MCQ Questions for Class 9 Maths Circles Quiz 9

If the length of the chord increase its perpendicular distance from the centre

0%
Increases
0%
Decreases
0%
Equal
0%
Constant

Explanation

$AB$ is a chord and its distance from centre

$O$ is

$OP_1$

$CD$ is a chord and its distance from centre

$O$ is

$OP_2$

$CD>AB\Rightarrow OP_2<OP_1$
As the length of the chord increases then the distance of the chord from the center decreases.
1838382_1867443_ans_4034aafb80d849f3a4600c3b90cb218a.png

1838382_1867443_ans_4034aafb80d849f3a4600c3b90cb218a.png

In the following, write true/false and if possible give reason for your answer.
If two chords

$AB$ and

$CD$ of a circle are at same distance

$4cm$ from center then

$AB=CD$

0%
True
0%
False

In the following, write true/false and if possible give reason for your answer.
Two chords of length

$10 cm$ and

$8 cm$ are at a distance from center

$8 cm$ and

$5 cm$ respectively.

0%
True
0%
False

Side

$AB$ of a quadrilateral is a diameter of its circumcircle and

$\angle ADC = 140^{0}$ , then

$\angle BAC$ equals:

0%
$80^{0}$
0%
$50^{0}$
0%
$40^{0}$
0%
$30^{0}$

Explanation

Here,

$ABCD$ is a cyclic quadrilateral.

$\therefore \angle D + \angle B = 180^{0}$ ....(Opposite angles of cyclic quadrilateral are supplementary)

$\implies$

$\angle B = 180° – \angle D$

$\implies$

$\angle B = 180° – 140°$

$\implies$

$\angle B = 40°$ .

$\angle ACB$ 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

0%
double
0%
triple
0%
half
0%
equal

In the given figure,

$r$ represents

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:

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.

1958774_1877809_ans_78feb039c7a24beda4ee9e03c88d3e67.png

In the figure, if

$A, B, C$ and

$D$ are vertices of cyclic quadrilateral, then

$\angle x$ will be:

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:

0%
$140^{0}$
0%
$120^{0}$
0%
$180^{0}$
0%
$275^{0}$

Explanation

$\Delta ABC, \angle A = 60^{0}$ ...[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 :

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 :

0%
$120^{0}$
0%
$60^{0}$
0%
$180^{0}$
0%
$90^{0}$

Explanation

We know that opposite 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 :

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

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.
1862675_1877843_ans_c6ce4ae523a44869aac897c05333cc55.png

The center of the circle lies

0%
in the interior of the circle
0%
in the exterior of the circle
0%
on the circle
0%
None of these

CBSE Questions for Class 9 Maths Circles Quiz 9 - MCQExams.com

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Practice Class 9 Maths Quiz Questions and Answers

CBSE Questions for Class 9 Maths Circles Quiz 9 - MCQExams.com

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Practice Class 9 Maths Quiz Questions and Answers

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