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MATHS M.C.Q

Circles · Rajasthan Board (RBSE) STD 9 (Rajasthan - English Medium). 249 questions.

Questions · Page 3 of 5

M.C.Q

Question 1011 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 $\angle\text{ABC}=$
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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Question 1021 Mark
In the given figure, ABCD is a cyclic quadrilateral, $\angle\text{CBQ}=48^\circ$ and a = 2b. Then, b is equal to:
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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Question 1031 Mark
If the length of a chord of a circle is $16\ cm$ and is at a distance of $15\ cm$ from the centre of the circle, then the radius of the circle is:
Answer
We will represent the given data in the figure

In the diagram$, AB$ is the given chord of $16\ cm$ 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 $\angle\text{OMA}=90^\circ$
$AO^2 = AM^2 + OM^2$
$AO^2 = 8^2 + 15^2$
$AO^2 = 64 + 225 = 289$
$\text{AO}=\sqrt{289}=17\text{cm}$
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Question 1041 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 = ?
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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Question 1061 Mark
Two circle are congruent if they have equal.
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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Question 1071 Mark
In the given figure if $OA = 5\ cm, AB = 8\ cm$ and $OD$ is perpendicular to $AB,$ then $CD$ is equal to:
Answer

$AC = 4\ cm$ and In triangle $\text{ACO},$
$AC^2 + OC^2 = AO^2$
$42 + OC^2 = 52$
$16 + OC^2 = 25$
$OC^2 = 25 - 16$
$OC^2 = 9$
$OC = 3$
Now $OD$ being the radius is $5\ cm$ and $OC$ is $3\ cm.$
So$, CD = OD - OC$
$ = 5 - 3 $
$= 2\ cm$
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Question 1081 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, $\angle\text{AOC}=?$
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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Question 1091 Mark
In the given figure, $\angle\text{AOB}=90^\circ$ and $\angle\text{ABC}=30^\circ.$ Then, $\angle\text{CAO}=?$
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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Question 1111 Mark
The greatest chord of a circle is called its:
Answer
  1. Diameter.
    Solution:
    The greatest chord of the circle is diameter of the circle.
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Question 1121 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, $\angle\text{BCD}=?$
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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Question 1131 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:
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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Question 1141 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:
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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Question 1151 Mark
AOB is the diameter of the circle. If $\angle\text{AOE}=150^\circ,$ then the measure of $\angle\text{CBE}$ is:
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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Question 1181 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:
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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Question 1201 Mark
In the given figure, O is the centre of a circle. If $\angle\text{OAC}=50^\circ$ then $\angle\text{ODB}=?$
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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Question 1211 Mark
Greatest chord of a circle is called its:
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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Question 1221 Mark
$AB$ and $CD$ are two parallel chords of a circle with centre $O$ such that $AB = 6\ cm$ and $CD = 12\ cm.$ The chords are on the same side of the centre and the distance between them is $3\ cm$. The radius of the circle, is:
Answer
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, $\text{OND},$
$x^2 + 36 = r^2 ....(i)$
Now, in right angled triangle $\text{AOM},$
$r^2 = 9 + (x + 3)^2 ....(ii)$
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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Question 1231 Mark
In the given figure, $CD$ is the diameter of a circle with centre $O$ and $CD$ is perpendicular to chord $AB. If AB = 12\ cm$ and $CE = 3\ cm,$ then radius of the circles is:
Answer
$OA = OC$
$\Rightarrow OA = OE + CE$
$\Rightarrow OA = OE + 3$
$\Rightarrow OE = OA - 3 ...(i)$
$\text{AE}=\frac{1}{2}\text{AB} [$Perpendicular drawn from the centre of a circle to the chord bisect the chord$]$
$=\frac{1}{2}(12)=6\text{cm}$
In right $\triangle\text{OEA},$
$OA^2 = OE^2 + AE^2$
$\Rightarrow OA^2 = (OA - 3)^2 + AE^2 [$From $(i)]$
$\Rightarrow OA^2 = OA^2 - 6OA + 9 + AE^2$
$\Rightarrow 6OA = 9 + 6^2$
$\Rightarrow 6OA = 9 + 36$
$\Rightarrow\ \text{OA}=\frac{45}{6}=7.5\text{cm}$
So, the radius of the circle is $7.5\ cm.$
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Question 1241 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, $\angle\text{CBD}=?$
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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Question 1261 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, $\angle\text{CBD}=?$
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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Question 1271 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:
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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Question 1291 Mark
A chord is at a distance of $8\ cm$ from the centre of a circle of radius $17\ cm.$ The length of the chord is:
Answer

Let O be the centre of the circle with radius $OA = 17\ cm.$
Since $\text{OC}\perp\text{AB}.$
In right $\triangle\text{OCA},$
$OA^2= OC^2 + AC^2\  [$By pythagoras theorem$]$
$AC^2= OA^2 - OC^2$
$\Rightarrow AC^2 = 17^2 - 8^2$
$\Rightarrow AC^2 = 289 - 64$
$\Rightarrow AC^2 = 225$
$\Rightarrow AC = 15\ cm$
We know that, the perpendicular drawn from the centre to the chord bisects the chord.
$\Rightarrow AB = 2AC = 2(15) = 30\ cm.$
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Question 1311 Mark
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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Question 1331 Mark
In the given figure$, CD$ is the diameter of a circle with centre $O$ and $CD$ is perpendicular to chord $AB.$ If $AB = 12\ cm$ and $CE = 3\ cm,$ then radius of the circles is:
Answer
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:
$OA^2 = OE^2+ AE^2\  [$Using paythagoras theorem$]$
$\Rightarrow (r)^2= (r - 3)^2 + 6^2$
$\Rightarrow r^2 = r^2 + 9 - 6r + 36$
$\Rightarrow 6r = 45$
$\Rightarrow\text{r}=\frac{45}{6}$
$=7.5\text{ cm}$
Hence, the required radius of the circle is $7.5\ cm.$
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Question 1341 Mark
In the given figure, O is the centre of a circle and $\angle\text{AOB}=140^\circ.$ Then, $\angle\text{ACB}=?$
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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Question 1351 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:
Answer

Both the circles pass through the centre of each other
$\Rightarrow O_1O_2 = r$
Common chord is $AB$
We know that perpendicular drawn from centre of circle to any chord bisects it.
$\Rightarrow P$ is the midpoint of $AB$
$\Rightarrow PA = PB$
$O_1A = r ($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 $O_1\ O_2)$
$\Rightarrow\text{AP}^2=\text{r}^2-\frac{\text{r}^2}{4}=\frac{\text{3r}^2}{4}$
$\Rightarrow\text{AP}=\frac{\sqrt3}{2}\text{r}$
Lenght of chord $\text{AP}=\text{2AP}=\sqrt{3}\text{r}$
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Question 1361 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:
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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Question 1371 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, $\angle\text{AOC}=?$
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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Question 1381 Mark
The radius of a circle is $6\ cm.$ The perpendicular distance from the centre oaf the circle to the chord which is $8\ cm$ in length, is:
Answer

$AB = 8\ cm$
$\Rightarrow AC = BC = 4\ cm$
Consider $\triangle\text{OCB},$ where $BC = 8\ cm,$
$OB = 6\ cm$
Now$, (OC)^2 + (BC)^2= (OB)^2$
$\Rightarrow (OC)^2 + 4^2 = 6^2$
$\Rightarrow (OC)^2 + 16 = 36$
$\Rightarrow (OC)^2 = 20$
$\Rightarrow\text{OC}=\sqrt{20}=2\sqrt{5}$
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Question 1391 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 $\angle\text{CBD}=?$
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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Question 1451 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 $\angle\text{BAC}=?$
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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Question 1461 Mark
$PS$ and $RS$ are two chord's of a circle such that $PQ = 10\ cm$ and $RS = 24\ cm$ and $PQ || RS$. The distance between $PQ$ and $RS$ is $17\ cm.$ Find the radius of circle.
Answer

Let $L$ and $M$ be the midpoints of $Rs$ and $PQ$ respectively.
Let $OM = x $thus$, OL = 17 - x$
Now in triangle $\text{RLO}, RL = 12$ and
$RL2 + OL2 = r2$
$122 + (17 - x)2 = r2 .....(i)$
Similarly,
In triangle $\text{OMP}, PM = 5$ 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 $13\ cm.$
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Question 1481 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:
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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Question 1491 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:
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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Question 1501 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:
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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M.C.Q - Page 3 - MATHS STD 9 Questions - Vidyadip