MCQ 11 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:
- A
$15\ cm.$
- B
$16\ cm.$
- ✓
$17\ cm.$
- D
$34\ cm.$
AnswerCorrect option: C. $17\ cm.$
$AB = 16\ cm$
$OC = 15\ cm$
$C$ is the mid-point of $AB$.
$\text{AC}=\text{BC}=\frac{16}{2}=8\text{cm}$
Consider $\triangle\text{OCA},$
$\text{OC}=\text{15cm},\ \text{AC}=\text{8cm}$
$\Rightarrow\text{OA}=\sqrt{(15)^2+(8)^2}$
$=\sqrt{225-64}$
$=\sqrt{289}$
$\Rightarrow\text{OA}=17\text{cm}$ View full question & answer→MCQ 21 Mark
In the given figure, if $\angle\text{ABC} = 45^\circ,$ then $\angle\text{AOC} =$
- A
$45^\circ $
- B
$60^\circ$
- C
$75^\circ$
- ✓
$90^\circ$
AnswerCorrect option: D. $90^\circ$
$\angle\text{AOC}$ is made by arc $\widehat{\text{AC}}$ at centre and $\angle\text{ABC}$ is made by $\widehat{\text{AC}}$ on circumference in major segment.
$\Rightarrow\angle\text{ABC}=\frac{1}{2}\angle\text{AOC}$
$\Rightarrow\angle\text{AOC}=2\times\angle\text{ABC}$
$=2\times45^\circ=90^\circ$ View full question & answer→MCQ 31 Mark
$ABC$ 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
$5\ cm.$
- ✓
$6\ cm.$
AnswerCorrect option: D. $6\ cm.$
$A D$ and $A C$ are radii of same circle and $C D$ is a chord.
Consider $\triangle \mathrm{ABC}$,
$B C^2=(A C)^2-(A B)^2$
$=5^2-4^2=25-16=9$
$\Rightarrow B C=3 \mathrm{~cm}$
Chord $C D=2 \times B C=6 \mathrm{~cm}$
View full question & answer→MCQ 41 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 $\angle\text{ABC} =$
- A
$60^\circ $
- ✓
$75^\circ$
- C
$90^\circ$
- D
$135^\circ$
AnswerCorrect option: B. $75^\circ$
$\angle\text{AOC}=\angle\text{AOB}+\angle\text{BOC}$
$=90^\circ+120^\circ=210^\circ$
$\angle\text{COA}=360^\circ-210^\circ=150^\circ$
If arc $\widehat{\text{COA}}$ makes $150^\circ$ at centre, then it will make half angle of the centre at circumference.
$\Rightarrow\angle\text{CBA}$ or $\angle\text{ABC}=\frac{150^\circ}{2}=75^\circ$ View full question & answer→MCQ 51 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:
AnswerCorrect option: C. $\sqrt{3}\text{r}$
Both the circles pass through the centre of each other
$\Rightarrow \mathrm{O}_1 \mathrm{O}_2=\mathrm{r}$
Common chord is $AB$
We know that perpendicular drawn from centre of circle to any chord bisects it.
$\Rightarrow P$ is the midpoint of $A B$
$\Rightarrow \mathrm{PA}=\mathrm{PB}$
$\mathrm{O}_1 \mathrm{~A}=\mathrm{r}$ (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 $\mathrm{O_1O_2}$)
$\Rightarrow\text{AP}^2=\text{r}^2-\frac{\text{r}^2}{4}=\frac{\text{3r}^2}{4}$
$\Rightarrow\text{AP}=\frac{\sqrt3}{2}\text{r}$
Lenght of chord $\text{AP}=\text{2AP}=\sqrt{3}\text{r}$
View full question & answer→MCQ 61 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
$\sqrt{7}\text{cm}.$
AnswerCorrect option: B. $2\sqrt{5}\text{cm}.$
$A B=8 \mathrm{~cm}$
$\Rightarrow A C=B C=4 \mathrm{~cm}$
Consider $\triangle \mathrm{OCB}$, where $\mathrm{BC}=8 \mathrm{~cm}$,
$O B=6 \mathrm{~cm}$
$\text { Now, }(O C)^2+(B C)^2=(O B)^2$
$\Rightarrow(O C)^2+4^2=6^2$
$\Rightarrow(O C)^2+16=36$
$\Rightarrow(O C)^2=20$
$\Rightarrow O C=\sqrt{20}=2 \sqrt{5}$
View full question & answer→MCQ 71 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 $\angle\text{BAO} =$
- A
$60^\circ$
- B
$45^\circ$
- ✓
$30^\circ$
- D
$15^\circ$
AnswerCorrect option: C. $30^\circ$
Let $\angle\text{BAO}=\theta$
Consider $\triangle\text{OAC},$
$\sin\theta=\frac{\text{OC}}{\text{OA}}=\frac{\frac{\text{r}}{2}}{\text{r}}$
$=\frac{1}{2}=\sin30^\circ$
$\Rightarrow\theta=30^\circ$ View full question & answer→MCQ 81 Mark
In a circle with centre $O, AB$ and $CD$ are two diameters perpendicular to each other. The length of chord $AC$ is:
AnswerCorrect option: D. $\frac{1}{\sqrt{2}}\text{AB}$
$OC = OA = r$ (radius)
$AB$ = Diameter $= 2r$
$\text{AC}=\sqrt{(\text{OA})^2+(\text{OC})^2}$
$=\sqrt{\text{r}^2+\text{r}^2}$
$=\sqrt{2}\text{r}$
$=\sqrt{2}\Big(\frac{\text{AB}}2{}\Big)$
$\Rightarrow\text{AC}=\frac{1}{\sqrt2}\text{AB}$ View full question & answer→MCQ 91 Mark
The greatest chord of a circle is called its:
AnswerThe greatest chord of the circle is diameter of the circle.
View full question & answer→MCQ 101 Mark
If two diameters of a circle intersect each other at right angles, then quadrilateral formed by joining their end points is a:
Answer
$AB$ and $CD$ are diameters of a circle and diameter makes $90^\circ$ 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. View full question & answer→MCQ 111 Mark
If $AB, BC$ and $CD$ are equal chords of a circle with $O$ as centre and $AD$ diameter, than $\angle\text{AOB} =$
- ✓
$60^\circ$
- B
$90^\circ$
- C
$120^\circ$
- D
AnswerCorrect option: A. $60^\circ$
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, $\angle\text{AOB}+\angle\text{BOC}+\angle\text{COD}=180^\circ$
$\Rightarrow\angle\text{AOB}+\angle\text{AOB}+\angle\text{AOB}=180^\circ$
$\Rightarrow3\angle\text{AOB}=180^\circ$
$\Rightarrow\angle\text{AOB}=60^\circ$ View full question & answer→MCQ 121 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:
AnswerCorrect option: A. In the interior of $S$.
$\text{m}\widehat{\text{AB}}=183^\circ$
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 $S$. View full question & answer→MCQ 131 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^\circ $
- B
$45^\circ$
- ✓
$55^\circ$
- D
$65^\circ$
AnswerCorrect option: C. $55^\circ$
$\angle\text{APC}+\angle\text{APB}=180^\circ$
$\Rightarrow\angle\text{APB}=180^\circ-90^\circ=90^\circ$
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.
$\Rightarrow\text{ABC}=\angle\text{ADC}$
$\Rightarrow\angle\text{ADC}=55^\circ$ View full question & answer→MCQ 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:
- A
$60^\circ$
- B
$75^\circ$
- C
$120^\circ$
- ✓
$150^\circ$
AnswerCorrect option: D. $150^\circ$
$\angle\text{AOB}=60^\circ$ $($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.
$\Rightarrow\angle\text{C}+\angle\text{D}=180^\circ$
$\Rightarrow\angle\text{ACB}+\angle\text{ADB}=180^\circ$
$\Rightarrow\angle\text{ACB}=180^\circ-30^\circ=150^\circ$ View full question & answer→MCQ 151 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:
- A
$6\text{cm}$
- B
$5\sqrt{2}\text{cm}$
- C
$7\text{cm}$
- ✓
$3\sqrt{5}\text{cm}$
AnswerCorrect option: D. $3\sqrt{5}\text{cm}$
$OB$ 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
$\text{r}^2-9=\text{r}^2-36+9+2\times3\sqrt{\text{r}^2-36}$
$\Rightarrow\text{r}^2-9=\text{r}^2-27+6\sqrt{\text{r}^2-36}$
$\Rightarrow18=6\sqrt{\text{r}^2-36}$
$\Rightarrow3=\sqrt{\text{r}^2-36}$
$\Rightarrow9=\text{r}^2-36$
$\Rightarrow\text{r}=\sqrt{45}=3\sqrt{5}\text{cm}$ View full question & answer→MCQ 161 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
$100^\circ$
- B
$75^\circ$
- ✓
$50^\circ$
- D
$25^\circ$
AnswerCorrect option: C. $50^\circ$
$\text{m}\big(\widehat{\text{AB}}\big)=260^\circ$
$\Rightarrow\text{m}\big(\widehat{\text{BA}}\big)=100^\circ$
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.
$\Rightarrow100^\circ=2\theta$
$\Rightarrow\theta=50^\circ$ View full question & answer→MCQ 171 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
$\angle\text{APB}=\angle\text{AQB}$
- ✓
$\angle\text{APB}+\angle\text{AQB}=180^\circ\ \text{or }\angle\text{APB}=\angle\text{AQB}$
- C
$\angle\text{APB}+\angle\text{AQB}=90^\circ$
- D
$\angle\text{APB}+\angle\text{AQB}=180^\circ$
AnswerCorrect option: B. $\angle\text{APB}+\angle\text{AQB}=180^\circ\ \text{or }\angle\text{APB}=\angle\text{AQB}$
$\angle\text{APB}$ 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)
$\angle\text{APB}+\angle\text{AQB}=180^\circ$ View full question & answer→MCQ 181 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:
- ✓
$90^\circ$ and $270^\circ $
- B
$90^\circ$ and $90^\circ$
- C
$270^\circ$ and $90^\circ $
- D
$60^\circ$ and $210^\circ $
AnswerCorrect option: A. $90^\circ$ and $270^\circ $
$\frac{\widehat{\text{AB}}\text{minor}}{\widehat{\text{AB}}\text{major}}=\frac{1}{3}=\frac{\angle\widehat{\text{AB}}\text{minor}}{\angle\widehat{\text{AB}}\text{major}}$
Let $\angle\widehat{\text{AB}}\text{minor}=\text{x}$
$\Rightarrow\angle\widehat{\text{AB}}\text{major}=\text{3x}$
Now we know $x + 3x = 360^\circ $
$\Rightarrow 4x = 360^\circ $
$\Rightarrow x = 90^\circ $
$\Rightarrow 3x = 270^\circ $ View full question & answer→MCQ 191 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
$23\ cm.$
- ✓
$30\ cm.$
AnswerCorrect option: D. $30\ cm.$
$\mathrm{PQ}=23 \mathrm{~cm}$
$\mathrm{AB}=16 \mathrm{~cm}$
$\Rightarrow B P=A P=8 \mathrm{~cm}$
$\mathrm{r}=17 \mathrm{~cm}$
$\Rightarrow E F=\text { diameter }=2 \mathrm{r}=34 \mathrm{~cm}$
Consider $\triangle \mathrm{OPB}$,
$r^2=O P^2+B P^2$
$\Rightarrow O P^2=(17)^2-(8)^2=289-64=225$
$\Rightarrow O P=15 \mathrm{~cm}$
$\Rightarrow O Q=23-15=8 \mathrm{~cm}$
Consider $\triangle \mathrm{OQD}$,
$r^2=O Q^2+Q D^2$
$\Rightarrow Q D^2=r^2-O Q^2=(17)^2-(8)^2=225$
$\Rightarrow O D=15 \mathrm{~cm}$
$\Rightarrow C D=2 \times Q D=30 \mathrm{~cm}$
View full question & answer→MCQ 201 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:
- ✓
$1 : 4$
- B
$3 : 4$
- C
$3 : 8$
- D
$1 : 2$
AnswerCorrect option: A. $1 : 4$
$ABC$ 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 $=\angle\text{AOC}=90^\circ$
$\Rightarrow\text{Required ratio}=\frac{\text{arc}\widehat{\text{ABC}}}{\text{circumference}}$
$=\frac{90^\circ}{360^\circ}=\frac{1}{4}=1:4$ View full question & answer→MCQ 211 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:
- A
$42^\circ$
- ✓
$48^\circ$
- C
$58^\circ$
- D
$52^\circ$
AnswerCorrect option: B. $48^\circ$
$\angle\text{ABC}=90^\circ$ ...(Diameter $AC$ makes $90^\circ$ at circumference)
$\angle\text{CDB}=\angle\text{CAB}$ ...(Angles on the same arc)
$\Rightarrow\angle\text{CAB}=42^\circ$
In $\triangle\text{ABC},$
$\angle\text{ACB}=180^\circ-90^\circ-42^\circ=48^\circ$ View full question & answer→MCQ 221 Mark
An equilateral triangle $ABC$ is inscribed in a circle with centre $O$. The measures of $\angle\text{BOC}$ is:
- A
$30^\circ$
- B
$60^\circ$
- C
$90^\circ$
- ✓
$120^\circ$
AnswerCorrect option: D. $120^\circ$
$\angle\text{BAC}=60^\circ$ (Angle of equilateral triangle)
Arc $\widehat{\text{BC}}$ makes angle $\angle\text{BAC}$ at circle and $\angle\text{BOC}$ at center of circle.
$\Rightarrow\angle\text{BAC}=\frac{1}{2}\angle\text{BOC}$
$\Rightarrow2\times\angle\text{BAC}=\angle\text{BOC}$
$\Rightarrow2\times60^\circ=\angle\text{BOC}$
$\Rightarrow\angle\text{BOC}=120^\circ$ View full question & answer→MCQ 231 Mark
In the given figure, if chords $AB$ and $CD$ of the circle intersect each other at right angles, then $x + y =$
- A
$45^\circ $
- B
$60^\circ $
- C
$75^\circ $
- ✓
$90^\circ $
AnswerCorrect option: D. $90^\circ $
$\angle\text{CAB}=\angle\text{CDB}=\text{x}^\circ$ ...(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^\circ = 180^\circ $
$\Rightarrow x + y = 90^\circ $ View full question & answer→MCQ 241 Mark
Angle formed in minor segment of a circle is:
Answer
Angle formed in a minor segment is always a obtuse angle.
View full question & answer→MCQ 251 Mark
Number of circles that can be drawn through three non-collinear points is:
Answer
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.
View full question & answer→MCQ 261 Mark
In the given figure, $O$ is the centre of the circle such that $\angle\text{AOC} = 130^\circ,$ then $\angle\text{ABC} =$
- A
$130^\circ$
- ✓
$115^\circ $
- C
$65^\circ$
- D
$165^\circ $
AnswerCorrect option: B. $115^\circ $
$\angle\text{ADC}=\frac{1}{2}\angle\text{AOC}$
$\big\{\angle\text{ADC}$ 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.
$\Rightarrow\angle\text{D}+\angle\text{B}=180^\circ$
$\Rightarrow\angle\text{ABC}=180^\circ-65^\circ=115^\circ$ View full question & answer→MCQ 271 Mark
One chord of a circle is known to be $10\ cm$. The radius of this circle must be:
AnswerCorrect option: B. Greater than $5\ cm.$
The longest chord of a circle is its diameter.
$\Rightarrow $ Diameter $> 10\ cm$
$\Rightarrow 2 \times $ Radius $> 10\ cm$
$\Rightarrow $ Radius $> 5\ cm$
View full question & answer→MCQ 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 $\angle\text{QRS} =$
- ✓
$41^\circ $
- B
$23^\circ$
- C
$67^\circ$
- D
$18^\circ$
AnswerCorrect option: A. $41^\circ $
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. $\angle\text{R}=\angle\text{QRS}=41^\circ$ View full question & answer→MCQ 291 Mark
$ABCD$ is a cyclic quadrilateral such that $\angle\text{ADB} = 30^\circ$ and $\angle\text{DCA} = 80^\circ,$ then $\angle\text{DAB} =$
- ✓
$70^\circ$
- B
$100^\circ$
- C
$125^\circ$
- D
$150^\circ$
AnswerCorrect option: A. $70^\circ$
$ABCD$ 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 $180^\circ$
$\Rightarrow\angle\text{A}+\angle\text{C}=180^\circ$
$\Rightarrow\angle\text{DAB}+\angle\text{BCD}=180^\circ$
$\Rightarrow\angle\text{DAB}=180^\circ-110^\circ=70^\circ$ View full question & answer→MCQ 301 Mark
- A
$30^{\circ}$
- B
$45^{\circ}$
- ✓
$60^{\circ}$
- D
$120^{\circ}$
AnswerCorrect option: C. $60^{\circ}$
(c) $60^{\circ}$
We observe that $\angle B A C$ is the angle in a semi-circle. Therefore, $\angle B A C=90^{\circ}$.
$\therefore \quad \angle O A C=\angle B A C-\angle O A B=90^{\circ}-60^{\circ}-30^{\circ}$
In $\triangle O A C$, we have
$O A=O C \Rightarrow \angle O A C=\angle O C A \Rightarrow \angle O C A=30^{\circ}$
Thus, in $\triangle O A C$, we have
$\begin{array}{ll}& \angle O A C=\angle O C A=30^{\circ} \\
\therefore \quad & \angle A O C=180^{\circ}-(\angle O A C+\angle O C A)=180^{\circ}
\left(30^{\circ}+30^{\circ}\right)=120^{\circ}\end{array}$
Clearly, arc $A C$ makes angle $\angle A O C=120^{\circ}$ at the centre $O$ and $\angle A D C$ at point on the remaining part of the circle.
$\therefore \quad \angle A D C=\frac{1}{2} \angle A O C=\frac{1}{2} \times 120^{\circ}=60^{\circ}$
View full question & answer→MCQ 311 Mark
- A
$30^{\circ}$
- B
$45^{\circ}$
- C
$90^{\circ}$
- ✓
$60^{\circ}$
AnswerCorrect option: D. $60^{\circ}$
(d) $60^{\circ}$
In $\triangle A O B$, we find that
$\begin{array}{ll}& O A=O B \\
\Rightarrow \quad & \angle O A B=\angle O B A\end{array}$
Using angle sum property in $\triangle O A B$, we obtain
$\begin{array}{ll}& \angle O A B+\angle O B A+\angle A O B=180^{\circ} \\
\Rightarrow \quad & 2 \angle O A B+90^{\circ}=180^{\circ} \Rightarrow \angle O A B=45^{\circ} \Rightarrow \angle O BA=45^{\circ}\end{array}$
Arc $A B$ subtends $\angle A O B=90^{\circ}$ at the centre $O$ and $\angle A C B$ at a point on the remaining part of the circle.
$\therefore \quad \angle A C B=\frac{1}{2} \angle A O B=\frac{1}{2} \times 90^{\circ}=45^{\circ}$
Thus, in $\triangle A B C$, we have $\angle A B C=30^{\circ}$ and $\angle A C B=45^{\circ}$.
$\begin{array}{ll}\therefore & \angle B A C=180^{\circ}-30^{\circ}-45^{\circ}=105^{\circ} \quad \text { [Using angle sum property in } \triangle A B C \text { ] } \\
\Rightarrow & \angle C A O+\angle O A B=105^{\circ} \Rightarrow \angle C A O+45^{\circ}=105^{\circ} \Rightarrow
\angle C A O=60^{\circ}\end{array}$
View full question & answer→MCQ 321 Mark
- A
$60^{\circ}$
- B
$50^{\circ}$
- ✓
$70^{\circ}$
- D
$80^{\circ}$
AnswerCorrect option: C. $70^{\circ}$
(c) $70^{\circ}$
In $\triangle A B D$, we have
$\begin{array}{l}\angle D A B=60^{\circ} \text { and } \angle A B D=50^{\circ} \\
\angle A D B=180^{\circ}-60^{\circ}-50^{\circ}=70^{\circ}\end{array}$
We find that $\angle A D B$ and $\angle A C B$ are angles made by the are $A B$ in the same segment.
$
∴\quad$ \angle A C B=\angle A D B \Rightarrow \angle A C B=70^{\circ}$
View full question & answer→MCQ 331 Mark
- ✓
$50^{\circ}$
- B
$40^{\circ}$
- C
$60^{\circ}$
- D
$70^{\circ}$
AnswerCorrect option: A. $50^{\circ}$
(a) $50^{\circ}$
In $\triangle O A B$, we have
$O A=O B \Rightarrow \angle O B A=\angle O A B \Rightarrow \angle O B A=40^{\circ}$
Using angle sum property in $\triangle A O B$, we obtain
$\begin{array}{ll}& \angle O A B+\angle O B A+\angle A O B=180^{\circ} \\
\Rightarrow & 40^{\circ}+40^{\circ}+\angle A O B=180^{\circ} \Rightarrow \angle A O B=100^{\circ}\end{array}$
Thus, $\operatorname{arc} A B$ subtends $\angle A O B=100^{\circ}$ at the centre and $\angle A C B$ at a point on the circumference.
$\therefore \quad \angle A C B=\frac{1}{2} \angle A O B=50^{\circ}$
View full question & answer→MCQ 341 Mark
- A
$30^{\circ}$
- B
$60^{\circ}$
- C
$90^{\circ}$
- ✓
$45^{\circ}$
AnswerCorrect option: D. $45^{\circ}$
(d) $45^{\circ}$
We find that angle in a semi-circle is a right angle. Therefore, $\angle A C B=90^{\circ}$.
It is given that
$A C=B C \Rightarrow \angle C A B=\angle C B A$
Using angle sum property in $\triangle A B C$, we obtain
$\angle A C B+\angle C A B+\angle C B A=180^{\circ} \Rightarrow 90^{\circ}+2 \angle C A B=180^{\circ} \Rightarrow\angle C A B=45^{\circ}$
View full question & answer→MCQ 351 Mark
- A
$20^{\circ}$
- ✓
$40^{\circ}$
- C
$60^{\circ}$
- D
$10^{\circ}$
AnswerCorrect option: B. $40^{\circ}$
(b) $40^{\circ}$
We observe that arc AC makes angle $\angle A B C=20^{\circ}$ at a point B on the circumference of the circle and $\angle A O C$ at the centre O.
$\therefore \quad \angle A O C=2 \angle A B C=2 \times 20^{\circ}=40^{\circ}$
View full question & answer→MCQ 361 Mark
If $A B=12 cm, B C=16 cm$ and $A B$ is perpendicular to $B C$, then the radius of the circle passing through the points A, B and C is
Answer(c) 10 cm
Perpendicular from the centre to a chord bisects the chord. Therefore, $L$ and $M$ are mid-points of $A B$ and $B C$ respectively. Thus, in right triangle $O L B$, we have
$O L=B M=\frac{1}{2} B C=8 cm \text { and } B L=\frac{1}{2} A B=6 cm$
Applying Pythagoras theorem in $\triangle O L B$, we obtain
$O B^2=O L^2+L B^2 \Rightarrow O B=\sqrt{8^2+6^2}=\sqrt{100}=10 cm$
View full question & answer→MCQ 371 Mark
AD is a diameter of a circle and AB is a chord. If AD = 34 cm, AB = 30 cm, the distance of AB from the centre of the circle is
Answer(d) 8 cm
Let O be the centre of the circle and $O L \perp A B$. Then, $L$ is the midpoint of $A B$. We have, $O A=17 cm$ and $A L=15 cm$.
Applying Pythagoras theorem in $\triangle O L A$, we obtain
$\begin{array}{ll}& O A^2=O L^2+A L^2 \\
\Rightarrow \quad & O L=\sqrt{O A^2-O L^2}=\sqrt{17^2-15^2}=\sqrt{289-225}=\sqrt{64}=8 cm\end{array}$
View full question & answer→MCQ 381 Mark
- A
$2: 1$
- B
$1: 2$
- ✓
$3: 1$
- D
$1: 3$
AnswerCorrect option: C. $3: 1$
(c) $3: 1$
$\frac{\operatorname{arc} A X B}{\operatorname{arc} A^{\prime} Y^{\prime} B}=\frac{m(\widehat{A X B})}{m\left(A^{\prime} X B^{\prime}\right)}=\frac{75^{\circ}}{25^{\circ}}=\frac{3}{1}$
View full question & answer→MCQ 391 Mark
- A
$30^{\circ}$
- ✓
$45^{\circ}$
- C
$15^{\circ}$
- D
$60^{\circ}$
AnswerCorrect option: B. $45^{\circ}$
(b) $15^{\circ}$
Equal chords of a circle are equidistant from the centre. Therefore, in $\triangle O P Q$, we have
$O P=O Q \Rightarrow \angle O P Q=\angle O Q P$
Using angle sum property in $\triangle O P Q$, we obtain
$\begin{array}{ll}& \angle O P Q+\angle O Q P+\angle P O Q=180^{\circ} \\
\Rightarrow & 2 \angle O P Q+150^{\circ}=180^{\circ} \Rightarrow \angle O P Q=15^{\circ} \\
\therefore & \angle A P Q=\angle O P A-\angle O P Q=90^{\circ}-15^{\circ}=75^{\circ}\end{array}$
View full question & answer→MCQ 401 Mark
- ✓
$30^{\circ}$
- B
$45^{\circ}$
- C
$60^{\circ}$
- D
$50^{\circ}$
AnswerCorrect option: A. $30^{\circ}$
(a) $30^{\circ}$
Given that $A B \| C D$ and $\angle B A D=30^{\circ}$. Therefore, $\angle A D C=30^{\circ}$.
AB is a diameter and D is a point on the circle. Therefore, $\angle A D B=90^{\circ}$.
$\therefore \quad \angle B D C=\angle A D B+\angle A D C=90^{\circ}+30^{\circ}=120^{\circ}$
$A B C D$ is a cyclic quadrilateral.
$\therefore \quad \angle B D C+\angle B A C=180^{\circ} \Rightarrow \angle B A C=180^{\circ}-120^{\circ}=60^{\circ}$
Hence, $\angle C A D=\angle B A C-\angle B A D=60^{\circ}-30^{\circ}=30^{\circ}$
View full question & answer→MCQ 411 Mark
- A
$25^{\circ}$
- B
$80^{\circ}$
- ✓
$50^{\circ}$
- D
$40^{\circ}$
AnswerCorrect option: C. $50^{\circ}$
(c) $50^{\circ}$
Let $E$ be a point on the circle. We observe that arc ABC makes $\angle A O C=100^{\circ}$ at O and $\angle A E C$ at E.
$\therefore \quad \angle A E C=\frac{1}{2} \angle A O C=\frac{1}{2} \times 100^{\circ}=50^{\circ}$
Using exterior angle property in cyclic quadrilateral $A B C E$, we obtain
$\text { ext } \angle C B D=\angle A E C=50^{\circ}$
View full question & answer→MCQ 421 Mark
- A
$95^{\circ}$
- B
$85^{\circ}$
- ✓
$105^{\circ}$
- D
$75^{\circ}$
AnswerCorrect option: C. $105^{\circ}$
(c) $105^{\circ}$
Given that $C F \| A B$ and $\angle A B C=85^{\circ}$.
$\therefore \angle B C F=85^{\circ} \Rightarrow \angle B C E=105^{\circ} \Rightarrow \angle B A D=105^{\circ} \quad$ [Exterior angle $=$ Interior opposite angle]
View full question & answer→MCQ 431 Mark
- A
$110^{\circ}$
- B
$100^{\circ}$
- C
$140^{\circ}$
- ✓
$130^{\circ}$
AnswerCorrect option: D. $130^{\circ}$
(d) $130^{\circ}$
In $\triangle A E D$, we find that
$\angle A E D=90^{\circ} \text { (angle in a semi-circle) and } \angle E A D=60^{\circ} \text {. Therefore, } \angle A DE=30^{\circ} \text {. }$
$A E D C$ is a cyclic quadrilateral such that
$\begin{array}{ll}& \angle E D C=\angle E D A+\angle A D C=30^{\circ}+70^{\circ}=100^{\circ} \\
\therefore \quad & \angle C A E=80^{\circ} \Rightarrow \angle D A C=20^{\circ}\end{array}$
Thus, $A B C D$ is a cyclic quadrilateral, such that $\angle D A B=20^{\circ}+30^{\circ}=50^{\circ}$
$\therefore \quad \angle B C D=180^{\circ}-50^{\circ}=130^{\circ}$
View full question & answer→MCQ 441 Mark
- A
$70^{\circ}$
- ✓
$110^{\circ}$
- C
$100^{\circ}$
- D
$130^{\circ}$
AnswerCorrect option: B. $110^{\circ}$
(b) $110^{\circ}$
ABCD is a cyclic quadrilateral such that $\angle A D C=70^{\circ}$.
$\therefore \quad \angle A B C+\angle A D C=180^{\circ} \Rightarrow \angle A B C+70^{\circ}=180^{\circ} \Rightarrow\angle A B C=110^{\circ}$
View full question & answer→MCQ 451 Mark
- A
$110^{\circ}$
- B
$50^{\circ}$
- C
$40^{\circ}$
- ✓
$80^{\circ}$
AnswerCorrect option: D. $80^{\circ}$
(d) $80^{\circ}$
We observe that DAEF is a cyclic quadrilateral.
$\therefore \quad \angle A D F=80^{\circ} \quad$ [Exterior angle $=$ Opposite interior angle]
Since ABCD is a parallelogram. Therefore,
$\angle A B C=\angle A D C=\angle A D F=80^{\circ}$
View full question & answer→MCQ 461 Mark
- A
$70^{\circ}$
- B
$75^{\circ}$
- C
$60^{\circ}$
- ✓
$105^{\circ}$
AnswerCorrect option: D. $105^{\circ}$
(d) $105^{\circ}$
Clearly, arc AC subtends $\angle A O C=105^{\circ}$ at the centre O and $\angle A B C$ at point B on the circumference. Therefore,
$\angle A B C=\frac{1}{2} \angle A O C=\frac{1}{2} \times 150^{\circ}=75^{\circ}$
Since $A B C D$ is a cyclic quadrilateral. Therefore,
$\angle A D C+\angle A B C=180^{\circ} \Rightarrow \angle A D C+75^{\circ}=180^{\circ} \Rightarrow \angle A DC=105^{\circ}$
View full question & answer→MCQ 471 Mark
- A
$43^{\circ}$
- B
$30^{\circ}$
- ✓
$34^{\circ}$
- D
$40^{\circ}$
AnswerCorrect option: C. $34^{\circ}$
(c) $34^{\circ}$
Since ABCD is a cyclic quadrilateral.
$\begin{array}{ll}\therefore & \angle A+\angle C=180^{\circ} \\
\Rightarrow & 2 x+40^{\circ}+4 x-64^{\circ}=180^{\circ} \\
\Rightarrow & 6 x=204^{\circ} \Rightarrow x=34^{\circ}\end{array}$
View full question & answer→MCQ 481 Mark
- A
$60^{\circ}$
- B
$70^{\circ}$
- ✓
$50^{\circ}$
- D
$80^{\circ}$
AnswerCorrect option: C. $50^{\circ}$
(c) $50^{\circ}$
we have, $\angle C P B=120^{\circ}$. Therefore, $\angle B P D=180^{\circ}-120^{\circ}=60^{\circ}$.
Arc $A D$ subtends $\angle A C D=70^{\circ}$ at $C$ and $\angle A B D$ at $B$ and angles in the same segment are equal.
Therefore, $\angle A B D=70^{\circ}$.
Using angle sum property in $\triangle P B D$, we obtain
$\angle B P D+\angle A B D+x^{\circ}=180^{\circ} \Rightarrow 60^{\circ}+70^{\circ}+x^{\circ}=180^{\circ} \Rightarrow x=50^{\circ}
$
In $\triangle A C P$, we have
$\begin{array}{rlrl}& \angle A C P =70^{\circ} \text { and } \angle A P C=60^{\circ} \\
\therefore \quad \angle C A B & =\angle C A P=180^{\circ}-70^{\circ}-60^{\circ}=50^{\circ}\end{array}$
Clearly, $C B$ makes angle $\angle C A B$ and $\angle C D B$ in the same segment.$
\therefore \quad \angle C D B=\angle C A B \Rightarrow \angle C D B=50^{\circ}$
View full question & answer→MCQ 491 Mark
- A
$60^{\circ}$
- B
$70^{\circ}$
- ✓
$80^{\circ}$
- D
$69^{\circ}$
AnswerCorrect option: C. $80^{\circ}$
(c) $80^{\circ}$
In $\triangle A B C$, we have
$\begin{array}{l}\angle B A C+\angle A B C+\angle A C B=180^{\circ} \\
\angle B A C+69^{\circ}+31^{\circ}=180^{\circ} \Rightarrow \angle B A C=80^{\circ}\end{array}$
We observe that $\angle B A C$ and $\angle B D C$ are angles in the same segment.$
\therefore \quad \angle B D C=\angle B A C \Rightarrow \angle B D C=80^{\circ}$
View full question & answer→MCQ 501 Mark
Answer(b) 10 cm
In triangles OEB and OFC, we have
$\begin{aligned}\angle O E B & =\angle O F C \\
O B & =O C\end{aligned}
$and, $\quad \angle B O E=\angle C O F$
So, by using $A A S$-congruence criterion, we obtain
$\begin{array}{ll}& \Delta B O E \cong \triangle C O F \\
\Rightarrow & O E=O F \\
\Rightarrow & \text { Chords } A B \text { and } C D \text { are equidistant from the centre. } \\
\Rightarrow \quad & A B=C D \Rightarrow C D=10 cm .\end{array}$
View full question & answer→
