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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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Increases
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Decreases
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Equal
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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.
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 $$AB=CD$$
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True
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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 $$10 cm$$ and $$8 cm$$ are at a distance from center $$8 cm$$ and $$5 cm$$ respectively.
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True
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False
Explanation
False, because larger chords are closer from the center.
Side $$AB$$ of a quadrilateral is a diameter of its circumcircle and $$\angle ADC = 140^{0}$$, then $$\angle BAC$$ equals:
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$$80^{0}$$
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$$50^{0}$$
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$$40^{0}$$
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$$30^{0}$$
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
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double
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triple
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half
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equal
In the given figure, $$r$$ represents
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Radius
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Diameter
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Tangent
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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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$$270^{0}$$
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$$360^{0}$$
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$$180^{0}$$
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$$90^{0}$$
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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$$70^{0}$$
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$$35^{0}$$
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$$110^{0}$$
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$$100^{0}$$
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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$$140^{0}$$
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$$120^{0}$$
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$$180^{0}$$
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$$275^{0}$$
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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$$110^{0}$$
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$$60^{0}$$
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$$40^{0}$$
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$$20^{0}$$
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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$$120^{0}$$
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$$60^{0}$$
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$$180^{0}$$
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$$90^{0}$$
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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$$80^{0}$$
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$$90^{0}$$
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$$110^{0}$$
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$$115^{0}$$
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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$$130^{0}$$
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$$230^{0}$$
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$$270^{0}$$
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$$360^{0}$$
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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