Circles Test

Result

Circles Test - 37

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Weekly Quiz Competition

Question 1
1 / -0

$$ABCD$$ is a cyclic quadrilateral such that $$AB$$ is a diameter of the circle circumscribing it and $$\displaystyle \angle ADC=140^{\circ}$$, then $$\displaystyle \angle BAC$$ is equal to:

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

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

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

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

Solution

Since $$ABCD$$ is a cyclic quadrilateral,
the sum of the opposite angles will be $$180$$ degrees.

Thus, $$∠CBA+∠ADC={ 180 }^{ ∘ }$$

$$\Rightarrow $$$$∠CBA={ 180 }^{ ∘ }-∠ADC$$

$$\Rightarrow $$$$∠CBA={ 180 }^{ ∘ }-{ 140 }^{ ∘ }$$

$$\Rightarrow $$$$∠CBA={ 40 }^{ ∘ }$$.

Also, angle in a semicircle in a right angle.
Therefore, $$∠ACB={ 90 }^{ ∘ }$$

Now, in $$\triangle ACB$$,

$$∠CBA+∠ACB+∠BAC={ 180 }^{ ∘ }$$ ...[angle property of a triangle]

$$\Rightarrow $$$${ 40 }^{ ∘ }+{ 90 }^{ ∘ }+∠BAC={ 180 }^{ ∘ }$$

$$\Rightarrow $$$$∠BAC={ 180 }^{ ∘ }-{ 130 }^{ ∘ }$$

$$\Rightarrow $$$$∠BAC=50^{ ∘ }$$.

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

Any cyclic parallelogram is a ______.

A
rectangle

B
rhombus

C
trapezium

D
square

Solution

We know that in a cyclic quadrilateral, sum of measures of opposite angles is $$180^o$$.

The given cyclic quadrilateral is a parallelogram.

Hence, opposite pairs of angles will be congruent.

$$\angle 1 + \angle 3 = 180^o$$ and $$\angle 2 + \angle 4 = 180^o$$

$$\Rightarrow 2 \times \angle 1 = 180^o$$ and $$2 \times \angle 2 = 180^o$$

$$\Rightarrow \angle 1 = \angle 2 = 90^o$$

Hence, $$\angle 3= \angle 4 = 90^o$$

Therefore, the given parallelogram has all its angles equal to $$90^o$$ i.e. a rectangle.

Hence, option $$A$$ is correct.
Question 3
1 / -0

In the given figure, ABCD is a cyclic quadrilateral and PQ is tangent to the circle with centre 'O'. BD is diameter, $$\angle DCQ = 40^o , \angle ABD = 60^o$$. The $$\angle BCP=$$ _______.

A
$$50^o$$

B
$$100^o$$

C
$$60^o$$

D
$$20^o$$

Solution
Question 4
1 / -0

In the diagram, O is the centre of the circle. Find the value of $$x$$.

A
$$111^o$$

B
$$123^o$$

C
$$69^o$$

D
$$49^o$$

Solution

By using circle theorem which states that the measure of an angle subtended by an arc at the center is twice the measure of an angle subtended by the same arc at the circumference of the circle.
The exterior angle at the center is $$360^{o}-138^{o}=222^{o}$$.
So, the value of $$X$$ is half of $$222^{o}$$ that is $$ \dfrac{222^{o}}{2}=111^{o}$$.
Hence, the answer is $$111^{o}$$.
Question 5
1 / -0

In the given figure, O is centre of the circle, then $$\angle XOZ $$ is ________.

A
$$ 2 \angle X$$

B
$$ 2 \angle Y$$

C
$$ 2 \angle Z$$

D
$$2[ \angle XYZ + \angle YXZ]$$
Question 6
1 / -0

In the give figure, ABCD is a cyclic quadrilateral, $$
\angle C B Q=48^{\circ}
$$ and a=2b. Then, b is equal to

A
$$

48^{\circ}

$$

B
$$

38^{\circ}

$$

C
$$

28^{\circ}

$$

D
none of these

Solution
Question 7
1 / -0

The circumcirlce of the quadrilateral formed by the lines $$x = a,x = 2a,y = - a,y = a$$ is

A
$${x^2} + {y^2} - 3ax - {a^2} = 0$$

B
$${x^2} + {y^2} + 3ax + {a^2} = 0$$

C
$${x^2} + {y^2}-3ax + {a^2} = 0$$

D
$${x^2} + {y^2} + 3ax - {a^2} = 0$$

Solution
Question 8
1 / -0

In the given figure, in $$ \square $$ pors.if $$ \angle R S P=80^{\circ} $$ then $$ \angle R Q T= $$____________

A
$$ 100^{\circ} $$

B
$$ 80^{\circ} $$

C
$$ 70^{\circ} $$

D
$$ 110^{\circ} $$

Solution
Question 9
1 / -0

Distance between chords of contact of the tangents of the circle $$ x ^ { 2 } + y ^ { 2 } + 2 y x + 27 y + c = 0 $$ from the origin and the point ($$g, f$$) is

A
$$ \sqrt { g ^ { 2 } + f ^ { 2 } } $$

B
$$ \frac { \sqrt { g ^ { 2 } + f ^ { 2 } - c } } { 2 } $$

C
$$ \frac { g ^ { 2 } + f ^ { 2 } - c } { 2 \sqrt { g ^ { 2 } + f ^ { 2 } } } $$

D
$$ \frac { \sqrt { g ^ { 2 } + f ^ { 2 } + c } } { 2 \sqrt { g ^ { 2 } + f ^ { 2 } } } $$

Solution
Question 10
1 / -0

Two tangents $$PQ$$ and $$PR$$ drawn to the circle $$x^2+y^2-2x-4y-20=0$$ from the point $$P(16,7)$$. If is the centre of the circle, area of quadrilateral $$PQCR$$ is

A
75

B
80

C
85

D
95

Solution