MCQ 11 Mark
In the given figure, $QR$ is a common tangent to the given circles touching externally at the point $T$. The tangent at $T$ meets $QR$ at $P$. If $PT = 3.8cm$, then the length of
$QR$ (in cm) is:
AnswerIt is given that $QR$ is a common tangent to the given circles touching externally at the point $T$.
Also, the tangent at $T$ meets $QR$ at $P$ such that $PT = 3.8cm$.
Now, $PQ$ and $PT$ are tangents drawn to the same circle from an external point.
$\therefore$ $PQ = PT = 3.8\ cm$ (Lengths of tangents drawn from an external point to a circle are equal)
$PR$ and $PT$ are tangents drawn to the same circle from an external point $T$.
$\therefore$ $PR = PT = 3.8\ cm$ (Lengths of tangents drawn from an external point to a circle are equal)
Now,
$QR = PQ + PR = 3.8\ cm + 3.8\ cm = 7.6\ cm$
Thus, the length of $QR$ is $7.6\ cm$.
View full question & answer→MCQ 21 Mark
In the figure, a quadrilateral $ABCD$ is drawn to circumscribe a circle such that its sides $AB, BC, CD$ and $AD$ touch the circle at $P, Q, R$ and $S$ respectively.
If $AB = x\ cm$$, BC= 7\ cm, CR = 3\ cm$ and $AS = 5\ cm$, then $x =$
AnswerIn the given figure,
$ABCD$ is a quadrilateral circumscribe a circle and its sides $AB, BC, CD$ and $DA$ touch the circle at $P, Q, R$ and $S$ respectively.
$AB = x\ cm, BC = 7\ cm, CR = 3\ cm, AS = 5\ cm$
$CR$ and $CQ$ are tangents to the circle from $C$
$CR = CQ = 3\ cm$
$BQ = BC – CQ = 7 – 3 = 4\ cm$
$BQ =$ and $BP$ are tangents from $B$
$BP = BQ = 4\ cm$
$AS$ and $AP$ are tangents from A$AP = AS = 5\ cm$
$AB = AP + BP = 5 + 4 = 9\ cm$
$x = 9\ cm$
View full question & answer→MCQ 31 Mark
In the given figure, if $AD, AE$ and $BC$ are tangents to the circle at $D, E$ and $F$ respectively, Then:
- A
$AD = AB + BC + CA$
- ✓
$2AD = AB + BC + CA$
- C
$3AD = AB + BC + CA$
- D
$4AD = AB + BC + CA$
AnswerCorrect option: B. $2AD = AB + BC + CA$
By the property of tangent
$AC = AB ($tangent from $A)...(i)$
$CD = CF ($tangent from $C)...(ii)$
$BF = BE ($tangent from $B)...(iii)$
Now taking $RHS,$
A$B + BC + CA = AB + BF + FC + CA$
$AB + BC + CA = AB + BE + CD + CA$ $[$ from $(ii)$ & $(iii) ]$
$AB + BC + CA = AE + AD$
$AB + BC + CA = 2AD$ View full question & answer→MCQ 41 Mark
The length of the tangent from a point $A$ at a circle, of radius $3\ cm,$ is $4\ cm$. The distance of $A$ from the centre of the circle is:
- A
$\sqrt{7}\text{cm}$
- B
$7\ cm$
- ✓
$5\ cm$
- D
$25\ cm$
AnswerCorrect option: C. $5\ cm$
We know, radius always perpendicular to tangent so we say $\triangle\text{OPA}$ is right angle triangle then $\angle\text{OPA}=90^{\circ}$
Now, we have to find OA
$ \Rightarrow O A^2=A P^2+O P^2 $
$ \Rightarrow O A^2=4^2+3^2$
$ \Rightarrow O A^2=16+9$
$\Rightarrow OA = \sqrt{25}$
$\Rightarrow OA = 5$
Hence, correct choice is $(c)$ View full question & answer→MCQ 51 Mark
If $PT$ is tahgent drawn froth a point P to a circle touching it at $T$ and $O$ is the centre of the circle, then $∠OPT + ∠POT =$
- A
$30^\circ$
- B
$60^\circ$
- ✓
$90^\circ$
- D
$180^\circ$
AnswerCorrect option: C. $90^\circ$
In the figure, $PT$ is the tangent to the circle with centre $O$.
$OP$ and $OT$ are joined
$PT$ is tangent and $OT$ is the radius
$OT ⊥ PT$
Now in right $\triangle\text{OTP}$
$\angle\text{OTP}=90^{\circ}$
$\angle\text{OPT}+\angle\text{POT}=180^{\circ}-90^{\circ}=90^{\circ}$ View full question & answer→MCQ 61 Mark
In the figure, a circle touches the side $DF$ of $\triangle\text{EDF}$ at $H$ and touches $ED$ and $EF$ produced at $K$ and $M$ respectively. If $EK = 9cm$, then the perimeter $\triangle\text{EDF}$ of is:
- ✓
$18\ cm$
- B
$13.5\ cm$
- C
$12\ cm$
- D
$9\ cm$
AnswerCorrect option: A. $18\ cm$
In $\triangle\text{DEF}$
DF touches the circle at H and circle touches $ED$ and $EF$ Produced at $K$ and $M$ respectively.
$EK = 9\ cm$
$EK$ and EM are the tangents to the circle.
$EM = EK = 9\ cm$
Similarly $DH$ and $DK$ are the tangent.
$DH = DK$ and $FH$ and $FM$ are tangents.
$FH = FM$
Now, perimeter of $\triangle\text{DEF}$$= ED + DF + EF$
$= ED + DH + FH + EF$
$= ED + DK + EM + EF$
$= EK + EM$
$= 9 + 9$
$= 18\ cm.$
View full question & answer→MCQ 71 Mark
In the given figure, there are two concentric circles with centre $O$. $PR$ and $PQS$ are tangents to the inner circle from point plying on the outer circle. If PR = $7.5\ cm$, then $PS$ is equal to:
- A
$10\ cm$
- B
$12\ cm$
- ✓
$15\ cm$
- D
$18\ cm$
AnswerCorrect option: C. $15\ cm$
Here, $PO = OS$(radius)
then $\triangle\text{POS}$ called isosceles triangle.
We know, In isosceles triangle line drawn from vertex to base, then line bisects the base in equal parts. so we say,
$PQ = QS ...(i)$
From the property of tangent
$PR = PQ = 7.5\ cm [$tangent from point $P] ...(ii)$
Now we have to find $PS,$
$PS = PQ + QS$
$\Rightarrow PS = PQ + PQ [$from eq.$(i)]$
$\Rightarrow PS = 7.5 + 7.5 [$fromeq.$(ii)]$
$\Rightarrow PS =15cm$ View full question & answer→MCQ 81 Mark
In the figure, if $PQR$ is the tangent to a circle at $Q$ whose centre is $O$, $AB$ is a chord parallel to $PR$ and $\angle\text{BQR}=70^\circ$ then $\angle\text{AQB}$ is equal to:
- A
$20^\circ $
- ✓
$40^\circ$
- C
$35^\circ$
- D
$45^\circ $
AnswerCorrect option: B. $40^\circ$
Given, $AB || PR$
$\angle\text{ABQ}=\angle\text{BQR}=70^\circ$ [alternate angles]
Also QD is perpendicular to AB and QD bisects AB.
In $\triangle\text{QDA}\ \text{and}\ \triangle\text{QDB}$
$\angle\text{QDA}= \angle\text{QDB}$ [each 90°]
$AD = BD$
$QD = QD$ [common side]
$\triangle\text{ADQ}=\angle\text{BDQ}$ [by SAS similarity criterion]
Then, $\angle\text{QAD}=\angle\text{QDA}\ ....(\text{i})$ [c.p.c.t.]
Also, $\angle\text{ABQ}=\angle\text{BQR}$ [alternate interior angle]
$\angle\text{ABQ}=70^\circ$$[\angle\text{BQR}=70^\circ]$
Hence, $\angle\text{QAB}=70^\circ [$from Eq. $(i)]$
Now, In $\triangle\text{ABQ},$
$\angle\text{A}+\angle\text{B}+\angle\text{Q}=180^\circ$
$\Rightarrow\angle\text{Q}=180^\circ-(70^\circ+70^\circ)=40^\circ$
View full question & answer→MCQ 91 Mark
If radii of two concentric circles are $4\ cm$ and $5\ cm$, then the length of each chord of one circle which is tangent to the other circle is:
- A
$3\ cm$
- ✓
$6\ cm$
- C
$9\ cm$
- D
$1\ cm$
AnswerCorrect option: B. $6\ cm$
Let $O$ be the centre of two concentric circles $C_1$ and $C_2$, whose radii are $r_1= 4$ cm and $r_2= 5cm.$
Now, we draw a chord $AC$ of circle $C_2$, which touches the circle $C_1$ at B.
Also, join $OB$, which is perpendicular to $AC$.
[Tangent at any point of circle is perpendicular to radius throughly the point of contact]
Now, in right angled $\triangle\text{OBC},$ by using Pythagoras theorem,
$ \Rightarrow \mathrm{OC}^2=\mathrm{BC}^2+\mathrm{BO}^2\left[(\text { hypotenuse })^2=(\text { base })^2+(\text { perpendicular })^2\right] $
$ \Rightarrow 5^2=\mathrm{BC}^2+4^2 $
$ \Rightarrow B C^2=25-16=9$
$⇒ BC = 3cm$
Length of chord $AC = 2 BC = 2 × 3 = 6cm.$
View full question & answer→MCQ 101 Mark
In the given figure, if $AB = 8\ cm$ and $PE = 3\ cm$, then $AE =$
- A
$11\ cm$
- B
$7\ cm$
- ✓
$5\ cm$
- D
$3\ cm$
AnswerCorrect option: C. $5\ cm$
We know that tangents drawn from the same external point will be equal in length.
Therefore,
$AB = AC$
It is given that,
$AB = 8\ cm$
Hence,
$AC = 8\ cm …… (1)$
Similarly,
$PE = CE$
It is given that,
$PE = 3\ cm$
Therefore,
$CE = 3\ cm …… (2)$
Subtracting equations $(1)$ and $(2)$, we get,
$AC − CE = 8 − 3$
From the figure we can see that,
$AC − CE = AE$
Therefore,
$AE = 8 − 3$
$AE = 5\ cm$ View full question & answer→MCQ 111 Mark
In the given figure, $RQ$ is a tangent to the circle with centre $O$. If $SQ = 6\ cm$ and $QR = 4\ cm,$ then $OR =$
- A
$8\ cm$
- B
$3\ cm$
- C
$2.5\ cm$
- ✓
$5\ cm$
AnswerCorrect option: D. $5\ cm$
In the figure, $0$ is the centre of the circle $QR$ is tangent to the circle and $QOS$ is a diameter $SQ = 6\ cm, QR = 4\ cm$
$\text{OQ}=\ \frac{1}{2}\ \text{QS}=\frac{1}{2}\times6=3\text{cm}$
$OQ$ is radius
$OQ ⊥ QR$
Now in right $\triangle\text{OQR}$
$O R^2=Q R^2+Q O^2=(3)^2+(4)^2=9+16=25=(5)^2$
$OR = 5cm$
View full question & answer→MCQ 121 Mark
If two tangents inclined at a angle of $60^\circ $ are drawn to a circle of radius $3\ cm$, then length of each tangent is equal to:
AnswerCorrect option: D. $3\sqrt{3}\text{cm}$
Let $P$ be an external point and a pair of tangents is drawn from point $P$ and angle between these two tangents is $60°$. Join $OA$ and $OP$.
Also, $OP$ is a bisector of line.
$\angle\text{APO}=\angle\text{CPO}=30^\circ$
Also, $OA ⊥ AP$
Tangent at any point of a circle is perpendicular to the radius through the point of contact.
In right angled $\triangle\text{OAP},\tan30^\circ=\frac{\text{OA}}{\text{AP}}=\frac{3}{\text{AP}}$
$\Rightarrow\frac{1}{\sqrt{3}}=\frac{3}{\text{AP}}$
$\Rightarrow\text{AP}=3\sqrt{3}\text{cm}$
Hence, the length of each tangent is $3\sqrt{3}\text{cm}$
View full question & answer→MCQ 131 Mark
If angle between two radii of a circle is $130^\circ$, the angle between the tangent at the ends of radii is:
- A
$90^\circ$
- ✓
$50^\circ$
- C
$70^\circ$
- D
$40^\circ$
AnswerCorrect option: B. $50^\circ$
Let $PQ$ and $RP$ be the radii of the circle with the centre $O.$
$\angle\text{ROQ}=130^\circ$
$\text{RP}\bot\text{OR}\ \text{and}\ \text{PQ}\bot\text{OQ}$ (Radii are perpendicular to the tangent)
In quadilateral $ROQP,$
$\angle\text{ORP}+\angle\text{RPQ}+\angle\text{PQO}+\angle\text{QOR}=360^\circ$
$\Rightarrow90^\circ+\angle\text{RPQ}+90^\circ+130^\circ=360^\circ$
$\Rightarrow\angle\text{RPQ}=50^\circ$
View full question & answer→MCQ 141 Mark
In the figure, if $PR$ is tangent to the circle at $P$ and $Q$ is the centre of the circle, then $∠POQ =$
- A
$110^\circ $
- B
$100^\circ $
- ✓
$120^\circ$
- D
$90^\circ$
AnswerCorrect option: C. $120^\circ$
We know, radius $OP$$\bot $ to tangent $PR$ then $\angle\text{OPR}=90^{\circ}$
Now,
$\angle\text{OPQ}=\angle\text{OPR}-\angle\text{QPR}$
$\angle\text{OPQ}=90^{\circ}-60^{\circ}$
$\angle\text{OPQ}=30^{\circ}$
In $\triangle\text{OPQ},$
$OP = OQ$ (radius of circle)
$\angle\text{OPQ}=\angle\text{OQP}=30^{\circ}$(opposite angle of same side)
we also know that sum of all angle of triangle is $180^\circ$, then
$\angle\text{OPQ}+\angle\text{OQP}+\angle\text{POQ}=180^{\circ}$
$\Rightarrow30^{\circ}+30^{\circ}+\angle\text{POQ}=180^{\circ}$
$\Rightarrow\angle\text{POQ}=120^{\circ}$ View full question & answer→MCQ 151 Mark
From a point $Q,$ the length of the tangent to a circle is $24\ cm $and the distance of $Q$ from the centre is $25\ cm$. The radius of the circle is:
- ✓
$7\ cm$
- B
$12\ cm$
- C
$15\ cm$
- D
$24.5\ cm$
AnswerCorrect option: A. $7\ cm$
We know, radius always perpendicular to tangent so we say $\triangle\text{OPQ}$ is right angle triangle then $\angle\text{OPQ}=90^{\circ}$
Now, we have to find $OP$
$ \Rightarrow O P^2=O Q^2-P Q^2 $
$ \Rightarrow O P^2=25^2-24^2$
$ \Rightarrow O P^2=625-576$
$\Rightarrow\text{OP}=\sqrt{49}$
$\Rightarrow OP = 7\ cm$
Hence, correct choice is $(A)$
View full question & answer→MCQ 161 Mark
From a point $P$ which is at a distance $13\ cm$ from the centre $O$ of a circle of radius $5\ cm,$ the pair of tangent $PQ$ and $PR$ to the circle are drawn. Then the area of the quadrilateral $PQOR$ is:
- ✓
$60 \mathrm{~cm}^2$
- B
$65 \mathrm{~cm}^2$
- C
$30 \mathrm{~cm}^2$
- D
$32.5 \mathrm{~cm}^2$
AnswerCorrect option: A. $60 \mathrm{~cm}^2$
Firstly, draw a circle of radius $5\ cm$ having centre $O$.
$P$ is a point at a distance of $13\ cm$ from $O$.
A pair of tangents $PQ$ and $PR$ are drawn.
Thus, quadrilateral $PQOR$ is formed.
$OQ ⊥ QP$ $[$since, $AP$ is a tangent line$]$
In right angled $\triangle\text{POQ}$
$ \Rightarrow O P^2=O Q^2+Q P^2 $
$ \Rightarrow 13^2=5^2+Q P^2 $
$\Rightarrow Q P^2=169-25=144=12^2$
$ \Rightarrow Q P=12 \mathrm{~cm}$
Now, $\text{area}\ \text{of}\triangle\text{OQP}=\frac{1}{2}\times\text{QP}\times\text{QO}=\frac{1}{2}\times12\times5=30\text{cm}^2$
Area of quadilateral $QORP$ = $2\triangle\text{OQP}=2\times30=60\text{cm}^2$ View full question & answer→MCQ 171 Mark
If $TP$ and $TQ $are two tangents to a circle with centre $O$ so that $\angle\text{POQ}=110^{\circ},$ then, $\angle\text{PTQ}$ is equal to:
- A
$60^\circ$
- ✓
$70^\circ$
- C
$80^\circ$
- D
$90^\circ$
AnswerCorrect option: B. $70^\circ$
$TP$ and $TQ$ are the tangents from $T$ to the circle with centre $O$ and $OP, OQ$ are joined and $\angle\text{POQ}=110^{\circ},$
But $\angle\text{POQ}+\angle\text{PTQ}=180^{\circ}$
$\Rightarrow110^{\circ}+\angle\text{PTQ}=180^{\circ}$
$\Rightarrow\angle\text{PTQ}=180^{\circ}-110^{\circ}=70^{\circ}$ View full question & answer→MCQ 181 Mark
In the figure, if $\angle\text{AOB}=125^\circ$ then $\angle\text{COD}$ is equal to:
- A
$45^\circ$
- B
$35^\circ$
- ✓
$55^\circ$
- D
$62\frac{1}{2}^\circ$
AnswerCorrect option: C. $55^\circ$
We know that, the opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle
$\angle\text{AOB}+\angle\text{COD}=180^\circ$
$\Rightarrow\angle\text{COD}=180^\circ-\angle\text{AOB}=180^\circ-125^\circ=55^\circ$
View full question & answer→MCQ 191 Mark
If tangents $PA$ and $PB$ from a point $P$ to a circle with centre $O$ are inclined to each other at
- ✓
$50^\circ$
- B
$60^\circ$
- C
$70^\circ$
- D
$80^\circ$
AnswerCorrect option: A. $50^\circ$
We know, radius is always perpendicular to tangent
then,
$\angle\text{OAP}=90^{\circ}(\text{OA}\bot\text{PA})$
$\angle\text{OBP}=90^{\circ}(\text{OB}\bot\text{PB})$
$\angle\text{APB}=80^{\circ}(\text{given})$
We also know that sum of all angles of a quadilateral is $360^\circ$ then,
$\angle\text{OAP}+\angle\text{OBP}+\angle\text{APB}+\angle\text{AOB}=360^{\circ}$
$\Rightarrow90^{\circ}+90^{\circ}+80^{\circ}+\angle\text{AOB}=360^{\circ}$
$\Rightarrow260^{\circ}+\angle\text{AOB}=360^{\circ}$
$\Rightarrow\angle\text{AOB}=100^{\circ}...(\text{i})$
Now, consider $\triangle\text{POA}\ \text{and}\ \triangle\text{POB},$
$OA = OB$ (Radius of circle)
$PA = PB$ (tangent grom external point $P$)
$OP = OP$ (commom)
So, By using SSS congurancy,
$\triangle\text{POA}\cong\triangle\text{POB}$
then $\angle\text{POA}=\triangle\text{POB}...(\text{ii})$
By eq. $(i)$ and $(ii)$
$\Rightarrow\angle\text{AOB}=100^{\circ}$
$\Rightarrow\angle\text{POA}+\angle\text{POB}=100^{\circ}$
$\Rightarrow2\angle\text{POA}=100^{\circ}$
$\Rightarrow\angle\text{POA}=50^{\circ}$
Hence, correct choice is $(A)$. View full question & answer→MCQ 201 Mark
The length of the tangent drawn from a point $8\ cm$ away from the centre of a circle of radius $6\ cm$ is:
- A
$\sqrt{7}\text{cm}$
- ✓
$2\sqrt{7}\text{cm}$
- C
$10\text{cm}$
- D
$5\text{cm}$
AnswerCorrect option: B. $2\sqrt{7}\text{cm}$
Let us first put the given data in the form of a diagram.
We know that the radius of a circle will always be perpendicular to the tangent at the point of contact.
Therefore, $OP$ is perpendicular to $QP$.
We can now use Pythagoras theorem to find the length of $QP$.
$QP^2= OQ^2- OP^2$
$QP^2= 8^2- 6^2$
$QP^2= 64 - 36$
$QP^2= 28$
QP = $\sqrt{28}$
QP = $2\sqrt{7}$
View full question & answer→MCQ 211 Mark
In the figure, $PR =$
- A
$20\ cm$
- ✓
$26\ cm$
- C
$24\ cm$
- D
$28\ cm$
AnswerCorrect option: B. $26\ cm$
In the figure, two circles with centre $O$ and $O’$ touch each other externally $PQ$ and $RS$ are the tangents drawn to the circles.
$OQ$ and $O’S$ are the radii of these circles and $OQ = 3\ cm, PQ = 4\ cm O’S = 5\ cm$ and $SR = 12\ cm$.
Now in right $\triangle\text{OQP}$
$ O P^2=(O Q)^2+P Q^2=(3)^2+(4)^2=9+16=25=(5)^2 $
$ O P=5 \mathrm{~cm}$
$\text { Similarly in right } \triangle \mathrm{RSO}$
$\left(O^{\prime} R\right)^2=(R S)^2+\left(O^{\prime} S\right)^2=(12)^2+(5)^2=144+25=169=(13)^2$
$O’R = 13cm$
Now $PR = OP + OO’ + O’R = 5 + (3 + 5) + 13 = 26cm.$
View full question & answer→MCQ 221 Mark
In the given figure, $PQ$ and $PR$ are tangents drawn from $P$ to a circle with centre $O$. If $\angle\text{OPQ}=35^\circ$, then:
- A
$a= 30^\circ , b= 60^\circ $
- ✓
$a= 35^\circ , b = 55^\circ $
- C
$a= 40^\circ , b = 50^\circ $
- D
$a= 45^\circ , b = 45^\circ $
AnswerCorrect option: B. $a= 35^\circ , b = 55^\circ $
We know, radius always $\bot$ $TP$ tangent
$\text{OQ}\bot\text{QP}$
$\text{OR}\bot\text{RP}$
From above eq. $\triangle\text{OQP}$ and $\triangle\text{ORP}$ is right angle triangle then,
$\angle\text{OQP}=\angle\text{ORP}=90^\circ$
$\triangle\text{OQP}\sim\triangle\text{ORP}$
then $\angle\text{QPO}=\angle\text{RPO}=35^\circ=\angle\text{a}$
sum of all angles in $\triangle\text{OQP}$ is $180^\circ $
$\angle\text{OQP}+\angle\text{QPO}+\angle\text{QOP}=180^\circ$
$\Rightarrow90^\circ+35^\circ+\angle\text{b}=180^\circ$
$\Rightarrow\angle\text{b}=55^\circ$ View full question & answer→MCQ 231 Mark
Two circles touch each other externally at P. $AB$ is a common tangent to the circle touching them at $A$ and $B$. The value of $\angle\text{APB}$ is:
- A
$30^\circ$
- B
$45^\circ$
- C
$60^\circ$
- ✓
$90^\circ$
AnswerCorrect option: D. $90^\circ$
It is given that two circles touch each other externally at $P$. $AB$ is a common tangent to the circle touching them at $A$ and $B$.
Draw a tangent to the circle at $P$, intersecting $AB$ at $T$.
Now, $TA$ and $TP$ are tangent drawn to the same circle from an external point $T$.
$\therefore$ $TA = TP$ (Length of tangents drawn from an external point to a circle are equal)
$TB$ and $TP$ are tangent drawn to the same circle from an external point T.
$\therefore$ $TB = TP$ (Length of tangents drawn from an external point to a circle are equal)
In $\triangle\text{ATP}$
$TA = TP$
$\therefore\angle\text{APT}=\angle\text{PAT}...(1)$(In a triangle, equal sides have equal angles opposite to them)
In $\triangle\text{BTP},$
$TB = TP$
$\therefore\angle\text{BPT}=\angle\text{PBT}...(2)$(In a triangle, equal sides have equal angles opposite to them)
Now, in $\triangle\text{APB},$
$\Rightarrow\angle\text{APB}+\angle\text{PAB}+\angle\text{PBA}=180^\circ$(Angle sum property)
$\Rightarrow\angle\text{APB}+\angle\text{APT}+\angle\text{BPT}=180^\circ$$[$From $(1)$ and $(2)]$
$\Rightarrow\angle\text{APB}+\angle\text{APB}=180^\circ$
$\Rightarrow2\angle\text{APB}=180^\circ$
$\Rightarrow\angle\text{APB}=90^\circ$
Thus, the value of $\angle\text{APB}\ \text{is}\ 90^\circ$ View full question & answer→MCQ 241 Mark
The pair of tangents $AP$ and $AQ$ drawn from an external point to a circle with centre $O$ are perpendicular to each other and length of each tangent is $5\ cm$. The radius of the circle is:
- A
$10\ cm$
- B
$7.5\ cm$
- ✓
$5\ cm$
- D
$2.5\ cm$
AnswerCorrect option: C. $5\ cm$
Given: $AP$ and $AQ$ are tangents to the ciecle with centre $O, AP$ $\bot$ $AQ$ and $AP = AQ = 5cm$
We know that radius of a circle is perpendicular to the tangent at the point of contact.
$\Rightarrow\text{OP}\bot\text{AP}\ \text{and}\ \text{OQ}\bot\text{AQ}$
Also sum of all angles of a quadilateral is $360^\circ $
$\Rightarrow\angle\text{O}+\angle\text{P}+\angle\text{A}+\angle\text{Q}=360^\circ$
$\Rightarrow\angle\text{O}+90^\circ+90^\circ=360^\circ$
$\Rightarrow\angle\text{O}=360^\circ-270^\circ=90^\circ$
Thus$\angle\text{O}=\angle\text{P}=\angle\text{A}=\angle\text{Q}=90^\circ$
$\Rightarrow OPAQ$ is a rectangle.
Since adjacent sides of $OPAQ$ i.e. $AP$ and $AQ$ are equal. Thus $OPAQ$ is a square radius $= OP = OQ = AP = AQ = 5cm$ View full question & answer→MCQ 251 Mark
If $PA$ and $PB$ are tangents to the circle with centre $O$ such that $\angle\text{APB}=50^\circ$ then $\angle\text{OAB}$ is equal to:
- ✓
$25^\circ$
- B
$30^\circ$
- C
$40^\circ$
- D
$50^\circ$
AnswerCorrect option: A. $25^\circ$
Given, $PA$ and $PB$ are tangent lines.
$PA = PB$ [Since, the length of tangents drawn from an$\angle\text{PAB}=\angle\text{PBA}=\theta$ [say]
In $\triangle\text{PAB},$
$\angle\text{P}+\angle\text{A}+\angle\text{B}=180^\circ$ [since, sum of angles of a triangle $= 180^\circ$
$50^\circ+\theta+\theta=180^\circ$
$2\theta=180^\circ-50^\circ=130^\circ$
$\theta=65^\circ$
Also, $OA ⊥ PA$ [Since, tangent at any point of a circle is perpendicular to the radius through the point of contact.]
$\angle\text{PAO}=90^\circ$
$\Rightarrow\angle\text{PAB}+\angle\text{BAO}=90^\circ$
$\Rightarrow65^\circ+\angle\text{BAO}=90^\circ$
$\Rightarrow\angle\text{BAO}=90^\circ-65=25^\circ$
View full question & answer→MCQ 261 Mark
In the given figure, a circle with centre $O$ is inscribed in a quadrilateral $ABCD$ such that, it touches sides $BC, AB, AD$ and $CD$ at points $P, Q, R$ and $S$ respectively.
If $AB$ $= 29cm, AD = 23cm$, $\angle\text{B}=90^\circ$ and $DS = 5cm,$ then the radius of the circle (in cm) is:
AnswerIn the figure, a circle touches the sides of a quadrilateral $ABCD$.
$\angle\text{B}=90^\circ,$ $OP = OQ = r$
$AB = 29cm, AD = 23cm, DS = 5cm$
$\angle\text{B}=90^\circ,$
$BA$ is tangent and $OQ$ is radius
$\angle\text{QPB}=90^\circ$
Similarly $OP$ is radius and $BC$ is tangents.
$\angle\text{OPB}=90^\circ$
But $\angle\text{B}=90^\circ,$ (given)
$PBQO$ is a square.
$DS = 5cm$
But $DS$ and $DR$ are tangents to the circles.
$DR = 5cm$
But $AD = 23cm$
$AR = 23 – 5= 18cm$
$AR = AQ$ (tangents to the circle from $A$.)
$AQ = 18cm$
But $AB = 29 cm$
$BQ = 29 – 18 = 11cm$
$OPBQ$ is a square.
$OQ = BQ = 11cm$
Radius of the circle $= 11cm$.
View full question & answer→MCQ 271 Mark
If four sides of a quadrilateral $ABCD$ are tangential to a circle, then:
- A
$AC + AD = BD + CD$
- ✓
$AB + CD = BC + AD$
- C
$AB + CD = AC + BC$
- D
$AC + AD = BC + DB$
AnswerCorrect option: B. $AB + CD = BC + AD$
A circle is inscribed in a quadrilateral $ABCD$ which touches the sides $AB, BC, CD$ and $DA at P, Q, R$ and $S$ respectively then the sum of two opposite sides is equal
to the sum of other two opposite sides
$AB + CD = BC + AD$
View full question & answer→MCQ 281 Mark
Two equal circles touch each other externally at $C$ and $AB$ is a common tangent to the circles. Then, $\angle\text{ACB}=$
- A
$60^\circ$
- B
$45^\circ$
- C
$30^\circ$
- ✓
$90^\circ$
AnswerCorrect option: D. $90^\circ$
we know, radius always $\bot$ to tangent then,
$\angle\text{OAB}=\angle\text{a}+\angle\text{b}=90^{\circ}[\because\text{OA}\bot\text{AB}]$
$\Rightarrow\angle\text{a}=90^{\circ}=\angle\text{b}...(\text{i})$
$\angle\text{O'BA}=\angle\text{C}+\angle\text{d}=90^{\circ}$
$\Rightarrow\angle\text{d}=90^{\circ}-\angle\text{c}...(\text{ii})$
Here, redius are equal.
$\angle\text{a}=\angle\text{e}$ (opposite angle of same side)
$\angle\text{d}=\angle\text{f}$ (opposite angle of same side)
Now,
$\Rightarrow\angle\text{ACB}=\angle\text{OCO'}-\angle\text{e}-\angle\text{f}$
$\Rightarrow\angle\text{ACB}=180^{\circ}-\angle\text{a}-\angle\text{d}$
Put the value from eq. $(i)$ and $(ii)$
$\Rightarrow\angle\text{ACB}=180^{\circ}-(90-\angle\text{b})-(90-\angle\text{c})$
$\Rightarrow\angle\text{ACB}=180^{\circ}-90-\angle\text{b}-90-\angle\text{c}$
$\Rightarrow\angle\text{ACB}=\angle\text{b}+\angle\text{c}...(\text{iii})$
Now In $\triangle\text{ACB}$
$\angle\text{b}+\angle\text{c}+\angle\text{ACB}=180^{\circ}$
from eq$....(iii)$
$\Rightarrow\angle\text{ACB}+\angle\text{ACB}=180^{\circ}$
$\Rightarrow2\angle\text{ACB}=180^{\circ}$
$\Rightarrow\angle\text{ACB}=90^{\circ}$ View full question & answer→MCQ 291 Mark
In the figure, if $AP = 10\ cm,$ then $BP =$
- A
$\sqrt{91}\text{cm}$
- ✓
$\sqrt{127}\text{cm}$
- C
$\sqrt{119}\text{cm}$
- D
$\sqrt{109}\text{cm}$
AnswerCorrect option: B. $\sqrt{127}\text{cm}$
In the figure,
$OA = 6cm, OB = 3cm$ and $AP = 10cm$
OA is radius and AP is the tangent
$OA ⊥ AP$
Now in right $\triangle\text{OAP}$
$O P^2=A P^2+O A^2=(10)^2+(6)^2=100+36=136$
$\text { Similarly } B P \text { is tangent and } O B \text { is radius }$
$ O P^2=O B^2+B P^2$
$ 136=(3)^2+B P^2$
$ 136=9+B P^2$
$\Rightarrow BP^2= 136 – 9 = 127$
BP = $\sqrt{127}\text{cm}$
View full question & answer→MCQ 301 Mark
In the figure, if tangents $PA$ and $PB$ are drawn to a circle such that $\angle\text{APB}=30^\circ$ and chord $AC$ is drawn parallel to the tangent $PB$, then $\angle\text{ABC}$ =
- A
$60^\circ$
- B
$90^\circ$
- ✓
$30^\circ$
- D
AnswerCorrect option: C. $30^\circ$
By property of tangent $PA = PB$ (tangent from P)
then, In $\triangle\text{ABP}$
$PA = PB$ and $\angle\text{PAB}=\angle\text{ABP}$
Sum of all angles of triangle APB is 180°
$\angle\text{PAB}+\angle\text{ABP}+\angle\text{APB}=180^\circ$
$\Rightarrow\angle\text{ABP}+\angle\text{ABP}+30^\circ=180^\circ$
$\Rightarrow2\angle\text{ABP}=150^\circ$
$\Rightarrow\angle\text{ABP}=75^\circ$
$\angle\text{ABP}=\angle\text{BAC}=75^\circ$(Alternate algles)
$\angle\text{ABP}=\angle\text{ACB}=75^\circ$(Alternate segment theorem)
Now, sum of all angles of $\triangle\text{ABC}$ 180°
$\Rightarrow\angle\text{BAC}+\angle\text{ACB}+\angle\text{ABC}=180^\circ$
$\Rightarrow75^\circ+75^\circ+\angle\text{ABC}=180^\circ$
$\Rightarrow\angle\text{ABC}=30^\circ$ View full question & answer→MCQ 311 Mark
In the given figure, if quadrilateral $PQRS$ circumscribes a circle, then $PD + QB =$
Answer
We know that tangents drawn to a circle from the same external point will be equal in length.
Therefore,
$PD = PA …… (1)$
$QB = QA …… (2)$
Adding equations $(1)$ and $(2)$, we get,
$PD + QB = PA + QA$
By looking at the figure we can say,
$PD + QB = PQ$. View full question & answer→MCQ 321 Mark
In a right triangle $ABC$, right angled at $B, BC = 12\ cm$ and $AB = 5\ cm$. The radius of the circle inscribed in the triangle (in cm) is:
Answer
$ \Rightarrow A C^2=A B^2+(B C)^2[\text { Pythagoras theorem }] $
$ \Rightarrow A C^2=25+144=169$
$\Rightarrow AC = 13cm$
$\text{ar}.\ \text{of}\ \triangle\text{ABC}$ $=\text{ar}.\ \text{of}\ \triangle\text{AOB}$ $+\text{ar}.\ \text{of}\ \triangle\text{BOC}$ $+\text{ar}.\ \text{of}\ \triangle\text{AOC}$
$\frac{5\times12}{2}=\frac{\text{AB}\times\text{r}}{2}+\frac{\text{BC}\times\text{r}}{2}+\frac{\text{AC}\times\text{r}}{2}$
$60=\text{r}(\text{AB}+\text{BC}+\text{AC})$ $[\because\text{Area}\ \text{of}\ \triangle=\frac{\text{Base}\times\text{Corr.alt}}{2}]$
$60=\text{r}(5+12+13)$
$60=30\text{r}\Rightarrow\text{r}=2\text{cm}$
View full question & answer→MCQ 331 Mark
$AB$ and $CD$ are two common tangents to circles which touch each other at $C$. If $D$ lies on $AB$ such that $CD = 4\ cm$, then $AB$ is equal to:
- A
$4\ cm$
- B
$6\ cm$
- ✓
$8\ cm$
- D
$12\ cm$
AnswerCorrect option: C. $8\ cm$
By property of tangent,
$AD = DC$ $($tangent from $D)$
$DB = DC$ $($tangent from $D)$
Given, $DC = 4\ cm$
Now, we have to find $AB$
$AB = AD + DB$
$\Rightarrow AB = DC + DC$
$\Rightarrow AB = 2DC$
$\Rightarrow AB = 2 \times 4$
$\Rightarrow AB = 8\ cm$ View full question & answer→MCQ 341 Mark
Two concentric circles of radii $3\ cm$ and $5\ cm$ are given. Then length of chord $BC$ which touches the inner circle at $P$ is equal to:
- A
$4\ cm$
- B
$6\ cm$
- ✓
$8\ cm$
- D
$10\ cm$
AnswerCorrect option: C. $8\ cm$
Here, radius $OQ$ $\bot$ to tangent $AB$ then we say,
$\angle\text{OQA}=\angle\text{OQB}=90^\circ$
and $\triangle\text{OQA}$ is right angle triangle then,
$ A Q^2=O A^2-O Q^2$
$ \Rightarrow A Q 2=5^2-3^2 $
$ \Rightarrow A Q^2=25-9$
$\Rightarrow\text{AQ}^2=\sqrt{16}$
$\Rightarrow AQ = 4cm$
By property of tangent.
$BQ = BP$$($tangent from point $B)$
$\because$ $OQ$ bisects $AB$ then $AQ = QB = 4cm$
$OP$ bisects $AB$ then $BP = PC = 4cm$
Now, we have to find $BC,$
$BC = BP + PC$
$\Rightarrow BC = 4 + 4$
$\Rightarrow BC = 8\ cm.$ View full question & answer→MCQ 351 Mark
At one end of a diameter $PQ$ of a circle of radius 5cm, tangent $XPY$ is drawn to the circle. The length of chord $AB$ parallel to $XY$ and at distance of $8\ cm$ from $P$ is:
- A
$5\ cm$
- B
$6\ cm$
- C
$7\ cm$
- ✓
$8\ cm$
AnswerCorrect option: D. $8\ cm$
In the figure, $PQ$ is diameter $XPY$ is tangent to the circle with centre $O$ and radius $5\ cm$ From $P$, at a distance of $8\ cm$ $AB$ is a chord drawn parallel to $XY$.
To find the length of $AB$ Join $OA$
$XY$ is tangent and $OP$ is the radius.
$OP ⊥ XY$ or $PQ ⊥ XY$$AB || XY$
$OQ$ is $⊥ AB$ which meets $AB$ at $R$
Now in right $\triangle\text{OAR}$
$ O A^2=O R^2+A R^2$
$(5)^2=(3)^2+A R^2$
$ 25=9+A R^2 $
$\Rightarrow A R^2=25-9=16=(4)^2$
$AR = 4cm$
But $R$ is mid-point of $AB$
$AB = 2 AR = 2 \times 4 = 8cm$
View full question & answer→MCQ 361 Mark
In the figure, if $TP$ and $TQ$ are tangents drawn from an external point $T$ to a circle with centre $O$ such that$\angle\text{TQP}=60^\circ,$ then:
- A
$25^\circ$
- ✓
$30^\circ$
- C
$40^\circ$
- D
$60^\circ$
AnswerCorrect option: B. $30^\circ$
In the figure, $TP$ and $TQ$ are the tangents drawn from $T$ to the circle with centre $O$.
$OP, OQ$ and $PQ$ are joined.
$\angle\text{TQP}=60^\circ$
$TP = TQ$ (Tangents from T to the circle)
$\angle\text{TPQ}=\angle\text{TQP}=60^\circ$
$\angle\text{PTQ}=180^\circ-(60^\circ+60^\circ)=180^\circ-120^\circ=60^\circ$
and $\angle\text{PTQ}=180^\circ-(60^\circ+60^\circ)=180^\circ-120^\circ=60^\circ$
But $OP = OQ$ (radii of the same circle.)
$\angle\text{OPQ}=\angle\text{OQP}$
But $\angle\text{OPQ}+\angle\text{OQP}=180^\circ-120^\circ=60^\circ$
But $\angle\text{OQP}=30^\circ$
View full question & answer→MCQ 371 Mark
In the given figure, $DE$ and $DF$ are tangents from an external point $D$ to a circle with centre $A$. If $DE = 5\ cm$ and $DE ⊥ DF$, then the radius of the circle is:
- A
$3\ cm$
- ✓
$5\ cm$
- C
$4\ cm$
- D
$6\ cm$
AnswerCorrect option: B. $5\ cm$
join $AE$ and $AF$.
Now, $DE$ is a tangent at $E$ and $AE$ is the radius through the point of contact E.
$\therefore\angle\text{AED}=90^\circ$(Tangent at any point of a circle is perpendicular to the radius through the point of contact)
Also, $DF$ is a tangent at F and AF is the radius through the point of contct F.)
$\therefore\angle\text{AFD}=90^\circ$(Tangent at any point of a circle is perpendicular to the radius through the point of contact)
$\therefore\angle\text{EDF}=90^\circ\ (\text{DE}\bot\text{DF})$
Also, $DF = DE$ (length of tangents drawn from an external point to a circle are equal)
so, $AEDF$ is a square.
$\therefore$ $AE = AF = DE = 5\ cm$ (sides if square are equal)
Thus, the radius of the circle is $5\ cm$. View full question & answer→MCQ 381 Mark
A tangent $PQ$ at a point $P$ of a circle of radius $5\ cm$ meets a line through the centre $O$ at a point $Q$ such that $OQ = 12\ cm$. Length $PQ$ is:
- A
$12\ cm$
- B
$13\ cm$
- C
$8.5\ cm$
- ✓
$\sqrt{119}\text{cm}$
AnswerCorrect option: D. $\sqrt{119}\text{cm}$
Let us first put the given data in the form of a diagram.
Given data is as follows:
$OQ = 12cm$
$OP = 5cm$
We have to find the length of QP.
We know that the radius of a circle will always be perpendicular to the tangent at the point of contact. Therefore, OP is perpendicular to QP. We can now use Pythagoras theorem to find the length of QP.
$ \mathrm{QP}^2=\mathrm{OQ}^2-\mathrm{OP}^2 $
$ \mathrm{QP}^2=12^2-5^2 $
$ \mathrm{QP}^2=144-25 $
$ \mathrm{QP}^2=119$
$\text{QP}\sqrt{119}$
Therefore the correct answer is choice $(d)$.
View full question & answer→MCQ 391 Mark
In the figure, if $AP = PB$, then:
- A
$AC = AB$
- ✓
$AC = BC$
- C
$AQ = QC$
- D
$AB = BC$
AnswerCorrect option: B. $AC = BC$
In the figure, $AP = PB$
But $AP$ and $AQ$ are the tangent from A to the circle.
$AP = AQ$
Similarly $PB = BR$
But $AP = PB$ (given)
$AQ = BR ….(i)$
But CQ and CR the tangents drawn from $C$ to the circle
$CQ = CR$
Adding in $(i)$
$AQ + CQ = BR + CR$
$AC = BC$
View full question & answer→MCQ 401 Mark
In the given figure, the sides $AB, BC$ and $CA$ of triangle $ABC$, touch a circle at $P, Q$ and $R$ respectively. If $PA = 4\ cm$, $BP = 3\ cm$ and $AC = 11\ cm$, then length of $BC$ is:
- A
$11\ cm$
- ✓
$10\ cm$
- C
$14\ cm$
- D
$15\ cm$
AnswerCorrect option: B. $10\ cm$
By property of tangent
$AP = AR = 4cm($tangent from $A) ...(i)$
$BP = BQ = 3cm($tangent from $B) ...(ii)$
$RC = QC($tangent from $C) ...(iii)$
$AC = 11cm($given$)$
Now, we have to find $BC$
$BC = BQ + QC$
$\Rightarrow BC = 3 + RC [$from eq. $(ii)$ and $(iii)]$
$\Rightarrow BC = 3 + (AC + AR) [$from fig$]$
$\Rightarrow BC = 3 + (11 - 4) [$from eq. $(i)]$
$\Rightarrow BC = 3 + 7$
$\Rightarrow BC = 10cm$ View full question & answer→MCQ 411 Mark
$PQ$ is a tangent drawn from a point $P$ to a circle with centre $O$ and $QOR$ is a diameter of the circle such that $\angle\text{POR}=120^{\circ}$ then $\angle\text{OPQ}$ is:
- A
$60^\circ$
- B
$45^\circ$
- ✓
$30^\circ$
- D
$90^\circ$
AnswerCorrect option: C. $30^\circ$
we know, radius always $\bot$ to tangent, then
$\angle\text{PQO}=90^{\circ}(\text{OQ}\bot\text{PQ})$
Given, $\angle\text{POR}=120^{\circ}$
then, $\angle\text{POQ}=\angle\text{QOR}-\angle\text{POR}$
$\Rightarrow\angle\text{POQ}=180^{\circ}-120^{\circ}$
$\Rightarrow\angle\text{POQ}=60^{\circ}$
Now, In $\triangle\text{OPQ}$
Sum of all angles are equal to $180^{\circ}$
then,
$\angle\text{OPQ}+\angle\text{POQ}+\angle\text{PQO}=180^{\circ}$
$\Rightarrow\angle\text{OPQ}+90^{\circ}+60^{\circ}=180^{\circ}$
$\Rightarrow\angle\text{OPQ}+150^{\circ}=180^{\circ}$
$\Rightarrow\angle\text{OPQ}=30^{\circ}$ View full question & answer→MCQ 421 Mark
$AP$ and $AQ$ are tangents drawn from a point $A$ to a circle with centre $O$ and radius $9\ cm$. If $OA = 15\ cm$, then $AP + AQ =$
- A
$12\ cm$
- B
$18\ cm$
- ✓
$24\ cm$
- D
$36\ cm$
AnswerCorrect option: C. $24\ cm$
By the property of tangent
$AP = AQ($tangent from $A)...(i)$
We know, radius always $\bot$ to tangent then $\triangle\text{OPQ}$ is right angle triangle the $\angle\text{OPA}=90^{\circ}$
$ \text { Now, } A P^2=O A^2-O P^2 $
$ \Rightarrow A P^2=15^2-9^2 $
$ \Rightarrow A P^2=225-81$
$\Rightarrow AP =$
$\sqrt{144}$
$\Rightarrow AP = 12cm$
we have to find $AP + AQ = 12 + 12 = 24cm$ [from(i)] View full question & answer→MCQ 431 Mark
At one end $A$ of a diameter $AB$ of a circle of radius $5\ cm$, tangent $XAY$ is drawn to the circle. The length of the chord $CD$ parallel to $XY$ and at a distance $8\ cm$ from $A$ is:
- A
$4\ cm$
- B
$5\ cm$
- C
$6\ cm$
- ✓
$8\ cm$
AnswerCorrect option: D. $8\ cm$
$XY$ is the tangent to the circle with centre $O$.
$CD$ is the chord.
$OA = OB = OD = 5cm$ (radii)
$PA = 8cm$
$PO = 3cm$
In $\triangle\text{POD},$
$ \Rightarrow P D^2+P O^2=O D^2$
$\Rightarrow 3^2+P D^2=5^2$
$ \Rightarrow P D^2=25-9=16$
$\Rightarrow PD = 4cm$
Hence, $CD = CP + PD = 4 + 4 = 8cm.$ View full question & answer→MCQ 441 Mark
In the adjacent figure, if $AB = 12\ cm$, $BC = 8\ cm$ and $AC = 10\ cm$, then $AD =$
- A
$5\ cm$
- B
$4\ cm$
- C
$6\ cm$
- ✓
$7\ cm$
AnswerCorrect option: D. $7\ cm$
Given,
$AB = AD + DB = 12cm...(i)$
$BC = BE + EC = 8cm...(ii)$
$CA + CF + FA = 10cm...(iii)$
from the property of tangent
$AD = AF ($ tangent from $A ) ...(iv)$
$DB = BE ($ tangent from $A ) ...(v)$
$CF = CE ($ tangent from $A ) ...(vi)$
Now, we have to find AD
By substracting eq.$(ii)$ from eq.$(i),$ then
$\Rightarrow AD + DB - (BE + EC) = 12 - 8$
$\Rightarrow AD + BE - BE - CF = 4 [$ from eq.$(v) ]$
$\Rightarrow AD - CF = 4$
$\Rightarrow AD - (10 - AF) = 4 [$ from eq,$(iii) ]$
$\Rightarrow AD - 10 + AF = 4$
$\Rightarrow AD - 10 + AD = 4$
$\Rightarrow 2AD = 14$
$\Rightarrow AD = 7$ View full question & answer→MCQ 451 Mark
Two circles of same radii r and centres $O$ and $O'$ touch each other at $P$ as shown in. If $O$ $O'$ is produced to meet the circle $C (O', r)$ at $A$ and $AT$ is a tangent to the
circle $C (O,r)$ such that $O'Q ⊥ AT$. Then $AO: AO' =$
Answer
From the given figure we have,
$AO = r + r + r$
$AO = 3r$
$AO’ = r$
Therefore,
$\frac{\text{AO}}{\text{AO'}}=\frac{3\text{r}}{\text{r}}$
$\frac{\text{AO}}{\text{AO'}}=3$
Also as $\text{O'Q}\parallel\text{OT}$ therefore $\frac{\text{AT}}{\text{AQ}}=\frac{\text{AO}}{\text{AO'}}$ View full question & answer→MCQ 461 Mark
In the figure, $AP$ is a tangent to the circle with centre $O$ such that $OP = 4\ cm$ and $\angle\text{OPA}=30^{\circ}$. Then, $AP =$ 
- A
$2\sqrt{2}\text{cm}$
- B
$2\text{cm}$
- ✓
$2\sqrt{3}\text{cm}$
- D
$3\sqrt{2}\text{cm}$
AnswerCorrect option: C. $2\sqrt{3}\text{cm}$
In the figure, AP is the tangent to the circle with centre $O$ such that $OP = 4\ cm$, $\angle\text{OPA}=30^{\circ}$
Join $OA$, let $AP = x$
$\cos30^{\circ}=\frac{\text{AP}}{\text{OP}}$
$\Rightarrow\frac{\sqrt{3}}{2}=\frac{\text{x}}{4}\Rightarrow\text{x}=\frac{4\times\sqrt{3}}{2}=2\sqrt{3}\text{cm}$ View full question & answer→MCQ 471 Mark
In the figure, two equal circles touch each other at $T$, if $QP = 4.5\ cm$, then $QR =$
- ✓
$9\ cm$
- B
$18\ cm$
- C
$15\ cm$
- D
$13.5\ cm$
AnswerCorrect option: A. $9\ cm$
In the figure, two equal circles touch, each other externally at $T$
$QR$ is the common tangent $QP = 4.5\ cm$
$PQ = PT ($tangents from $P$ to the circle$)$
Similarly $PT = PR$
$PQ = PT = PR$
Now $QR = PQ + PR = 4.5 + 4.5 = 9\ cm$
View full question & answer→MCQ 481 Mark
$ABC$ is a right angled triangle, right angled at $B$ such that $BC = 6\ cm$ and $AB = 8\ cm$. A circle with centre $O$ is inscribed in $\triangle ABC$. The radius of the circle is:
- A
$1\ cm$
- ✓
$2\ cm$
- C
$3\ cm$
- D
$4\ cm$
AnswerCorrect option: B. $2\ cm$
In a right $\triangle\text{ABC},$ $\angle\text{B}=90^{\circ}$
$BC = 6\ cm, AB = 8\ cm$
$A C^2=A B^2+B C^2$ (Pythagoras Theorem)
$=(8)^2+(6)^2=64+36=100=(10)^2$
$AC = 10cm$
An incircle is drawn with centre 0 which touches the sides of the triangle $ABC$ at $P, Q$ and $R ,OP, OQ$ and $OR$ are radii and $AB, BC$ an $CA$ are the tangents to the circle.
$OP ⊥ AB, OQ ⊥ BC$ and $OR ⊥ CA$
$OPBQ$ is a square.
Let $r$ be the radius of the incircle.
$PB = BQ = r$
$AR = AP = 8 – r,$
$CQ = CR = 6 – r$
$AC = AR + CR$
$\Rightarrow 10 = 8 – r + 6 – r$
$\Rightarrow 10 = 14 – 2r$
$\Rightarrow 2r = 14 – 10 = 4$
$\Rightarrow r = 2$
Radius of the incircle $= 2cm.$
View full question & answer→MCQ 491 Mark
In the figure, $APB$ is a tangent to a circle with centre $O$ at point $P$. If $\angle\text{QPB}=50^\circ$ then the measure of $\angle\text{POQ}$ is: 
- ✓
$100^\circ$
- B
$120^\circ$
- C
$140^\circ$
- D
$150^\circ$
AnswerCorrect option: A. $100^\circ$
In the figure, $APB$ is a tangent to the circle with centre $O$.
$\angle\text{QPB}=50^\circ$
$OP$ is radius and $APB$ is a tangent.
$OP ⊥ AB$
$\Rightarrow\text{OPB}=90^\circ$
$\Rightarrow\angle\text{OPQ}+\angle\text{QPB}=90^\circ$
$\angle\text{OPQ}+50^\circ=90^\circ$
$\Rightarrow\angle\text{OPQ}=90^\circ-50^\circ=40^{\circ}$
But $OP = OQ$
$\angle\text{OPQ}=\angle\text{OQP}=40^{\circ}$
$\angle\text{POQ}=180^{\circ}-(40^{\circ}+40^{\circ})=180^{\circ}-80^{\circ}=100^{\circ}$
View full question & answer→MCQ 501 Mark
In the figure, $PQ$ and $PR$ are two tangents to a circle with centre $O$. If $\angle\text{QPR}=46^\circ$ then $\angle\text{QOR}$ equals:
- A
$67^\circ$
- ✓
$134^\circ$
- C
$44^\circ$
- D
$46^\circ$
AnswerCorrect option: B. $134^\circ$
$\angle\text{OQP}=90^\circ$ $[$Tangent is $\perp$ to the radius through the point of contact$]$
$\angle\text{ORP}=90^\circ$
$\angle\text{OQP}+\angle\text{QPR}+\angle\text{ORP}+\angle\text{QOR}=360^\circ$ $[$Angle sum property of a quad.$]$
$90^\circ+46^\circ+90^\circ+\angle\text{QOR}=360^\circ$
$\angle\text{QOR}=360^\circ-90^\circ-46^\circ-90^\circ=134^\circ$
View full question & answer→