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Maths M.C.Q

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

Questions · Page 3 of 5

M.C.Q

MCQ 1011 Mark
AD is diameter of a circle, O being the centre and AB is a chord. Let the centre of AB be denoted by M, then find OM.
  • A
    7cm
  • B
    8cm
  • C
    6cm
  • D
    5cm
Answer
  1. 7cm
    Solution:
    Join OM. OM will be perpendicular to AB. Since the line joining the midpoint of a chord to the centre is always perpendicular to the chord.
    AB = 48cm, so, $\text{AM}=\frac{48}{2}=24\text{cm}$ (O is the midpoint of AD)
    Now, applying pythagoras theorem, we get:-
    OA2 = AM2 + OM2
    252 = 242 + OM2
    OM2 = 252 - 242
    OM2 = 625 - 576 = 49
    OM = 7cm
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MCQ 1031 Mark
In the given figure, AOB is a diameter of a circle and CD || AB. If $\angle\text{BAD}=30^\circ,$ then $\angle\text{CAD}=?$
  • A
    45º
  • B
    30º
  • C
    50º
  • D
    60º
Answer
  1. 30º
    Solution:
    $\angle\text{ADC}=\angle\text{BAD}=30^\circ$ (Alternate angles)
    $\angle\text{ADB}=90^\circ$ (Angle in semicircle)
    $\therefore\angle\text{CDB}=(90^\circ+30^\circ)=120^\circ$
    But ABCD being a cyclic quadrilateral, we have:
    $\angle\text{BAC}+\angle\text{CDB}=180^\circ$
    $\Rightarrow\angle\text{BAD}+\angle\text{CAD}+\angle\text{CDB}=180^\circ$
    $\Rightarrow30^\circ+\Rightarrow\angle\text{CAD}+120^\circ=180^\circ$
    $\Rightarrow\angle\text{CAD}=(180^\circ-150^\circ)=30^\circ$
    $\Rightarrow\angle\text{CAD}=30^\circ$
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MCQ 1041 Mark
If the angles subtended by two chords of a circle at the centre are equal then the chords are:
  • A
    Angle equal.
  • B
    Equal.
  • C
    None of these.
  • D
    Not equal.
Answer
  1. Equal.
    Solution:
    If the angles subtended by two chords of a circle at the centre are equal then the chords are equal. It's a theorem.
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MCQ 1051 Mark
  • A
    15cm
  • B
    12.5cm
  • C
    10cm
  • D
    5cm
Answer
  1. 10cm
    Solution:
    Draw $\text{OE}\perp\text{AB}$ and $\text{OF}\perp\text{CD}.$
    In $\triangle\text{OEB}$ and $\triangle\text{OFC},$ we have:
    OB = OC (Radius of a circle)
    $\angle\text{BOE}=\angle\text{COF}$ (Vertically opposite angles)
    $\angle\text{OEB}=\angle\text{OFC}$ (90º each)
    $\therefore\triangle\text{OEB}\cong\triangle\text{OFC}$ (By ASA congruency rule)
    $\therefore\text{OE}=\text{OF}$ {by cpct}
    Chords equidistance from the centre is equal.
    $\therefore$ CD = AB = 10cm
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MCQ 1071 Mark
In a circle with centre O, AB and CD are two diameters perpendicular to each other. The length of chord AC, is:
  • A
    $\frac{1}{\sqrt{2}\text{ AB}}$
  • B
    $\sqrt{2}$
  • C
    $\frac{1}{2\text{AB}}$
  • D
    $\text{2AB}$
Answer
  1. $\frac{1}{\sqrt{2}\text{ AB}}$
    Solution:
    We are given a circle with centre at O and two perpendicular diameters AB and CD.
    We need to find the length of AC
    We have the following corresponding figure:

    Since, AB = CD (Diameter of the same circle)
    Also, $\angle\text{AOC}=90^\circ$
    And, $\text{AO}=\frac{\text{AB}}{2}$
    Here, AO = OC (radius)
    In $\triangle\text{AOC}$ is right angled triangle,
    $\text{AC}^2=\text{AO}^2+\text{OC}^2=\text{AO}^2+\text{AO}^2$
    $=\Big(\frac{\text{AB}}{2}\Big)^2+\Big(\frac{\text{AB}}{2}\Big)^2$
    $\text{AC}2=\frac{\text{AB}^2}{2}$
    $\text{AC}=\frac{\text{AB}}{\sqrt{2}}$
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MCQ 1091 Mark
In the given figure, AD is the diameter of the circle and AE = DE. If, $\angle\text{ABC}=115^\circ,$ then the measure of $\angle\text{CAE}$ is:
  • A
    70º
  • B
    60º
  • C
    80º
  • D
    90º
Answer
  1. 70º
    Solution:

    Since, AE = DE,
    therefore, $\angle\text{DAE}=\angle\text{ADE}=45^\circ$ (In $\triangle\text{AED},\angle\text{E}$ And other two angles are equal.)
    Now, BADC is a cyclic quadrilateral,
    $\Rightarrow\angle\text{B}+\angle\text{D}=180^\circ$
    $\Rightarrow115^\circ+\angle\text{D}=180^\circ$
    $\Rightarrow\angle\text{CDA}=\angle\text{D}=65^\circ$
    Also, $\angle\text{ACD}=90^\circ$ (Angle in a semicircle)
    So, we have:-
    In $\triangle\text{ACD},$
    $\Rightarrow\angle\text{CAD}+\angle\text{ACD}+\angle\text{CDA}=180^\circ$
    $\Rightarrow\angle\text{CAD}+90^\circ+65^\circ=180^\circ$
    $\Rightarrow\angle\text{CAD}=25^\circ$
    Finally,
    $\angle\text{CAE}=\angle\text{CAD}+\angle\text{DAE}=25^\circ+45^\circ=70^\circ$
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MCQ 1101 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
    80°
  • B
    100°
  • C
    50°
  • D
    40°

Answer
  1. 100°
    Solution:
    Since AB || DC,
    $\angle\text{ADC}+\angle\text{BAD}=180^\circ$
    $\Rightarrow\ 100^\circ+\angle\text{BAD}=180^\circ$
    $\Rightarrow\ \angle\text{BAD}=80^\circ$
    We know that the opposite angles of a quadrilateral are supplementary.
    $\angle\text{BAD}+\angle\text{ABC}=180^\circ$
    $\Rightarrow\ 80^\circ+\angle\text{ABC}=180^\circ$
    $\Rightarrow\ \angle\text{ABC}=100^\circ$
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MCQ 1141 Mark
In the given figure, O is the centre of a circle and $\angle\text{AOC}=120^\circ.$ Then, $\angle\text{BDC}=?$
  • A
    45º
  • B
    30º
  • C
    15º
  • D
    60º
Answer
  1. 30º
    Solution:
    $\angle\text{COB}=180^\circ-120^\circ=60^\circ$ (Linear pair)
    Now, arc BC subtends $\angle\text{COB}$ at the centre and $\angle\text{BDC}$ at the point D of the remaining part of the circle.
    $\therefore\angle\text{COB}=2\angle\text{BDC}$
    $\Rightarrow\angle\text{BDC}=\frac{1}{2}\angle\text{COB}=(\frac{1}{2}\times60^\circ)=30^\circ$
    $\Rightarrow\angle\text{BDC}=30^\circ$
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MCQ 1151 Mark
Circle having same centre are said to be:
  • A
    Secant
  • B
    Chord
  • C
    Concentric
  • D
    Circle
Answer
  1. Concentric
    Solution:
    Concentric circles are those circle that are drawn with same point as centre but different radii.
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MCQ 1181 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
    90º
  • B
    70º
  • C
    60º
  • D
    80º
Answer
  1. 80º
    Solution:
    We have:
    $\angle\text{AEB}+\angle\text{CEB}=180^\circ$ (Linear pair angles)
    $\Rightarrow110^\circ+\angle\text{CEB}=180^\circ$
    $\Rightarrow\angle\text{CEB}=(180^\circ-110^\circ)=70^\circ$
    $\Rightarrow\angle\text{CEB}=70^\circ$
    In $\triangle\text{CEB},$ we have:
    $\angle\text{CEB}+\angle\text{EBC}+\angle\text{ECB}=180^\circ$ (Angle sum property of a triangle)
    $\Rightarrow70^\circ+30^\circ+\angle\text{ECB}=180^\circ$
    $\Rightarrow\angle\text{ECB}=(180^\circ-100^\circ)=80^\circ$
    The angles in the same segment are equal.
    Thus, $\angle\text{ADB}=\angle\text{ECB}=80^\circ$
    $\Rightarrow\angle\text{ADB}=80^\circ$
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MCQ 1221 Mark
Greatest chord of a circle is called its:
  • A
    Secant
  • B
    Radius
  • C
    Chord
  • D
    Diameter
Answer
  1. Diameter
    Solution:
    Since diameter is the longest segment that can be drawn in a circle(touching the circle at both ends), therefore it is the longest possible chord also.
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MCQ 1231 Mark
Write the correct answer in the following:
AD is a diameter of a circle and AB is a chord. If AD = 34cm, AB = 30cm, the distance of AB from the centre of the circle is:
  • A
    17cm.
  • B
    15cm.
  • C
    4cm.
  • D
    8cm.
Answer
  1. 8cm.
    Solution:
    Draw $\text{OP}\bot\text{AB}.$
    As perpendicular from the centre to a chord bisect the chord,
    So, $\text{AP}=\frac{1}{2}\times\text{AB}=\frac{1}{2}\times30=15\text{cm}$
    Radius $=\text{OA}=\frac{1}{2}\times34=17\text{cm}$
    In right $\triangle\text{OPA},$ we have
    $\text{OP}=\sqrt{\text{OA}^2-\text{AP}^2}=\sqrt{(17)^2-(15)^2}$
    $=\sqrt{289-225}=\sqrt{64}=8\text{cm}$
    Hence, (d) is the correct answer.
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MCQ 1241 Mark
What is a diameter in terms of the radius?
  • A
    $\text{d}=2\pi\text{r}$
  • B
    r = 2d
  • C
    d = r
  • D
    d = 2r
Answer
  1. d = 2r
    Solution:
    Diameter is twice of radius.
    thus, d = 2r.
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MCQ 1251 Mark
The angle in a semicircle measures:
  • A
    45°
  • B
    60°
  • C
    90°
  • D
    36°
Answer
  1. 90°
    Solution:
    The angle in a semicircle measures 90°.
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MCQ 1261 Mark
A circle is drawn. It divides the plane into:
  • A
    No Parts
  • B
    4 Parts
  • C
    5 Parts
  • D
    3 Parts
Answer
  1. 3 Parts
    Solution:
    A circle divides the plane into 3 parts namely, the points outside the circle, the points inside the circle and the points on the circle.
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MCQ 1281 Mark
In the given figure, O is the centre of a circle in which $\angle\text{AOC}=100^\circ.$ Side AB of quad. OABC has been produced to D. Then, $\angle\text{CBD}=?$
  • A
    80º
  • B
    50º
  • C
    40º
  • D
    25º
Answer
  1. 50º
    Solution:
    Take a point E on the remaining part of the circumference.
    Join AE and CE.
    Then, $\angle\text{AEC}=\frac{1}{2}\angle\text{AOC}=(\frac{1}{2}\times100^\circ)=50^\circ$
    Now, side AB of the cyclic quadrilateral ABCE has been produced to D.
    $\therefore$ Exterior $\angle\text{CBD}=\angle\text{AEC}=50^\circ$
    $\Rightarrow\angle\text{CBD}=50^\circ$
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MCQ 1291 Mark
In a circle, the major arc is 3 times the minor arc. The corresponding central angles and the degree measures of two arcs are:
  • A
    90° and 270°
  • B
    90° and 90°
  • C
    270° and 90°
  • D
    60° and 210°
Answer
  1. 90° and 270°
    Solution:

    $\frac{\widehat{\text{AB}}\text{minor}}{\widehat{\text{AB}}\text{major}}=\frac{1}{3}=\frac{\angle\widehat{\text{AB}}\text{minor}}{\angle\widehat{\text{AB}}\text{major}}$
    Let $\angle\widehat{\text{AB}}\text{minor}=\text{x}$
    $\Rightarrow\angle\widehat{\text{AB}}\text{major}=\text{3x}$
    Now we know x + 3x = 360°
    ⇒ 4x = 360°
    ⇒ x = 90°
    ⇒ 3x = 270°
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MCQ 1301 Mark
In the given figure, O is the centre of a circle and $\angle\text{OAB}=50^\circ.$ Then, $\angle\text{CDA}=?$
  • A
    40°
  • B
    50°
  • C
    75°
  • D
    25°

Answer
  1. 50°
    Solution:
    OA = OB [Radii of the same circle]
    $\Rightarrow\ \angle\text{OBA}=\angle\text{OAB}=50^\circ$
    Since angles in the same segment are equal, $\angle\text{ABC}=\angle\text{CDA}.$
    That is, $\angle\text{ABO}=\angle\text{CDA}=50^\circ$
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MCQ 1321 Mark
In the given figure, sides AB and AD of quad. ABCD are produced to E and F respectively. If $\angle\text{CBE}=100^\circ$ then $\angle\text{CDF}=?$
  • A
    100°
  • B
    80°
  • C
    130°
  • D
    90°

Answer
  1. 80°
    Solution:
    Since ABCD is a cyclic equilateral,
    $\angle\text{CBE}=\angle\text{ADC}=100^\circ$
    Since ADF is a straight line,
    $\angle\text{CDF}+\angle\text{ADC}=180^\circ$
    $\Rightarrow\ \angle\text{CDF}+100^\circ=180^\circ$
    $\Rightarrow\ \angle\text{CDF}=80^\circ$
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MCQ 1351 Mark
In the given figure, O is the centre of the circle and $\angle\text{SPQ}=50^\circ.$ Then, the measure of $\angle\text{SRQ}$ is:
  • A
    130º
  • B
    100º
  • C
    110º
  • D
    120º
Answer
  1. 130º
    Solution:

    $\angle\text{OQP}=\text{OPQ}=50^\circ$
    $\angle\text{POQ}=80^\circ$ (From angle sum property)
    $\angle\text{SOQ}=180^\circ-80^\circ=100^\circ$ (From linear pair)
    Completing the cyclic quadrilateral, QRSL, (L being any point on the circumference)
    $\angle\text{SLQ}=50^\circ$
    From cyclic quadrilateral, we have
    $\angle\text{SRQ}=180^\circ-50^\circ=130^\circ$
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MCQ 1361 Mark
In a circle, the major arc is 3 times the minor arc. The corresponding central angles and the degree measures of two arcs are:
  • A
    60º and 210º
  • B
    90º and 90º
  • C
    90º and 27º
  • D
    270º and 90º
Answer
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MCQ 1371 Mark
In the given figure, A and B are the centres of two circles having radii 5cm and 3cm respectively and intersecting at points P and Q respectively. If AB = 4cm, then the length of common chord PQ is:
  • A
    3cm
  • B
    6cm
  • C
    7.5cm
  • D
    9cm

Answer
  1. 6cm
    Solution:
    We know that, the line joining centres is the perpendicular bisector of the common chord.
    Then,
    AP = 5cm, BP = 3cm and AB = 4cm
    AP2 = 52 = 25
    BP2 + AB2 = 32 + 42 = 25
    In $\triangle\text{ABP},$
    Since AP2 = BP2 + AB2
    $\triangle\text{ABP},$ is a right-angled triangle and PQ = 2BP = 2(3) = 6cm.
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MCQ 1381 Mark
One chord of a circle is known to be 10cm. The radius of this circle must be:
  • A
    Greater than 5cm.
  • B
    5cm.
  • C
    Greater than or equal to 5cm.
  • D
    Less than 5cm.
Answer
  1. Greater than 5cm.
    Solution:
    We are given length of a chord to be 10cm and we have to give information about the radius of the circle.
    Since in any circle, the diameter of the circle is greater than any chord.
    So diameter > 10
    ⇒ 2r > 10
    ⇒ r > 5cm
    Radius is greater than 5cm.
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MCQ 1411 Mark
In the given figure, AB is a diameter of the circle APBR. APQ and RBQ are straight lines. If $\angle\text{A}=35^\circ$ and then the measure of $\angle\text{PBR}$ is:
  • A
    135º
  • B
    155º
  • C
    165º
  • D
    115º
Answer
  1. 115º
    Solution:

    $\angle\text{APB}=\angle\text{BPQ}=90^\circ$
    Now,
    In $\triangle\text{APB},$
    $\angle\text{BAP}+\angle\text{APB}+\angle\text{ABP}=180^\circ$
    $35^\circ+90^\circ+\angle\text{ABP}=180^\circ$
    $\angle\text{ABP}=55^\circ$
    Again,
    In $\triangle\text{BPQ}$
    $\Rightarrow\angle\text{BPQ}+\angle\text{PQB}+\angle\text{PBQ}=180^\circ$
    $\Rightarrow90^\circ+25^\circ+\angle\text{PBQ}=180^\circ$
    $\Rightarrow\angle\text{PBQ}=65^\circ$
    Since, RBQ is a straight line,
    $\angle\text{RBA}+\angle\text{ABP}+\angle\text{PBQ}=180^\circ$
    $\angle\text{RBA}+55^\circ+65^\circ=180^\circ$
    $\angle\text{RBA}=60^\circ$
    Finally,
    $\angle\text{PBR}=\angle\text{ABP}+\angle\text{RBA}$
    $=55^\circ+60^\circ=115^\circ$
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MCQ 1421 Mark
Write the correct answer in the following:
ABCD is a cyclic quadrilateral such that AB is a diameter of the circle circumscribing it and $\angle\text{ADC}=140^\circ,$ then $\angle\text{BAC}$ is equal to:
  • A
    80°.
  • B
    50°.
  • C
    40°.
  • D
    30°.
Answer
  1. 50°.
    Solution:
    Given, ABCD is a cyclic quadrilateral and $\angle\text{ADC}=140^\circ.$

    We know that, sum of the opposite angles in a cyclic quadrilateral is 180°.
    $\angle\text{ADC}+\angle\text{ABC}=180^\circ$
    $\Rightarrow\ 140^\circ+\angle\text{ABC}=180^\circ$
    $\Rightarrow\ \angle\text{ABC}=180^\circ-140^\circ$
    $\therefore\angle\text{ABC}=40^\circ$
    Since, $\angle\text{ACB}$ is an angle in a semi-circle.
    $\therefore\angle\text{ACB}=90^\circ$
    In $\triangle\text{ABC},\ \ \angle\text{BAC}+\angle\text{ACB}+\angle\text{ABC}=180^\circ$ [by angle sum property of a triangle]
    $\Rightarrow\angle\text{BAC}+90^\circ+40^\circ=180^\circ$
    $\Rightarrow\angle\text{BAC}=180^\circ-130^\circ=50^\circ$
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MCQ 1431 Mark
In the given figure, if $\angle\text{AOB}=80^\circ$ and $\angle\text{ABC}=30^\circ,$ then $\angle\text{CAO}$ is equal to:
  • A
    60º
  • B
    40º
  • C
    80º
  • D
    30º
Answer
  1. 60º
    Solution:

    In triangle AOB, OA = OB (Radii)
    $\Rightarrow\angle\text{OAB}=\angle\text{OBA}$
    Now,
    $\Rightarrow\angle\text{OAB}+\angle\text{ABO}+\angle\text{AOB}=180^\circ$
    $\Rightarrow2\angle\text{OAB}=80^\circ=180^\circ$
    $\Rightarrow\angle\text{OAB}=50^\circ$
    Now, $\angle\text{CAB}=\frac{1}{2}\angle\text{AOB}=\frac{80^\circ}{2}=40^\circ$
    Again, in triangle ABC,
    $\angle\text{CAB}+\angle\text{ABC}+\angle\text{BCA}=180^\circ$
    $\angle\text{CAB}+30^\circ+40^\circ=180^\circ$
    $\angle\text{CAB}=110^\circ$
    Thus,
    $\angle\text{CAO}=\angle\text{CAB}-\angle\text{OAB }$
    $=110^\circ-50^\circ=60^\circ$
    Hence, proved.
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MCQ 1441 Mark
AOB is the diameter of the circle. ABCD is a cyclic trapezium in which AB || DC. If $\angle\text{BED}=65^\circ,$ then $\angle\text{BDC}$ is equal to:
  • A
    40º
  • B
    25º
  • C
    65º
  • D
    75º
Answer
  1. 25º
    Solution:

    AB is diameter and ABCD is semi-circle, so $\angle\text{ADB}=90^\circ$
    Also, $\angle\text{BED}=\angle\text{BAD}=65^\circ$ {Angles in same segment}
    so, in $\triangle\text{ABD},\angle\text{BAD}+\angle\text{ABD}+\angle\text{ADB}=180^\circ$
    $\angle\text{ADB}=180-155=25^\circ$
    Now, since, AB || CD
    So, $\angle\text{ADB}=\angle\text{BDC}=25^\circ$ {alternate interior angles}
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MCQ 1451 Mark
A chord of length 14cm is at a distance of 6cm from the centre of a circle. The length of another chord at a distance of 2cm from  he centre of the circle is:
  • A
    12cm
  • B
    18cm
  • C
    14cm
  • D
    16cm
Answer
  1. 18cm
    Solution:
    We are given the chord of length 14cm and perpendicular distance from the centre to the chord is 6cm. We are asked to find the length of another chord at a distance of 2cm from the centre.
    We have the following figure

    We are given AB = 14cm, OD = 6cm, MO = 2cm, PQ = ?
    Since, perpendicular from centre to the chord divide the chord into two equal parts
    Therefore
    AO2 = AD2 + OD2 [using paythagoras theorem]
    = 72 + 62
    = 49 + 36
    $\text{AO}=\sqrt{85}$
    Now consider the $\triangle\text{OPQ}$ in which OM = 2cm
    So using Pythagoras theorem in $\triangle\text{OPM}$
    PM2 = OP2 - OM2
    $=(\sqrt{85})^2-2^2$ ($\because$ OP = AO = radius)
    PM2 = 81
    PM = 9cm
    Hence PQ = 2PM
    = 2 × 9
    PQ = 18cm
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MCQ 1461 Mark
The given figure shows two intersecting circles. If $\angle\text{ABC}=75^\circ,$ then the measure of $\angle\text{PAD}$ is:
  • A
    75º
  • B
    105º
  • C
    150º
  • D
    125º
Answer
  1. 105º
    Solution:

    Since $\angle\text{PQC}+\angle\text{PBC}=180^\circ$ (Opposite angles of a cyclic quadrilateral))
    $\angle\text{PQC}=180^\circ-75^\circ=105^\circ$
    Again, $\angle\text{DPQ}+\angle\text{PQC}=180^\circ$ (Linear Pair)
    so, $\angle\text{DPQ}=75^\circ$
    Also, $\angle\text{PAD}+\angle\text{DQP}=180^\circ$ (Opposite angles of a cyclic quadrilateral)
    $\angle\text{PAD}=105^\circ$
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MCQ 1471 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
    65°
  • B
    70°
  • C
    110°
  • D
    90°

Answer
  1. 70°
    Solution:
    BC = BD [Given]
    $\angle\text{BDC}=\angle\text{CBD}=35^\circ$ [Angle opposite equal sides are equal]
    In $\triangle\text{BCD},$
    $\angle\text{BCD}+\angle\text{BDC}+\angle\text{CBD}=180^\circ$ [Angle sum property]
    $\Rightarrow\ \angle\text{BCD}+35^\circ+35^\circ=180^\circ$
    $\Rightarrow\ \angle\text{BCD}=110^\circ$
    Since ABCD is a cyclic quadrilateral,
    $\angle\text{BAD}+\angle\text{BCD}=180^\circ$
    $\Rightarrow\ \angle\text{BAD}+110^\circ=180^\circ$
    $\Rightarrow\ \angle\text{BAD}=70^\circ$
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MCQ 1481 Mark
In the given figure PQ and RS are two equal chords of a circle with centre O. OA and OB are perpendiculars on chords PQ and RS, respectively. If $\angle\text{AOB}=140^\circ,$ then $\angle\text{PAB}$ is equal to.
  • A
    70º
  • B
    50º
  • C
    60º
  • D
    40º
Answer
  1. 70º
    Solution:

    In triangle ABO, AO = BO
    So, $\angle\text{BAO}=\angle\text{ABO}=\text{x}$
    $\text{x}+\text{x}+140^\circ=180^\circ$
    $\Rightarrow2\text{x}=40^\circ$
    $\text{x}=20^\circ$
    Now $\angle\text{QAO}+\angle\text{BAO}+\angle\text{PAB}=180^\circ$
    Substituting the values we get:
    $90^\circ+20^\circ+\angle\text{PAB}=180^\circ$
    $\angle\text{PAB}=70^\circ$
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MCQ 1491 Mark
In the given figure, O is the centre of a circle in which $\angle\text{OBA}=20^\circ$ and $\angle\text{OCA}=30^\circ.$ Then, $\angle\text{BOC}=?$
  • A
    50°
  • B
    90°
  • C
    100°
  • D
    130°

Answer
  1. 100°
    Solution:
    In $\triangle\text{OAB},$
    OA = OB [Radii of the same circle]
    $\Rightarrow\ \angle\text{OBA}=\angle\text{OAB}=20^\circ$ [Angle opposite equal sides are equal]
    In $\triangle\text{OAC},$
    OA = OC [Radii of the same circle]
    $\Rightarrow\ \angle\text{OCA}=\angle\text{OAC}=30^\circ$ [Angle opposite equal sides are equal]
    Now, $\angle\text{BAC}=\angle\text{BAO}+\angle\text{CAO}$
    $=20^\circ+30^\circ$
    $=50^\circ$
    $\angle\text{BOC}=2\angle\text{BAC}=2(50^\circ)=100^\circ.$
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MCQ 1501 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
    60º
  • B
    75º
  • C
    120º
  • D
    150º
Answer
  1. 150º
    Solution:
    We are given that the chord is equal to its radius.
    We have to find the angle subtended by this chord at the minor arc.
    We have the corresponding figure as follows:

    We are given that
    AO = OB = AB
    So,
    $\triangle\text{AOB}$ is an equilateral triangle.
    Therefore, we have
    $\angle\text{AOB}=60^\circ$
    Since the angle subtended by any chord at the centre is twice the angle subtended at any point on the circle.
    So, $\angle\text{AQB}=\frac{\angle\text{AOB}}{2}=\frac{60}{2}=30^\circ$
    $\Rightarrow\angle\text{AQB}=30^\circ$
    Take a point P on the minor arc
    Since APBQ is a cyclic quadrilateral
    So, opposite angles are supplementary. That is
    $\angle\text{APB}+\angle\text{AQB}=180^\circ$
    $\angle\text{APB}+130^\circ=180^\circ\ [\angle\text{AQB}=30^\circ]$
    $\angle\text{APB}=180^\circ-30^\circ=150^\circ$
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