Circles Test

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Circles Test - 28

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Weekly Quiz Competition

Question 1
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One of the base angle a cyclic trapezium is double the other. What is the measure of the larger angle?

A
$$60^0$$

B
$$80^0$$

C
$$75^0$$

D
$$120^0$$

Solution

In cyclic trapezium,
Sum of opposite angle $$=180^o$$ ...[opposite angles of a cyclic quadrilateral are supplementary].

Let the smaller angle be $$x$$
and larger angle $$=2x$$.

Then, $$2x+x=180^o \implies 3x=180^o \implies x=60^o$$.

Therefore, Larger angle $$=2x=2\times 60^o=120^o$$.

Hence, option $$D$$ is correct.
Question 2
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Find $$x.$$

A
$$40^{\circ}$$

B
$$50^{\circ}$$

C
$$30^{\circ}$$

D
$$60^{\circ}$$

Solution

ABCD is a quadrilateral inside the circle,
$$\angle D + \angle B = 180^o$$
$$130^o + \angle B = 180^o$$
$$\angle B = 50^o$$
AB is diameter $$\angle C = 90^o$$ [angle in a semicircle]
In $$\Delta ABC$$
$$\angle A + \angle B + \angle C = 180^o$$
$$\angle A + 50^o + 90^o = 180^o$$
$$\Rightarrow \angle A + 140^o = 180^o$$
$$\Rightarrow \angle A = 180^o - 140^o$$
$$=40^o$$
$$\therefore \angle CAB = 40^o \Rightarrow x = 40^o$$
option (A) is correct.
Question 3
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If $$O$$ is the centre of the circle and $$A, B$$ and $$C$$ are points on its circumference and $$\angle AOC = 130^{\circ}$$, find $$\angle ABC$$.

A
$$105^{\circ}$$

B
$$115^{\circ}$$

C
$$110^{\circ}$$

D
$$120^{\circ}$$

Solution

Take any point $$P$$ on the circumference of the circle as shown.
Join $$AP$$ and $$CP$$.

$$\because ABC$$ subtends $$\angle AOC$$ at centre $$O$$ and $$\angle APC$$ at any point $$P$$ on the circumference of the circle.
$$\therefore \angle AOC = 2 \angle APC$$ [$$\because$$ Angle subtended by an arc at the centre is twice the angle subtended by the same arc at any point on the circumference]
$$\displaystyle \implies \angle APC = \frac{1}{2} \angle AOC ~~~[ \because AOC = 130^{\circ}]$$
$$\implies$$ $$\displaystyle \frac{1}{2} \times 130^{\circ} = 65^{\circ}$$.

$$\because$$ $$ABCP$$ is a cycle quadrilateral,
$$\implies \angle APC + \angle ABC = 180^{\circ}$$ ...[$$\because$$ sum of opposite angles of a cyclic quadrilateral is $$180^{\circ}$$]
$$\implies 65^{\circ} + \angle ABC = 180^{\circ}$$
$$\therefore \angle ABC = 180^{\circ} - 65^{\circ} = 115^{\circ}$$.

Hence, option $$B$$ is correct.
Question 4
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In the given figure, AB=AC and $$\displaystyle \angle ABC=68^{\circ},$$ what is the value of $$\displaystyle \angle BPC$$?

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

B
$$\displaystyle 42^{\circ}$$

C
$$\displaystyle 44^{\circ}$$

D
$$\displaystyle 48^{\circ}$$

Solution

Theorem: Chord of a circle subtends equal angles at different points on the circumference of the circle on same side

$$\Rightarrow \angle BAP=\angle BPC$$

$$ AB=AC\Rightarrow \angle B=\angle C=68^{°}$$ (Opposite angles of equal sides are equal)
In $$\triangle ABC $$
$$\angle A+ \angle B + \angle C =180^o$$ (Sum of angles in a triangle )

$$68^o+68^o+ \angle A = 180^o$$

$$136^o+\angle A = 180^o$$

$$\Rightarrow \angle A=44^{°}$$

$$\Rightarrow \angle A=\angle BAC =\angle BPC =44^{°}$$ (Angles in same segment are equal )
Question 5
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In the given figure, $$BD=DC$$ and $$\displaystyle \angle DBC=30^{\circ}$$ what is the value of $$\displaystyle \angle BAC$$?

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

B
$$\displaystyle 50^{\circ}$$

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

D
$$\displaystyle 65^{\circ}$$

Solution

As $$BD=DC$$ ...... (given)
$$\Rightarrow \angle DBC=\angle DCB=30^{\circ}$$

$$\Rightarrow \angle BDC =180^\circ -30^\circ-30^\circ=120^{\circ}$$

As $$ABCD$$ is a cyclic quadrilateral ; $$\angle BAC+\angle BDC =180^{\circ}$$

$$\Rightarrow \angle BAC =180^{\circ}-120^{\circ} =60^{\circ}$$
Question 6
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$$ABCD$$ is a cyclic quadrilateral, in which $$BC$$ is parallel to $$AD$$, $$\displaystyle\angle ADC=\displaystyle 110^{\circ}$$ and $$\displaystyle \angle BAC=50^{\circ}$$. What is the value of $$\displaystyle \angle DAC$$?

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

B
$$\displaystyle 50^{\circ}$$

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

D
$$\displaystyle 64^{\circ}$$

Solution

$$ABCD$$ is a cyclic quadrilateral.
Then, $$\angle ABC+\angle ADC=180^o$$ ...[Opposite angles of a cyclic quadrilateral are supplementary]
$$\Rightarrow$$ $$\angle ABC =180^{\circ}-110^{\circ} =70^{\circ}$$.

In $$\triangle ABC$$,
$$\angle ABC+\angle ACB+\angle BAC=180^o$$ ...[Angle sum property]
$$\Rightarrow$$ $$70^o+\angle ACB+50^o=180^o$$
$$\Rightarrow$$ $$ \angle ACB=180^{\circ}-(50^{\circ}+70^{\circ})=60^{\circ}$$.

Given, $$BC$$ and $$AD$$ are parallel,
$$\Rightarrow \angle ACB=\angle DAC =60^{\circ}$$ ...[Alternate interior angles].

Hence, option $$C$$ is correct.
Question 7
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What is the value of $$ \displaystyle \angle ABC $$ ?

A
$$ \displaystyle 105^{o} $$

B
$$ \displaystyle 110^{o} $$

C
$$ \displaystyle 115^{o} $$

D
$$ \displaystyle 120^{o} $$

Solution

Extending A and C onto the circumference to D as shown in the figure

Theorem: Angle subtended by an arc at the centre is double the angle subtended by the same arc at the circumference of a circle

$$\Rightarrow \angle AOC=2\times \angle ADC $$

$$\Rightarrow \angle ADC=\dfrac {130}{2}=65^o$$

As $$ABCD$$ is a cyclic quadrilateral, sum of its opposite interior angles are equal to $$180^o$$

$$\Rightarrow \angle ADC+\angle CBA =180^o$$

$$\Rightarrow \angle ABC =180^o - 65^o=115^o$$
Question 8
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ABCD is a cyclic trapezium and $$ \displaystyle AB\parallel CD $$. If $$AB$$ is the diameter of the circle and $$ \displaystyle \angle CAB=30^{\circ} $$, then the value of $$ \displaystyle \angle ADC $$ is:

A
$$ \displaystyle 110^{\circ}$$

B
$$ \displaystyle 115^{\circ}$$

C
$$ \displaystyle 120^{\circ}$$

D
$$ \displaystyle 125^{\circ}$$

Solution

Here, $$\angle CAB=30^o$$ ...[Given].
Also, $$\angle ACB=90^o$$ ...[Angle inscribed in a semi-circle].

Now, in $$\triangle ACB$$,
$$\angle ACB+\angle CAB+\angle ABC=180^o$$ ...[Angle sum property]
$$\Rightarrow$$ $$90^o+30^o+\angle ABC=180^o$$
$$\Rightarrow$$  $$\angle ABC=180^o-120^o$$
$$\Rightarrow$$  $$\angle ABC=60^o$$.

Here, $$ABCD$$ is a cyclic quadrilateral.
We know that, sum of opposite angles of a cyclic quadrilateral is $$180^o$$.
$$\therefore$$ $$\angle ABC+\angle ADC=180^o$$
$$\Rightarrow$$  $$60^o+\angle ADC=180^o$$
$$\Rightarrow$$  $$\angle ADC=180^o-60^o$$
$$\Rightarrow$$  $$\angle ADC=120^o.$$

Hence, option $$D$$ is correct.
Question 9
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In the given figure $$\text{O}$$ is the center of the circle and measure of $$\angle \text{AOC} $$ is $$ \displaystyle 100^{\circ} $$ what is the value of $$ \displaystyle \angle \text{ADC} $$ ?

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

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

C
$$ \displaystyle 50^{\circ} $$

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

Solution

The angle subtended by an arc at the center of the circle is twice the angle subtended by it on the circumference of the circle.

$$\implies \angle \text{ADC}=\dfrac{1}{2}\angle \text{AOC}$$

$$=\dfrac{1}{2}\times 100$$

$$=50^{\circ}$$
Question 10
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ABCD is cyclic quadrilateral. The tangents to a circle at A and C meet at P. If $$\angle$$ APC = 50$$^{\circ} $$, then the value of $$\angle$$ ADC is:

A
$$ \displaystyle 65^{\circ} $$

B
$$ \displaystyle 70^{\circ} $$

C
$$ \displaystyle 80^{\circ} $$

D
none of these

Solution

Given $$\angle APC=50^0 $$

As the radius is always perpendicular to tangent at the point of contact,
$$\angle OCP=\angle OAP=90^0$$

As $$AOCP $$ is a quadrilateral, sum of all its interior angles are equal to $$360^0$$

$$\therefore\ \angle AOC=130^0\quad \quad [360^0-(90^0+90^0+50^0)]$$

We know that, angle subtended by an arc at the center of a circle is twice the angle subtended by the same arc on the circumference of the circle.

$$\therefore \ \angle ADC=\dfrac{130^0}{2}$$
$$=65^0$$

Hence, $$\angle ADC=65^0$$.