MCQ 511 Mark
In the given figure, a circle is inscribed in a quadrilateral $\text{ABCD}$ touching its sides $AB, BC, CD$ and $AD$ at $P, Q, R$ and $S$ respectively. If the radius of the circle is $10\ cm, BC = 38\ cm, PB = 27\ cm$ and $\text{AD} \perp \text{CD}$ then the length of $CD$ is :
- A
$11\ cm$
- B
$15\ cm$
- C
$20\ cm$
- ✓
$21\ cm$
AnswerCorrect option: D. $21\ cm$
We know that tangles from an external point to the circle are equal.
$BQ = PB = 27\ cm$
So $, CQ = BC - BQ $
$= 38 - 27 = 11\ cm$
$\Rightarrow CR = CQ = 11\ cm$
In quad. $\text{SORD},$
$\angle\text{SDR}=90^\circ....(\therefore\text{AD}\perp\text{CD})$
$\Rightarrow\angle\text{OSD}=\angle\text{ORD}=90^\circ$
Also $, OS = OR$ and $SD = SR$
So, quad. $\text{SORD}$ is a square.
Thus $, DR = SO = 10\ cm$
Hence $, CD = DR + CR $
$= 10 + 11 = 21\ cm.$
View full question & answer→MCQ 521 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 531 Mark
In the given figure $,O$ is the centre of a circle. $\text{AOC}$ is its diameter, such that $\angle\text{ACB} = 50^\circ.$ If $AT$ is the tangent to the circle at the point $A,$ then $\angle \text{BAT} = ?$
- A
$40^\circ$
- ✓
$50^\circ$
- C
$60^\circ$
- D
$65^\circ$
AnswerCorrect option: B. $50^\circ$
Construction : Join $OC$.
Since $AC$ is a diameter of the circle.
$\angle\text{ABC}=90^\circ ....($angle in a semicircle is $90^\circ )$
In $\triangle\text{ABC},$
$\angle\text{ABC}+\angle\text{BCA}+\angle\text{BAC}=180^\circ ......($Angle Sum Property$)$
$\Rightarrow90^\circ+50^\circ+\angle\text{BAC}=180^\circ$
$\Rightarrow\angle\text{BAC}=40^\circ$
Since $AC$ is a tangent to the inner circle.
$\Rightarrow\angle\text{OAT}=90^\circ .....($Tangent is perpendicular to the radius of a circle$)$
$\Rightarrow\angle\text{BAO}+\angle\text{BAT}=90^\circ$
$\Rightarrow40^\circ+\angle\text{BAT}=90^\circ$
$\Rightarrow\angle\text{BAT}=50^\circ$
View full question & answer→MCQ 541 Mark
$PQ$ is a tangent to a circle with centre $O$ at the point $P$. If $\triangle\text{OPQ}$ is an isosceles triangle, then $\angle\text{OQP}$ is equal to :
- A
$30^\circ$
- ✓
$45^\circ$
- C
$60^\circ$
- D
$90^\circ$
AnswerCorrect option: B. $45^\circ$
Given that $\triangle\text{PQO}$ is an isosceles triangle.
Since $PQ$ is a tangent to the circle at $P$.
Sunce $PQ$ is a tangent to the circle at $P.$
$\angle\text{OPQ}=90^\circ .....($tangent is perpendicular to the radius of a circle$)$
In $\triangle\text{OPQ},$
$OP = OQ$
$\Rightarrow\angle\text{OQP}=\angle\text{POQ}$
Using Angle Sum Property,
$\angle\text{OQP}+\angle\text{POQ}+\angle\text{OPQ}=180^\circ$
$\Rightarrow\angle\text{OQP}+\angle\text{OQP}+90^\circ=180^\circ$
$\Rightarrow2\angle\text{OQP}=90^\circ$
$\Rightarrow\angle\text{OQP}=45^\circ$ View full question & answer→MCQ 551 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 $\text{PQOR}$ is :
- ✓
$60\ cm^2$
- B
$65\ cm^2$
- C
$30\ cm^2$
- D
$32.5\ cm^2$
AnswerCorrect option: A. $60\ cm^2$
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 $\text{PQOR}$ is formed.
$OQ \perp QP \ [$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 $
$\Rightarrow QP = 12\ cm$
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 $\text{QORP} = 2\triangle\text{OQP}=2\times30=60\text{ cm}^2$ View full question & answer→MCQ 561 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 571 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 581 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,
$\angle\text{OAP}=90^{\circ}(\text{OA}\bot\text{PA})$
$\angle\text{OBP}=90^{\circ}(\text{OB}\bot\text{PB})$
$\angle\text{APB}=80^{\circ}\ ($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 $\triangle\text{POA}$ and $ \triangle\text{POB},$
$OA = OB \ ($Radius of circle$)$
$PA = PB\ ($tangent grom external point $P)$
$OP = OP\ ($commom$)$
So, By using $\text{SSS}$ 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 591 Mark
Each question consists of two statements, namely, Assertion $(A)$ and Reason $(R)$. for selecting the correct answer, use the following code :
|
Assertion $(A)$
|
Reason $(R)$
|
In the given figure, a quad. $\text{ABCD}$ is drawn to circumscribe a given circle as shown. Then $, AB + BC = AD + DC.$
 |
In two concentric circles, the chord of the larger circle, which to uches the smaller circle, is bisected at the point of contact. |
- A
Both Assertion $(A)$ and Reason $(R)$ are true and Reason $(R)$ is a correct explanation of Assertion $(A)$.
- B
Both Assertion $(A)$ and Reason $(R)$ are true but Reason $(R)$ is not a correct explanation of Assertion $(A).$
- C
Assertion $(A)$ is true and Reason $(R)$ is false.
- ✓
Assertion $(A)$ is false and Reason $(R)$ is true.
AnswerCorrect option: D. Assertion $(A)$ is false and Reason $(R)$ is true.
The Assertion $(A)$ is false since sum of the opposite sides of a quadrilateral circumscribing a circle are equal, and not the adjacent sides.
The chord of the larger circle is the tangent to the smaller circle.
We know that the perpendicular drawn from the centre to the chord
So, the Reason $(R)$ is true.
But is not the correct explanation for the Assertion $(A).$
View full question & answer→MCQ 601 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 $QP.$
$ \mathrm{QP}^2=\mathrm{OQ}^2-\mathrm{OP}^2 $
$ \mathrm{QP}^2=8^2-6^2 $
$ \mathrm{QP}^2=64-36 $
$ \mathrm{QP}^2=28 $
$QP = \sqrt{28}$
$QP = 2\sqrt{7}$ View full question & answer→MCQ 611 Mark
$AB $ is a chord of circle and $\text{AOC}$ is its diameter such that $\angle\text{ACB} = 40^\circ$. If $AT$ is the tangent to the circle at the point $A,$ then $\angle\text{BAT}$ is equal to :
- A
$60^\circ$
- B
$55^\circ$
- C
$45^\circ$
- ✓
$40^\circ$
AnswerCorrect option: D. $40^\circ$
Here $\angle\text{B}=90^\circ \ [$ Angle of semicircle$]$
Now, in triangle $\text{ABC},\angle\text{CAB}+\angle\text{B}+\angle\text{C}=180^\circ$
$\Rightarrow\angle\text{CAB}+90^\circ+40^\circ=180^\circ$
$\Rightarrow\angle\text{CAB}=180^\circ-130^\circ=50^\circ$
Now,$\angle\text{CAT}=90^\circ\ [$Angle between tangent and redius through the point of contact$]$
$\therefore\angle\text{CAB}+\angle\text{BAT}=90^\circ$
$\Rightarrow50+\angle\text{BAT}=90^\circ$
$\Rightarrow\text{BAT}=40^\circ$
View full question & answer→MCQ 621 Mark
If $PQ = 28\ cm,$ then the perimeter of $\triangle\text{PLM}$ is :
- A
$48\ cm$
- ✓
$56\ cm$
- C
$42\ cm$
- D
$28\ cm$
AnswerCorrect option: B. $56\ cm$
We know that $ ,PQ =\frac{1}{2}(\text{perimeter of }\triangle \text{ PLM})$
$\Rightarrow 28$
$=\frac{1}{2}$
$(\text{Perimeter of }\triangle\text{PLM)}$
$\Rightarrow (\text{Perimeter of}\triangle\text{PLM)} = 28 \times 2 = 56 \text{ cm}$
View full question & answer→MCQ 631 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 $\triangle\text{OQP}$
$OP^2= (OQ)^2+ PQ^2= (3)^2+ (4)^2$
$= 9 + 16 = 25 = (5)^2$
$OP = 5\ cm$
Similarly in right $\triangle\text{RSO}$
$(O’R)^2= (RS)^2+ (O’S)^2= (12)^2+ (5)^2$
$= 144 + 25 = 169 = (13)^2$
$O’R = 13\ cm$
Now $PR = OP + OO’ + O’R$
$ = 5 + (3 + 5) + 13 = 26\ cm.$ View full question & answer→MCQ 641 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 651 Mark
If the angle between two radii of a circle is $130^\circ ,$ then the angle between the tangent at the ends of the radii is :
- A
$65^\circ$
- B
$40^\circ$
- ✓
$50^\circ$
- D
$90^\circ$
AnswerCorrect option: C. $50^\circ$
In quad. $\text{AOBP}$
$\angle\text{PAO}+\angle\text{PBO}+\angle\text{AOB}+\angle\text{APB}=360^\circ....($Angle Sum Property$)$
$\Rightarrow90^\circ+90^\circ+130^\circ+\angle\text{APB}=360^\circ....($Since radius of a circle is perpendicular to the tangent$)$
$\Rightarrow\angle\text{APB}=50^\circ$ View full question & answer→MCQ 661 Mark
In figure, if $\angle\text{AOB}=125^\circ,$ then $\angle\text{COD}$ is equal to :
- A
$62.5^\circ$
- B
$45^\circ$
- C
$35^\circ$
- ✓
$55^\circ$
AnswerCorrect option: D. $55^\circ$
we know that, the opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
i.e., $\angle\text{AOB}+\angle\text{COD}=180^\circ$
$\Rightarrow\ \angle\text{COD}=180^\circ-\angle\text{AOB}$
$=180^\circ - 125^\circ = 55^\circ$
View full question & answer→MCQ 671 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 681 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 $AP = AQ = 5\ cm$
We know that radius of a circle is perpendicular to the tangent at the point of contact.
$\Rightarrow\text{OP}\bot\text{AP}$ 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$\angle\text{O}=\angle\text{P}=\angle\text{A}=\angle\text{Q}=90^\circ$
$\Rightarrow \text{OPAQ}$ is a rectangle.
Since adjacent sides of $\text{OPAQ}$
i.e. $AP$ and $AQ$ are equal.
Thus $\text{OPAQ}$ is a square radius $= OP = OQ = AP = AQ = 5\ cm$ View full question & answer→MCQ 691 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, $OA \perp PA \ [$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 701 Mark
In the given figure, $O$ is the centre of a circle, $\text{BOA}$ is its diameter and the tangent at the point $P$ meets $BA$ extended at $T. \angle\text{PBO} = 30^\circ$ then $\angle\text{PTA} =?$
- A
$60^\circ$
- ✓
$30^\circ$
- C
$15^\circ$
- D
$45^\circ$
AnswerCorrect option: B. $30^\circ$
In $\triangle\text{OBP},$
$\text{OB} = \text{OP} ....($radii of the circle$)$
$\Rightarrow\angle\text{OBP}=\angle\text{OPB}=30^\circ ....($angles opposite equal sides are equal$)$
Since $PT$ is a tangent,
$\angle\text{OPT}=90^\circ$
In $\triangle\text{BPT},$
$\angle\text{BPT}+\angle\text{PBT}+\angle\text{PTB}=180^\circ....($Angle Sum Property$)$
$\Rightarrow(30^\circ+90^\circ)+30^\circ+\angle\text{PTB}=180^\circ$
$\Rightarrow\angle\text{PTB}=30^\circ$
That is, $\angle\text{PTA}=30^\circ$
View full question & answer→MCQ 711 Mark
$O$ is the centre of a circle of radius $5\ cm$. At a distance of $13\ cm$ from $O,$ a point $P$ is taken. From this point, two tangents $PQ$ and $PR$ are drawn to the circle. Then, the area of quadrilateral $\text{PQOR}$ is :
- ✓
$60 \mathrm{~cm}^2 $
- B
$ 32.5 \mathrm{~cm}^2 $
- C
$ 65 \mathrm{~cm}^2 $
- D
$ 30 \mathrm{~cm}^2 $
AnswerCorrect option: A. $60 \mathrm{~cm}^2 $
In $\triangle\text{OPQ}$ and $\triangle\text{ORP},$
$\angle\text{OQP}=\angle\text{ORP}=90^\circ ....($Since $OP$ and $RP$ are tangent to the circle$)$
$\text{OP}=\text{OP} ....($common side$)$
$\text{OQ}=\text{OR} ....($radii of the same circle$)$
$\Rightarrow\triangle\text{OPQ}\cong\triangle\text{ORP} ....(\text{RHS}$ congruence criterion$)$
So, the areas of both the triangle will be the swame.
In $\triangle\text{OPQ},$
By Pythagoras theorem,
$\text{OP}^2=\text{OQ}^2+\text{PQ}^2$
$\Rightarrow\text{PQ}^2=\text{OP}^2-\text{OQ}^2$
$\Rightarrow\text{PQ}^2=\text{13}^2-\text{5}^2$
$\Rightarrow\text{PQ}^2=144$
$\Rightarrow\text{PQ}=12\text{ cm}$
$\text{ar}(\triangle\text{OPQ})=\frac{1}{2}\times\text{PQ}\times\text{OQ}$
$=\frac{1}{2}\times\text{12}\times5$
$=30\text{ cm}^2$
$\text{ar}(\text{quad PQOR})=\text{ar}(\triangle\text{OPQ})+\text{ar}(\triangle\text{ORP})$
$\Rightarrow\text{ar}(\text{quad PQOR})=30\text{ cm}^2+30\text{ cm}^2$
$\Rightarrow\text{ar}(\text{quad PQOR})=60\text{ cm}^2$
View full question & answer→MCQ 721 Mark
In the given figure, $PA$ and $PB$ are two tangents to the circle with centre $O$. If $\angle\text{APB}=60^\circ$ then $\angle\text{OAB}$ is :
- A
$15^\circ$
- ✓
$30^\circ$
- C
$60^\circ$
- D
$90^\circ$
AnswerCorrect option: B. $30^\circ$
We know that tangent from an external point to a circle are equal.
So,
$\text{PA}=\text{PB}$
$\Rightarrow\angle\text{PAB}+\angle\text{PBA} ....($angles opposite equal sides are equal$)$
Now in $\triangle\text{PAB},$
$\angle\text{PAB}+\angle\text{PBA}+\angle\text{APB}=180^\circ ...($angles Sum Property$)$
$\Rightarrow\angle\text{PAB}+\angle\text{PAB}+60^\circ=180^\circ$
$\Rightarrow2\angle\text{PAB}+60^\circ=180^\circ$
$\Rightarrow2\angle\text{PAB}=120^\circ$
$\Rightarrow\angle\text{PAB}=60^\circ$
Since $AP$ is a tangent to the circle,
$\angle\text{OAP}=90^\circ$
$\Rightarrow\angle\text{OAB}+\angle\text{PAB}=90^\circ$
$\Rightarrow\angle\text{OAB}+60^\circ=90^\circ$
$\Rightarrow\angle\text{OAB}=30^\circ$
View full question & answer→MCQ 731 Mark
Directions : In the following questions, a statement of assertion $(A)$ is followed by a statement of reason $(R)$. Mark the correct choice as:
Assertion : Figure, $\text{AOB}$ is a diameter of a circle with centre $O$ and $AC$ is a tangent to the circle at $A$ .If $\angle\text{BOC}=125^\circ,$ then $\angle\text{ACO}=35^\circ$
Reason : $\angle\text{ACO}$ and $\angle\text{BOC}$ form a linear pair.
- A
Assertion and Reason both are correct statements and Reason is the correct explanation of Assertion.
- B
Assertion and Reason both are correct statements but Reason is not the correct explanation of Assertion.
- ✓
Assertion is correct statement but Reason is wrong statement.
- D
Assertion is wrong statement but Reason is correct statement.
AnswerCorrect option: C. Assertion is correct statement but Reason is wrong statement.
$\angle\text{BOC}=125^\circ$
Since, $AC$ is a tangent to the circle at $A$
$\therefore\angle\text{OAC}=90^\circ \ [\because$ Radius is perpendicular to the tangent at point of contact$]$
Now, $\angle\text{AOC} + \angle\text{BOC} = 180^\circ \ [$Linear pair$]$
In $\triangle\text{AOC}, \angle\text{AOC} +\angle\text{ACO} +\angle\text{OAC} = 180^\circ \ [$By angle sum property$]$
$\Rightarrow55^\circ+\angle\text{ACO}+90^\circ=180^\circ$
$\Rightarrow\angle\text{ACO} = 180^\circ - 55^\circ - 90^\circ = 35^\circ$
$\therefore$ Assertion is correct but Reason is wrong.
View full question & answer→MCQ 741 Mark
In the given figure, point $P$ is $26\ cm$ away from the centre $O$ of a circle and the length $PT$ of the tangent drawn from $P$ to the circle is $24\ cm$. Then the radius ofthe circle is :
- ✓
$10\ cm$
- B
$12\ cm$
- C
$13\ cm$
- D
$15\ cm$
AnswerCorrect option: A. $10\ cm$
Construction : Join $OT$.
$PT = 24\ cm$
$OP = 26\ cm$
Sunce $PT$ is a tangent to the circle at $T.$
$\angle\text{PTO}=90^\circ .....($tangent is perpendicular to the radius of a circle$)$
In $\triangle\text{PTO},$
By Pythagoras theorem,
$ \mathrm{OP}^2=\mathrm{PT}^2+\mathrm{OT}^2 $
$ \Rightarrow \mathrm{OT}^2=\mathrm{OP}^2-\mathrm{PT}^2 $
$ \Rightarrow \mathrm{OT}^2=26^2-24^2 $
$ \Rightarrow \mathrm{OT}^2=676-576 $
$ \Rightarrow \mathrm{OT}^2=100 $
$ \Rightarrow \mathrm{OT}^2=10 \mathrm{~cm} $
View full question & answer→MCQ 751 Mark
In the given figure, $AD$ and $AE$ are the tangents to a circle with centre $O$ and $BC$ touches the circle at $F$. If $AE = 5\ cm,$ then perimeter of $\triangle\text{ABC}$ is :
- A
$15\ cm$
- ✓
$10\ cm$
- C
$22.5\ cm$
- D
$20\ cm$
AnswerCorrect option: B. $10\ cm$
We know that tangent from an external point to the circle equal.
So,
$AE = AD = 5\ cm$
$BF = BE$
$CF = CD$
Perimeter of $\triangle\text{ABC}$
$= AB + BC + AC$
$= AB + (BE + DC) + AC$
$= AB + (BE + DC) + AC$
$= (AB + BE) + (AC + DC)$
$= AE + AD$
$= 5 + 5$
$= 10\ cm$
View full question & answer→MCQ 761 Mark
In the given figure, a circle with centre $O$ is inscribed in a quadrilateral $\text{ABCD}$ such that, it touches sides $BC, AB, AD$ and $CD$ at points $P, Q, R $ and $S$ respectively. If $AB$
$= 29\ cm, AD = 23\ cm, \angle\text{B}=90^\circ$ and $DS = 5\ cm,$ then the radius of the circle $($in $cm)$ is :
AnswerIn the figure, a circle touches the sides of a quadrilateral $\text{ABCD}.$$\angle\text{B}=90^\circ, OP = OQ = r$
$AB = 29\ cm, AD = 23\ cm, DS = 5\ cm$
$\angle\text{B}=90^\circ,$
$BA$ 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$)$
$\text{PBQO}$ is a square.
$DS = 5\ cm$
But $DS$ and $DR$ are tangents to the circles.
$DR = 5\ cm$
But $AD = 23\ cm$
$AR = 23 – 5= 18\ cm$
$AR = AQ\ ($tangents to the circle from $A.)$
$AQ = 18\ cm$
But $AB = 29 \ cm$
$BQ = 29 – 18 = 11\ cm$
$\text{OPBQ}$ is a square.
$OQ = BQ = 11\ cm$
Radius of the circle $= 11\ cm.$
View full question & answer→MCQ 771 Mark
If four sides of a quadrilateral $\text{ABCD}$ 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 $\text{ABCD}$ 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 781 Mark
Which of the following statements is not true?
- A
A line which intersects a circle at two points, is called a secant of the circle.
- B
A line intersecting a circle at one point only, is called a tangent to the circle.
- C
The point at which a line touches the circle, is called the point of contact.
- ✓
A tangent to the circle can be drawn from a point inside the circle.
AnswerCorrect option: D. A tangent to the circle can be drawn from a point inside the circle.
Options $(a), (b)$ and $(c)$ are all true.
However, Option $(d)$ is false since it is not possible to draw a tangent from a point inside a circle.
View full question & answer→MCQ 791 Mark
In the given figure, $AB$ and $AC$ are tangents to a circle with centre $O$ and radius $8\ cm$. If $OA = 17\ cm,$ then the length of $AC ($in $\ cm)$ is :
- A
$9$
- ✓
$15$
- C
$\sqrt{353}$
- D
$25$
Answer
Construction : Join $OC.$
Since $AC$ is a tangent to the circle.
$\angle\text{OCA}=90^\circ ....($tangent is perpendicular to the radius of a circle$)$
In $\triangle\text{OCA},$
By Pythagoras theorem,
$ O A^2=O C^2+A C^2 $
$ \Rightarrow 17^2=8^2+A C^2 $
$ \Rightarrow A C^2=289-64 $
$ \Rightarrow A C^2=225 $
$\Rightarrow AC = 15\ cm$ View full question & answer→MCQ 801 Mark
If $PT$ is a tangent to the circle with centre $O,$ then $x + y$ is equal to :
- ✓
$90^\circ$
- B
$60^\circ$
- C
$75^\circ$
- D
$100^\circ$
AnswerCorrect option: A. $90^\circ$
Here $\angle\text{T} = 90^\circ \ [$Angle between tangent and radius through the point of contact$]$
Now, in triangle $\text{OPT}$, we know that
$\angle\text{O} + \angle\text{P} + \angle\text{T} = 180^\circ$
$[$Angle sum property of a triangle$]$
$\Rightarrow x + y + 90^\circ = 180^\circ $
$\Rightarrow x + y = 180^\circ $
$\Rightarrow x + y = 90^\circ $
View full question & answer→MCQ 811 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 821 Mark
In a circle of radius $7\ cm,$ tangent $PT$ is drawn from a point $P,$ such that $PT = 24\ cm$. If $O$ is the centre of the circle, then $OP =?$
- A
$30\ cm$
- B
$28\ cm$
- ✓
$25\ cm$
- D
$18\ cm$
AnswerCorrect option: C. $25\ cm$
$PT = 24\ cm, OT = 7\ cm.$
Since $PT$ is a tangent to the circle at $T$.
$\angle\text{PTO}=90^\circ ....($tangent is perpendicular to the radius of a circle$)$
In $\triangle\text{PTO},$
By using Pythagoras theorem,
${OP}^2={PT}^2+{OT}^2$
$ \Rightarrow {OP}^2=24^2+7^2 $
$ \Rightarrow O P^2=576+49 $
$ \Rightarrow O P^2=625$
$\Rightarrow OP = 25\ cm.$
View full question & answer→MCQ 831 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,
$OA = 6\ cm, OB = 3\ cm$ and $AP = 10\ cm$
$OA$ is radius and $AP$ is the tangent
$OA \perp AP$
Now in right $\triangle\text{OAP}$
$O P^2=A P^2+O A^2=(10)^2+(6)^2$
$=100+36=136$
Similarly $BP$ is tangent and $OB$ 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 841 Mark
In the given figure, $AB = 8\ cm$. If $PE = 3\ cm,$ then the measure of $AE$ is :
- ✓
$5\ cm$
- B
$7\ cm$
- C
$9\ cm$
- D
$4.5\ cm$
AnswerCorrect option: A. $5\ cm$
Since Tangents from an external point to a circle are equal.
$\therefore PE = EC = 3\ cm$ and $AB = AE = 8\ cm$
Therefore $, AE = AC - EC = 8 - 3 = 5\ cm$
View full question & answer→MCQ 851 Mark
Choose the correct option and give justification. 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. $7 \ cm.$
$\because\angle\text{OPQ}=90^\circ$
$[$The tangent at any point of a circle is $\perp$ to the radius through the point of contact$]$
$\therefore$ In right triangle $\text{OPQ},$
$O Q^2=O P^2+P Q^2$
$[$By Pythagoras theorem$]$
$ \Rightarrow(25)^2=O P^2+(24)^2 $
$ \Rightarrow 625=O P^2+576$
$ \Rightarrow O P^2=625-576=49$
$\Rightarrow OP = 7\ cm.$ View full question & answer→MCQ 861 Mark
In the given figure, two circles touch each other at $C$ and $AB$ is a tangent to both the circles. The measure of $\angle\text{ACB}$ is :
- A
$45^\circ$
- B
$60^\circ$
- ✓
$90^\circ$
- D
$120^\circ$
AnswerCorrect option: C. $90^\circ$
Draw a tangent to the circle at point $C$ meet $AB$ at $P$.
Then,
$PA = PC$
$\Rightarrow \angle\text{PAC}=\angle\text{PCA}$
And $PB = PC$
$\Rightarrow\angle\text{PBC}=\angle\text{PCB}$
$\therefore\angle\text{PAC}+\angle\text{PBC}=\angle\text{PCA}+\angle\text{PCB}=\angle\text{ACB}$
$\Rightarrow\angle\text{PAC}+\angle\text{PBC}+\angle\text{ACB}=2\angle\text{ACB}$
$\Rightarrow180^\circ=2\angle\text{ACB}$
$\Rightarrow\angle\text{ACB}=90^\circ$ View full question & answer→MCQ 871 Mark
In the given figure $,O$ is the centre of the circle, $PQ$ is a chord and $PT$ is the tangent at $P$. If $ \angle\text{POQ} = 70^\circ$ then $\angle\text{TPQ} $ is equal to :
- ✓
$35^\circ$
- B
$45^\circ$
- C
$55^\circ$
- D
$70^\circ$
AnswerCorrect option: A. $35^\circ$
In $\triangle\text{OPQ},$
$\text{OP}=\text{OQ} ....($radii of the same circle$)$
$\Rightarrow\angle\text{OQP}=\angle\text{OPQ} ....($angles opposite equal sides are equal$)$
$\triangle\text{OPQ},$
$\angle\text{OQP}+\angle\text{OPQ}+\angle\text{POQ}=180^\circ ......($Angle Sum Property$)$
$\Rightarrow\angle\text{OPQ}+\angle\text{OPQ}+70^\circ=180^\circ$
$\Rightarrow2\angle\text{OPQ}=110^\circ$
$\Rightarrow\angle\text{OPQ}=55^\circ$
Since $PT$ is a tangent to the inner circle.
$\Rightarrow\angle\text{OPT}=90^\circ .....($Tangent is perpendicular to the radius of a circle$)$
$\Rightarrow\angle\text{OPQ}+\angle\text{TPQ}=90^\circ$
$\Rightarrow55^\circ+\angle\text{TPQ}=90^\circ$
$\Rightarrow\angle\text{TPQ}=35^\circ$
View full question & answer→MCQ 881 Mark
The number of tangents that can be drawn from an external point to a circle is :
Answer
We can draw only two tangents from an external point to a circle.
View full question & answer→MCQ 891 Mark
$TP$ and $TQ$ are tangents from an external point $T,$ to a circle with centre $\text{O} \angle\text{TPQ} = 60^\circ$ then the measure of $ \angle\text{OPQ}$ is :
- A
$40^\circ$
- B
$90^\circ$
- ✓
$30^\circ$
- D
$60^\circ$
AnswerCorrect option: C. $30^\circ$
Here $\angle\text{QPT} = 60\ [$Angles opposite to equal sides$]$
And$ \angle\text{PTQ} = 180^\circ - (60^\circ + 60^\circ) = 60^\circ\ [$Angle sum property of a triangle$]$
$\angle\text{POQ} = 180^\circ - 60^\circ = 120^\circ$
Let $\angle\text{OPQ} = \text{OQP} = \text{x}\ [$Angles opposite to equal sides $($Radii$)]$
$\therefore$ In triangle $\text{OPQ},$
$\angle\text{POQ} +\text{x}+\text{x} =180^\circ$
$\Rightarrow 120^\circ + 2\text{x} = 180^\circ$
$\Rightarrow 2\text{x} = 60^\circ$
$\Rightarrow \text{x} = 30^\circ$
Therefore $,\angle\text{OPQ} = 30^\circ$
View full question & answer→MCQ 901 Mark
In the given figure, if $\angle\text{AOD}=135^\circ$ then $\angle\text{BOC}$ is equal to :
- A
$25^\circ$
- ✓
$45^\circ$
- C
$52.5^\circ$
- D
$62.5^\circ$
AnswerCorrect option: B. $45^\circ$
We know that sum of the angles subtended by opposite sides of a quadrilateral having a circumscribed circle is $180^\circ .$
$\Rightarrow\angle\text{AOD}+\angle\text{BOC}=180^\circ$
$\Rightarrow135^\circ+\angle\text{BOC}=180^\circ$
$\Rightarrow\angle\text{BOC}=45^\circ$
View full question & answer→MCQ 911 Mark
In the given figure, the length of $BC$ is :
- A
$7\ cm$
- ✓
$10\ cm$
- C
$14\ cm$
- D
$15\ cm$
AnswerCorrect option: B. $10\ cm$
We know that tangent from an external point to a circle are equal.
So,
$AF = AE = 4\ cm$
$\Rightarrow EC = AC - AE $
$= 11 - 4 = 7\ cm$
Now,
$CD = CE = 7\ cm$
and $BF = BD = 3\ cm$
$BD = BD + CD$
$\Rightarrow BD = 3 + 7$
$\Rightarrow BD = 10\ cm$
View full question & answer→MCQ 921 Mark
Choose the correct option and give justification. In Fig., if $TP$ and $TQ$ are the two tangents to a circle with centre $O$ so that $\angle POQ = 110^\circ ,$ then $\angle PTQ$ is equal to :
- A
$60^\circ$
- ✓
$70^\circ$
- C
$80^\circ$
- D
$90^\circ$
AnswerCorrect option: B. $70^\circ$
$\angle POQ = 110^\circ , \angle OPT = 90^\circ $ and $\angle OQT = 90^\circ$
$[$The tangent at any point of a circle is $\perp$ to the radius through the point of contact$]$
In quadrilateral $\text{OPTQ},$
$\angle POQ + \angle OPT + \angle OQT + \angle PTQ = 360^\circ$
$[$Angle sum property of quadrilateral$]$
$\Rightarrow 110^\circ + 90^\circ + 90^\circ + \angle PTQ = 360^\circ$
$\Rightarrow 290^\circ + \angle PTQ = 360^\circ$
$\Rightarrow \angle PTQ = 360^\circ - 290^\circ$
$\Rightarrow \angle PTQ = 70^\circ$
View full question & answer→MCQ 931 Mark
In the given figure $,PQ$ is a tangent to a circle with centre $O. A$ is the point of contact. If $\angle\text{PAB} = 67^\circ$ then the measure of $\angle\text{AQB} $ is :
- A
$73^\circ$
- B
$64^\circ$
- C
$53^\circ$
- ✓
$44^\circ$
AnswerCorrect option: D. $44^\circ$
Since $\angle\text{BAC}$ is inscribed in a semicircle, $\angle\text{BAC}=90^\circ.$
Since $\text{PAQ}$ is a straight line,
$\angle\text{PAB}+\angle\text{BAC}+\angle\text{CAQ}=180^\circ $
$\Rightarrow67^\circ+90^\circ+\angle\text{CAQ}=180^\circ$
$\Rightarrow\angle\text{CAQ}=23^\circ$
We know that angles that subtend the same arc are equal.
$\Rightarrow\angle\text{CBA}=\angle\text{CAQ}=23^\circ$
In $\triangle\text{BAQ},$
$\angle\text{BAQ}+\angle\text{QBA}+\angle\text{AQB}=180^\circ ....($Angle Sum Property$)$
$\Rightarrow(90^\circ+23^\circ)+23^\circ+\angle\text{AQB}=180^\circ$
$\Rightarrow\angle\text{AQB}=44^\circ$
View full question & answer→MCQ 941 Mark
Which of the following statment is not true ?
- A
If a point $P$ lies inside a circle, no tangent can be drawn to the circle passing through $P$.
- B
If a point $P$ lies on a circle, then one and only one tangent can be drawn to the circle at $P$.
- C
If a point $P$ lies outside a circle, then only two tangents can be drawn to the circle from $P$.
- ✓
A circle can have more than two parallel tangents parallel to a given line.
AnswerCorrect option: D. A circle can have more than two parallel tangents parallel to a given line.
Options $(a), (b)$ and $(c)$ are all true.
However, Option $(d)$ is false since we can draw only parallel tamngent on either side of the diameter,
which would be parallel to a given line.
View full question & answer→MCQ 951 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 $\text{APB}$ is $180^\circ $
$\angle\text{PAB}+\angle\text{ABP}+\angle\text{APB}=180^\circ$
$\Rightarrow\angle\text{ABP}+\angle\text{ABP}+30^\circ=180^\circ$
$\Rightarrow2\angle\text{ABP}=150^\circ$
$\Rightarrow\angle\text{ABP}=75^\circ$
$\angle\text{ABP}=\angle\text{BAC}=75^\circ\ ($Alternate algles$)$
$\angle\text{ABP}=\angle\text{ACB}=75^\circ\ ($Alternate segment theorem$)$
Now, sum of all angles of $\triangle\text{ABC} 180^\circ $
$\Rightarrow\angle\text{BAC}+\angle\text{ACB}+\angle\text{ABC}=180^\circ$
$\Rightarrow75^\circ+75^\circ+\angle\text{ABC}=180^\circ$
$\Rightarrow\angle\text{ABC}=30^\circ$ View full question & answer→MCQ 961 Mark
In the given figure, if quadrilateral $\text{PQRS}$ 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 971 Mark
In figure, $AB$ is a chord of the circle and $\text{AOC}$ is its diameter such that $\angle\text{ACB}=50^\circ.$ If $AT$ is the tangent to the circle at the point $A,$ then $\angle\text{BAT}$ is equal to :
- A
$65^\circ$
- B
$60^\circ$
- ✓
$50^\circ$
- D
$40^\circ$
AnswerCorrect option: C. $50^\circ$
In figure, $\text{AOC}$ is a diameter of the circle.
We know that, diameter subtends an angle $90^\circ$ at the circle.
So, $\angle\text{ABC}=90^\circ$
In $\triangle\text{ACB},\ \angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
$[$Since, sum of all angles of a triangle is $180^\circ ]$
$\Rightarrow\ \angle\text{A}+90^\circ+50^\circ=180^\circ$
$\Rightarrow\ \angle\text{A}+140=180$
$\Rightarrow\ \angle\text{A}=180^\circ-140^\circ=40^\circ$
$\angle\text{A}\text{ or }\angle\text{OAB}=40^\circ$
Now, $AT$ is the tangent to the circle at point $A$.
So, $OA$ is perpendicular to $AT.$
$\therefore\ \angle\text{OAT}=90^\circ\ \ [\text{from figure}]$
$\Rightarrow\ \angle\text{OAB}+\angle\text{BAT}=90^\circ$
On putting $\angle\text{OAB}=40^\circ,$ we get
$\Rightarrow\ \angle\text{BAT}=90^\circ-40^\circ=50^\circ$
Hence, the value of $\angle\text{BAT}$ is $50^\circ .$
View full question & answer→MCQ 981 Mark
In Figure, if $\text{PQR}$ 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 $, AB \| PR$
$\therefore\ \angle\text{ABQ}=\angle\text{BQR}=70^\circ \ [$alternate angles$]$
Also, $QD$ is perpendicular to $AB$ and $QD$ bisects $AB$.
In $\triangle\text{QDA } $ and $\triangle\text{QDB},\ \ \angle\text{QDA}=\angle\text{QDB}\ \ [\text{each }90^\circ]$
$AD = BD$
$QD = QD\ [$common side$]$
$\therefore\ \triangle\text{ADQ}\cong\triangle\text{BDQ} \ [$by $\text{SAS}$ similarity criterion$]$
Then $\angle\text{QAD}=\angle\text{QBD}\ \ [\text{CPCT}] ...(\text{i})$
Also $\angle\text{ABQ}=\angle\text{BQR} \ [$alternate interior angle$]$
$\therefore\ \angle\text{ABQ}=70^\circ\ \ [\because\angle\text{BQR}=70^\circ]$
Hence, $\angle\text{QAB}=70^\circ\ \ [\text{from Eq. (i})]$
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 991 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 1001 Mark
To draw a pair of tangents to a circle, which are inclined to each other at angle of $45^\circ,$ we have to draw the tangents at the end points of those two radii, the angle between which is :
- A
$105^\circ$
- ✓
$135^\circ$
- C
$140^\circ$
- D
$145^\circ$
AnswerCorrect option: B. $135^\circ$
Since $AB$ and $AC$ are the tangent to the circle.
$\angle\text{OBA}=\angle\text{OCA}=90^\circ ($tangent is perpendicular to the radius of a circle$)$
In $\text{ACOB},$
$\angle\text{OBA}+\angle\text{BAC}+\angle\text{OCA}+\angle\text{BOC}=360^\circ$
$\Rightarrow90^\circ+45^\circ+45^\circ+\angle\text{BOC}=360^\circ$
$\Rightarrow\angle\text{BOC}=135^\circ$ View full question & answer→