = 180^{\circ}$$ [Opposite angles of a cyclic quadrilateral are supplementary]
$$\therefore \ 70^{\circ} + \angle BCD = 180^{\circ}$$
$$\therefore \angle BCD = 180^{\circ} - 70^{\circ} = 110^{\circ}$$
Also, $$\angle BCD + \angle DCP = 180^{\circ}$$ [Linear pair]
$$\Rightarrow 110^{\circ} + y = 180^{\circ}$$
$$\therefore \ y = 70^{\circ}$$

Hence, option B is correct.
Question 9
1 / -0

In a cyclic quadrilateral $$ABDC,\,\angle CAB=80^{\circ}$$ and $$\angle ABC=40^{\circ}$$. The measure of the $$\angle ADB$$ will be?

A
$$120^{\circ}$$

B
$$80^{\circ}$$

C
$$60^{\circ}$$

D
$$40^{\circ}$$

Solution

In $$\Delta ACB$$,
$$\because\;\angle ACB=180^{\circ}-(80^{\circ}+40^{\circ})=60^{\circ}$$
$$\therefore\;\angle ADB=\angle ACB$$
$$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=60^{\circ}$$
Question 10
1 / -0

In fig, a is the centre of the circle. If $$\angle$$ BOD 160$$^{\circ}$$ find the values of x and y.

A
80$$^{\circ}$$, 100$$^{\circ}$$

B
135$$^{\circ}$$, 45$$^{\circ}$$

C
140$$^{\circ}$$, 40$$^{\circ}$$

D
120$$^{\circ}$$, 60$$^{\circ}$$

Solution

In the cyclic quadrilateral ABCD
$$\displaystyle \angle BCD = \frac{1}{2} \angle BOD$$
[Angle made by an arc at the centre is double the angle made by it, at any other point on the remaining part of the circle]
$$\displaystyle \therefore \angle x = \frac{1}{2} \times 160^{\circ} = 80^{\circ}$$
$$\therefore x = 80^{\circ}$$
$$\angle x + \angle y = 180^{\circ}$$
[opp. angles of a cyclic quadrilateral]
$$\therefore \angle y = 180^{\circ} - \angle x$$
$$\angle y = 180^{\circ} - 80^{\circ} = 100^{\circ}$$
$$y = 100^{\circ}$$