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MATHS M.C.Q

Circles · Rajasthan Board (RBSE) STD 9 (Rajasthan - English Medium). 29 questions.

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M.C.Q

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29 questions · timed · auto-graded

Question 11 Mark
Number of circles that can be drawn through three non-collinear points is:
  1. 1
  2. 0
  3. 2
  4. 3
Answer
  1. 1
Solution:

Three non-collinear points make a triangle and there is only one circle that can pass through all three points,
i.e. circumcircle of that triangle.
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Question 21 Mark
In the given figure, if $\angle\text{ABC} = 45^\circ,$ then $\angle\text{AOC} =$
  1. 45°
  2. 60°
  3. 75°
  4. 90°
Answer
  1. 90°
Solution:

$\angle\text{AOC}$ is made by arc $\widehat{\text{AC}}$ at centre and $\angle\text{ABC}$ is made by $\widehat{\text{AC}}$ on circumference in major segment.
$\Rightarrow\angle\text{ABC}=\frac{1}{2}\angle\text{AOC}$
$\Rightarrow\angle\text{AOC}=2\times\angle\text{ABC}$
$=2\times45^\circ=90^\circ$
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Question 31 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 $\angle\text{ABC} =$
  1. 60°
  2. 75°
  3. 90°
  4. 135°
Answer
  1. 75°
Solution:

$\angle\text{AOC}=\angle\text{AOB}+\angle\text{BOC}$
$=90^\circ+120^\circ=210^\circ$
$\angle\text{COA}=360^\circ-210^\circ=150^\circ$
If arc $\widehat{\text{COA}}$ makes 150° at centre, then it will make half angle of the centre at circumference.
$\Rightarrow\angle\text{CBA}$ or $\angle\text{ABC}=\frac{150^\circ}{2}=75^\circ$
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MCQ 41 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
    $\sqrt{2}\text{r}\text{AB}$
  • ✓
    $\sqrt{3}\text{r}$
  • D
    $\frac{\sqrt3}{2}$
Answer
Correct option: C.
$\sqrt{3}\text{r}$

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 $O_1O_2)$
$\Rightarrow\text{AP}^2=\text{r}^2-\frac{\text{r}^2}{4}=\frac{\text{3r}^2}{4}$
$\Rightarrow\text{AP}=\frac{\sqrt3}{2}\text{r}$
Lenght of chord $\text{AP}=\text{2AP}=\sqrt{3}\text{r}$
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MCQ 51 Mark
The radius of a circle is $6\ cm$. The perpendicular distance from the centre of the circle to the chord which is $8\ cm$ in length, is:
  • A
    $\sqrt{5}\text{cm}.$
  • ✓
    $2\sqrt{5}\text{cm}.$
  • C
    $2\sqrt{7}\text{cm}.$
  • D
    $\sqrt{7}\text{cm}.$
Answer
Correct option: B.
$2\sqrt{5}\text{cm}.$

$AB = 8\ cm$
$\Rightarrow AC = BC = 4\ cm$
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 61 Mark
If O is the centre of a circle of radius r and AB is a chord of the circle at a distance $\frac{\text{r}}2{}$ from O, then $\angle\text{BAO} =$
  1. 60°
  2. 45°
  3. 30°
  4. 15°
Answer
  1. 30°
Solution:

Let $\angle\text{BAO}=\theta$
Consider $\triangle\text{OAC},$
$\sin\theta=\frac{\text{OC}}{\text{OA}}=\frac{\frac{\text{r}}{2}}{\text{r}}$
$=\frac{1}{2}=\sin30^\circ$
$\Rightarrow\theta=30^\circ$ 
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Question 71 Mark
In a circle with centre O, AB and CD are two diameters perpendicular to each other. The length of chord AC is:
  1. $2\text{AB}$
  2. $\sqrt{2}$
  3. $\frac{1}{2}\text{AB}$
  4. $\frac{1}{\sqrt{2}}\text{AB}$
Answer
  1. $\frac{1}{\sqrt{2}}\text{AB}$
Solution:

OC = OA = r (radius)
AB = Diameter = 2r
$\text{AC}=\sqrt{(\text{OA})^2+(\text{OC})^2}$
$=\sqrt{\text{r}^2+\text{r}^2}$
$=\sqrt{2}\text{r}$
$=\sqrt{2}\Big(\frac{\text{AB}}2{}\Big)$
$\Rightarrow\text{AC}=\frac{1}{\sqrt2}\text{AB}$
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Question 81 Mark
The greatest chord of a circle is called its:
  1. Radius.
  2. Secant.
  3. Diameter.
  4. None of these.
Answer
  1. Diameter.
Solution:
The greatest chord of the circle is diameter of the circle.
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Question 91 Mark
If two diameters of a circle intersect each other at right angles, then quadrilateral formed by joining their end points is a:
  1. Rhombus.
  2. Rectangle.
  3. Parallelogram.
  4. Square.
Answer
  1. Square.
Solution:

AB and CD are diameters of a circle and diameter makes 90° at any point on circle.
$\Rightarrow\angle\text{CAD}=\angle\text{CBD}=\angle\text{BCA}=\angle\text{ADB}=90^\circ$
Also, diagonals AB and CD are perpendicular to each other.
Thus, ABCD is a square.
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Question 101 Mark
Angle formed in minor segment of a circle is:
  1. Acute.
  2. Obtuse.
  3. Right angle.
  4. None of these.
Answer
  1. Obtuse.
Solution:

Angle formed in a minor segment is always a obtuse angle.
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Question 111 Mark
If the length of a chord of a circle is 16cm and is at a distance of 15cm from the centre of the circle, then the radius of the circle is:
  1. 15cm.
  2. 16cm.
  3. 17cm.
  4. 34cm.
Answer
  1. 17cm.
Solution:

AB = 16cm
OC = 15cm
C is the mid-point of AB.
$\text{AC}=\text{BC}=\frac{16}{2}=8\text{cm}$
Consider $\triangle\text{OCA},$
$\text{OC}=\text{15cm},\ \text{AC}=\text{8cm}$
$\Rightarrow\text{OA}=\sqrt{(15)^2+(8)^2}$
$=\sqrt{225-64}$
$=\sqrt{289}$
$\Rightarrow\text{OA}=17\text{cm}$
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Question 121 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 $\angle\text{ADC}=$
  1. 35°
  2. 45°
  3. 55°
  4. 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 131 Mark
$\text{ABC}$ is a triangle with $B$ as right angle, $AC = 5\ cm$ and $AB = 4\ cm$. A circle is drawn with $A$ as centre and $AC$ as radius. The length of the chord of this circle passing through $C$ and $B$ is:
  • A
    $3\ cm.$
  • B
    $4\ cm.$
  • C
    $5\ cm.$
  • ✓
    $6\ cm.$
Answer
Correct option: D.
$6\ cm.$

$AD$ and $AC$ are radii of same circle and $CD$ is a chord.
Consider $\triangle\text{ABC},$
$BC^2 = (AC)^2 - (AB)^2$
$=5^2 - 4^2 = 25 - 16 = 9$
$\Rightarrow BC = 3\ cm$
Chord $CD = 2 \times BC = 6\ cm$
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Question 141 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:
  1. 60°
  2. 75°
  3. 120°
  4. 150°
Answer
  1. 150°
Solution:

$\angle\text{AOB}=60^\circ$ $($Since $ \triangle\text{AOB}$ is equilateral triangle$)$
Now, $\angle\text{ADB}=30^\circ$
(Since chord AB makes 60 at centre, same chord will make half of the angle at circumference of angle made at centre)
Now $\angle\text{ACB}$ is angle made by chord at minor arc of circle.
ABCD is cyclic Quadrilateral.
$\Rightarrow\angle\text{C}+\angle\text{D}=180^\circ$
​​​​​​​$\Rightarrow\angle\text{ACB}+\angle\text{ADB}=180^\circ$
$\Rightarrow\angle\text{ACB}=180^\circ-30^\circ=150^\circ$
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Question 151 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:
  1. $6\text{cm}$
  2. $5\sqrt{2}\text{cm}$
  3. $7\text{cm}$
  4. $3\sqrt{5}\text{cm}$
Answer
  1.  $3\sqrt{5}\text{cm}$
Solution:

OB and OD are the radii of a circle.
In $\triangle\text{OED},$
$\text{r}^2=\text{OE}^2+\text{ED}^2=\text{OE}^2+(6)^2$
$\Rightarrow\text{OE}=\sqrt{\text{r}^2-36}\dots(1)$
In $\triangle\text{OFB},$
$\text{r}^2=\text{OF}^2+\text{BF}^2=\text{OF}^2+(3)^2$
$\Rightarrow\text{OF}=\sqrt{\text{r}^2-9}\dots(2)$
$\text{OF}-\text{OE}=3\text{cm}$(given)
$\sqrt{\text{r}^2-9}-\sqrt{\text{r}^2-36}=3$
$\sqrt{\text{r}^2-9}=\sqrt{\text{r}^2-36}+3\dots(3)$
Squaring equation (3), we have
$\text{r}^2-9=\text{r}^2-36+9+2\times3​​\sqrt{\text{r}^2-36}$
$\Rightarrow\text{r}^2-9=\text{r}^2-27+6​​\sqrt{\text{r}^2-36}$
$\Rightarrow18=6​​\sqrt{\text{r}^2-36}$
$\Rightarrow3=\sqrt{\text{r}^2-36}$
$\Rightarrow9=\text{r}^2-36$
$\Rightarrow\text{r}=\sqrt{45}=3\sqrt{5}\text{cm}$
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Question 161 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:
  1. In the interior of S.
  2. In the exertior of S.
  3. On the segment AB.
  4. 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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Question 171 Mark
If AB, BC and CD are equal chords of a circle with O as centre and AD diameter, than $\angle\text{AOB} =$
  1. 60°
  2. 90°
  3. 120°
  4. None of these.
Answer
  1. 60°
Solution:

Chord AB = Chord BC = Chord CD
$\Rightarrow\angle\text{AOB}=\angle\text{BOC}=\angle\text{COD}$ (equal chords subtend equal angles at the center)
Now, $\angle\text{AOB}+\angle\text{BOC}+\angle\text{COD}=180^\circ$
$\Rightarrow\angle\text{AOB}+\angle\text{AOB}+\angle\text{AOB}=180^\circ$
$\Rightarrow3\angle\text{AOB}=180^\circ$
$\Rightarrow\angle\text{AOB}=60^\circ$
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Question 181 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:
  1. 100°
  2. 75°
  3. 50°
  4. 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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Question 191 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:
  1. $\angle\text{APB}=\angle\text{AQB}$
  2. $\angle\text{APB}+\angle\text{AQB}=180^\circ\ \text{or }\angle\text{APB}=\angle\text{AQB}$
  3. $\angle\text{APB}+\angle\text{AQB}=90^\circ$
  4. $\angle\text{APB}+\angle\text{AQB}=180^\circ$
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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Question 201 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:
  1. 90° and 270°
  2. 90° and 90°
  3. 270° and 90°
  4. 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 211 Mark
In a circle of radius $17\ cm$, two parallel chords are drawn on opposite side of a diameter. The distance between the chords is $23\ cm.$ If the length of one chord is $16\ cm$, then the length of the other is:
  • A
    $34\ cm.$
  • B
    $15\ cm.$
  • C
    $23\ cm.$
  • ✓
    $30\ cm.$
Answer
Correct option: D.
$30\ cm.$

$PQ = 23\ cm$
$AB = 16\ cm$
$\Rightarrow BP = AP = 8\ cm$
$r = 17\ cm$
$\Rightarrow EF $= diameter $= 2r = 34\ cm$
Consider $\triangle\text{OPB},$
$r^2 = OP^2 + BP^2$
$\Rightarrow OP^2= (17)^2 - (8)^2 $
$= 289 - 64 $
$= 225$
$\Rightarrow OP = 15\ cm$
$\Rightarrow OQ = 23 - 15 = 8\ cm$
Consider $\triangle\text{OQD},$
$r^2 = OQ^2 + QD^2$
$\Rightarrow QD^2= r^2 - OQ^2 $
$= (17)^2 - (8)^2$
$= 225$
$\Rightarrow OD = 15\ cm$
$\Rightarrow CD = 2 \times QD = 30\ cm$
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Question 221 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:
  1. 1 : 4
  2. 3 : 4
  3. 3 : 8
  4. 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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Question 231 Mark
In the given figure, O is the centre of the circle and $\angle\text{BDC} = 42^\circ.$ The measure of $\angle\text{ACB}$ is:
  1. 42°
  2. 48°
  3. 58°
  4. 52°
Answer
  1. 48°
Solution:

$\angle\text{ABC}=90^\circ$ ...(Diameter AC makes 90° at circumference)
$\angle\text{CDB}=\angle\text{CAB}$ ...(Angles on the same arc)
$\Rightarrow\angle\text{CAB}=42^\circ$
In $\triangle\text{ABC},$
$\angle\text{ACB}=180^\circ-90^\circ-42^\circ=48^\circ$
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Question 241 Mark
An equilateral triangle ABC is inscribed in a circle with centre O. The measures of $\angle\text{BOC}$ is:
  1. 30°
  2. 60°
  3. 90°
  4. 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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Question 251 Mark
In the given figure, if chords AB and CD of the circle intersect each other at right angles, then x + y =
  1. 45°
  2. 60°
  3. 75°
  4. 90°
Answer
  1. 90°
Solution:

$\angle\text{CAB}=\angle\text{CDB}=\text{x}^\circ$ ...(Both are on the same arc)
Consider $\triangle\text{ODB},$
$\angle\text{DOB}=90^\circ,\ \angle\text{OBD}=\text{y},\ \angle\text{ODB}=\text{x}$
In $\triangle\text{ODB},$
x + y + 90° = 180°
⇒ x + y = 90°
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Question 261 Mark
In the given figure, O is the centre of the circle such that $\angle\text{AOC} = 130^\circ,$ then $\angle\text{ABC} =$
  1. 130°
  2. 115°
  3. 65°
  4. 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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Question 271 Mark
One chord of a circle is known to be 10cm. The radius of this circle must be:
  1. 5cm.
  2. Greater than 5cm.
  3. Greater than or equal to 5cm.
  4. Less than 5cm.
Answer
  1. Greater than 5cm.
Solution:
The longest chord of a circle is its diameter.
⇒ Diameter > 10cm
⇒ 2 × Radius > 10cm
⇒ Radius > 5cm
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Question 281 Mark
PQRS is a cyclic quadrilateral such that PR is a diameter of the circle. If $\angle\text{QPR} = 67^\circ$ and $ \angle\text{SPR} = 72^\circ,$ then $\angle\text{QRS} =$
  1. 41°
  2. 23°
  3. 67°
  4. 18°
Answer
  1. 41°
Solution:

In a cyclic quadrilateral, Opposite angles are supplementary.
$\Rightarrow\angle\text{P}+\angle\text{R}=180^\circ$
Now, $\angle\text{P}=67^\circ+72^\circ=139^\circ$
Thus, $\angle\text{R}=180^\circ-139^\circ=41^\circ$
i.e. $\angle\text{R}=\angle\text{QRS}=41^\circ$
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Question 291 Mark
ABCD is a cyclic quadrilateral such that $\angle\text{ADB} = 30^\circ$ and $\angle\text{DCA} = 80^\circ,$ then $\angle\text{DAB} =$
  1. 70°
  2. 100°
  3. 125°
  4. 150°
Answer
  1. 70°
Solution:

ABCD is a cyclic Quadrilateral.
Consider $\triangle\text{ABD}$ and $\triangle\text{ABC}.$
Both are on the same base AB and $\angle\text{ADB}$ and $\angle\text{ACB}$ are the angles in the same segment AB.
$\Rightarrow\angle\text{ADB}=\angle\text{ACB}=30^\circ$
$\Rightarrow\angle\text{BCD}=80^\circ+30^\circ=110^\circ$
In a cyclic Quadrilateral, sum of opposite angles is 180°
$\Rightarrow\angle\text{A}+\angle\text{C}=180^\circ$
$\Rightarrow\angle\text{DAB}+\angle\text{BCD}=180^\circ$
$\Rightarrow\angle\text{DAB}=180^\circ-110^\circ=70^\circ$
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