Question 14 Marks
Prove that the line joining the mid-point of a chord to the centre of the circle passes through the mid-point of the corresponding minor arc.
AnswerGiven:
$C$ is the mid-point of chord $AB$.
To prove: $D$ is the mid-point of arc $AB$.
Proof: In $\triangle\text{OAC}$ and $\triangle\text{OBC}$
$OA = OB$ [Radius of circle]
$OC = OC$ [Common]
$AC = BC$ [$C$ is the mid-point of $AB$]
Then $\triangle\text{OAC}\cong\triangle\text{OBC}$ [By $SSS$ condition]
$\therefore\angle\text{AOC}=\angle\text{BOC}$
$\Rightarrow\text{m}\overline{\text{A}}\text{D}\cong\text{m}\overline{\text{B}}\text{D}$
$\Rightarrow\overline{\text{A}}\text{D}\cong\overline{\text{B}}\text{D}$
Hence, $D$ is the mid-point of arc $AB$. View full question & answer→Question 24 Marks
Suppose you are given a circle. Give a construction to find its centre.
Answer
Steps of Construction:
$1.$ Take three points $A, B$ and $C$ on the given circle.
$2.$ Join $AB$ and $BC$.
$3.$ Draw the perpendicular bisectors of chord $AB$ and $BC$ which intersect each other at $O$.
$4.$ Point $O$ will be the required centre of the circle because we know that the perpendicular bisector of the chord always passes through the centre. View full question & answer→Question 34 Marks
Two chords $AB, CD$ of lengths $5\ cm, 11\ cm$ respectively of a circle are parallel. If the distance between $AB$ and $CD$ is $3\ cm$, find the radius of the circle.
AnswerLet $AB$ and $CD$ be two parallel chord of the circle with center $o$ such that $AB = 5\ cm$ and $CD = 11\ cm$. let the radius of the circle be $r\ cm$. 
Draw $\text{OP}\perp\text{AB}$ and $\text{OQ}\perp\text{CD}$ as well as point $O, Q$ and $P$ are collinear.
Clearly, $PQ = 3\ cm$ Let $OQ = x$ then $OP = x + 3$ In $\triangle\text{OAP}$ and $\triangle\text{OCQ}$
we have $\text{OA}^2=\text{OP}^2+\text{AP}^2$
$\Rightarrow\text{r}^2=(\text{x}+3)^2+\Big(\frac{5}{2}\Big)^2\dots(1)$ And $\text{OC}^2=\text{OQ}^2+\text{CQ}^2$
$\Rightarrow\text{r}^2=\text{x}^2+\Big(\frac{11}{2}\Big)^2\dots(2)$ From $(1)$ and $(2)$
we get $(\text{x}+3)^2+\Big(\frac{5}{2}\Big)^2=\text{x}^2+\Big(\frac{11}{2}\Big)^2$
$\Rightarrow\text{x}^2+\text{6x}+9+\frac{25}{4}=\text{x}^2+\frac{121}{4}$
$\Rightarrow\text{6x}+\frac{61}{4}=\frac{121}{4}$
$\Rightarrow\text{6x}=\frac{121-61}{4}$
$\Rightarrow\text{6x}=\frac{60}{4}$
$\Rightarrow\text{x}=\frac{5}{2}$ Putting the value of $x$ in $(2)$ we get,
$\text{r}^2=\Big(\frac{5}{2}\Big)^2+\Big(\frac{11}{2}\Big)^2$
$=\frac{25}{4}+\frac{121}{4}$
$=\frac{146}{4}$
$\Rightarrow\text{r}=\sqrt{\frac{146}{4}}$
$\text{r}=\sqrt{\frac{146}{4}}\text{cm}$ View full question & answer→Question 44 Marks
In a cyclic quadrilateral $ABCD$, if $\angle\text{A}-\angle\text{C}=60^\circ,$ prove that the smaller of two is $60^\circ$
AnswerWe have $\angle\text{A}-\angle\text{C}=60^\circ\dots(1)$
Since, $ABCD$ is a cyclic quadrilateral Then
$\angle\text{A}+\angle\text{C}=180^\circ\dots(2)$ Add equations $(1)$ and $(2)$
$\angle\text{A}-\angle\text{C}+\angle\text{A}+\angle\text{C}=60^\circ+180^\circ$
$\Rightarrow2\angle\text{A}=240^\circ$
$\Rightarrow\angle\text{A}=\frac{240^\circ}{2}=120^\circ$
Put value of $\angle\text{A}$ in equation $(2)$
$120^\circ+\angle\text{C}=180^\circ$
$\Rightarrow\angle\text{C}=180^\circ-120^\circ=60^\circ$
View full question & answer→Question 54 Marks
In the given figure, $ABCD$ is a quadrilateral inscribed in a circle with centre $O$. $CD$ is produced to $E$ such that $\angle\text{AED} = 95^\circ$ and $\angle\text{OBA} = 30^\circ$ Find $\angle\text{OAC.}$
AnswerWe are given $ABCD$ is a quadrilateral with center $O$, $\angle\text{ADE}=95^\circ$ and $\angle\text{OBA}=30^\circ$
We need to find $\angle\text{OAC}$
We are given the following figure:
Since $\angle\text{ADE}=95^\circ$
$\Rightarrow\angle\text{ADC}=180^\circ-95^\circ=85^\circ$
Since squo; $ABCD$ is cyclic quadrilateral
This means
$\angle\text{ABC}+\angle\text{ADC}=180^\circ$
$\Rightarrow\angle\text{ABO}+\angle\text{OBC}+\angle\text{ADC}=180^\circ$
$\Rightarrow30^\circ+\angle\text{OBC}+85^\circ=180^\circ$
$\Rightarrow\angle\text{OBC}=180^\circ-115^\circ=65^\circ$
Since $OB = OC$ (radius)
$\Rightarrow\angle\text{OBC}+\angle\text{OCB}=65^\circ$
In $\triangle\text{OBC}$
$\angle\text{BOC}+\angle\text{OBC}+\angle\text{OBC}=180^\circ$
$\angle\text{BOC}+2\angle\text{OBC}=180^\circ$
$\angle\text{BOC}+2\times65^\circ=180^\circ$
$\angle\text{BOC}=180^\circ-130^\circ$
$\angle\text{BOC}=50^\circ$
Since $DBAC$ and $DBOC$ are formed on the same base which is chord.
So,
$\angle\text{BAC}=\frac{\angle\text{BOC}}{2}$
$=\frac{50^\circ}{2}$
$\angle\text{BAC}=25^\circ$
Consider $\triangle\text{BOA}$ which is isosceles triangle.
$\angle\text{OAB}=30^\circ$
$\Rightarrow\angle\text{OAC}+\angle\text{BAC}=30^\circ$
$\Rightarrow\angle\text{OAC}+25^\circ=30^\circ$
$\Rightarrow\angle\text{OAC}=5^\circ$ View full question & answer→Question 64 Marks
A chord of a circle is equal to the radius of the circle. Find the angle subtended by the chord at a point on the minor arc and also at a point on the major arc.
AnswerWe have,
Radius $OA =$ Chord $AB $
$\Rightarrow OA = OB = AB $
Then triangle $OAB$ is an equilateral triangle.
$\therefore\angle\text{AOB}=60^\circ$ [one angle of equilateral triangle] By degree measure theorem
$\angle\text{AOB}=2\angle\text{APB}$
$\Rightarrow60^\circ=2\angle\text{APB}$
$\Rightarrow\angle\text{APB}=\frac{60^\circ}{2}=30^\circ$
Now, $\angle\text{APB}+\angle\text{AQB}=180^\circ$ [opposite angles of cyclic quadrilateral]
$\Rightarrow300+\angle\text{AQB}=180^\circ$
$\Rightarrow\angle\text{AQB}=180^\circ-30^\circ=150^\circ$
Therefore, Angle by chord $AB$ at minor arc $= 150^\circ$
Angle by chord $AB$ at major arc $= 30^\circ$ View full question & answer→Question 74 Marks
Three girls Ishita, Isha and Nisha are playing a game by standing on a circle of radius $20\ m$ drawn in a park. Ishita throws a ball to Isha, Isha to Nisha and Nisha to Ishita. If the distance between Ishita and Isha and between Isha and Nisha is $24\ m$ each, what is the distance between Ishita and Nisha.
AnswerLet $R, S$ and $M$ be the position of Ishita, Isha and Nisha respectively.
$\text{AR}=\text{AS}=\frac{24}{2}=12\text{cm}$
$OR = OS = OM = 20\ cm$ [Radii of circle] In
$\triangle\text{OAR,}$ $OA^2 + AR^2 = OR^2\ OA^2 + 12^2 = 20^2\ OA^2 = 400 - 144 = 256\ m^2 OA = 16m$
We know that, in an isosceles triangle altitude divides the base.
So in $\triangle\text{RSM},\angle\text{RCS}=90^\circ$ and $RC = CM$
Area of $\triangle\text{ORS}=\frac{1}{2}\times\text{OA}\times\text{RS}$
$\Rightarrow\frac{1}{2}\times\text{RC}\times\text{OS}$
$=\frac{1}{2}\times16\times24$ $\Rightarrow\text{RC}\times20=16\times24$
$\Rightarrow\text{RC}=19.2$
$\Rightarrow\text{RM}=2(19.2)=38.4\text{m}$
So, the distance between Ishita and Nisha is $38.4m.$ View full question & answer→Question 84 Marks
In the given figure, $ABCD$ is a cyclic quadrilateral in which $AC$ and $BD$ are its diagonals. If $\angle\text{DBC}=55^\circ$ and $\angle\text{BAC}=45^\circ,$ find $\angle\text{BCD}.$
AnswerSince angles in the same segment of a circle are equal.
$\therefore\angle\text{CAD}=\angle\text{DBC}=55^\circ$
$\therefore\angle\text{DAB}=\angle\text{CAD} +\angle\text{BAC}=55^\circ +45^\circ=100^\circ$
But, $\angle\text{DAB}+\angle\text{BCD}=180^\circ$ [Opposite angles of a cyclic quadrilateral]
$\therefore\angle\text{BCD}=180^\circ-100^\circ=80^\circ$
View full question & answer→Question 94 Marks
Prove that the line segment joining the mid-point of the hypotenuse of a right triangle to its opposite vertex is half of the hypotenuse.
Answer
Let $\triangle\text{ABC}$ be a right triangle right angled at $B$.
Let $P$ be the mid-point of hypotenuse $AC$.
Draw a circle with centre at $P$ and $AC$ as a diameter.
Since, $\angle\text{ABC}=90^\circ.$
Therefore, the circle passes through $B$.
$\therefore$ $BP$ = Radius Also, $AP = CP$ = Radius
$\therefore$ $AP = BP = CP$
Hence, $\text{BP}=\frac{1}2{}\text{AC}$ View full question & answer→Question 104 Marks
$ABCD$ is a cyclic qudrilateral in which: $\text{BC}\parallel\text{AD},\ \angle\text{ADC}=110^\circ$ and $\angle\text{BAC}=50^\circ.$ Find $\angle\text{DAC}.$
Answer
Since, $ABCD$ is a cyclic quadrilateral.
Then, $\angle\text{ABC}+\angle\text{ADC}=180^\circ$
$\Rightarrow\angle\text{ABC}+110^\circ=180^\circ$
$\Rightarrow\angle\text{ABC}=180^\circ-110^\circ=70^\circ$
Since, $AD || BC$ Then, $\angle\text{DAB}+\angle\text{ABC}=180^\circ$ [Co-interior angles]
$\Rightarrow\angle\text{DAC}+50^\circ+70^\circ=180^\circ$
$\Rightarrow\angle\text{DAC}=180^\circ-50^\circ-70^\circ=60^\circ$ View full question & answer→Question 114 Marks
In the given figure, $\angle\text{BAD}=78^\circ,\angle\text{DCF}=\text{x}^\circ$ and $\angle\text{DEF}=\text{y}^\circ.$ Find the values of $x$ and $y$.
AnswerWe have, $\angle\text{BAD}=78^\circ,\angle\text{DCF}=\text{x}^\circ$ and $\angle\text{DEF}=\text{y}^\circ.$
Since, $ABCD$ is a cyclic quadrilateral.
Then, $\angle\text{BAD}+\angle\text{BCD}=180^\circ$
$\Rightarrow78^\circ+\angle\text{BCD}=180^\circ$
$\Rightarrow\angle\text{BCD}=180^\circ-78^\circ=102^\circ$
Now, $\angle\text{BCD}+\angle\text{DCF}=180^\circ$ [Linear pair of angles]
$\Rightarrow102^\circ=\text{x}^\circ=180^\circ$
$\Rightarrow\text{x}=180^\circ-102^\circ=78^\circ$
Since, $DCEF$ is a cyclic quadrilateral
Then, $x + y = 180^\circ$
$\Rightarrow 78^\circ+ y = 180^\circ$
$\Rightarrow y = 180^\circ - 78^\circ = 102^\circ$
View full question & answer→Question 124 Marks
In the given figure, if $O$ is the circumcentre of $\angle\text{ABC},$ then find the value of $\angle\text{OBC} + \angle\text{BAC.}$
Answer
Since, $O$ is the circumcentre of $\triangle\text{ABC,}$
So, $O$ would be centre of the circle passing through points $A, B$ and $C$.
$\angle\text{ABC}=90^\circ$ $($Angle in the semicircle is $90^\circ)$
$\Rightarrow\angle\text{OAB}+\angle\text{OBC}=90^\circ\dots(1)$
As $OA = OB$ (Radii of the same circle)
$\therefore\angle\text{OAB}=\angle\text{OBA}$ (Angle opposite to equal sides are equal)or,
$\angle\text{BAC}=\angle\text{OBA}$
From $(1)$
$\angle\text{BAC}+\angle\text{OBC}=90^\circ$ View full question & answer→Question 134 Marks
$ABCD$ is a cyclic qudrilateral in which: $\angle\text{BCD}=100^\circ$ and $\angle\text{ABD}=70^\circ$ find $\angle\text{ADB}.$
Answer
Since, $ABCD$ is a cyclic quadrilateral.
Then, $\angle\text{BAD}+\angle\text{BCD}=180^\circ$
$\Rightarrow\angle\text{BAD}+100^\circ=180^\circ$
$\Rightarrow\angle\text{BAD}=180^\circ-100^\circ=80^\circ$ In by angle sum property
$\angle\text{ABD}+\angle\text{ADB}+\angle\text{BAD}=180^\circ$
$\Rightarrow70^\circ+\angle\text{ADB}+80^\circ=180^\circ$
$\Rightarrow\angle\text{ADB}=180^\circ-70^\circ-80^\circ=30^\circ$ View full question & answer→Question 144 Marks
A line segment $AB$ is of length $5\ cm$. Draw a circle of radius $4\ cm$ passing through $A$ and $B$. Can you draw a circle of radius $2\ cm$ passing through $A$ and $B$? Give reason in support of your answer.
Answer$1.$ Draw a line segment $AB$ of $5\ cm$.
$2.$ Draw the perpendicular bisectors of $AB$.
$3.$ With centre $A$ and radius of $4\ cm$, draw an arc which intersects the perpendicular bisector at point $O$.
The point $O$ will be the required centre.
$4.$ Join $OA$.
$5.$ With centre $O$ and radius $OA$, draw a circle.
No, we cannot draw a circle of radius $2\ cm$ passing through $A$ and $B$ because when we draw an arc of radius $2\ cm$ with centre $A$, the arc will not intersect the perpendicular bisector and we will not find the centre.
View full question & answer→Question 154 Marks
$\text{ABCD}$ ia a cyclic quadrilateral in which $BA$ and $CD$ when produced meet in $E$ and $EA = ED$. Prove that:
$i. AD \| BC$.
$ii. EB = EC$.
Answer
Given $ABCD$ is a cyclic quadrilateral in which $EA = ED$ To prove:
$i. AD \| BC$.
$ii. EB = EC$.
Proof:
$i.$ Since $EA = ED$
Then, $\angle\text{EAD}=\angle\text{EDA}\dots(1) [$Oppo. angles to equal sides$]$
Since, $\text{ABCD}$ is a cyclic quadrilateral
Then, $\angle\text{ABC}+\angle\text{ADC}=180^\circ$
But $\angle\text{ABC}+\angle\text{EBC}=180^\circ [$Linear pair of angles$]$
Then, $\angle\text{ADC}=\angle\text{EBC}\dots(2)$
Compare equations $(1)$ and $(2)$
$\angle\text{EAD}=\angle\text{EBC}\dots(3)$
Since, corresponding angles are equal
Then, $BC \| AD$
$ii.$ From equation $(3)$
$\angle\text{EAD}=\angle\text{EBC}\dots(3)$
Similarly $\angle\text{EDA}=\angle\text{ECB}\dots(4)$
Compare equations $(1)(3)$ and $(4)$
$\angle\text{EBC}=\angle\text{ECB}$
$\Rightarrow \text{EB}=\text{EC} [$Opposite angles to equal sides$]$ View full question & answer→Question 164 Marks
Two chords $AB$ and $CD$ of lengths $5\ cm$ and $11\ cm$ respectively of a circle are parallel to each other and are opposite side of its centre. If the distance between $AB$ and $CD$ is $6\ cm$, find the radius of the circle.
AnswerDraw $\text{OM}\perp\text{AB}$ and $\text{ON}\perp\text{CD}.$
Join $OB$ and $OD$.
$\text{BM}=\frac{\text{AB}}{2}=\frac{5}{2}$ [Perpendicular from the centre bisects the chord]
$\text{ND}=\frac{\text{CD}}{2}=\frac{11}{2}$
Let $ON$ be $x$, so $OM$ will be $6 - x$.$\triangle\text{MOB}$
$\text{OM}^2+\text{MB}^2=\text{OB}^2$
$(6-\text{x})^2+\Big(\frac{5}{2}\Big)^2+\text{OB}^2$
$36+\text{x}^2-12\text{x}+\frac{25}{4}=\text{OB}^2\dots(\text{i})$
In $\triangle\text{NOD}$
$\text{ON}^2+\text{ND}^2=\text{OD}^2$
$\text{x}^2+\Big(\frac{11}{2}\Big)^2=\text{OD}^2$
$\text{x}^2+\frac{121}{4}=\text{OD}^2\dots(\text{ii})$
We have $OB = OD$. [Radii of same circle] So, from equation $(i)$ and $(ii)$.
$36+\text{x}^2-\text{12x}+\frac{25}{4}=\text{x}^2+\frac{121}{2}$
$\Rightarrow\text{12x}=36+\frac{25}{4}-\frac{121}{4}$
$=\frac{144+25-121}{4}$
$=\frac{48}{4}=12$
$\text{x}=1$
From equation $(ii)$$(1)^2+\Big(\frac{121}{4}\Big)=\text{OD}^2$
$\text{OD}^2=1+\frac{121}{4}=\frac{125}{4}$
$\text{OD}=\frac{5\sqrt{5}}2{}$
So, the radius of the circle is found to be $55\sqrt{2}\text{cm}.$ View full question & answer→Question 174 Marks
In the given figure, $O$ is the center of the circle. Find $\angle\text{CBD}.$
AnswerWe have, $\angle\text{AOC}=100^\circ$ By degree measure theorem
$\angle\text{AOC}=2\angle\text{APC}$
$\Rightarrow100^\circ=2\angle\text{APC}$
$\Rightarrow\angle\text{APC}=\frac{100^\circ}{2}=50^\circ$
$\therefore\angle\text{APC}+\angle\text{ABC}=180^\circ$ [Opposite angles of cyclic quad.]
$\Rightarrow50^\circ+\angle\text{ABC}=180^\circ$
$\Rightarrow\angle\text{ABC}=180^\circ-50^\circ=130^\circ$
$\therefore\angle\text{ABC}+\angle\text{CBD}=180^\circ$ [Linear pair of angles]
$\Rightarrow130^\circ+\angle\text{CBD}=180^\circ$
$\Rightarrow\angle\text{CBD}=180^\circ-130^\circ=50^\circ$
View full question & answer→Question 184 Marks
$ABCD$ is a cyclic trapezium with $AD || BC$. If $\angle\text{B}=70^\circ,$determine other three angles of the trapezium.
Answer
We have $ABCD$ is a cyclic trapezium with $AD || BC$ and $\angle\text{B}=70^\circ.$
Since, $ABCD$ is a cyclic quadrilateral Then,
$\angle\text{B}+\angle\text{D}=180^\circ$
$\Rightarrow70^\circ+\angle\text{D}=180^\circ$
$\Rightarrow\angle\text{D}=180^\circ-70^\circ=110^\circ$
Since, $AD || BC$ Then, $\angle\text{A}+\angle\text{B}=180^\circ$ [Co-interior angles]
$\Rightarrow\angle\text{A}+70^\circ=180^\circ$
$\Rightarrow\angle\text{A}=180^\circ-70^\circ=110^\circ$
Since, ABCD is a cyclic quadrilateral Then, $\angle\text{A}+\angle\text{C}=180^\circ$
$\Rightarrow110^\circ+\angle\text{C}=180^\circ$ View full question & answer→Question 194 Marks
Give a method to find the centre of a given circle.
Answer
Steps of Construction:
$1.$ Take three points $A, B$ and $C$ on the given circle.
$2.$ Join $AB$ and $BC$.
$3.$ Draw the perpendicular bisectors of the chord $AB$ and $BC$ which intersect each other at $O$.
$4.$ Point $O$ will give the required circle because we know that, the Perpendicular bisectors of chord always pass through the centre. View full question & answer→Question 204 Marks
If the two sides of a pair of opposite sides of a cyclic quadrilateral ae equal, prove that its diagonals are equal.
Answer
Given $ABCD$ is a cyclic quadrilateral in which $AB = DC$
To prove $AC = BD$ Proof In $\triangle\text{PAB}$ and $\triangle\text{PDC}$
$\text{AB}=\text{DC}$ [Given]
$\angle\text{BAP}=\angle\text{CDP}$ [Angles in the same segment]
$\angle\text{PBA}=\angle\text{PCD}$ [Angles in same segment]
Then, $\triangle\text{PAB}\cong\triangle\text{PDC}$ [By $ASA$ condition]
$\therefore\text{PA}=\text{PD}\dots(1)$ $[C.P.C.T.]$ and $\text{PC}=\text{PB}\dots(2)$
$[C.P.C.T.]$ Add equation $(1)$ and $(2) PA + PC = PD + PB \Rightarrow AC = BD$ View full question & answer→Question 214 Marks
In a cyclic quadrilateral $ABCD$ if $\text{m}\angle\text{A}=\big(\text{m}\angle\text{C}\big).$ Find $\text{m}\angle\text{A}.$
Answer
We have, $\angle\text{A}=3\angle\text{C}$
Let $\angle\text{C}=\text{x}$ Then, $\angle\text{A}=\text{3x}$
$\therefore\angle\text{A}+\angle\text{C}=180^\circ$ [Opposite angles of cyclic quad.]
$\Rightarrow\text{3x}+\text{x}=180^\circ$
$\Rightarrow\text{4x}=180^\circ$
$\Rightarrow\text{x}=\frac{180^\circ}{4}=45^\circ$
$\therefore\angle\text{A}=\text{3x}$
$=3\times45^\circ$
$=135^\circ$ View full question & answer→Question 224 Marks
Prove that the angle in a segment shorter than a semicircle is greater than a right angle.
Answer
Given: $\angle\text{ACB}$ is an angle in minor segment.
To prove: $\angle\text{ACB}>90^\circ$
Proof: By degree measure theorem Reflex
$\angle\text{AOB}=2\angle\text{ACB}$ And reflex
$\angle\text{AOB}>180^\circ$ Then, $2\angle\text{ACB}>180^\circ$
$\Rightarrow\angle\text{ACB}>\frac{180^\circ}{2}$
$\Rightarrow\angle\text{ACB}>90^\circ$ View full question & answer→Question 234 Marks
In figure, $O$ is the centre of the circle, then prove that $\angle\text{x}=\angle\text{y}+\angle\text{z}.$
AnswerWe have, $\angle3=\angle4$ [Angles in same segment]
$\therefore\angle\text{x}=2\angle3$ [By degree measure theorem]
$\Rightarrow\angle\text{X}=\angle3+\angle3$
$\Rightarrow\angle\text{X}=\angle3+\angle4\dots(\text{i})$
$[\angle3=\angle4]$ But $\angle\text{y}=\angle3+\angle1$ [By exterior angle property]
$\Rightarrow\angle3=\angle\text{y}-\angle1\dots(\text{ii})$ From $(i)$ and $(ii)$
$\angle\text{x}=\angle\text{y}-\angle1+\angle4$
$\Rightarrow\angle\text{x}=\angle\text{y}+\angle4-\angle1$
$\Rightarrow\angle\text{x}=\angle\text{y}+\angle\text{z}+\angle1-\angle1$ [By exterior angle property] $\Rightarrow\angle\text{x}=\angle\text{y}+\angle\text{z}$
View full question & answer→Question 244 Marks
In the given figure, $O$ is the center of the circle. If $\angle\text{CEA}=30^\circ,$ find the value of $x, y$ and $z$.
AnswerWe have, $\angle\text{CEA}=30^\circ$ Since, quad.
$ABCE$ is a cyclic quadrilateral.
Then, $\angle\text{ABC}+\angle\text{CEA}=180^\circ$
$\Rightarrow\text{x}+30^\circ=180^\circ$
$\Rightarrow\text{x}=180^\circ-30^\circ=150^\circ$ By degree measure theorem
$\angle\text{AOC}=2\angle\text{CEA}$
$\Rightarrow\text{y}=2\times30^\circ=60^\circ$
$\therefore\angle\text{ADC}=\angle\text{CEA}$ [Angle in same segment]
$\Rightarrow\text{z}=30^\circ$
View full question & answer→Question 254 Marks
In the given figure, $\triangle\text{PQR}$ is an isosceles triangle with $PQ = PR$ and $\text{m}\angle\text{PQR}=35^\circ.$ Find $\text{m}\angle\text{QSR}$ and $\text{m}\angle\text{QTR}.$ 
AnswerWe have, $\angle\text{PQR}=35^\circ$ Since, $\triangle\text{PQR}$ is an isosceles triangle with $PQ = PR$.
Then, $\angle\text{PQR}=\angle\text{PRQ}=35^\circ$ In
$\triangle\text{PQR},$ by angle sum property $\angle\text{p}+\angle\text{PQR}+\angle\text{PRQ}=180^\circ$
$\Rightarrow\angle\text{p}+35^\circ+35^\circ=180^\circ$
$\Rightarrow\angle\text{p}=180^\circ-35^\circ-35^\circ=110^\circ$
$\therefore\angle\text{QSR}=\angle\text{p}=110^\circ$ [Angles in same segment]
Now, $\angle\text{QSR}+\angle\text{QTR}=180^\circ$ [Opposite angles of cyclic quad.]
$\Rightarrow110^\circ+\angle\text{QTR}=180^\circ$
$\Rightarrow\angle\text{QTR}=180^\circ-110^\circ=70^\circ$
View full question & answer→Question 264 Marks
Find the length of a chord which is at a distance of $4\ cm$ from the centre of a circle of radius $6\ cm.$
AnswerGiven that ,
Distance $(OC) = 4\ cm$ Radius of the circle $(OA) = 6\ cm$ In $\triangle\text{OCA,}$
by Pythagoras theorem
$AC^2+ OC^2= OA^2$
$\Rightarrow AC^2 + 4^2 = 6^2$
$\Rightarrow AC^2 = 36 - 16$
$\Rightarrow AC^2 = 20$
$\Rightarrow\text{AC}=\sqrt{20}$
$\Rightarrow AC = 4.47cm$
We know that the perpendicular distance from centre to chord bisects the chord.
$AC = BC = 4.47cm$
Then $AB = 4.47 + 4.47 = 8.94cm$ View full question & answer→Question 274 Marks
The lengths of two parallel chords of a circle are $6\ cm$ and $8\ cm$. If the smaller chord is at a distance of $4\ cm$ from the centre, what is the distance of the other chord from the centre?
AnswerDistance of smaller chord $AB$ from centre of circle $= 4\ cm, OM = 4\ cm$
$\text{MB}=\frac{\text{AB}}2{}=\frac{6}{2}=3\text{cm}$
In $\triangle\text{OMB }$ $OM^2 + MB^2 = OB^2 4^2 + 9^2 = OB^2 16 + 9 = OB^2$
$\text{OB}=\sqrt{25}$
$OB = 5\ cm$ In $\triangle\text{OND }$ $OD = OB = 5\ cm$ [Radii of same circle]
$\text{ND}=\frac{\text{CD}}{2}=\frac{8}{2}=4\text{cm}$
$ON^2 + ND^2 = OD^2 ON^2 + 4^2 = 5^2 ON^2 = 25 - 16$
$\text{ON}=\sqrt{9}$
$ON = 3\ cm$
So, the distance of bigger chord from the circle is $3\ cm.$ View full question & answer→Question 284 Marks
Prove that a diameter of a circle which bisects a chord of the circle also bisects the angle subtended by the chord at the centre of the circle.
AnswerGiven:
$PQ$ is a diameter of circle which bisects the chord $AB$ at $C$.
To Prove: $PQ$ bisects $\angle\text{AOB}$
Proof: In $\angle\text{AOC}$ and
$\angle\text{BOC}$ $OA = OB$ [Radius of circle] $OC = OC$ [Common] $AC = BC$ [Given]
Then $\triangle\text{AOC}\cong\triangle\text{BOC}$ [By $SSS$ condition] $\angle\text{AOC}=\angle\text{BOC}$ $[C.P.C.T]$
Hence $PQ$ bisects $\angle\text{AOB}.$ View full question & answer→Question 294 Marks
On a semi-circle with $AB$ as diameter, a point $C$ is taken, so that $\text{m}\big(\angle\text{CAB}\big)=30^\circ.$ Find $\text{m}\big(\angle\text{ACB}\big)$ and $\text{m}\big(\angle\text{ABC}\big).$
Answer
We have, $\angle\text{CAB}=30^\circ$
$\therefore\angle\text{ACB}=90^\circ$ [Angle in semicircle] In
$\triangle\text{ABC},$ by angle sum property
$\angle\text{CAB}+\angle\text{ACB}+\angle\text{ABC}=180^\circ$
$\Rightarrow30^\circ+90^\circ+\angle\text{ABC}=180^\circ$
$\Rightarrow\angle\text{ABC}=180^\circ-90^\circ-30^\circ=60^\circ$ View full question & answer→Question 304 Marks
Prove that the angle in a segment greater than a semi-circle is less than a right angle.
Answer
Given: $\angle\text{ACB}$ is an angle in minor segment.
To prove: $\angle\text{ACB}<90^\circ$
Proof: By degree measure theorem
$\angle\text{AOB}=2\angle\text{ACB}$
And $\angle\text{AOB}<180^\circ$
Then, $2\angle\text{ACB}<180^\circ$
$\Rightarrow\angle\text{ACB}<\frac{180^\circ}{2}$
$\Rightarrow\angle\text{ACB}<90^\circ$ View full question & answer→Question 314 Marks
A circular park of radius 40m is situated in a colony. Three boys Ankur, Amit and Anand are sitting at equal distance on its boundary each having a toy telephone in his hands to talk to each other. Find the length of the string of each phone.
AnswerGiven that $AB = BC = CA$
So, $ABC$ is an equilateral triangle $OA$ (radius) $= 40m$ Medians of equilateral triangle pass through the circum centre $(O)$ of the equilateral triangle $ABC$.
We also know that median intersect each other at the ratio $2 : 1$ As $AD$ is the median of equilateral triangle $ABC$, we can write:$\frac{\text{OA}}{\text{OD}}=\frac{2}{1}$
$\Rightarrow\frac{\text{4OM}}{\text{OD}}=\frac{2}{1}$
$\Rightarrow\text{OD}=20\text{m}$
Therefore, $AD = OA + OD = (40 + 20)m = 60m$ In $\triangle\text{ADC}$
By using Pythagoras theorem$\text{AC}^2 = \text{AD}^2 + \text{DC}^2$
$\text{AC}^2=60^2+\big(\text{AC}^2\big)^2$
$\text{AC}^2=3600+\frac{\text{AC}^2}{4}$
$\Rightarrow\frac{3}{4}\text{AC}^2=3600$
$\Rightarrow\text{AC}^2=4800$
$\Rightarrow\text{AC}=40\sqrt{3}\text{m}$
So, length of string of each phone will be $40\sqrt{3}\text{m}.$ View full question & answer→Question 324 Marks
In the given figure, two congruent circles with centres $O$ and $O'$ intersect at $A$ and $B$. If $\angle\text{AOB} = 50^\circ,$ then find $\angle\text{APB.}$ 
AnswerSince both the circles are congruent, they will have equal radii.
Let their radii be $‘r’$ So, from the given figure we have, $OA = OB = O'A = O'B = r$
Now, since all the sides of the quadrilateral $OBO’A$ are equal it has to be a rhombus.
One of the properties of a rhombus is that the opposite angles are equal to each other.
So, since it is given that $\angle\text{AO}'\text{B}=50^\circ$ we can say that the angle opposite it,
that is to say that $\angle\text{AO}\text{B}$ should also have the same value.
Hence we get $\angle\text{AO}\text{B}=50^\circ$ Now, consider the first circle with the centre $‘O’$ alone.
$‘AB’$ forms a chord and it subtends an angle of $50^\circ$ with its centre, that is $\angle\text{AO}\text{B}=50^\circ.$
A property of a circle is 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.
This means that, $\angle\text{APB}=\frac{\angle\text{AOB}}{2}$
$=\frac{50^\circ}{2}$ $=25^\circ$
Hence the measure of $\angle\text{APB}$ is $25^\circ$. View full question & answer→Question 334 Marks
In the given figure, two circles intersect at $A$ and $B$. The centre of the smaller circle is $O$ and it lies on the circumference of the larger circle. If $\angle\text{APB} = 70^\circ,$ find $\angle\text{ACB.}$
AnswerConsider the smaller circle whose centre is given as $‘O’$.
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.
So, here we have $\angle\text{AOB}=2\angle\text{APB}$ $=2(70^\circ)$ $\angle\text{AOB}=140^\circ$
Now consider the larger circle and the points $‘A’, ‘C’, ‘B’$ and $‘O’$ along its circumference.
$‘ACBO’$ form a cyclic quadrilateral.
In a cyclic quadrilateral it is known that the opposite angles are supplementary, meaning that the opposite angles add up to $180^\circ $.
$\angle\text{AOB}+\angle\text{ACB}=180^\circ$
$\angle\text{ACB}=180^\circ-\angle\text{AOB}$
$=180^\circ-140^\circ$ $\angle\text{ACB}=40^\circ$
Hence ,the measure of $\angle\text{ACB}$ is $40^\circ.$ View full question & answer→Question 344 Marks
Find the length of a chord which is at a distance of $5\ cm$ from the centre of a circle of radius $10\ cm.$
AnswerGiven that,
Distance $(OC) = 5\ cm$ Radius of the circle $(OA) = 10\ cm$ In $\triangle\text{OCA,}$ by
Pythagoras theorem $OC^2+ AC^2= OA^2$
$\Rightarrow 5^2 + AC^2 = 10^2$
$\Rightarrow 25 + AC^2 = 100$
$\Rightarrow AC^2 = 100 - 25$
$\Rightarrow AC^2 = 75$
$\Rightarrow\text{AC}=\sqrt{75}$
$\Rightarrow AC = 8.66cm$
We know that, the perpendicular from the centre to chord bisects the chord
Therefore, $AC = BC = 8.66\ cm$
Then the chord $AB = 8.66 + 8.66 = 17.32cm$ View full question & answer→Question 354 Marks
$O$ is the circumference of the triangle $ABC$ and $OD$ is perpendicular on $BC$. Prove that $\angle\text{BOD}=\angle\text{A}.$
AnswerGiven $O$ is the circum centre of triangle $ABC$ and $\text{OD}\perp\text{BC}$
To prove $\angle\text{BOD}=2\angle\text{A}$
Proof: In $\triangle\text{OBD}$ and $\triangle\text{OCD}$ $\angle\text{ODB}=\angle\text{ODC}$ [Each 90°]
$OB = OC [$Radius of circle$]$
$OD = OD [$Common$]$
Then $\triangle\text{OBD}\cong\triangle\text{OCD}$ $[$By $RHS$ Condition$]$.
$\therefore\angle\text{BOD}=\angle\text{COD}\dots(\text{i})$ $[PCT]$.
By degree measure theorem$\angle\text{BOC}=2\angle\text{BAC}$
$\Rightarrow2\angle\text{BOD}=2\angle\text{BAC}$ $[$By using $(i)]$
$\Rightarrow\angle\text{BOD}=\angle\text{BAC}$ View full question & answer→Question 364 Marks
In figure, $O$ is the centre of the circle, $BO$ is the bisector of $\angle\text{ABC}.$ Show that $AB = AC$.
AnswerGiven, $BO$ is the bisector of $\angle\text{ABC}$
To prove $AB = BC$ Proof: Since, $BO$ is the bisector of $\angle\text{ABC}$
Then, $\angle\text{ABO}=\angle\text{CBO}\dots(\text{i})$
Since, $OB = OA$ [Radius of circle] Then, $\angle\text{ABO}=\angle\text{DAB}\dots(\text{ii})$ [opposite angles to equal sides]
Since $OB = OC$ [Radius of circle] Then, $\angle\text{OAB}=\angle\text{OCB}\dots(\text{iii})$ [opposite angles to equal sides]
Compare equations $(i), (ii)$ and $(iii)$ $\angle\text{OAB}=\angle\text{OCB}\dots(\text{iii})$ In
$\triangle\text{OAB}$ and $\triangle\text{OCB}$
$\angle\text{OAB}=\angle\text{OCB}$ [From $(iv)$]
$\angle\text{OBA}=\angle\text{OBC}$ [Given] $OB = OB$ [Common]
Then, $\triangle\text{OAB}\cong\triangle\text{OCB}$ [By $AAS$ condition]
$\therefore\text{AB}=\text{BC}$ $[C.P.C.T.]$
View full question & answer→Question 374 Marks
In the given figure, $ABCD$ is a cyclic quadrilateral. Find the value of $x$.
Answer$\angle\text{EDC}+\angle\text{CDA}=180^\circ$ [Linear pair of angles]
$\Rightarrow80^\circ+\angle\text{CDA}=180^\circ$
$\Rightarrow\angle\text{CDA}=180^\circ-80^\circ=100^\circ$
Since, $ABCD$ is a cyclic quadrilateral.
$\angle\text{ADC}+\angle\text{ABC}=180^\circ$
$\Rightarrow100^\circ+\angle\text{ABC}=180^\circ$
$\Rightarrow\angle\text{ABC}=180^\circ-100^\circ=80^\circ$
Now, $\angle\text{ABC}+\angle\text{ABF}=180^\circ$ [Linear pair of angles]
$\Rightarrow80^\circ+\text{x}^\circ=180^\circ$
$\Rightarrow\text{x}=180^\circ-80^\circ=100^\circ$
View full question & answer→Question 384 Marks
In figure, $O$ is the centre of a circle and $PQ$ is a diameter. If $\angle\text{ROS} = 40^\circ, $ find $\angle\text{RTS}.$ 
AnswerSince $PQ$ is diameter Then, $\angle\text{PRQ} = 90^\circ$ [Angle in semicircle]
$\therefore\angle\text{PRQ}+\angle\text{TRQ}=180^\circ$ [Linear pair of angle]
$900+\angle\text{TRQ}=180^\circ$
$\angle\text{TRQ}=180^\circ-90^\circ=90^\circ$
By degree measure theorem
$\angle\text{ROS}=2\angle\text{RQS}$
$\Rightarrow40^\circ=2\angle\text{RQS}$
$\Rightarrow\angle\text{RQS}=\frac{40^\circ}{2}=20^\circ$ In
$\triangle\text{RQT},$ by Angle sum property $\angle\text{RQT}+\angle\text{QRT}+\angle\text{RTS}=180^\circ$
$\Rightarrow20^\circ+90^\circ+\angle\text{RTS}=180^\circ$
$\Rightarrow\angle\text{RTS}=180^\circ-20^\circ-90^\circ=70^\circ$
View full question & answer→Question 394 Marks
An equilateral triangle of side $9\ cm$ is inscribed in a circle. Find the radius of the circle.
AnswerLet $ABC$ be an equilateral triangle of side $9\ cm$ and let $AD$ is one of its median.
Let $G$ be the centroid of $\triangle\text{ABC}$ Then $AG : GD = 2 : 1$
We know that in an equilateral triangle, centroid coincides with the circum centre.
Therefore, $G$ is the centre of the circumference with circum radius $GA$.
Also $G$ is the centre and $GD$ is perpendicular to $BC.$
Therefore, In right triangle $ADB,$ we have $AB^2 = AD^2 + DB^2$
$\Rightarrow 9^2 = AD^2 + DB^2$
$\Rightarrow\text{AD}=\sqrt{81-\frac{81}{4}}=\frac{9\sqrt3}{2}\text{cm}$
$\therefore\text{Radius}=\text{AG}=\frac{2}{3}\text{AD}$
$=3\sqrt3\text{cm}.$ View full question & answer→Question 404 Marks
In a cyclic quadrilateral $ABCD$ if $AB || CD$ and $\angle\text{B}=70^\circ,$ find the remaining angles.
Answer
We have, $\angle\text{B}=70^\circ$
Since, $ABCD$ is a cyclic quadrilateral
Then, $\angle\text{B}+\angle\text{D}=180^\circ$
$\Rightarrow70^\circ+\angle\text{D}=180^\circ$
$\Rightarrow\angle\text{D}=180^\circ-70^\circ=110^\circ$
Since, $AB || DC$
Then, $\angle\text{B}+\angle\text{C}=180^\circ$ [Co-interior angles]
$\Rightarrow70^\circ+\angle\text{C}=180^\circ$
$\Rightarrow\angle\text{C}=180^\circ-70^\circ=110^\circ$
Now, $\angle\text{A} +\angle\text{C}=180^\circ$ [Opposite angles of cyclic quad.]
$\Rightarrow\angle\text{A}+110^\circ=180^\circ$
$\Rightarrow\angle\text{A}=180^\circ-110^\circ=70^\circ$ View full question & answer→Question 414 Marks
In figure, $O$ is the centre of the circle. If $\angle\text{APB}=50^\circ,$ find $\angle\text{AOB}$ and $\angle\text{OAB}.$
Answer$\angle\text{APB}=50^\circ$ By degree measure theorem
$\angle\text{AOB}=2\angle\text{APB}$
$\Rightarrow\angle\text{APB}=2\times50^\circ=100^\circ$ since $OA = OB$ [Radius of circle]
Then $\angle\text{OAB}=\angle\text{OBA}$ [Angles opposite toequalsides]
Let $\angle\text{OAB}=\text{x}$
In $\triangle\text{OAB},$ by angle sum property
$\angle\text{OAB}+\angle\text{OBA}+\angle\text{AOB}=180^\circ$
$\Rightarrow\text{x}+\text{x}+100^\circ=180^\circ$
$\Rightarrow\text{2x}=180^\circ-100^\circ$
$\Rightarrow\text{2x}=80^\circ$
$\Rightarrow\text{x}=40^\circ$
$\angle\text{OAB}=\angle\text{OBA}=40^\circ$
View full question & answer→Question 424 Marks
Draw different pairs of circles. How many points does each pair have in common? What is the maximum number of common points?
AnswerEach pair of circles have $0, 1$ or $2$ points in common. The maximum number of points in common is $‘2’$.
View full question & answer→Question 434 Marks
In figure, $O$ is the centre of the circle. Find $\angle\text{BAC}.$
AnswerWe have $\angle\text{AOB}=80^\circ$ And $\angle\text{AOC}=110^\circ$
Therefore, $\angle\text{AOB}+\angle\text{AOC}+\angle\text{BOC}=360^\circ$ [Complete angle]
$\Rightarrow80^\circ+100^\circ+\angle\text{BOC}=360^\circ$
$\Rightarrow\angle\text{BOC}=360^\circ-80^\circ-110^\circ$
$\Rightarrow\angle\text{BOC}=170^\circ$
By degree measure theorem$\angle\text{BOC}=2\angle\text{BAC}$
$\Rightarrow170^\circ=2\angle\text{BAC}$
$\Rightarrow\angle\text{BAC}=\frac{170^\circ}{2}=85^\circ$
View full question & answer→Question 444 Marks
In the given figure, $O$ and $O'$ are centers of two circles intersecting at $B$ and $C$. $ACD$ is a straight line, find $x$. 
AnswerBy degree measure theorem$\angle\text{AOB} = 2\angle\text{ACB}$
$\Rightarrow130^\circ=2\angle\text{ACB}$
$\Rightarrow\angle\text{ACB}=\frac{130^\circ}{2}=65^\circ$ [Liner a pair of angles]
$\Rightarrow65^\circ+\angle\text{BCD}=180^\circ$
$\Rightarrow\angle\text{BCD}=180^\circ-65^\circ=115^\circ$
By degree measure theorem reflex$\angle\text{BOD}=2\angle\text{BCD}$
$\Rightarrow\text{reflex }\angle\text{BOD}=2\times115^\circ=230^\circ$
Now, reflex $\angle\text{BOD}+\angle\text{BO}'\text{D}=360^\circ$ [Complex angle]
$\Rightarrow230^\circ+\text{x}=360^\circ$
$\Rightarrow\text{x}=360^\circ-230^\circ$
$\therefore\text{x}=130^\circ$
View full question & answer→Question 454 Marks
Given an arc of a circle, complete the circle.
Answer
Steps of Construction:
$1.$ Take three points $A, B$ and $C$ on the given arc.
$2.$ Join $AB$ and $BC$.
$3.$ Draw the perpendicular bisectors of chords $AB$ and $BC$ which intersect each other at point $O$.
Then $O$ will be the required centre of the required circle.
$4.$ Join $OA$.
$5.$ With centre $O$ and radius $OA$, complete the circle. View full question & answer→Question 464 Marks
Circles are described on the sides of a triangle as diameters. Proved that the circle on any two sides intersect each other on the third side (or third side produced).
Answer
Since, $AB$ is a diameter Then, $\angle\text{ADB}=90^\circ\dots(1)$ [Angles in semicircle]
Since, $AC$ is a diameter Then, $\angle\text{ADC}=90^\circ\dots(2)$ [Angles in semicircle]
Add equations $(1)$ and $(2)$
$\angle\text{ADB}+\angle\text{ADC}=90^\circ+90^\circ$
$\Rightarrow\angle\text{BDC}=180^\circ$
Then, $BDC$ is a line Hence, the circles on any two sides intersect each other on the third side. View full question & answer→Question 474 Marks
Prove that the circles described on the four sides of a rhombus as diameters, pass through the point of intersection of its diagonals.
Answer
Let $ABCD$ be a rhombus such that its diagonals $AC$ and $BD$ intersect at $O$.
Since, the diagonals of a rhombus intersect at right angle.
$\therefore\angle\text{AOB}=\angle\text{BOC}$
$\angle\text{COD}=\angle\text{DOA}=90^\circ$
Now, $\angle\text{AOB}=90^\circ\Rightarrow$ circle described on $AB$ as diameter will pass through $O$.
Similarly, all the circles described on $BC, AD$ and $CD$ as diameters pass through $O$. View full question & answer→