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Maths M.C.Q

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

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M.C.Q

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MCQ 11 Mark
In the given figure, O is the centre of a circle. Then, $\angle\text{OAB}=?$
  • A
    50°
  • B
    60°
  • C
    55°
  • D
    65°

Answer
  1. 65°
    Solution:
    OA = OB [Radii of the same circle]
    $\Rightarrow\ \angle\text{OAB}=\angle\text{OBA}$
    In $\triangle\text{OAB},$
    $\angle\text{BOA}+\angle\text{OAB}+\angle\text{OBA}=180^\circ$ [Angle sum property]
    $\Rightarrow\ 50^\circ+\angle\text{OAB}+\angle\text{OAB}=180^\circ$
    $\Rightarrow\ 2\angle\text{OAB}=130^\circ$
    $\Rightarrow\ \angle\text{OAB}=65^\circ$
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MCQ 21 Mark
In the given figure, O is the centre of a circle and $\angle\text{AOB}=140^\circ.$ Then, $\angle\text{ACB}=?$
  • A
    70°
  • B
    80°
  • C
    110°
  • D
    40°

Answer
  1. 110°
    Solution:
    Minor $\angle\text{AOB}=140^\circ$
    Major $\angle\text{AOB}=360^\circ-140^\circ$
    ⇒ Major $\angle\text{AOB}=220^\circ$
    Since $\angle\text{ACB}=\frac{1}{2}\text{major}\angle\text{AOB}$
    $\Rightarrow\ \angle\text{ACB}=\frac{1}{2}(220^\circ)$
    $\Rightarrow\ \angle\text{ACB}=110^\circ$
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MCQ 41 Mark
The greatest chord of a circle is called its:
  • A
    Radius.
  • B
    Secant.
  • C
    Diameter.
  • D
    None of these.
Answer
  1. Diameter.
    Solution:
    The greatest chord of the circle is diameter of the circle.
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MCQ 61 Mark
In the given figure, O is the centre of a circle. If $\angle\text{OAC}=50^\circ$ then $\angle\text{ODB}=?$
  • A
    40°
  • B
    50°
  • C
    60°
  • D
    75°

Answer
  1. 50°
    Solution:
    Since angles in the same segment of a circle are equal.
    $\angle\text{CDB}=\angle\text{BAC}$
    That is , $\angle\text{ODB}=\angle\text{OAC}=50^\circ.$
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MCQ 71 Mark
Greatest chord of a circle is called its:
  • A
    Chord
  • B
    Diameter
  • C
    Secant
  • D
    Radius
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 81 Mark
One half of the whole arc of a circle.
  • A
    Semi-circle
  • B
    Sector
  • C
    Circumference
  • D
    Segment
Answer
  1. Semi-circle
    Solution:
    A semi-circle is half the circle. In other words, half of the total length of the circle makes the semicircle.
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MCQ 91 Mark
In the given figure, O is the centre of a circle in which $\angle\text{AOC}=100^\circ.$ Side AB of quadrilateral OABC has been produced to D. Then, $\angle\text{CBD}=?$
  • A
    50°
  • B
    40°
  • C
    25°
  • D
    80°

Answer
  1. 50°
    Solution:

    Construction: Let E be a point on the reamaining part of the circumference of the circle.
    Join AE and CE.
    $\angle\text{AEC}=\frac{1}{2}\angle\text{AOC}$
    $\Rightarrow\ \angle\text{AEC}=\frac{1}{2}(100^\circ)=50^\circ$
    Since AECB forms a cyclic quadrilateral.
    $\Rightarrow\ \angle\text{CBD}=\angle\text{AEC}=50^\circ$
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MCQ 101 Mark
The given figure shows two congruent circles with centre O and O’ intersecting at A and B. If $\angle\text{AO}'\text{B}=50^\circ,$ then the measure of $\angle\text{APB}$ is:
  • A
    25º
  • B
    50º
  • C
    45º
  • D
    40º
Answer
  1. 25º
    Solution:

    Since both the triangles are congruent,
    So, OA = O'A,
    OB = O'B
    AB = AB  (Common)
    Hence, $\triangle\text{AOB}\cong\triangle\text{AO}'\text{B}$
    Thus, $\angle\text{AOB}=\angle\text{AO}'\text{B}=50^\circ$
    Since, PB is a straight line, therefore:-
    $\angle\text{AOP}+\angle\text{AOB}=180^\circ$
    $\Rightarrow\angle\text{AOP}=180^\circ-50^\circ=130^\circ$
    Again, In triangle OPA,
    $\Rightarrow\angle\text{P}=\angle\text{A}$
    $\Rightarrow\angle\text{A}+\angle\text{P}+\angle\text{O}=180^\circ$
    $\Rightarrow2\angle\text{P}+130^\circ=180^\circ$
    $\Rightarrow\angle\text{P}=\frac{50^\circ}{2}=25^\circ$
    Thus, $\angle\text{OPA}=25^\circ$
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MCQ 111 Mark
AB and CD are two parallel chords of a circle with centre O such that AB = 6cm and CD = 12cm. The chords are on the same side of the centre and the distance between them is 3cm. The radius of the circle, is:
  • A
    $7\text{cm}$
  • B
    $6\text{cm}$
  • C
    $3\sqrt{5}\text{cm}$
  • D
    $5\sqrt{2}\text{cm}$
Answer
  1. $3\sqrt{5}\text{cm}$
    Solution:
    Let the distance between the center and the chord CD be x cm and the radius of the circle is r cm.
    We have to find the radius of the following circle:

    In right angled triangle, OND,
    x2 + 36 = r2 ....(i)
    Now, in right angled triangle AOM,
    r2 = 9 + (x + 3)2 ....(ii)
    From (i) and (ii), we have
    $\text{r}^2=9+((\sqrt{\text{r}})^2-36+3)^2$
    $\Rightarrow\text{r}^2=9+\text{r}^2-36+9+6\sqrt{\text{r}^2-36}$
    $\Rightarrow3=\sqrt{\text{r}^2-36}$
    $\Rightarrow9=\text{r}^2-36$ [squaring both the sides]
    $\Rightarrow\text{r}^2=45\Rightarrow\text{r}=3\sqrt{5}\text{cm}$
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MCQ 121 Mark
In the given figure, CD is the diameter of a circle with centre O and CD is perpendicular to chord AB. If AB = 12cm and CE = 3cm, then radius of the circles is:
  • A
    6cm
  • B
    9cm
  • C
    7.5cm
  • D
    8cm

Answer
  1. 7.5cm
    Solution:
    OA = OC
    ⇒ OA = OE + CE
    ⇒ OA = OE + 3
    ⇒ OE = OA - 3 ...(i)
    $\text{AE}=\frac{1}{2}\text{AB}$ [Perpendicular drawn from the centre of a circle to the chord bisect the chord]
    $=\frac{1}{2}(12)=6\text{cm}$
    In right $\triangle\text{OEA},$
    OA2 = OE2 + AE2
    ⇒ OA2 = (OA - 3)2 + AE2 [From (i)]
    ⇒ OA2 = OA2 - 6OA + 9 + AE2
    ⇒ 6OA = 9 + 62
    ⇒ 6OA = 9 + 36
    $\Rightarrow\ \text{OA}=\frac{45}{6}=7.5\text{cm}$
    So, the radius of the circle is 7.5cm.
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MCQ 141 Mark
A chord is at a distance of 8cm from the centre of a circle of radius 17cm. The length of the chord is:
  • A
    25cm
  • B
    12.5cm
  • C
    30cm
  • D
    9cm
Answer
  1. 30cm
    Solution:

    Let O be the centre of the circle with radius OA = 17cm.
    Since $\text{OC}\perp\text{AB}.$
    In right $\triangle\text{OCA},$
    OA2= OC2 + AC2 [By pythagoras theorem]
    AC2 = OA2 - OC2
    ⇒ AC2 = 172 - 82
    ⇒ AC2 = 289 - 64
    ⇒ AC2 = 225
    ⇒ AC = 15cm
    We know that, the perpendicular drawn from the centre to the chord bisects the chord.
    ⇒ AB = 2AC = 2(15) = 30cm.
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MCQ 151 Mark
Circle having same centre are said to be:
  • A
    Secant
  • B
    Concentric
  • C
    Chord
  • D
    Circle
Answer
  1. Concentric
    Solution:
    Concentric circles are those circle that is drawn with the same point as a centre but different radii.
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MCQ 171 Mark
  • A
    7.5cm
  • B
    9.5cm
  • C
    10.5cm
  • D
    8.5cm
Answer
  1. 8.5cm
    Solution:
    Join AC.
    Then AE : CE = DE : BE (Intersecting secant theorem)
    $\therefore$ AE × BE = DE × CE ....(i)
    Let CD = x cm
    Then AE = (AB + BE) = (11 + 3)cm = 14cm;
    BE = 3cm; CE = (x + 3.5)cm; DE = 3.5cm
    $\therefore$ 14 × 3 = (x + 3.5) × 3.5 [FROM (1)]
    $\Rightarrow\text{x}+3.5=\frac{14\times3}{3.5}=\frac{42}{3.5}=12$
    ⇒ x = (12 - 3.5)cm = 8.5cm
    Hence, CD = 8.5cm
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MCQ 191 Mark
  • A
    9cm
  • B
    6cm
  • C
    8cm
  • D
    7.5cm
Answer
  1. 7.5cm
    Solution:
    Let OA = OC = r cm.
    Then OE = (r - 3)cm and $\text{AE}=\frac{1}{2}\text{AB}=6\text{cm}$
    Now, in right $\triangle\text{OAE},$ we have:
    OA2 = OE2 + AE2 [Using paythagoras theorem]
    ⇒ (r)2 = (r - 3)2 + 62
    ⇒ r2 = r2 + 9 - 6r + 36
    ⇒ 6r = 45
    $\Rightarrow\text{r}=\frac{45}{6}=7.5\text{cm}$
    Hence, the required radius of the circle is 7.5cm.
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MCQ 201 Mark
In the given figure, O is the centre of a circle and $\angle\text{AOB}=140^\circ.$ Then, $\angle\text{ACB}=?$
  • A
    110º
  • B
    70º
  • C
    80º
  • D
    40º
Answer
  1. 110º
    Solution:
    Let, D on any point on circumference and join AD and BD,
    Now, $\angle\text{ADB}=\frac{\angle\text{AOB}}{2}$
    $\angle\text{ADB}=\frac{140}{2}=70^\circ$
    Now, in cyclic quadrilateral ADBC
    $\Rightarrow\angle\text{ADB}+\angle\text{ACB}=180^\circ$
    $\Rightarrow\angle\text{ACB}=180^\circ-\angle\text{ADB}$
    $\Rightarrow\angle\text{ACB}=180^\circ-70^\circ$
    $\Rightarrow\angle\text{ACB}=110^\circ$
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MCQ 211 Mark
Two equal circles of radius r intersect such that each passes through the centre of the other. The length of the common chord of the circle is:
  • A
    $\sqrt{\text{r}}$
  • B
    $\sqrt{2}\text{r}\text{AB}$
  • C
    $\sqrt{3}\text{r}$
  • D
    $\frac{\sqrt3}{2}$
Answer
  1. $\sqrt{3}\text{r}$
    Solution:

    Both the circles pass through the centre of each other
    ⇒ O1O2 = r
    Common chord is AB
    We know that perpendicular drawn from centre of circle to any chord bisects it.
    ⇒ P is the midpoint of AB
    ⇒ PA = PB
    O1A = r (radius of circle)
    Consider $\triangle\text{O}_1\text{PA}$
    $\big(\text{O}_1\text{A}\big)^2=\text{AP}^2+\text{O}_1\text{P}^2$
    $\Rightarrow\text{r}^2=\text{AP}^2+\Big(\frac{\text{r}}{2}\Big)^2$ ...(P is also mid-point of O1O2)
    $\Rightarrow\text{AP}^2=\text{r}^2-\frac{\text{r}^2}{4}=\frac{\text{3r}^2}{4}$
    $\Rightarrow\text{AP}=\frac{\sqrt3}{2}\text{r}$
    Lenght of chord $\text{AP}=\text{2AP}=\sqrt{3}\text{r}$
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MCQ 221 Mark
In the given figure, P and Q are centers of two circles intersecting at B and C. ACD is a straight line. Then, the measure of $\angle\text{BQD}$ is:
  • A
    115º
  • B
    150º
  • C
    105º
  • D
    130º
Answer
  1. 150º
    Solution:

    $\angle\text{APB}=150^\circ,$ so, $\angle\text{ACB}=75^\circ$ {Angle subtended by an arc at centre is twice the angle subtended at any point on circumference}
    Now, ACD is straight line, so, $\angle\text{ACB}+\angle\text{DCB}=180^\circ$
    $\angle\text{DCB}=180-75=105^\circ$
    Now, angle subtended by arc BD on centre is twice of $\angle\text{DCB}=2\times105=210^\circ$
    Now, $\angle\text{BQD}=360^\circ-210^\circ=150^\circ$
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MCQ 231 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
    50°
  • B
    70°
  • C
    20°
  • D
    60°

Answer
  1. 60°
    Solution:
    OA = OC [Radii of the same circle]
    $\Rightarrow\ \angle\text{OBA}=\angle\text{OAB}=20^\circ$
    In $\triangle\text{OAB,}$
    $\angle\text{OBA}+\angle\text{OAB}+\angle\text{AOB}=180^\circ$ [Angle sum property]
    $\Rightarrow\ 20^\circ+20^\circ+\angle\text{AOB}=180^\circ$
    $\Rightarrow\ \angle\text{AOB}=140^\circ$
    Now,
    OB = OC [Radii of the same circle]
    $\Rightarrow\ \angle\text{OBC}=\angle\text{OCB}=50^\circ$
    In $\triangle\text{OCB},$
    $\angle\text{OBC}+\angle\text{OCB}+\angle\text{COB}=180^\circ$ [Angle sum property]
    $\Rightarrow\ 50^\circ+50^\circ+\angle\text{COB}=180^\circ$
    $\Rightarrow\ \angle\text{COB}=80^\circ$
    So,
    $\angle\text{AOB}=\angle\text{AOC}+\angle\text{COB}$
    $\Rightarrow\ \angle\text{AOC}=\angle\text{AOB}-\angle\text{COB}$
    $\Rightarrow\ \angle\text{AOC}=140^\circ-80^\circ$
    $\Rightarrow\ \angle\text{AOC}=60^\circ$
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MCQ 251 Mark
In the given 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
    80º
  • B
    70º
  • C
    60º
  • D
    50º
Answer
  1. 60º
    Solution:
    We have:
    $\angle\text{CDB}=\angle\text{CAB}=40^\circ$ (Angles in the same segment of a circle)
    In $\triangle\text{CBD},$ we have:
    $\angle\text{CDB}+\angle\text{BCD}+\angle\text{CBD}=180^\circ$ (Angle sum property of a triangle)
    $\Rightarrow40^\circ+80^\circ+\angle\text{CBD}=180^\circ$
    $\Rightarrow\angle\text{CBD}=(180^\circ-120^\circ)=60^\circ$
    $\Rightarrow\angle\text{CBD}=60^\circ$
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MCQ 271 Mark
Let C be the mid-point of an arc AB of a circle such that m AB^ = 183º. If the region bounded by the arc ACB and line segment AB is denoted by S, then the centre O of the circle lies.
  • A
    In the interior of S.
  • B
    On the segment AB.
  • C
    On AB and bisect AB.
  • D
    In the exterior of S.
Answer
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MCQ 301 Mark
In the given figure, O is the centre of a circle. If $\angle\text{AOB}=100^\circ$ and $\angle\text{AOC}=90^\circ$ then $\angle\text{BAC}=?$
  • A
    85°
  • B
    80°
  • C
    95°
  • D
    75°

Answer
  1. 85°
    Solution:
    $\angle\text{BOA}+\angle\text{AOC}+\angle\text{BOC}=360^\circ$ [Angles around a point are 360°]
    $\Rightarrow\ 100^\circ+90^\circ+\angle\text{BOC}=360^\circ$
    $\Rightarrow\ \angle\text{BOC}=170^\circ$
    Now,
    $\angle\text{BAC}=\frac{1}{2}(\angle\text{BOC})=\frac{1}{2}(170^\circ)=85^\circ$
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MCQ 311 Mark
PS and RS are two chord's of a circle such that PQ = 10cm and RS = 24cm and PQ || RS. The distance between PQ and RS is 17cm. Find the radius of circle.
  • A
    13cm
  • B
    15cm
  • C
    None of these.
  • D
    10cm
Answer
  1. 13cm
    Solution:

    Let L and M be the midpoints of Rs and PQ respectively.
    Let OM = x thus, OL = 17 - x
    Now in triangle RLO, RL = 12 and
    RL2 + OL2 = r2
    122 + (17 - x)2 = r2 .....(i)
    Similarly,
    In triangle OMP, PM = 5 and
    PM2 + OM2 = OP2
    (x)2 + 52 = r2 .....(ii)
    Equating (i) and (ii), we get :-
    122 + (17 - x)2 = x2 + 52
    144 + 289 - 34x + x2 = x2 + 25 {using, (a - b)2}
    On solving, we get:
    $\text{x}=\frac{408}{34}=12$
    So, 17 - x = 17 - 12 = 5
    Thus, from(i),
    RL2 + OL2 = r2
    122 + 52 = r2
    144 + 25 = r2
    r2 = 169
    so, radius $=\sqrt{169}$
    Hence, the radius is 13cm.
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MCQ 331 Mark
Two equal circles of radius r intersect such that each passes through the centre of the other. The length of the common chord of the circles, is:
  • A
    $\sqrt{3}\text{r}$
  • B
    $\sqrt{2}\text{rAB}$
  • C
    $\frac{\sqrt{3}}{2}\text{r}$
  • D
    $\sqrt{\text{r}}$
Answer
  1. $\sqrt{3}\text{r}$
    Solution:
    We need to find a common chord.
    We have the corresponding figure as follows:

    AO = AO' = r (radius)
    And OO' = r
    So, $\triangle\text{OAO}'$ is an equilateral triangle.
    We know that the attitude of an equilateral triangle with side r is given by $\frac{2}{\sqrt{3}}$
    That is $\text{AM}=\frac{\sqrt{3}}{2\text{r}}$
    We know that the line joining centre of the circles divides the common chord into two equal parts.
    So we have
    $\text{AB}=2\text{AM}=2.\frac{\sqrt{3}}{2\text{r}}$
    $\text{AB}=\sqrt{3}\text{r}$
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MCQ 341 Mark
In the figure, O is the centre of the circle. If $\angle\text{OPQ}=25^\circ$ and $\angle\text{ORQ}=20^\circ,$ then the measures of $\angle\text{POR}$ and $\angle\text{PQR}$ are respectively:
  • A
    90º, 45º
  • B
    60º, 30º
  • C
    120º, 60º
  • D
    None of these.
Answer
  1. 90º, 45º
    Solution:
    Here, given
    OP = OQ and OR = OQ (Radius of circle)
    So, {angles opposite to equal sides are also equal}
    Hence,
    PQR = 25º + 20º = 45º
    and PQR = 2 PQR = 2 45º = 90º
    {Angle subtended by same sides on centre is double the angle at opposite vertex}
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MCQ 351 Mark
Let C be the mid-point of an arc AB of a circle such that $\text{m}\widehat{\text{AB}}=183^\circ.$ If the region bounded by the arc ACB and the line segment AB is denoted by S, then the centre O of the circle lies:
  • A
    In the interior of S.
  • B
    In the exertior of S.
  • C
    On the segment AB.
  • D
    On AB and bisects AB.
Answer
  1. In the interior of S.
    Solution:

    $\text{m}\widehat{\text{AB}}=183^\circ$
    O is the center of the circle and AB is a chord.
    The region bounded by arc and line segment AB is shaded.
    We can see, 'O', the center, always lie in the interior of S.
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MCQ 371 Mark
An equilateral triangle ABC is inscribed in a circle with centre O. The measures of $\angle\text{BOC}$ is:
  • A
    60º
  • B
    90º
  • C
    30º
  • D
    120º
Answer
  1. 120º
    Solution:
    We are given that an equilateral $\triangle\text{ABC}$ is inscribed in a circle with centre O. We need to find $\angle\text{BOC}.$
    We have the following corresponding figure.

    We are given AB = BC = AC
    Since the sides, AB, BC, and AC are these equal chords of the circle.
    Hence,
    $\angle\text{AOB}+\angle\text{BOC}+\angle\text{AOC}=360$
    $\Rightarrow\angle\text{BOC}+\angle\text{BOC}+\angle\text{BOC}=360$
    $\Rightarrow3\angle\text{BOC}=360$
    $\Rightarrow\angle\text{BOC}=\frac{360}{3}$
    $\Rightarrow\angle\text{BOC}=120^\circ$
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MCQ 381 Mark
AOB is a diameter of the circle and C, D, E are any three points on the semicircle. Then $\angle\text{AED}+\angle\text{BCD}$ is equal to:
  • A
    260º
  • B
    280º
  • C
    250º
  • D
    270º
Answer
  1. 270º
    Solution:

    Join OE, OD, OC.
    We get four triangles, namely, $\triangle\text{AOE},\triangle\text{EOD},\triangle\text{DOC},\triangle\text{COB}$
    Now, all these triangles are isosceles triangles.
    So,
    $\angle1=\angle2$
    $\angle3=\angle4$
    $\angle5=\angle6$
    $\angle7=\angle8$
    Now, required angle $\angle\text{AED}+\angle\text{BCD}=\angle2+\angle3+\angle6+\angle7$
    Now sum of all the angles must be 720º (as they are the angles of four triangles $\therefore\text{Sum}=180^\circ\times4=720^\circ$)
    Therefore,
    $\angle1+\angle2+\angle3+\angle4+\angle5+\angle6+\angle7+\angle8+\angle9+\angle10+\angle11+\angle12=720^\circ$
    $\Rightarrow2\angle2+2\angle3+2\angle6+2\angle7+\angle9+\angle10+\angle11+\angle12=720^\circ$
    $\Rightarrow2(\angle2+\angle3+\angle6+\angle7)+\angle9+\angle10+\angle11+\angle12=720^\circ$
    $\Rightarrow2(\angle2+\angle3+\angle6+\angle7)+180^\circ=720$
    $\Rightarrow2(\angle2+\angle3+\angle6+\angle7)=540^\circ$
    $\Rightarrow(\angle2+\angle3+\angle6+\angle7)=\frac{540^\circ}{2}=270^\circ$
    Thus, the required angle is equal to 270º
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MCQ 391 Mark
In the given figure, chords AD and BC intersect each other at right angles at a point P. If $\angle\text{DAB} = 35^\circ,$ then $\angle\text{ADC}=$
  • A
    35°
  • B
    45°
  • C
    55°
  • D
    65°
Answer
  1. 55°
    Solution:

    $\angle\text{APC}+\angle\text{APB}=180^\circ$
    $\Rightarrow\angle\text{APB}=180^\circ-90^\circ=90^\circ$
    In $\triangle\text{APB},$
    $\angle\text{ABP}=180^\circ-\angle\text{APB} -\angle\text{BAP}$
    $180^\circ-90^\circ-35^\circ=55^\circ$
    Now Arc $\widehat{\text{AC}}$ makes $\angle\text{ABC}$ and $\angle\text{ADC}$ on circle.
    $\Rightarrow\text{ABC}=\angle\text{ADC}$
    $\Rightarrow\angle\text{ADC}=55^\circ$
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MCQ 411 Mark
In the given figure, BOC is a diameter of a circle and AB = AC. Then, $\angle\text{ABC}=?$
  • A
    30°
  • B
    45°
  • C
    60°
  • D
    90°

Answer
  1. 45°
    Solution:
    Since BOC is a diameter of a circle, $\angle\text{BAC}$ is 90°.
    Given that AB = AC.
    $\Rightarrow\ \angle\text{ABC}=\angle\text{ACB}$
    In $\triangle\text{BAC},$
    $\angle\text{ABC}+\angle\text{ACB}+\angle\text{BAC}=180^\circ$ [Angle sum property]
    $\Rightarrow\ \angle\text{ABC}+\angle\text{ABC}+90^\circ=180^\circ$
    $\Rightarrow\ 2\angle\text{ABC}=90^\circ$
    $\Rightarrow\ \angle\text{ABC}=45^\circ$
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MCQ 431 Mark
If A and B are two points on a circle such that $\text{m}\big(\widehat{\text{AB}}\big)=260^\circ.$ A possible value for the angle subtended by arc BA at a point on the circle is:
  • A
    100°
  • B
    75°
  • C
    50°
  • D
    25°
Answer
  1. 50°
    Solution:

    $\text{m}\big(\widehat{\text{AB}}\big)=260^\circ$
    $\Rightarrow\text{m}\big(\widehat{\text{BA}}\big)=100^\circ$
    Now Let $\widehat{\text{BA}}$ subtend an angle $\theta$ at a point C on circle.
    Now, we know that angle subtend by an arc at the center is double the angle subtended at any point on the circle.
    $\Rightarrow100^\circ=2\theta$
    $\Rightarrow\theta=50^\circ$
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MCQ 451 Mark
In the given figure, O is the centre of a circle and diameter AB bisects the chord CD at a point E such that CE = ED = 8cm and EB = 4cm. The radius of the circle is:
  • A
    10cm
  • B
    12cm
  • C
    6cm
  • D
    8cm

Answer
  1. 10cm
    Solution:
    Construction: Join OD.

    OB = OD
    ⇒ OB = OE + EB
    ⇒ OB = OE + 4
    ⇒ OE = OB - 4 ...(i)
    In right $\triangle\text{OED},$
    OD2 = OE2 + DE2
    ⇒ OD2 = (OB - 4)2 + DE2 [From (i)]
    ⇒ OD2 = OB2 - 80A + 16 + 82
    ⇒ 8OD = 16 + 64
    ⇒ 8OD = 80
    $\Rightarrow\ \text{OD}=\frac{80}{8}=10\text{cm}$
    So, the radius of the circle is 10cm.
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MCQ 471 Mark
If AB is a chord of a circle, P and Q are the two points on the circle different from A and B, then:
  • A
    $\angle\text{APB}=\angle\text{AQB}$
  • B
    $\angle\text{APB}+\angle\text{AQB}=180^\circ\ \text{or }\angle\text{APB}=\angle\text{AQB}$
  • C
    $\angle\text{APB}+\angle\text{AQB}=90^\circ$
  • D
    $\angle\text{APB}+\angle\text{AQB}=180^\circ$
Answer
  1. $\angle\text{APB}+\angle\text{AQB}=180^\circ$
    Solution:

    $\angle\text{APB}$ and $\angle\text{AQB}$ are on the same arc.
    $\Rightarrow\angle\text{APB}=\angle\text{AQB}$
    But, if AB = diameter, then $\angle\text{APB}=\angle\text{AQB}=90^\circ$
    (Because diameter makes Right angle at any point on circumference of circle)
    $\angle\text{APB}+\angle\text{AQB}=180^\circ$
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MCQ 481 Mark
If ABC is an arc of a circle and $\angle\text{ABC}=135^\circ,$ then the ratio of arc ABC^ to the circumference, is:
  • A
    1 : 4
  • B
    1 : 2
  • C
    3 : 8
  • D
    3 : 4
Answer
  1. 3 : 8
    Solution:
    The length of an arc subtending an angle $\theta$ in a circle of radius r is given by the formula,
    Length of the arc $=\frac{\theta}{360^\circ}2\pi\text{r}$
    Here, it is given that the are subtends an angle of 135º with its centre. So the length of the given arc in a circle with radius r is given as
    Length of the arc $=\frac{135^\circ}{360^\circ}2\pi\text{r}\ ....(\text{i})$
    The circumference of the same circle with radius $\text{r}=2\pi\text{r}.\ ....(\text{ii})$
    The ratio between the lengths of the arc and the circumference of the circle will be
    $\frac{\text{Length of the arc}}{\text{Cirrumference of the circle}}=\frac{135^\circ(2\pi\text{r})}{360^\circ(2\pi\text{r})}=\frac{135^\circ}{360^\circ}=\frac{3}{8}$ [FROM (i) and (ii)]
    RATIO = 3 : 8
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MCQ 491 Mark
In the given figure, O is the centre of a circle. If $\angle\text{AOB}=100^\circ$ and $\angle\text{AOC}=90^\circ,$ then $\angle\text{BAC}=?$
  • A
    95º
  • B
    85º
  • C
    75º
  • D
    80º
Answer
  1. 85º
    Solution:
    We have:
    $\angle\text{BOC}+\angle\text{BOA}+\angle\text{AOC}=360^\circ$
    $\Rightarrow\angle\text{BOC}+100^\circ+90^\circ=360^\circ$
    $\Rightarrow\angle\text{BOC}=(360^\circ-190^\circ)=170^\circ$
    $\therefore\angle\text{BAC}=(\frac{1}{2}\times\angle\text{BOC})=(\frac{1}{2}\times170^\circ)=85^\circ$
    $\Rightarrow\angle\text{BAC}=85^\circ$
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MCQ 501 Mark
In the given figure, equilateral $\triangle\text{ABC}$ is inscribed in a circle and ABCD is a quadrilateral, as shown. Then, $\angle\text{BDC}=?$
  • A
    90°
  • B
    60°
  • C
    120°
  • D
    150°

Answer
  1. 120°
    Solution:
    Since $\triangle\text{BDC}$ is an equilateral traingle, $\angle\text{BAC}=60^\circ.$
    Since ABCD is a cyclic equilateral,
    $\angle\text{BAC}+\angle\text{BDC}=180^\circ$
    $\Rightarrow\ 60^\circ+\angle\text{BDC}=180^\circ$
    $\Rightarrow\ \angle\text{BDC}=120^\circ$
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M.C.Q - Maths STD 9 Questions - Vidyadip