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CBSE Questions for Class 9 Maths Circles Quiz 9 - MCQExams.com
CBSE
Class 9 Maths
Circles
Quiz 9
If the length of the chord increase its perpendicular distance from the centre
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0%
Increases
0%
Decreases
0%
Equal
0%
Constant
Explanation
A
B
is a chord and its distance from centre
O
is
O
P
1
C
D
is a chord and its distance from centre
O
is
O
P
2
C
D
>
A
B
⇒
O
P
2
<
O
P
1
As the length of the chord increases then the distance of the chord from the center decreases.
In the following, write true/false and if possible give reason for your answer.
If two chords
A
B
and
C
D
of a circle are at same distance
4
c
m
from center then
A
B
=
C
D
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0%
True
0%
False
Explanation
True, because equal chords are equidistant from the center.
In the following, write true/false and if possible give reason for your answer.
Two chords of length
10
c
m
and
8
c
m
are at a distance from center
8
c
m
and
5
c
m
respectively.
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0%
True
0%
False
Explanation
False, because larger chords are closer from the center.
Side
A
B
of a quadrilateral is a diameter of its circumcircle and
∠
A
D
C
=
140
0
, then
∠
B
A
C
equals:
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0%
80
0
0%
50
0
0%
40
0
0%
30
0
Explanation
Here,
A
B
C
D
is a cyclic quadrilateral.
∴
∠
D
+
∠
B
=
180
0
....(
Opposite angles of cyclic quadrilateral are supplementary)
⟹
\angle B = 180° – \angle D
\implies
\angle B = 180° – 140°
\implies
\angle B = 40°
.
\angle ACB
is an angle in a semicircle.
\therefore \angle ACB = 90°
.
Now in
\Delta ABC
,
\angle ABC + \angle ACB + \angle BAC = 180^{0}
....(Angle sum property of a triangle)
\implies
40° + 90° + \angle BAC = 180°
\implies
\angle BAC= 180° – (40° + 90°)
\implies
\angle BAC= 180° – 130°
\implies
\angle BAC = 50°
.
Thus, option
B
is correct.
Chords equidistant to each other from center of circle are
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0%
double
0%
triple
0%
half
0%
equal
In the given figure,
r
represents
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0%
Radius
0%
Diameter
0%
Tangent
0%
Chord
Explanation
{\textbf{Step -1: Find r.}}
{\text{The distance from the center point to any point on the circle is called the radius of a circle}}
{\text{Here, r is the radius in the figure.}}
{\textbf{Hence, the correct answer is option A.}}
Sum of opposite angles in a cyclic quadrilateral is:
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0%
270^{0}
0%
360^{0}
0%
180^{0}
0%
90^{0}
Explanation
Given:
ABCD
is a cyclic quadrilateral of a circle with centre
O
.
To prove : Sum of opposite angles in a cyclic quadrilateral is
180^0
.
Construction : Two diagonals
AC
and
BD
are drawn.
Proof:
\angle ADB=\angle ACB
[angles in the same segment of a circle]
=\angle BAC=\angle BDC
Again,
\angle ADC=\angle ADB+\angle BDC
=\angle ACB=\angle BAC
\therefore \angle ADC+\angle ABC=
\angle ACB+\angle BAC+\angle ABC
\therefore \angle ADC+\angle ABC=2
right angle [Sum of angles in a triangle is
180^0
]
Similarly,
\angle BAD+\angle BCD=2
right angle, i.e.
180^0
.
Hence, option
C
is correct.
In the figure, if
A, B, C
and
D
are vertices of cyclic quadrilateral, then
\angle x
will be:
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0%
70^{0}
0%
35^{0}
0%
110^{0}
0%
100^{0}
Explanation
We know,
\angle CBA + 70^{0} = 180^{0}
(linear pair)
Then,
\angle CBA = 180^{0} – 70^{0} = 110^{0}
.
\because ABCD
is cyclic quadrilateral.
We know, the sum of opposite angles of a cyclic quadrilateral is
180^{o}
.
Thus,
\angle CBA + \angle CDA = 180^{0}
\implies
110^{0} + \angle CDA = 180^{0}
\implies
\angle CDA = 180^{0} – 110^{0}
= 70^{0}
.
But
\angle x + \angle CDA = 180^{0}
(linear pair)
\implies
\angle x + 70^{0} = 180^{0}
\implies
\angle x = 180^{0} – 70^{0}
= 110^{0}
.
Thus, option
C
is correct.
If in the given figure,
\Delta ABC
is an equilateral triangle, then
\angle BCD
will be:
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0%
140^{0}
0%
120^{0}
0%
180^{0}
0%
275^{0}
Explanation
In
\Delta ABC, \angle A = 60^{0}
...[Equilateral triangle].
In cyclic quadrilateral
ABCD
,
\angle A + \angle D = 180^{0}
....[
Opposite angles of cyclic quadrilateral are supplementary]
\implies
\angle D = 180^{0} – \angle A
= 180^{0} – 60^{0}
\implies
\angle BDC = 120^{0}
.
Thus, option
B
is correct.
In the given figure,
ABCD
is a cyclic quadrilateral whose diagonals intersect each other at
O
. If
\angle ACB = 50^{0}
and
\angle ABC = 110^{0}
, then
\angle BDC
will be :
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0%
110^{0}
0%
60^{0}
0%
40^{0}
0%
20^{0}
Explanation
\because
Angles formed on same base
AB
,
\therefore \angle ADB = \angle ACB
\implies
\angle ADB = 50^{0}
...[because
\angle ACB = 50^{0}
].
We know that sum of opposite angles of cyclic quadrilateral is
180^{0}
\therefore \angle ADC + \angle ABC = 180^{0}
\implies
\angle ADC + 110^{0} = 180^{0}
....[because
\angle ABC = 110^{0}]
\implies
\angle ADC = 180^{0} – 110^{0} = 70^{0}
.
Now
\angle ADC = \angle ADB + \angle BDC
\implies
70^{0} = 50^{0} + \angle BDC
\implies
70^{0} – 50^{0} = \angle BDC
\implies
20^{0} = \angle BDC
.
Thus,
\angle BDC = 20^{0}
.
Therefore, option
D
is correct.
In fig,
ABCD
is a cyclic quadrilateral. if
\angle D = 120^{0}
, then
\angle CBE
will be :
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0%
120^{0}
0%
60^{0}
0%
180^{0}
0%
90^{0}
Explanation
We know that o
pposite angles of cyclic quadrilateral are supplementary.
Then,
\angle ADC+\angle ABC=180^o
\implies
120^o+\angle ABC=180^o
\implies
\angle ABC=180^o-120^o=60^o
.
Also,
\angle ABC+ \angle CBE = 180^o
...(Linear pair)
\implies
\angle 60^o+ \angle CBE = 180^o
\implies
\angle CBE=180^o-60^o=120^o
Thus, option
A
is correct.
In figure,
D
is cyclic quadrilateral in which
AC
and
BD
are diagonals. If
\angle DBC = 55^{0}
and
\angle BAC = 45^{0}
, then
\angle BCD
will be :
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0%
80^{0}
0%
90^{0}
0%
110^{0}
0%
115^{0}
Explanation
Here,
\angle CAD = \angle DBC = 55^{0}
(angles formed by same arc in the same segment are equal)
We know,
\angle DAB = \angle CAD + \angle BAC
=55^{0}+ 45^{0} = 100^{0}
.
But
\angle DAB + \angle BCD = 180^{0}
(opposite angles of cyclic quadrilateral)
\implies
\angle BCD = 180^{0} – 100^{0} = 80^{0}
.
Thus, option
A
is correct.
In figure,
O
is center of circle and
\angle PQR = 115^{0}, \angle POR
will be
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0%
130^{0}
0%
230^{0}
0%
270^{0}
0%
360^{0}
Explanation
Major angle
\angle POR = 2\angle PQR
x = 2 \times115^o
x = 230^{0}
Sum of angles formed at center of circle is
360^{0}
\therefore \angle POR + x = 360^{0}
\angle POR + 230^{0} = 360^{0}
\angle POR = 360^{0} – 230^{0}
\angle POR = 130^{0}
Thus, (A) is correct.
The center of the circle lies
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in the interior of the circle
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in the exterior of the circle
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on the circle
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None of these
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