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Maths M.C.Q

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

Questions · Page 2 of 5

M.C.Q

MCQ 511 Mark
In the given figure, $\angle\text{BPC}=19^\circ,$ arc AB = arc BC = arc CD, Then, the measure of $\angle\text{APD}$ is:
  • A
    38º
  • B
    76º
  • C
    57º
  • D
    59º
Answer
  1. 57º
    Solution:
    Equal arcs subtend equal angles at the centre and the angle subtended by them at the circumference would be double the angle subtended by them at the centre. As the angle subtended at centre were same so the angle subtended at the circumference would also become same. Thus each arc would make an angle of 19º. Thus the total length of all the three angles would be thrice 19º that is 57º.
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MCQ 521 Mark
In the given figure, AOB is a diameter and ABCD is a cyclic quadrilateral. If $\angle\text{ADC}=120^\circ$ then $\angle\text{BAC}=?$
  • A
    60°
  • B
    30°
  • C
    20°
  • D
    45°

Answer
  1. 30°
    Solution:
    We know that the opposite angles of a quadrilateral are supplementary.
    $\angle\text{ADC}+\angle\text{ABC}=180^\circ$
    $\Rightarrow\ 120^\circ+\angle\text{ABC}=180^\circ$
    $\Rightarrow\ \angle\text{ABC}=60^\circ$
    Since BOC is a diameter $\angle\text{ACB}=90^\circ.$
    In $\triangle\text{CAB},$
    $\angle\text{ABC}+\angle\text{BAC}+\angle\text{ACB}=180^\circ$ [Angle sum property]
    $\Rightarrow\ 60^\circ+\angle\text{BAC}+90^\circ=180^\circ$
    $\Rightarrow\ \angle\text{BAC}=30^\circ$
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MCQ 561 Mark
In the given figure, O is the centre of a circle and $\angle\text{OAB}=50^\circ.$ Then, $\angle\text{BOD}=?$
  • A
    80º
  • B
    100º
  • C
    130º
  • D
    50º
Answer
  1. 100º
    Solution:
    OA = OB (Radii of a circle)
    $\Rightarrow\angle\text{OBA}=\angle\text{OAB}=50^\circ$
    In $\triangle\text{OAB},$ we have:
    $\angle\text{OAB}+\angle\text{OBA}+\angle\text{AOB}=180^\circ$ (Angle sum property of a triangle)
    $\Rightarrow50^\circ+50^\circ+\angle\text{AOB}=180^\circ$
    $\Rightarrow\angle\text{AOB}=(180^\circ-100^\circ)=80^\circ$
    Since $\angle\text{AOB}+\angle\text{BOD}=180^\circ$ (Linear pair)
    $\therefore\angle\text{BOD}=(180^\circ-80^\circ)=100^\circ$
    $\Rightarrow\angle\text{BOD}=100^\circ$
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MCQ 571 Mark
PQRS is a cyclic quadrilateral such that PR is a diameter of the circle. If 67º and $\angle\text{SPR}=72^\circ,$ then $\angle\text{QRS}=$
  • A
    41º
  • B
    23º
  • C
    67º
  • D
    18º
Answer
  1. 41º
    Solution:
    Here we have a cyclic quadrilateral PQRS with PR being a diameter of the circle. Let the centre of this circle be O.
    We are given that $\angle\text{QPR}$ and $\angle\text{SPR}=72^\circ.$

    So, we see that,
    $\angle\text{QPS}=\angle\text{QPR}+\angle\text{RPS}$
    $=67^\circ+72^\circ$
    $=139^\circ$
    In a cyclic quadrilateral, it is known that the opposite angles as supplementary.
    $\angle\text{QPS}+\angle\text{QRS}=180^\circ$
    $\angle\text{QRS}=180^\circ-\angle\text{QPS}$
    $\angle\text{QRS}=180^\circ-139^\circ=41^\circ$
    $\angle\text{QRS}=41^\circ$
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MCQ 581 Mark
In the give figure, AB and CD are two intersecting chords of a circle. If $\angle\text{CAB}=40^\circ$ and $\angle\text{BCD}=80^\circ$then $\angle\text{CBD}=?$
  • A
    80°
  • B
    60°
  • C
    50°
  • D
    70°

Answer
  1. 60°
    Solution:
    Since angles in the same segment are equal,
    $\angle\text{BDC}=\angle\text{BAC}=40^\circ$
    In $\triangle\text{BDC},$
    $\angle\text{BDC}+\angle\text{BCD}+\angle\text{CBD}=180^\circ$ [Angle sum property]
    $\Rightarrow\ 40^\circ+80^\circ+\angle\text{CBD}=180^\circ$
    $\Rightarrow\ \angle\text{CBD}=60^\circ$
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MCQ 591 Mark
In the given figure, ABCD is a cyclic quadrilateral in which DC is produced to E and CF is drawn parallel to AB such that $\angle\text{ADC}=95^\circ$ and $\angle\text{ECF}=20^\circ.$ Then, $\angle\text{EAD}=?$
  • A
    95°
  • B
    85°
  • C
    105°
  • D
    75°

Answer
  1. 105°
    Solution:
    Since CF || AB, $\angle\text{ABC}=\angle\text{BCF}=85^\circ$
    $\angle\text{BAD}=\angle\text{BCE}$
    $\Rightarrow\ \angle\text{BAD}=\angle\text{BCF}+\angle\text{ECF}$
    $\Rightarrow\ \angle\text{BAD}=105^\circ$
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MCQ 601 Mark
Write the correct answer in the following: In Fig. if $\angle\text{DAB} = 60^\circ, \angle\text{ABD} = 50^\circ,$ then $\angle\text{ACB}$ is equal to:
  • A
    60°.
  • B
    50°.
  • C
    70°.
  • D
    80°.
Answer
  1. 70°.
    Solution:
    In $\triangle\text{ADB},$ we have
    $\angle\text{A}+\angle\text{B}+\angle\text{D}=180^\circ$ [Angle sum property of a triangle]
    $\Rightarrow60^\circ+50^\circ+\angle\text{D}=180^\circ$
    $\Rightarrow\angle\text{D}=180^\circ-110^\circ=70^\circ$
    i.e., $\angle\text{ABD}=70^\circ$
    Now, $\angle\text{ACB}=\angle\text{ADB}=70^\circ$
    [$\because$ Angles in the same segment of a circle are equal]
    Hence, (c) is the correct answer.
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MCQ 611 Mark
If a chord of a circle is equal to its radius, then the angle subtended by this chord in major segment is:
  • A
    30º
  • B
    90º
  • C
    45º
  • D
    60º
Answer
  1. 30º
    Solution:
    Since the chord is equal to the radius therefore, it will form an equilateral triangle inside the circlewith the third vertex being the centre of the circle.
    So the chord will make an angle of 60º at the centre. As the angle made by the chord at any other point of the circumfrence would be half.
    So, we have that angle made at the major segment would be 30º.
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MCQ 621 Mark
In the given figure, O is the centre of a circle and $\angle\text{OAB}=50^\circ.$ Then, $\angle\text{BOD}=?$
  • A
    130°
  • B
    50°
  • C
    100°
  • D
    80°

Answer
  1. 100°
    Solution:
    OA = OB [Radii of the same circle]
    $\Rightarrow\ \angle\text{OAB}=\angle\text{OBA}=50^\circ$ [Angle opposite equal sides are equal]
    $\angle\text{BOD}=\angle\text{OAB}+\angle\text{OBA}$
    $\Rightarrow\ \angle\text{BOD}=50^\circ+50^\circ$
    $\Rightarrow\ \angle\text{BOD}=100^\circ$
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MCQ 641 Mark
If ABC is an arc of a circle and $\angle\text{ABC} = 135^\circ,$ then the ratio of arc $\widehat{\text{ABC}}$ to the circumference is:
  • A
    1 : 4
  • B
    3 : 4
  • C
    3 : 8
  • D
    1 : 2
Answer
  1. 1 : 4
    Solution:

    ABC is an arc of circle.
    Take point D in the altrenative segment and join AD and CD.
    $\angle\text{ABC}=135^\circ$ (Given)
    $\angle\text{ABC}+\angle\text{ADC}=180^\circ$ (Sum of opposite angles of cyclic quadrilateral is 180°)
    $\Rightarrow\angle\text{ADC}=180^\circ-\angle\text{ABC}=180^\circ-135^\circ=45^\circ$
    Now, $\angle\text{AOC}=2\times\angle\text{ADC}=2\times45^\circ=90^\circ$
    $\widehat{\text{ABC}}=$ measure of the central angle $=\angle\text{AOC}=90^\circ$
    $\Rightarrow\text{Required ratio}=\frac{\text{arc}\widehat{\text{ABC}}}{\text{circumference}}$
    $=\frac{90^\circ}{360^\circ}=\frac{1}{4}=1:4$
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MCQ 651 Mark
An equilateral triangle ABC is inscribed in a circle with centre O. The measures of $\angle\text{BOC}$ is:
  • A
    30°
  • B
    60°
  • C
    90°
  • D
    120°
Answer
  1. 120°
    Solution:

    $\angle\text{BAC}=60^\circ$ (Angle of equilateral triangle)
    Arc $\widehat{\text{BC}}$ makes angle $\angle\text{BAC}$ at circle and $\angle\text{BOC}$ at center of circle.
    $\Rightarrow\angle\text{BAC}=\frac{1}{2}\angle\text{BOC}$
    $\Rightarrow2\times\angle\text{BAC}=\angle\text{BOC}$
    $\Rightarrow2\times60^\circ=\angle\text{BOC}$
    $\Rightarrow\angle\text{BOC}=120^\circ$
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MCQ 661 Mark
If P and Q are any two Points on a circle then PQ is called a:
  • A
    Radius
  • B
    Diameter
  • C
    Secant
  • D
    Chord
Answer
  1. Chord
    Solution:
    A chord is a line formed by any two points on a circle.
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MCQ 681 Mark
If a chord of a circle is equal to its radius, then the angle subtended by this chord in major segment is:
  • A
    90º
  • B
    60º
  • C
    45º
  • D
    30º
Answer
  1. 30º
    Solution:
    Since the chord is equal to the radius therefore, it will form an equilateral triangle inside the circle with the third vertex being the centre of the circle.
    So the chord will make an angle of 60º at the centre. As the angle made by the chord at any other point of the circumference would be half.
    So, we have that angle made at the major segment would be 30º.
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MCQ 691 Mark
The angle in a semicircle measures.
  • A
    90º
  • B
    36º
  • C
    60º
  • D
    45º
Answer
  1. 90º
    Solution:
    The angle in a semicircle measures 90º.
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MCQ 711 Mark
In the given figure, O is the centre of a circle and $\angle\text{AOC}=130^\circ.$ Then, $\angle\text{ABC}=?$
  • A
    50°
  • B
    65°
  • C
    115°
  • D
    130°

Answer
  1. 115°
    Solution:
    Minor $\angle\text{AOC}=130^\circ$
    Major $\angle\text{AOC}=360^\circ-130^\circ$
    ⇒ Major $\angle\text{AOC}=230^\circ$
    Since $\angle\text{ABC}=\frac{1}{2}\text{major}\angle\text{AOC}$
    $\Rightarrow\ \angle\text{ABC}=\frac{1}{2}(230^\circ)$
    $\Rightarrow\ \angle\text{ABC}=115^\circ$
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MCQ 741 Mark
In the given figure, ABCD is a cyclic quadrilateral in which DC is produced to E and CF is drawn parallel to AB such that $\angle\text{ADC}=95^\circ$ and $\angle\text{ECF}=20^\circ.$ Then, $\angle\text{BAD}=?$
  • A
    95º
  • B
    85º
  • C
    75º
  • D
    105º
Answer
  1. 105º
    Solution:
    We have:
    $\angle\text{ABC}+\angle\text{ADC}=180^\circ$
    $\Rightarrow\angle\text{ABC}+95^\circ=180^\circ$
    $\Rightarrow\angle\text{ABC}=(180^\circ-95^\circ)=85^\circ$
    Now, CF || AB and CB is the transversal.
    $\therefore\angle\text{BCF}=\angle\text{ABC}=85^\circ$ (Alternate interior angles)
    $\Rightarrow\angle\text{BCE}=(85^\circ+20^\circ)=105^\circ$
    $\Rightarrow\angle\text{DCB}=(180^\circ-105^\circ)=75^\circ$
    $\Rightarrow\angle\text{DCB}=75^\circ$
    Now, $\angle\text{BAD}+\angle\text{BCD}=180^\circ$
    $\Rightarrow\angle\text{BAD}+75^\circ=180^\circ$
    $\Rightarrow\angle\text{BAD}=(180^\circ-75^\circ)$
    $\Rightarrow\angle\text{BAD}=105^\circ$
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MCQ 751 Mark
AB and CD are two equal chords of a circle with centre O such that $\angle\text{AOB}=80^\circ$ then $\angle\text{COD}=?$
  • A
    100°
  • B
    80°
  • C
    120°
  • D
    40°

Answer
  1. 80°
    Solution:
    Given that AB = CD.
    Since equal chord, subtend equal angles at the centre,
    $\angle\text{COD}=\angle\text{AOB}=80^\circ.$
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MCQ 761 Mark
AB is a chord of a circle with radius r. if P is any point on the circle such that $\angle\text{APB}$ is a right angle, then AB is equal to:
  • A
    r2
  • B
    3r
  • C
    r
  • D
    2r
Answer
  1. 2r
    Solution:
    Since the angle subtended by the diameter or the semicircle is a right angle, therefore the AB is nothing but the diameter. Hence, it is equal to 2r, r being the radius.
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MCQ 771 Mark
In the given figure, O is the centre of the circle. If $\angle\text{CAB}=40^\circ$ and $\angle\text{CBA}=110^\circ,$ the value of x is:
  • A
    55º
  • B
    80º
  • C
    60º
  • D
    50º
Answer
  1. 60º
    Solution:

    In $\triangle\text{ABC}$
    $\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
    $\Rightarrow\angle\text{C}=180^\circ-40^\circ-110=30^\circ$
    Since AB is a chord and angle made by a chord at the centre is twice the angle made by it on any point on the circumference, therefore:-
    x = 2 × 30º = 60º
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MCQ 801 Mark
In the given figure, BOC is a diameter of a circle and AB = AC. Then, $\angle\text{ABC}=?$
  • A
    90º
  • B
    30º
  • C
    60º
  • D
    45º
Answer
  1. 45º
    Solution:
    Since an angle in a semicircle is a right angle, $\angle\text{BAC}=90^\circ$
    $\therefore\angle\text{ABC}+\angle\text{ACB}=90^\circ ....(\text{i})$
    Now, AB = AC (Given)
    $\Rightarrow\angle\text{ABC}=\angle\text{ACB}=45^\circ\ ....(\text{ii})$
    $\Rightarrow\angle\text{ABC}+\angle\text{ABC}=90^\circ$ [From (i) and (ii)]
    $\Rightarrow2\angle\text{ABC}=90$
    $\Rightarrow\angle\text{ABC}=45^\circ$
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MCQ 811 Mark
  • A
    20cm
  • B
    16cm
  • C
    10cm
  • D
    8cm
Answer
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MCQ 821 Mark
In the given figure, O is the centre of the circle such that $\angle\text{AOC} = 130^\circ,$ then $\angle\text{ABC} =$
  • A
    130°
  • B
    115°
  • C
    65°
  • D
    165°
Answer
  1. 115°
    Solution:

    $\angle\text{ADC}=\frac{1}{2}\angle\text{AOC}$
    $\big\{\angle\text{ADC}$ and $\angle\text{AOC}$ are made by same $\widehat{\text{AC}}$ on centre and cricumference$\big\}$
    $\Rightarrow\angle\text{ADC}=\frac{1}{2}\times130^\circ=65^\circ$
    ADCB is a cyclic Quadrilateral.
    $\Rightarrow\angle\text{D}+\angle\text{B}=180^\circ$
    $\Rightarrow\angle\text{ABC}=180^\circ-65^\circ=115^\circ$
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MCQ 831 Mark
  • A
    It is 2 : 1
  • B
    It is 3 : 1
  • C
    It is 1 : 3
  • D
    It is 1 : 2
Answer
  1. It is 3 : 1
    Solution:
    Since, circles are congruent thus we can consider the two arcs as in the same circle. Now the length of the arcs is directly proportional the angle subtended by the arcs. Therefore the lengths of the given arcs would be same in ratio as the ratio of the given angles.
    Hence the required ratio is $\frac{75}{25}=\frac{3}{1}.$
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MCQ 841 Mark
In the given figure, O is the centre of a circle and $\angle\text{AOC}=120^\circ.$ Then, $\angle\text{BDC}=?$
  • A
    60°
  • B
    45°
  • C
    30°
  • D
    15°

Answer
  1. 30°
    Solution:
    Since BOA is a diameter.
    $\angle\text{AOC}+\angle\text{BOC}=180^\circ$
    $\Rightarrow\ 120^\circ+\angle\text{BOC}=180^\circ$
    $\Rightarrow\ \angle\text{BOC}=60^\circ$
    So, $\angle\text{BDC}=\frac{1}{2}\angle\text{BOC}=\frac{1}{2}(60^\circ)=30^\circ$
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MCQ 851 Mark
The given figure shows two intersecting circles. If $\angle\text{ABC}=75^\circ$ then the measure of $\angle\text{PAD}$ is:
  • A
    105º
  • B
    150º
  • C
    125º
  • D
    75º
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, $\angle\text{DQP}+\angle\text{PQC}=180^\circ$ (Linear Pair)
    So, $\angle\text{DQP}=75^\circ$
    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 871 Mark
In the given figure, AB || CD and O is the centre of the circle. If $\angle\text{ADC}=25^\circ,$ then the measure of $\angle\text{AEB}$ is:
  • A
    40º
  • B
    80º
  • C
    25º
  • D
    80º
Answer
  1. 40º
    Solution:

    Here, AB || CD and $\angle\text{ADC}=25^\circ,$
    So, $\angle\text{DAB}=25^\circ,$ (opposite interior angles are equal)
    Now, $\angle\text{ADC}=25^\circ,$ so, $\angle\text{AOC}=50^\circ$ (Angle subtended by arc AC at centre is twice the angle subtended at circumference)
    Similarly, $\angle\text{DAB}=25^\circ,$ so, $\angle\text{DOB}=25^\circ$ (Angle subtended by arc BD at centre is twice the angle subtended at circumference)
    $\angle\text{AOB}+\angle\text{DOB}+\angle\text{AOC}=180^\circ$ (All lie in straight line)
    $\angle\text{AOB}=180-50-50=80^\circ$
    Now, $\angle\text{AEB}=40^\circ$ (Angle subtended by arc AB at centre is twice the angle subtended at circwnference)
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MCQ 891 Mark
In the given, AB is side of regular five sided polygon and AC is a side of a regular six sided polygon inscribed in the circle with centre O. AO and CB intersect at P, then $\angle\text{APB}$ is equal to:
  • A
    72º
  • B
    90º
  • C
    86º
  • D
    96º
Answer
  1. 96º
    Solution:

    Here AC is side of hexagon, so it will subtend 60º angle at centre and also sides & radius are equal.
    Thus,  AC = OC = OA  and $\angle\text{COA}=\angle\text{OAC}=\angle\text{ACO}=60^\circ$
    AB is side of pentagon, so it would subtend angle of $\frac{360}{5}=72^\circ$ angle at centre.
    So, $\angle\text{BOP}=72^\circ$
    So, $\angle\text{COB}=72+60=132^\circ$
    Also since, $\text{OC},\text{OB},\angle\text{OCP}=\text{OBP}$
    $\triangle\text{COB}\angle\text{COB}+\angle\text{OBC}+\angle\text{OCB}=180^\circ$
    $2\angle\text{OBC}=180-(132)=48^\circ$
    $\angle\text{OBC}=24^\circ$
    Now, $\triangle\text{BOP}\angle\text{BOP}+\angle\text{OPB}+\angle\text{PBO}=180^\circ$
    $\angle\text{OPB}=180-(24+72)=180-96=84^\circ$
    Now,
    $\angle\text{APB}$ and $\angle\text{OPB}$ lie on straight line, so they are supplementry angles.
    $\angle\text{APB}=180-\angle\text{OPB}=180-84=96^\circ$
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MCQ 901 Mark
If P and Q are any two points on a circle then PQ is called a:
  • A
    Chord
  • B
    Diameter
  • C
    Secant
  • D
    Radius
Answer
  1. Chord
    Solution:
    A chord is a line formed by any two points on a circle.
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MCQ 911 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
    50°
  • B
    60°
  • C
    70°
  • D
    80°

Answer
  1. 70°
    Solution:
    Since angles in the same segment of a circle are equal.
    $\angle\text{BAC}=\angle\text{BDC}=60^\circ.$
    In $\triangle\text{BDC},$
    $\angle\text{BDC}+\angle\text{DBC}+\angle\text{BCD}=180^\circ.$
    $\Rightarrow\ 60^\circ+50^\circ+\angle\text{BCD}=180^\circ$
    $\Rightarrow\ \angle\text{BCD}=70^\circ$
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MCQ 921 Mark
The constant distance of a point on a circle from the centre of the circle is called.
  • A
    Diameter
  • B
    Circle
  • C
    Radius
  • D
    Centre
Answer
  1. Radius
    Solution:
    Radius is the fixed distance of a fixed point from a point on the circle.
    Also more precisely, a circle is the locii or the path of a point that moves maintaing a fixed distance from a given point.
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MCQ 931 Mark
In the given figure, ABCD is a quadrilateral inscribed in circle with centre O. CD is produced to E. If $\angle\text{ADE}=95^\circ$ and $\angle\text{OBA}=30^\circ,$ then $\angle\text{OAC}$ is equal to:
  • A
    10º
  • B
    20º
  • C
    15º
  • D
    5º
Answer
  1. 5º
    Solution:

    Here, $\angle\text{ADC}$ and $\angle\text{ADE}$ are supplementary to each other,
    So, $\angle\text{ADC}=180-95=85^\circ$
    Also, ABCD is also a cyclic quadrilateral so, $\angle\text{ADC}$ and $\angle\text{ABC}$ are supplementary
    So, $\angle\text{ABC}=\angle\text{ADC}=180^\circ$
    $\angle\text{ABC}=180-95=85^\circ$
    angle subtended at centre is double the angle suntended at circumference.
    So, $\angle\text{AOC}=170^\circ$
    Now, in triangle AOC, OA = OC(Radii) so, $\angle\text{OAC}=\text{OCA}$
    $\angle\text{OCA}+\angle\text{OAC}+\angle\text{AOC}=180^\circ$
    $2\angle\text{OAC}=180-170=10^\circ$
    $\angle\text{OAC}=5^\circ$
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MCQ 941 Mark
ABCD is a cyclic quadrilateral such that $\angle\text{ADB}=30^\circ$ and $\angle\text{DCA}=80^\circ,$ then $\angle\text{DAB}=$
  • A
    125º
  • B
    70º
  • C
    150º
  • D
    100º
Answer
  1. 70º
    Solution:
    It is given that ABCD is cyclic quadrilateral $\angle\text{ADB}=90^\circ$ and $\angle\text{DCA}=80^\circ.$ We have to find $\angle\text{DAB}.$
    We have the following figure regarding the given information

    $\angle\text{BDA}=\angle\text{BCA}=30^\circ$ (Angle in the same segment are equal)
    Now, since ABCD is a cyclic quadrilateral
    So, $\angle\text{DAB}+\angle\text{BCD}=180^\circ$
    $\angle\text{DAB}+\angle\text{BCA}+\angle\text{DCA}=180^\circ\ [\angle\text{BCD}=\angle\text{BCA}+\angle\text{DCA}]$
    $\angle\text{DAB}+30^\circ+80^\circ=180^\circ$
    $\angle\text{DAB}=180^\circ-110^\circ$
    $\angle\text{DAB}=70^\circ$
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MCQ 971 Mark
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
    40º
  • B
    30º
  • C
    75º
  • D
    50º
Answer
  1. 50º
    Solution:
    ln the given quadrilateral,
    $\angle\text{ABC}+\angle\text{ADC}=180^\circ$
    $140^\circ+\angle\text{ABC}=180^\circ$
    $\angle\text{ABC}=40^\circ$
    Since, AB is diameter so ABCD lies in semi-circle.
    thus, $\angle\text{BCA}=90^\circ$
    In triangle, ABC,
    $\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
    $\angle\text{BAC}=180^\circ-40^\circ-90^\circ=180^\circ-130^\circ=50^\circ$
    $\angle\text{BAC}=50^\circ$
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MCQ 981 Mark
if A and B are two points on a circle such that m (AB^) = 260º. A possible value for the angle subtended by arc BA at a point on the circle is:
  • A
    50º
  • B
    100º
  • C
    75º
  • D
    25º
Answer
  1. 50º
    Solution:
    We are given m(AB^) = 260º

    Suppose point P is on the circle.
    Since m(AB^) = 260º
    So, angle AOB = 360º - 260º = 100º
    We know that angle subtended by chord AB at the centre is twice that of subtended at the point P
    So, $\angle\text{APB}=\frac{\angle\text{AOB}}{2}=\frac{100}{2}=50^\circ$
    $\Rightarrow\angle\text{APB}=50^\circ$
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MCQ 1001 Mark
In Fig., AB and CD are two equal chords of a circle with centre O. OP and OQ are perpendiculars on chords AB and CD, respectively. If $\angle\text{POQ}=150^\circ,$ then $\angle\text{APQ}$ is equal to:
  • A
    60º
  • B
    30º
  • C
    15º
  • D
    75º
Answer
  1. 75º
    Solution:

    As AB = CD
    So, OP = OQ (equal chords are equidistant from the centre)
    $\angle1=\angle2$ (angles opposite to equal sides are equal)
    $\angle1+\angle2+\angle\text{POQ}=180^\circ$
    $\angle1+\angle1+150^\circ=180^\circ$
    $\therefore\angle1=15^\circ$
    Since APB is a line segement
    $\therefore\angle\text{BPO}+\angle1+\angle\text{APQ}=180^\circ$
    $90^\circ+15^\circ+\angle\text{APQ}=180^\circ$
    $\therefore\angle\text{APQ}=75^\circ$
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M.C.Q - Page 2 - Maths STD 9 Questions - Vidyadip