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Maths M.C.Q

Circles · Gujarat Board (GSEB) STD 9 (GSEB - English Medium). 244 questions.

Questions · Page 5 of 5

M.C.Q

MCQ 2021 Mark
PQRS is a cyclic quadrilateral such that PR is a diameter of the circle. If $\angle\text{QPR} = 67^\circ$ and $ \angle\text{SPR} = 72^\circ,$ then $\angle\text{QRS} =$
  • A
    41°
  • B
    23°
  • C
    67°
  • D
    18°
Answer
  1. 41°
    Solution:

    In a cyclic quadrilateral, Opposite angles are supplementary.
    $\Rightarrow\angle\text{P}+\angle\text{R}=180^\circ$
    Now, $\angle\text{P}=67^\circ+72^\circ=139^\circ$
    Thus, $\angle\text{R}=180^\circ-139^\circ=41^\circ$
    i.e. $\angle\text{R}=\angle\text{QRS}=41^\circ$
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MCQ 2061 Mark
In the given figure ABCD is a cyclic quadrilateral in which AB || DC and $\angle\text{BAD}=100^\circ.$ Then, $\angle\text{ABC}=?$
  • A
    100º
  • B
    40º
  • C
    50º
  • D
    80º
Answer
  1. 100º
    Solution:
    Since ABCD is a cyclic quadrilateral, we have:
    $\angle\text{BAD}+\angle\text{BCD}=180^\circ$ (Opposite angles of a cyclic quadrilateral)
    $\Rightarrow100^\circ+\angle\text{BCD}=180^\circ$
    $\Rightarrow\angle\text{BCD}=(180^\circ-100^\circ)=80^\circ$
    Now, AB || DC and CB is the transversal.
    $\therefore\angle\text{ABC}+\angle\text{BCD}=180^\circ$
    $\Rightarrow\angle\text{ABC}+80^\circ=180^\circ$
    $\Rightarrow\angle\text{ABC}=(180^\circ-80^\circ)=100^\circ$
    $\Rightarrow\angle\text{ABC}=100^\circ$
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MCQ 2071 Mark
The figure shows two circles which intersect at A and B. The centre of the smaller circle is O and it lies on the circumference of the larger circle. If $\angle\text{APB}=70^\circ,$ then the measure of $\angle\text{AGB}$ is:
  • A
    40º
  • B
    50º
  • C
    70º
  • D
    50º
Answer
  1. 40º
    Solution:

    Since, AB is a chord and makes $\angle\text{APB}=70^\circ$ at the circumference, so $\angle\text{AOB}=140^\circ$
    Now, as AOBC is a cyclic quadrilateral then, sum of opposite angles must be 180º.
    $\angle\text{AOB}+\angle\text{ACB}=180^\circ$
    $\Rightarrow140^\circ+\angle\text{ACB}=180^\circ$
    $\Rightarrow\angle\text{ACB}=40^\circ$
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MCQ 2081 Mark
In the give figure, ABCD is a cyclic quadrilateral in which BC = CD and $\angle\text{CBD}=35^\circ,$ Then, $\angle\text{BAD}=?$
  • A
    70º
  • B
    90º
  • C
    65º
  • D
    110º
Answer
  1. 70º
    Solution:
    BC = CD (given)
    $\Rightarrow\angle\text{BDC}=\angle\text{CBD}=35^\circ$
    In $\triangle\text{BCD},$ we have:
    $\angle\text{BCD}+\text{BDC}+\angle\text{CBD}=180^\circ$ (Angle sum property of a triangle)
    $\Rightarrow\angle\text{BCD}+35^\circ+35^\circ=180^\circ$
    $\Rightarrow\angle\text{BCD}=(180^\circ-70^\circ)=110^\circ\Rightarrow\angle\text{BCD}=110^\circ$
    In cyclic quadrilateral ABCD, we have:
    $\angle\text{BAD}+\angle\text{BCD}=180^\circ$
    $\Rightarrow\angle\text{BAD}+110^\circ=180^\circ$
    $\therefore\angle\text{BAD}=(180^\circ-110^\circ)=70^\circ$
    $\Rightarrow\angle\text{BAD}=70^\circ$
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MCQ 2101 Mark
In the given figure, O is the centre of a circle and $\angle\text{AOB}=130^\circ.$ Then, $\angle\text{ACB}=?$
  • A
    50°
  • B
    65°
  • C
    115°
  • D
    155°

Answer
  1. 115°
    Solution:
    Minor $\angle\text{AOB}=130^\circ$
    Major $\angle\text{AOB}=360^\circ-130^\circ$
    ⇒ Major $\angle\text{AOB}=230^\circ$
    Since $\angle\text{ACB}=\frac{1}{2}\text{major}\angle\text{AOB}$
    $\Rightarrow\ \angle\text{ACB}=\frac{1}{2}(230^\circ)$
    $\Rightarrow\ \angle\text{ACB}=115^\circ$
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MCQ 2111 Mark
Write the correct answer in the following:
If AB = 12cm, BC = 16cm and AB is perpendicular to BC, then the radius of the circle passing through the points A, B and C is:
  • A
    6cm.
  • B
    8cm.
  • C
    10cm.
  • D
    12cm.
Answer
  1. 10cm.
    Solution:
    AB is perpendicular to BC, therefore ABC is a right triangle.
    In right $\triangle\text{ABC},$ we have
    $\text{AC}=\sqrt{\text{AB}^2+\text{BC}^2}$
    $=\sqrt{(12)^2+(16)^2}$
    $\sqrt{144+256}$
    $\text{AC}=20\text{cm}$
    $\therefore\text{Radius}=\frac{1}{2}\times\text{diameter}=\frac{1}{2}\times20\text{cm}=10\text{cm}$
    Hence, (c) is the correct answer.
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MCQ 2131 Mark
In the given figure, O is the centre of a circle and chords AC and BD intersect at E. If $\angle\text{AEB}=110^\circ$ and $\angle\text{CBE}=30^\circ,$ then $\angle\text{ADB}=?$
  • A
    70°
  • B
    60°
  • C
    80°
  • D
    90°

Answer
  1. 80°
    Solution:
    $\angle\text{AED}=\angle\text{ECB}+\angle\text{EBC}$
    $\Rightarrow\ 110^\circ=\angle\text{ECB}+30^\circ$
    $\Rightarrow\ \angle\text{ECB}=80^\circ$
    Since angles in the same segment are equal,
    $\angle\text{ADB}=\angle\text{ECB}=80^\circ$
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MCQ 2141 Mark
The figure shows two circles which intersect at A and B. The centre of the smaller circle is O and it lies on the circumference of the larger circle. If $\angle\text{APB}=70^\circ,$ then the measure of $\angle\text{ACB}$ is:
  • A
    40º
  • B
    50º
  • C
    60º
  • D
    70º
Answer
  1. 40º
    Solution:

    Since, AB is a chord and makes $\angle\text{APB}=70^\circ$ at the circumference, so $\angle\text{AOB}=140^\circ$
    Now, as AOBC is a cyclic quadrilateral then, sum of opposite angles must be 180º
    $\angle\text{AOB}+\angle\text{ACB}=180^\circ$
    $\Rightarrow140^\circ+\angle\text{ACB}=180^\circ$
    $\Rightarrow\angle\text{ACB}=40^\circ$
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MCQ 2161 Mark
In the given figure, O is the centre of a circle and $\angle\text{ACB}=30^\circ.$ Then, $\angle\text{AOB}=?$
  • A
    30°
  • B
    15°
  • C
    60°
  • D
    90°

Answer
  1. 60°
    Solution:
    We know that, the angle subtended by an arc of a circle at the centre is double the angle subtended by it at any point on the remaining part of the circle.
    So,
    $\angle\text{AOB}=2\angle\text{ACB}$
    $=2(30^\circ)$
    $=60^\circ$
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MCQ 2181 Mark
In the given figure, AB is a chord of a circle with centre O and BOC is a diameter. If $\text{OD}\perp\text{AB}$ such that OD = 6cm, then AC = ?
  • A
    9cm
  • B
    12cm
  • C
    15cm
  • D
    7.5cm

Answer
  1. 12cm
    Solution:
    In $\triangle\text{BOD}$ and $\triangle\text{CAB},$
    Since BOC is the diameter, $\angle\text{CAB}=90^\circ.$
    Also, $\angle\text{ODB}=90^\circ.$
    So, $\angle\text{DBO}=\angle\text{ABC}$ [Common angles]
    $\Rightarrow\ \triangle\text{BOD}\sim\triangle\text{BCA}$ [AA congruence criterion]
    $\Rightarrow\ \frac{\text{OD}}{\text{CA}}=\frac{\text{BO}}{\text{BA}}$
    $\Rightarrow\ \frac{\text{OD}}{\text{CA}}=\frac{1}{2}$ [Since radius = 2 diameter]
    $\Rightarrow\ \frac{6}{\text{CA}}=\frac{1}{2}$
    ⇒ CA = 12cm that is, AC = 12cm.
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MCQ 2211 Mark
Two point on a circle makes the:
  • A
    Secant
  • B
    Chord
  • C
    Diameters
  • D
    Diameter
Answer
  1. Chord
    Solution:
    A chord is the line joining any two points on the circle.
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MCQ 2221 Mark
The line which meet a circle in two points is called a:
  • A
    Radius of circle.
  • B
    Secant of circle.
  • C
    Chord of circle.
  • D
    Diameter of circle.
Answer
  1. Secant of circle.
    Solution:
    A line that meets a circle at any two points is called secant of that circle and secant intercepted between these two points is chord of that circle.
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MCQ 2261 Mark
If A, B, C are three points on a circle with centre O such that $\angle\text{AOB}=90^\circ$ and $\angle\text{BOC}=120^\circ,$ then $\angle\text{ABC}=$
  • A
    60º
  • B
    90º
  • C
    75º
  • D
    135º
Answer
  1. 75º
    Solution:
    To solve this problem we need to know that the angle subtended by an arc at the centre of the circle is double the angle subtended by the arc in the remaining part of the circle.

    Here we are given that A, B, C, are three points on a circle with centre O such that
    $\angle\text{AOB}=90^\circ$ and $\angle\text{BOC}=120^\circ.$
    From
    $\angle\text{AOC}=360^\circ-\angle\text{AOB}-\angle\text{BOC}$
    $\Rightarrow\angle\text{AOC}=360^\circ-90^\circ-120^\circ$
    $\Rightarrow\angle\text{AOC}=360^\circ-210^\circ$
    $\Rightarrow\angle\text{AOC}=150^\circ$
    Now, as seen earlier, the angle made by the arc AC with the centre of the circle will be twice the angle it makes in any point in the remaining part of the circle.
    Since the point C lies on the remaining part of the circle, the angle the arc AC makes with this point has to be half of the angle 'AC' makes with the centre. Therefore we have,
    $\angle\text{ABC}=\frac{\angle\text{AOC}}{2}=\frac{150^\circ}{2}=75^\circ$
    $\Rightarrow\angle\text{ABC}=75^\circ$
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MCQ 2271 Mark
In the given figure, ABCD is a cyclic quadrilateral, $\angle\text{CBQ}=48^\circ$ and a = 2b. Then, b is equal to:
  • A
    48º
  • B
    28º
  • C
    38º
  • D
    18º
Answer
  1. 28º
    Solution:

    Here, $\angle\text{ABC}$ is supplementary to $\angle\text{CBQ}$
    So, $\angle\text{ABC}=180-48=132^\circ.$
    Since, ABCD is cyclic quadrilateral, $\angle\text{B}+\angle\text{D}=180^\circ$ {opposite angles are supplementery}
    So, $\angle\text{D}=180-132=48^\circ$
    Now, in triangle PDC, $\angle\text{P}+\angle\text{D}+\angle\text{C}=180^\circ$
    = a + 48º + (48º + b) = 180º {since, $\angle\text{C}$ is external angle to B and b, and sum of two opposite interior angles is equal to external angle}
    = 3b + 96 = 180º
    3b = 180 - 96 = 84
    b = 28º
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MCQ 2281 Mark
If the length of a chord of a circle is 16cm and is at a distance of 15cm from the centre of the circle, then the radius of the circle is:
  • A
    15cm
  • B
    34cm
  • C
    17cm
  • D
    16cm
Answer
  1. 17cm
    Solution:
    We will represent the given data in the figure

    In the diagram, AB is the given chord of 16cm length and OM is the perpendicular distance from the centre to AB.
    We know that perpendicular from the centre to any chord divides it into two equal parts.
    So, $\text{AM}=\text{MB}=\frac{16}{2}=8\text{cm}.$
    Now consider right triangle $\triangle\text{OMA}$ and by using Pythagoras theorem $\angle\text{OMA}=90^\circ$
    AO2 = AM2 + OM2
    AO2 = 82 + 152
    AO2 = 64 + 225 = 289
    $\text{AO}=\sqrt{289}=17\text{cm}$
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MCQ 2291 Mark
In the given figure, BOC is a diameter of a circle with centre O. If AB and CD are two chords such that AB || CD. If AB = 10cm, then CD = ?
  • A
    5cm
  • B
    12.5cm
  • C
    15cm
  • D
    10cm

Answer
  1. 10cm
    Solution:
    In $\triangle\text{BEO}$ and $\triangle\text{CFO},$
    OB = OC [Radii of the same circle]
    $\angle\text{OBE}=\angle\text{OCF}$ [Alternate angles since AB || CD]
    $\angle\text{BOE}=\angle\text{COF}$ [Vertically angles]
    $\Rightarrow\ \triangle\text{BEO}\cong\triangle\text{CFO}$ [ASA congruence criterion]
    ⇒ OE = OF [C.P.C.T.]
    Since chord are equidistant from the centre are equal, AB = CD = 10 cm.
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MCQ 2301 Mark
In the given figure, $\triangle\text{ABC}$ and $\triangle\text{DBC}$ are inscribed in a circle such that $\angle\text{BAC}=60^\circ$ and $\angle\text{DBC}=50^\circ.$ Then, $\angle\text{BCD}=?$
  • A
    60º
  • B
    50º
  • C
    80º
  • D
    70º
Answer
  1. 70º
    Solution:
    $\angle\text{BDC}=\angle\text{BAC}=60^\circ$ (Angles in the same segment of a circle)
    In $\triangle\text{BDC},$ we have
    $\angle\text{DBC}+\angle\text{BDC}+\angle\text{BCD}=180^\circ$ (Angle sum property of a triangle)
    $\therefore50^\circ+60^\circ+\angle\text{BCD}=180^\circ$
    $\Rightarrow\angle\text{BCD}=180^\circ-(50^\circ+60^\circ)=(180^\circ-110^\circ)=70^\circ$
    $\Rightarrow\angle\text{BCD}=70^\circ$
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MCQ 2311 Mark
The chord of a circle is equal to its radius. The angle subtended by this chord at the minor arc of the circle is:
  • A
    60°
  • B
    75°
  • C
    120°
  • D
    150°
Answer
  1. 150°
    Solution:

    $\angle\text{AOB}=60^\circ$ $($Since $ \triangle\text{AOB}$ is equilateral triangle$)$
    Now, $\angle\text{ADB}=30^\circ$
    (Since chord AB makes 60 at centre, same chord will make half of the angle at circumference of angle made at centre)
    Now $\angle\text{ACB}$ is angle made by chord at minor arc of circle.
    ABCD is cyclic Quadrilateral.
    $\Rightarrow\angle\text{C}+\angle\text{D}=180^\circ$
    $\Rightarrow\angle\text{ACB}+\angle\text{ADB}=180^\circ$
    $\Rightarrow\angle\text{ACB}=180^\circ-30^\circ=150^\circ$
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MCQ 2321 Mark
Write the correct answer in the following: In Fig. $\angle\text{AOB}=90^\circ$ and $\angle\text{ABC}=30^\circ,$ then $\angle\text{CAO}$ is equal to:
  • A
    30°.
  • B
    45°.
  • C
    90°.
  • D
    60°.
Answer
  1. 60°.
    Solution:
    In $\triangle\text{OAB},$ we have
    $\text{OA}=\text{OB}$
    [Radii of the same circle]
    $\therefore\angle\text{OAB}=\angle\text{OBA}$
    In triangle OAB, we have
    $\angle\text{OAB}+\angle\text{OBA}+\angle\text{AOB}=180^\circ$
    $\therefore2\angle\text{OAB}=(180^\circ-\angle\text{AOB})$
    $=(180^\circ-90^\circ)=90^\circ$ [$\because$ sum of angles of $\triangle$ is 180°]
    $\Rightarrow\angle\text{OAB}=\frac{1}{2}\times90^\circ=45^\circ$
    Also, $\angle\text{ACB}=\frac{1}{2}\angle\text{AOB}=\frac{1}{2}\times90^\circ=45^\circ$
    Now, in $\triangle\text{CAB},$ we have
    $\angle\text{CAB}=180^\circ-(\angle\text{ABC}+\angle\text{ACB})$
    $=180^\circ-(30^\circ+45^\circ)=105^\circ$
    Now, $\angle\text{CAO}=\angle\text{CAB}-\angle\text{OAB}$
    $\Rightarrow\angle\text{CAO}=105^\circ-45^\circ=60^\circ$
    Hence, (d) is the correct answer.
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MCQ 2331 Mark
AOB is the diameter of the circle. If $\angle\text{AOE}=150^\circ,$ then the measure of $\angle\text{CBE}$ is:
  • A
    125º
  • B
    120º
  • C
    105º
  • D
    115º
Answer
  1. 105º
    Solution:

    Here, AOB is diameter,
    So, $\angle\text{BOE}=180-150=30^\circ$ {Angles lie in straight line}
    Now, OE & OB are radius so, OE = OB .i.e $\angle\text{OEB}=\angle\text{OBE}$
    In $\triangle\text{BOE},\angle\text{BOE}+\angle\text{OBE}+\angle\text{BEO}=180^\circ$
    $=30+2\angle\text{OBE}=180^\circ$
    $=2\angle\text{OBE}=180-30=150^\circ$
    $=\angle\text{OBE}=75^\circ$
    Now, $\angle\text{OBE}$ and $\angle\text{CBE}$ lie on staright line
    so, $\angle\text{OBE}+\angle\text{CBE}=180^\circ$
    $\angle\text{CBE}=180-75=105^\circ$
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MCQ 2341 Mark
  • A
    $\frac{1}{3}$ of the circle.
  • B
    $\frac{1}{2}$ of the circle.
  • C
    $\frac{1}{4}$ of the circle.
  • D
    $\frac{1}{5}$ of the circle.
Answer
  1. $\frac{1}{3}$ of the circle.
    Solution:
    Complete the cyclic quadrilateral PQRS, with S being a point on a point on the major arc. Then $\angle\text{S}=60^\circ$ (Opposite angles of a cyclic quadrilateral)
    Then Major $\angle\text{POR}=120^\circ$
    Thus fraction the minor arc $=\frac{120^\circ}{130^\circ}=\frac{1}{3}$
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MCQ 2361 Mark
ABCD is a parallelogram. A circle passes through A and D and cuts AB at E and DC at F. If $\angle\text{BEF}=80^\circ,$ then $\angle\text{ABC}$ is equal to:
  • A
    75º
  • B
    80º
  • C
    100º
  • D
    120º
Answer
  1. 80º
    Solution:

    $\angle\text{AEF}+80^\circ=180^\circ$ (Linear Pair)
    $\angle\text{AEF}=100^\circ$
    $\angle\text{ADF}+\angle\text{AEF}=180^\circ$ (Opposite angles of a cyclic quadrilateral)
    $\angle\text{ADF}=180^\circ-100^\circ=80^\circ$
    $\angle\text{ADF}=\angle\text{ABC}=80^\circ$ (Opposite angles of a parallelogram)
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MCQ 2381 Mark
Two circle are congruent if they have equal.
  • A
    Radius
  • B
    Diameter
  • C
    Secant
  • D
    Chord
Answer
  1. Radius
    Solution:
    Equal radius would generate two same circles that are exact copy of each other, hence making them congruent.
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MCQ 2391 Mark
Angles in the same segment of a circle area are:
  • A
    Equal
  • B
    Complementary
  • C
    Supplementary
  • D
    None of these
Answer
  1. Equal
    Solution:
    Angles in a the same segment of a circle area are equal.
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MCQ 2401 Mark
ABCD is a parallelogram. A circle passes through A and D and cuts AB at E and DC at F. If $\angle\text{BEF}=80^\circ,$ then $\angle\text{ABC}$ is equal to:
  • A
    80º
  • B
    75º
  • C
    120º
  • D
    100º
Answer
  1. 80º
    Solution:

    $\angle\text{AEF}+80^\circ=180^\circ$
    $\angle\text{AEF}=100^\circ$
    $\angle\text{ADF}+\angle\text{AEF}=180^\circ$ (Opposite angles of a cyclic quadrilateral)
    $\angle\text{ADF}=180^\circ-100^\circ=80^\circ$
    $\angle\text{ADF}=\angle\text{ABC}=80^\circ$ (opposite angles of a parallelogram)
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MCQ 2421 Mark
In the given figure, O is the centre of a circle in which $\angle\text{OAB}=20^\circ$ and $\angle\text{OCB}=50^\circ.$ Then, $\angle\text{AOC}=?$
  • A
    20º
  • B
    60º
  • C
    70º
  • D
    50º
Answer
  1. 60º
    Solution:
    $\text{OA}=\text{OB}\Rightarrow\angle\text{OBA}=\angle\text{OAB}=20^\circ.$
    In $\triangle\text{OAB},$
    $\angle\text{OAB}+\angle\text{OBA}+\angle\text{AOB}=180^\circ$
    $\Rightarrow20^\circ+20^\circ+\angle\text{AOB}=180^\circ$
    $\Rightarrow\angle\text{AOB}=140^\circ.$
    $\text{OB}=\text{OC}\Rightarrow\angle\text{OBC}=\angle\text{OCB}=50^\circ.$
    In $\triangle\text{OCB},$
    $\angle\text{OCB}+\angle\text{OBC}+\angle\text{COB}=180^\circ$
    $\Rightarrow50^\circ+50^\circ+\angle\text{COB}=180^\circ$
    $\Rightarrow\angle\text{COB}=80^\circ.$
    $\angle\text{AOB}=140^\circ\Rightarrow\angle\text{AOC}+\angle\text{COB}=140^\circ$
    $\Rightarrow\angle\text{AOC}+80^\circ=140^\circ$
    $\Rightarrow\angle\text{AOC}=140^\circ-80^\circ$
    $\Rightarrow\angle\text{AOC}=60^\circ.$
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MCQ 2431 Mark
In the given figure, $\angle\text{AOB}=90^\circ$ and $\angle\text{ABC}=30^\circ.$ Then, $\angle\text{CAO}=?$
  • A
    30°
  • B
    45°
  • C
    60°
  • D
    90°

Answer
  1. 60°
    Solution:
    $\angle\text{AOB}=2\angle\text{ACB}$
    $\Rightarrow\ \angle\text{ACB}=\frac{1}{2}\angle\text{AOB}$
    $\Rightarrow\ \angle\text{ACB}=\frac{1}{2}(90^\circ)$
    $\Rightarrow\ \angle\text{ACB}=45^\circ$
    $\angle\text{COA}=2\angle\text{CBA}=2(30^\circ)=60^\circ$
    Since AOD is a straight line,
    $\therefore\ \angle\text{COD}+\angle\text{AOC}=180^\circ$
    $\therefore\ \angle\text{COD}+60^\circ=180^\circ$
    $\therefore\ \angle\text{COD}=120^\circ$
    $\Rightarrow\ \angle\text{CAO}=\frac{1}{2}\angle\text{COD}=\frac{1}{2}\times120^\circ=60^\circ$
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