Circles Test

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Circles Test - 19

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Weekly Quiz Competition

Question 1
1 / -0

In the figure, $$O$$ is the centre of the circle. If $$\angle {ADC}={140}^{o}$$, then what is the value of $$x$$?

A
$${35}^{o}$$

B
$${55}^{o}$$

C
$${60}^{o}$$

D
$${50}^{o}$$

Solution

$$ OA = OC \Rightarrow \angle OAC = \angle ACO = X $$
$$ \angle ADC+\angle AMC = 180^{\circ} $$ (cyclic quadrilateral )
$$ \angle AMC = 180^{\circ} -140^{\circ} = 40^{\circ} $$
$$ \angle ADC = 2\angle AMC = 2\times 40^{\circ} = 80^{\circ} $$
In $$ \triangle AOC $$
$$ \angle ADC+\angle OAC+\angle OCA = 180^{\circ} $$
$$ x+80^{\circ}+x = 180^{\circ} $$
$$ 2x = 100^{\circ} $$
$$ x = 50^{\circ} $$
Question 2
1 / -0

The geometric shape that has no corners is _____.

A
square

B
circle

C
rectangle

D
hexagon

Solution

The geometric shape which has no corners is circle.
Question 3
1 / -0

We can draw a circle using given three points if given points are

A
collinear

B
non collinear

C
no condition on points

D
none of these

Solution

A circle is defined by any three non-collinear points. This means that, given any three points that are not on the same line, you can draw a circle that passes through them. It is possible to construct this circle using only a compass and straightedge.
Question 4
1 / -0

The perpendicular distance between the biggest chord and the centre is

A
Zero

B
Equal

C
$$9 cm $$

D
$$10 cm $$

Solution

Zero
The biggest chord of the circle is the diameter.
Now the perpendicular distance between the biggest chord (diameter) and the centre of the the circle is 0. (Since the chord passes through the centre)
Question 5
1 / -0

Is it possible to draw a circle with given two points?

A
Yes

B
No

C
Can't say anything

D
None of these

Solution

Infinite number of circles can be drawn from two points
starting from the two points as a diameter we can draw circle
as the circle is moving up it becomes a chord to the next circle with bigger diameter. as such a way we can draw infinite number of circles passing through two points.
Question 6
1 / -0

If chord $$AOB $$ is diameter of the circle and $$AC = BC$$, then $$\angle$$ $$ACB$$ is equal to:

A
$$30$$

B
$$60$$

C
$$90$$

D
$$45$$

Solution

$$\angle ACB = 90^o$$ as it is inscribed in a semi circle.
So option C is the correct answer.
Question 7
1 / -0

Two chords $$AB$$ and $$AC$$ of a circle subtends angles equal to $$ { 90 }^{ o } $$ and $$ { 150 }^{ o } $$, respectively at the centre. Find $$ \angle BAC $$, if $$AB$$ and $$AC$$ lie on the opposite sides of the centre.

A
$$ { 120 }^{ o } $$

B
$$ { 240 }^{ o } $$

C
$$ { 60 }^{ o } $$

D
$$ { 30 }^{ o } $$

Solution

Given- $$ AB$$ and $$AC$$ are the chords of a circle with centre $$O$$.
$$ \angle AOB={ 90 }^{ o }$$ and $$ \angle AOC={ 150 }^{ o }$$.
To find out- $$ \angle BAC=?$$
Solution-
$$BC$$ is joined.
Now $$ \angle BOC={ 360 }^{ o }-\left( \angle AOB+\angle AOC \right) = { 360 }^{ o }-({ 90 }^{ o }+{ 150 }^{ o })={ 120 }^{ o }$$.
So, $$ \angle BAC=\dfrac { 1 }{ 2 } \angle BOC=\dfrac { 1 }{ 2 } \times { 120 }^{ o }={ 60 }^{ o }$$ ....(The angle at the centre, subtended by a chord of a circle is double of that a the circumference.)
Question 8
1 / -0

$$ABCD$$ is a cyclic quadrilateral such that $$ A={ 90 }^{ o }, B={ 70 }^{ o }, C={ 95 }^{ o } $$ and $$D={ 105 }^{ o }.$$
The statement is:

A
true.

B
false.

C
true in special condition.

D
Data insufficeint

Solution

Given, $$ABCD$$ is a cyclic quadrilateral where $$ \angle A={ 90 }^{ o } , \angle B={ 70 }^{ o }, \angle C={ 95 }^{ o }$$ & $$\angle D={ 105 }^{ o }.$$
Since $$ABCD$$ is a cyclic quadrilateral,
then $$\angle A+\angle C ={ 180 }^{ o }$$ and $$ \angle B+\angle D={ 180 }^{ o }$$......$$(i)$$(The sum of the opposite angles of a cyclic quadrilateral is $${ 180 }^{ o })$$
But here, $$ \angle A+\angle C={ 90 }^{ o }+95^{ o }=185^{ o }$$ and $$\angle B+\angle D={ 70 }^{ o }+{ 10 }5^{ o }=175^{ o }.$$
This is a contradiction to $$(i)$$.
That is, the statement is false.
Hence, option $$B$$ is correct.
Question 9
1 / -0

If arcs $$AXB$$ and $$CYD$$ of a circle are congruent, then the ratio of $$AB$$ and $$CD$$ is

A
$$1:1$$

B
$$2:1$$

C
$$1:2$$

D
$$0$$

Solution

$$ Given-\quad \\ AXB\quad \& \quad CYD\quad are\quad congruent\quad arcs\quad of\quad a\quad circle.\\ to\quad find\quad out-\\ the\quad ratio\quad AB\quad :\quad CD=?\\ Solution-\\ We\quad join\quad AB\quad \& \quad CD.\quad \\ AXB\quad \& \quad CYD\quad are\quad congruent\quad arcs\quad of\quad the\quad circle.\\ \therefore \quad Their\quad repective\quad chords\quad will\quad be\quad equal.\\ So\quad AB=CD.\\ \therefore AB:CD=1:1.\\ Ans-\quad Option\quad A.\\ $$
Question 10
1 / -0

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

A
$$ { 60 }^{ o\quad } $$

B
$$ { 50 }^{ o\quad } $$

C
$$ { 40 }^{ o\quad } $$

D
$$ { 30 }^{ o\quad } $$

Solution

Given:
$$ABCD$$ is a cyclic quadrilateral with $$AB$$ as the diameter of the circle.
Also, $$ \angle ADC={ 140 }^{ o }.$$
We have to find: $$ \angle BAC$$.

Then,
$$ \angle ADC+\angle ABC={ 180 }^{ o }$$ ....(since the sum of the opposite angles of a quadrilateral is $${ 180 }^{ o } $$).
$$ \therefore \angle ABC={ 180 }^{ o }-{ 140 }^{ o }={ 40 }^{ o }.$$
Also, $$ \angle ACB={ 90 }^{ o }$$ ....(since the angle subtended by a diameter at the circumference of the circle, is $${ 90 }^{ o }$$).
$$ \therefore$$ In $$\Delta ABC$$ we have,
$$ \angle ACB={ 90 }^{ o }$$ and $$\angle ABC={ 40 }^{ o }.$$
So, $$\angle CAB={ 180 }^{ o }-(\angle ACB+\angle ABC)$$
$$={ 180 }^{ o }-({ 90 }^{ o }+{ 40 }^{ o })$$
$$\therefore \angle BAC ={ 50 }^{ o }$$ .

Hence, option $$B$$ is correct.