Download App

Maths 3 Marks Question

Circles · Jharkhand Board (JAC) STD 10 (Jharkhand - English Medium). 29 questions.

Questions

3 Marks Question

Take a timed test

29 questions · self-marked practice — reveal the answer and mark yourself.

Question 13 Marks
In the given figure, there are two concentric circles with centre $O$ of radii 5cm and $ 3\ cm$. From an external point $P$, tangent $PA$ and $PB$ are drawn to these circles.
If $AP = 12\ cm$, find the length of $BP.$
Answer

Given, $O A=5 cm$ and $O B=3 cm$
$AP=12 cm$
From the property of tangent we know radius of the circle is always perpendicular to tangent at the point of contact. so, $\triangle\text{OAP}$ is right angle triangle.
$OP^2= AP^2 + OA^2$
$\Rightarrow OP^2 = (12)^2 + (5)^2$
$\Rightarrow OP^2 = 144 + 25$
$\Rightarrow\text{OP}=\sqrt{169}$
$\Rightarrow OP = 13cm$
Now consider $\triangle\text{OBP},$ is also right angle triangle.
$OP^2= BP^2 + OB^2$
$13^2= AP^2 + (3)^2$
$\Rightarrow BP^2= 169 - 9$
$ \Rightarrow\text{BP}=\sqrt{160}$
$\Rightarrow\text{BP}=4\sqrt{10}\ \text {cm}$
Hence, length of $\text{BP}=4\sqrt{10} \text{cm.}$
View full question & answer→
Question 23 Marks
Prove that a diameter AB of a circle bisects all those chords which are parallel to the tangent at the point A.
Answer
Given, AB is a diameter of the circle.
A tangent is drawn from point A.
Draw a chord CD parallel to the tangent MAN.

So, CD is a chord of the circle and OA is a radius of the circle.
$\angle\text{MAO}=90^{\circ}$
[Tangent at any point of a circle is perpendicular to the radius through the point of contact.]
$\angle\text{CEO}=\angle\text{MAO}$ [corresponding angles.]
$\angle\text{CEO}=90^{\circ}$
Thus, OE bisects CD
[perpendicular from centre of circle to chord bisects the chord.]
Similarly, the diameter AB bisects all. Chord which are parallel to the tangent at the point A.
View full question & answer→
Question 33 Marks
If PA and PB are tangents from an outside point P such that PA = 10cm and $\angle\text{APB}=60^{\circ}.$ Find the length of chord AB.
Answer
PA and PB are the tangents from a point PQ outside the circle with centre O. PA = 10cm and $\angle\text{APB}=60^{\circ}$
Tangents drawn from a point outside the circle are equal.
$\text{PA}+\text{PB}=10\text{cm}\ \angle\text{PAB}=\angle\text{PBA}$ (Angles opposite to equal sides)
But in $\triangle\text{APB},$
$\angle\text{APB}+\angle\text{PAB}+\angle\text{PBA}=180^{\circ}$ (Angles of a triangle)
$\Rightarrow60^{\circ}+\angle\text{PAB}+\angle\text{PAB}=180^{\circ}$
$\Rightarrow 2\angle\text{PAB}=180^{\circ}-60^{\circ}=120^{\circ}$
$\angle\text{PAB}=60^{\circ}$
$\angle\text{PBA}=\angle\text{PAB}=60^{\circ}$
PA = PB = AB = 10cm
Hence length of chord AB = 10cm.
View full question & answer→
Question 43 Marks
$A$ is a point at a distance $13 \ cm$ from the centre $O$ of a circle of radius $5 \ cm$ . $A P$ and $A Q$ are the tangents to the circle at $P$ and $Q$. If a tangent $B C$ is drawn at a point $R$ lying on the minor arc $P Q$ to intersect $A P$ at $B$ and $A Q$ at $C$, find the perimeter of the $\triangle A B C$.
Answer
Given: Two tangents are drawn from an external point A to the circle with centre O, Tangent BC is drawn at a point R, radius of circle equals to 5cm.
To find: Perimeter of $\triangle\text{ABC}.$
Proof: $\angle\text{OPA}=90^{\circ}$[Tangent at any point of a circle is perpendicular to the radius through the point of contact]
$OA^2 = OP^2 + PA^2$​​​​​​​ [by Pythagoras Theorem]
$(13)^2 = 5^2 + PA^2$
$\Rightarrow PA^2 = 144^2 = 12^2$
$\Rightarrow PA = 12cm$
Now, perimeter of $\triangle\text{ABC}.$
$= AB + BC + CA$
$​​​​​​​= (AB + BR) + (RC + CA)$
$= AB + BP + CQ + CA [BR = BP, RC = CQ$ [tangents from internal point to a circle are equal]
$= AP + AQ = 2AP = 2 × (12) = 24cm [AP = AQ$ tangent from internal point to a circle are equal]
Hence, the perimeter of $\triangle\text{ABC}.$ = 24cm.
View full question & answer→
Question 53 Marks
In the figure, PQ is a chord of a circle and PT is the tangent at P such that $\angle\text{QPT}=60^{\circ}.$ Then, find $\angle\text{PRQ}$.
Answer

Construction: Take any point on major arc PQ and name it S. join SQ and SP.
In the given fig. PT is the tangent. SO, $\text{PT}\bot\text{PO}.$
$\angle\text{QPT}=60^{\circ}$
Thus, $\angle\text{OPQ}=90^{\circ}-60^{\circ}=30^{\circ}$
OQ = OP (Radii of the circle.)
$\angle\text{OQP}=\angle\text{OPQ}=30^{\circ}$
In $\triangle\text{OPQ},$
$\angle\text{OPQ}+\angle\text{OQP}+\angle\text{POQ}=180^{\circ}$
$\Rightarrow30^{\circ}+30^{\circ}+\angle\text{POQ}=180^{\circ}$
$\Rightarrow\angle\text{POQ}=120^{\circ}$
Now,
$\angle\text{PSQ}=\frac{1}{2}\angle\text{POQ}=\frac{1}{2}\times120=60^{\circ}$
PSQR is a cyclic quadilateral. Thus,
$\angle\text{PSQ}+\angle\text{PRQ}=180^{\circ}$
$\Rightarrow60^{\circ}+\angle\text{PRQ}=180^{\circ}$
$\Rightarrow\angle\text{PRQ}=120^{\circ}$
View full question & answer→
Question 63 Marks
AB and CD are common tangents to two circles of equal radii. Prove that AB = CD.
Answer

Given: Two circles with center O and O'. AB and CD are common tangents to the circles which intersrcts in P.
To prove:
AP = PC (length of tangents drawn from an external point to the circle are equal) ...(i)
PB = PD (length of tangents drawn from an external point to the circle are equal) ...(ii)
Adding (i) and (ii), we get
AP + PB = PC + PD
⇒ AB = CD
Hence proved
View full question & answer→
Question 73 Marks
Two concentric circles are of diameters 30cm and 18cm. Find the length of the chord of the larger circle which touches the smaller circle.
Answer
Let R be the radius of outer circle and r be the radius if small circle of two concentric circle AB is the chord of the outer circle and touches the smaller circle at P Join OP, OA. $\therefore\text{OP}\bot\text{AB}$ and bisects it at P OA = R and OP = r $\therefore\text{OA}=\frac{30}{2}=15\text{cm},\text{OP}=\frac{18}{2}=9\text{cm}$$\therefore$ in Right $\triangle\text{OAP}$
$\text{AP}=\sqrt{\text{OA}^2-\text{OP}^2}$ $=\sqrt{{15^2}-{9^2}}$ $=\sqrt{225-81}$ $\sqrt{144}=12\text{cm}$ But AB = 2AP = 2 × 12cm = 24cm.
View full question & answer→
Question 83 Marks
$O$ is the centre of a circle of a radius $8 \ cm$ the tangent at a point $A$ on the circle cuts a line through $O$ at $B$ such that $A B$ $=15 cm$. Find $O B$.
Answer
Radius $OA = 8\ cm$, $ST$ is the tangent to the circle at A and $AB = 15\ cm$

OA ⊥ tangent TS In right
$\triangle \text{OAB},$
$\Rightarrow OB^2 = OA^2 + AB^2​​​​​​​$​​​​​​​ (Pythagoras Theorem)
$\Rightarrow OB^2 = (8)^2 + (15)^2$
$\Rightarrow OB^2 = 64 + 225$
$​​​​​​​\Rightarrow OB^2 = 289 = (17)^2 OB = 17cm.$
View full question & answer→
Question 93 Marks
Prove that the intercept of a tangent between two parallel tangents to a circle subtends a right angle at the centre.
Answer
consider circle with center 'O' and has two parallel tangents through A & B at ends of diameter.

Let tangents through M intersects the tangents parallel at P and Q required to prove is that $\text{POQ}=90^\circ$
From fig. It is clear that ABQP is a quadilateral.
$\angle{\text{A}}+\angle\text{B}=90^\circ+90^\circ=180^\circ$[At point of contact tangent & radius are perpendicular]
$\angle{\text{A}}+\angle{\text{B}}+\angle{\text{P}}+\angle{\text{Q}}=360^\circ$ [Angle sum property]
$\angle{\text{P}}+\angle{\text{Q}}=360^\circ-180^\circ=180^\circ\dots(\text{i})$
$\text{At P&Q }\angle\text{APO}=\angle\text{OPQ}=\frac12\angle{\text{P}}$
$\angle{\text{BQO}}=\angle\text{PQO}=\frac12\angle{\text{Q}}\text{ in (i)}$
$2\angle\text{OPQ}+2\angle\text{PQO}=180^\circ$
$\angle\text{OPQ}+\angle\text{POQ}=90^\circ\dots{\text{(ii)}}$
In $\triangle\text{OPQ},\angle\text{OPQ}+\triangle\text{PQO}+\triangle\text{POQ}=180^\circ$ [Angle sum property]
$90^\circ+\angle\text{POQ}=180^\circ$ [from(ii)]
$\angle\text{OPQ}=180^\circ-90^\circ=90^\circ$
$\therefore\angle{\text{POQ}}=90^\circ$
View full question & answer→
Question 103 Marks
If the tangent at a point $P$ to a circle with centre $O$ cuts a line through $O$ at $Q$ such that $P Q=24 cm$ and $O Q=25 cm$. Find the radius of the circle.
Answer


Given, $PQ = 15cm$
$OQ = 25cm$
$OP = xcm$
we know,
radius OP is always perpendicular to tangent $PQ$
By using pythagouras, we find $OB$
$OQ = OP^2 + PQ^2$
$\Rightarrow (25)^2 = (x)^2 + (24)^2$
$\Rightarrow 625 = x^2+ 576$
$\Rightarrow x^2= 625 - 576$
$\Rightarrow\text{x}=\sqrt{49} $
Hence, length of OP is 7cm.
View full question & answer→
Question 113 Marks
In the given figure, common tangents PQ and RS to two circles intersect at A. Prove that PQ = RS.
Answer
The figure given in the question is,
We know from the property of tangents that the length of two tangents drawn from a common external point will be equal.
Therefore,
PA = RA …… (1)
AQ = AS …… (2)
Let us add equation (1) and (2)
PA + AQ = RA + AS
PQ = RS
Thus we have proved that PQ = RS.
View full question & answer→
Question 123 Marks
The common tangents AB and CD to two circles with centres O and O’ intersect at E between their centres. Prove that the points O, E and O’ are collinear.
Answer
Joint AO, OC and O’D, O’B
Now, in $\triangle\text{EO}'\text{D}$ and,
O’D = O’B [radius]
O’E = O’E [common side]
ED = EB [since, tangentdrawn from an external point to the circle are equal in the length.]

$\therefore\triangle\text{EO'D}\cong\triangle\text{EO'B}$ [by SSS congruence rule]
$\angle\text{EO'D}=\angle\text{EO'B}$
O'E is the angle bisects of $\angle\text{DEB}\ ...\text{(i)}$
Similarily, OE is the angle bisects of $\angle\text{AEC}$
Now, in quadilateral DEBO'
$\angle\text{OD'E}=\angle\text{O'BE}=90^{\circ}$[since, CED is the tangent to the circle and O'D is the radius, i.e,$\text{O'D}\bot\text{CED}$]
$\Rightarrow\angle\text{O'DE}+\angle\text{O'BE}=180^{\circ}$
$\therefore\angle\text{DEB}+\angle\text{DO'B}=180^{\circ}\ ...\text{(ii)}$
[since, DEBo' is cyclic quadilateral]
since, AB is a straight line.
$\therefore\angle\text{AED}+\text{DEB}=180^{\circ}$
$\Rightarrow\angle\text{AED}+180^{\circ}-\angle\text{DO'B}=180^{\circ}$
View full question & answer→
Question 133 Marks
Out of the two concentric circles, the radius of the outer circle is $5 \ cm$ and the chord $C D$ of length 8 cm is a tangent to the inner circle. Find the radius of the inner circle.
Answer


Let the centre of the two concentric circles be $0$ .
$CD$ is the tangent to the inner circle.
$OP$ joins the centre of the circle to the tangent at the point of contant.
$OP \perp CD$ and $OP$ bisects $C D$.
Thus, $PD =4 cm$
In $\triangle OPD$,
$\Rightarrow OP^2 + PD^2 = OD^2$
$\Rightarrow OP^2 = OD^2 - PD^2$
$^\Rightarrow OP^2 = (5)^2 - (4)^2 = 25 - 16 = 9$
$\Rightarrow OP = 3cm$
Hence, the radius of the inner circle $=3 \ cm$
View full question & answer→
Question 143 Marks
In the given figure, a circle touches all the four sides of a quadrilateral ABCD with AB = 6cm, BC = 7cm and CD = 4cm. Find AD.
Answer
A circle touches the sides AB, BC, CD and DA of a quadrilateral ABCD at P, Q, R and S respectively. AB = 6cm, BC = 7cm, CD = 4cm

Let AD = x
AP and AS are the tangents to the circle
AP = AS
Similarly,
BP = BQ
CQ = CR
and DR = DS
AB + CD = AD + BC
⇒ 6 + 4 = 7 + x
⇒ 10 = 7 + x
⇒ x = 10 – 7 = 3
AD = 3cm
View full question & answer→
Question 153 Marks
From an external point P, tangents PA = PB are drawn to a circle with centre O. If $\angle\text{PAB}=50^{\circ}$, then find $\angle\text{AOB}$.
Answer

A = PB [tangents drawn from external point are equal]
$\angle\text{PAB}=\angle\text{PAB}=50^{\circ}$[angles equal to opposite sides]
$\angle\text{APB}=180^{\circ}-50^{\circ}=80^{\circ}$[angle-sum property of a A]
In cyclic quad. OAPB
$\angle\text{AOB}+\angle\text{APB}=180^{\circ}$[sum of opposite angles of a cyclic quadrilateral is 180°]
$\angle\text{AOB}+80^{\circ}=180^{\circ}$
$\angle\text{AOB}=180^{\circ}-80^{\circ}=100^{\circ}.$
View full question & answer→
Question 163 Marks
In the given figure, PQ is tangent at a point R of the circle with centre O. If $\angle\text{TRQ}=30^{\circ},$ find m $\angle\text{PRS}$.
Answer

Given, $\angle\text{TRQ}=30^{\circ}$
ST is diameter and right angle present in the semi circle.
Then, $\angle\text{SRT}=90^{\circ}$
Now,
$\angle\text{TRQ}+\angle\text{SRT}+\angle\text{PRS}=180^{\circ}$
$\Rightarrow 30^{\circ}+90^{\circ}+\angle\text{PRS}=180^{\circ}$
$\Rightarrow 120^{\circ}+\angle\text{PRS}=180^{\circ}$
$\angle\text{PRS}=180^{\circ}-120^{\circ}$
Hence,
$\angle\text{PRS}=60^{\circ}$
View full question & answer→
Question 173 Marks
Two concentric circles are of radii $5\ cm$ and $3\ cm$. Find the length of the chord of the larger circle which touches the smaller circle.
Answer
Let us first draw whatever is given in the problem so that we can understand the problem better.
We have to find the length of $A B$, which is the chord of the larger circle which touches the smaller circle. Clearly, $OC$ is the radius of the smaller circle and is touching the tangent $A B$. We know that the radius of the circle will always form a right angle with the tangent at the point of contact. We have draw $OA$ in order to complete the triangle $OAC$ which will be a right triangle. From the figure it is very clear that $OA$ is the radius of the larger circle which is $5 cm$ . We can now find $A C$ using Pythagoras theorem.
We have, $AC^{2}$
$\Rightarrow OA^2 - OC^2 AC^{2}$
$\Rightarrow 5^2 - 3^2 AC^{2}$
$\Rightarrow 25 - 9 AC^2$
$\Rightarrow 16$
$\text{AC}=\sqrt{16}$
$= 4$ Similarly we can find $CB.$
We have,$CB^2$
$\Rightarrow OB^2 - OC^2CB^{2}$
$^\Rightarrow 5^2 - 3^2CB^2$
$\Rightarrow 25 - 9CB^2$
$\Rightarrow 16$
$C B=\sqrt{16}=4$ From the figure we can see that, $A B=A C+C B$ Since we have found the values of $A C$ and $C B$, let us substitute the values in the above equation. We get, $A B=4+4 A B=8$ Therefore, the length of the chord of the larger circle which touches the smaller circle is $8 \ cm$ .
View full question & answer→
Question 183 Marks
In given fig. tangents PQ and PR are drawn from an external point to a circle with centre O, such that $\angle\text{RPQ}=30^{\circ}$. A chord RS is drawn parallel to the tangent PQ.Find $\angle\text{RQS}$
Answer

It is given that tangents PQ and PR are drawn from an external point P to a circle with centre O, such that $\angle\text{RPQ}=30^{\circ}$. Also, $\text{RS}\parallel\text{PQ}$.
join OR and OS. PQ and PR are tangents drawn from an external point P to a circle.
$\therefore$ PQ = PR (lengths of tangents frawn from an external point to a circle are equal)
In $\triangle\text{PQR},$
PQ = PR
$\therefore\angle\text{PRQ}=\angle\text{PQR}$ (in a triangle, equal sides have equal anles opposite to them)
Now,
$\angle\text{PRQ}+\angle\text{PQR}+\angle\text{RPQ}=180^{\circ}$ (Angle sum property)
$\Rightarrow2\angle\text{PRQ}=180^{\circ}-30^{\circ}=150^{\circ}$
$\Rightarrow\angle\text{PRQ}=75^{\circ}$
$\therefore\angle\text{PQR}=\angle\text{PRQ}=75^{\circ}...(1)$
Now, PR is the tangent and OR is the radius through the point of contact R.
$\therefore\angle\text{ORP}=90^{\circ}$ (Tangent at any point of a circle is perpendicular to the radius through the point of a contact)
$\Rightarrow\angle\text{ORQ}=\angle\text{ORQ}-\angle\text{PRQ}=90^{\circ}-75^{\circ}=15^{\circ}$
$\therefore\angle\text{ORS}=\angle\text{SRQ}-\angle\text{ORQ}=75^{\circ}-15^{\circ}=60^{\circ}$
In $\triangle\text{SOR},$ OR = OS (Radii of the circle)
$\therefore\angle\text{ORS}+\angle\text{OSR}+\angle\text{ROS}=180^{\circ}$(Angle sum property)
$\Rightarrow\angle\text{ROS}=180^{\circ}-120^{\circ}=60^{\circ}$
we know that the angle subtended by an arc at the center is twice the angle subtended by it any point on the remaining part of the circle.
$\therefore\angle\text{ROS}=2\angle\text{RQS}$
$\Rightarrow\angle\text{RQS}=\frac{1}{2}\angle\text{ROS}=\frac{1}{2}\times60^{\circ}=30^{\circ}$
View full question & answer→
Question 193 Marks
In the given fig. PA and PB are tangents to the circle from an external point P. CD is another tangent touching the circle at Q. if PA = 12cm, QC = QD = 3cm, then find PC + PD.
Answer
Given: PA and PB are the tangents to the circle.
PA = 12cm
QC = QD = 3cm
To find: PC + PD
PA = PB = 12cm (The length of the tangents drawn from an external point to a circle are equal.)
Similarly, QC = AC = 3cm.
and QD = BD = 3cm.
Now, PC = PA - AC = 12 - 3 = 9cm
similarily, PD = PB - BD = 12 - 3 = 9cm
Hence, PC + PD = 9 + 9 = 18cm.
View full question & answer→
Question 203 Marks
AB is a chord of a circle with centre O, AOC is a diameter and AT is the tangent at A as shown in the figure. Prove that $\angle\text{BAT}=\angle\text{ACB}.$
Answer
Since, AC is a diameter line, so angle in semicircle makes an angle $90^{\circ}$.
$\angle\text{ABC}=90^{\circ}$ [by property]
In $\triangle\text{ABC}.$
$\angle\text{CAB}+\angle\text{ABC}+\angle\text{ACB}=180^{\circ}$ [sum of all interior angles of any triangle is 180°]
$\Rightarrow\angle\text{CAB}+\angle\text{ACB}=180^{\circ}-90^{\circ}=90^{\circ}\ ...\text{(i)}$
Since, diameter of a circle is perpendicular to the tangent.
i.e. $\text{CA}\bot\text{AT}$
$\angle\text{CAT}=90^{\circ}$
$\Rightarrow\angle\text{CAB}+\angle\text{BAT}=90^{\circ}\ ...\text{(ii)}$
From Eqs. (i) and (ii),
$\angle\text{CAB}+\angle\text{ACB}=\angle\text{CAB}+\angle\text{BAT}$
$\Rightarrow\angle\text{ACB}=\angle\text{BAT}$
Hence proved.
View full question & answer→
Question 213 Marks
The length of three concesutive sides of a quadilateral circumscribing a circle are 4cm, 5cm, 7cm respectively. Determine the length of thefourth side.
Answer
Let us first put the given date in the form of a diagram.

From the property of tangents we know that the length of two tangents drawn from the same external will be equal. Therefore we have,
AR = SA
Let us represent AR and SA by 'a'.
Similarily,
QB = RB
Let us represent SD and DP by 'b'.
PC = CQ
Let us represent PC and PQ by 'c'.
SD = DP
Let us represent QB and RB by 'd'.
View full question & answer→
Question 223 Marks
In the given figure, a circle is inscribed in a quadrilateral ABCD in which $\angle\text{B}=90^{\circ}.$ It AD = 23cm, AB = 29cm and DS = 5cm, find the radius r of the circle.
Answer
In the figure, O is the centre of the circle inscribed in a quadrilateral ABCD and $\angle\text{B}=90^{\circ}$ AD = 23cm, AB = 29cm, DS = 5cm.

OP = OQ (radii of the same circle)
AB and BC are tangents to the circle and OP and OQ are radii
OP ⊥ BC and OQ ⊥ AB
$\angle\text{OPB}=\angle\text{OQB}=90^{\circ}$
PBQO is a square
DS and DR are tangents to the circle
DR = DS = 5cm
AR = AD – DR = 23 – 5 = 18cm
AR and AQ are the tangents to the circle
AQ = AR = 18cm But AB = 29cm
View full question & answer→
Question 233 Marks
In the given figure, PA and PB are tangents from an external point P to a circle with centre O. LN touches the circle at M. Prove that PL + LM = PN + MN.
Answer
Given: In the figure, PA and PB are the tangents to the circle with centre O from a point P outside it LN touches it at M.

To prove: PL + LM = PN + MN
Prove: PA and PB are tangents to the circle from P
PA = PB
Similarly from L, LA and LM are tangents
LA = LM
Similarly NB = NM
Now PA = PB ⇒ PL + LA = PN + NB
PL + LM = PN + NM
Hence proved.
View full question & answer→
Question 243 Marks
Prove that the centre of a circle touching two intersecting lines lies on the angle bisector of the lines.
Answer
Given: Two tangents PQ and PR are drawn from an external point P to a circle with centre O.
To prove: Centre of a circle touching two intersecting line s lies on the angle bisector of the lines.

Construction: Join OR, and OQ.
In $\triangle\text{POR}\ \text{and}\ \triangle\text{PQO}$
$\angle\text{PRO}=\angle\text{PQO}=90^{\circ}$ [tangent at any point of a circle is perpendicular to the radius through the point of contact]
OR = OQ [radii of same circle]
Since, OP is common.
$\triangle\text{PRO}\cong\triangle\text{PQO}$ [SAS]
Hence, $\angle\text{RPO}=\angle\text{QPO}$ [by CPCT]
Thus, O lies on angle bisector of PR and PQ.
Hence proved.
View full question & answer→
Question 253 Marks
In the given figure, two tangents AB and AC are drawn to a circle with centre O such that $\angle\text{BAC}=120^{\circ}$. Prove that OA = 2AB.
Answer
Cosider $\triangle\text{OAB}$ and $\triangle\text{OAC}. $

We have,
OB = OC (Since they are radii of the same circle.)
AB = AC (Since length of two tangents drawn from an external point will be equal.)
OA is the common side.
Therefore by SSS congruency, we can say that$\triangle\text{OAB}$ and $\triangle\text{OAC}$ are congruent triangles.
Therefore,
$\angle\text{OAB}=\angle\text{OAC}$
It is given that,
$\angle\text{OAB}+\angle\text{OAC}=120^{\circ}$
$2\angle\text{OAB}=120^{\circ}$
$\angle\text{OAB}=60^{\circ}$
We know that,
$\cos\angle\text{OAB}=\frac{\text{AB}}{\text{OA}}$
$\cos60^{\circ}=\frac{\text{AB}}{\text{OA}}$
we know that,
$\cos60^{\circ}=\frac{1}{2}$
Therefore,
$\frac{1}{2}=\frac{\text{AB}}{\text{OA}}$
OA = 2AB
View full question & answer→
Question 263 Marks
If $d_1, d_2 (d_2 > d_1)$ be the diameters of two concentric circles and c be the length of a chord of a circle which is tangent to the other circle, prove that . $\text{d}^2_2=\text{c}^2+\text{d}^2_2$
Answer

Let $O$ be the center of two concentric circles and $PQ$ be the tangent to the inner circle that touches the circle at $R$.
Now, $\text{OQ}=\frac{1}{2}\text{d}_2$ and $\text{OR}=\frac{1}{2}\text{d}_1$
Also, $PQ = c$
As, $PQ$ is the tangent to the circle.
$\Rightarrow\text{OR}\bot\text{PQ}$
$\Rightarrow\text{QR}=\frac{1}{2}\text{PQ}=\frac{1}{2}\text{c}$
in triangle $OQR,$
$\therefore$ By pythagoras throrem,
$(OQ)^2 = (OR)^2 + (RQ)^2$​​​​​​​
$\Rightarrow\Big(\frac{\text{d}_2}{2}\Big)^2=\Big(\frac{\text{d}_1}{2}\Big)^2+\Big(\frac{\text{c}}{2}\Big)^2$
$\Rightarrow \text{(d}_2)^{2} = \big(\text{d}_{1}\big)^{2} + \text{(c})^{2} $
View full question & answer→
Question 273 Marks
In the figure, the tangent at a point C of a circle and a diameter AB when extended itersect at P. If $\angle\text{PCA}=110^{\circ},$ find $\angle\text{CBA}.$
Answer

$\angle\text{ACB}=90^{\circ}$(Angle inscribed in a semi-circle)
$\angle\text{PCO}=90^{\circ}$(PC is a tangent at C)
Now, $\angle\text{PCA}=\angle\text{PCO}+\angle\text{OCA}$
$\Rightarrow110^{\circ}=90^{\circ}+\angle\text{OCA}$
$\Rightarrow\angle\text{OCA}=20^{\circ}$ Since, OC = OA (radii of the circle.)
$\angle\text{OCA}=\angle\text{OAC}=20^{\circ}$
In $\triangle\text{ABC,}$
$\angle\text{BAC}+\angle\text{ACB}+\angle\text{CBA}=180^{\circ}$
$\Rightarrow90^{\circ}+20^{\circ}+\angle\text{CBA}=180^{\circ}$
$\angle\text{CBA}=70^{\circ}$
View full question & answer→
Question 283 Marks
From an external point P, tangents PA and PB are drawn to a circle with centre O. If CD is the tangent to the circle at a point E and PA = 14cm, find the perimeter of $\triangle\text{PCD}.$
Answer

We have to find perimeter of $\triangle\text{PCD}$
From property of tangent
AP = PB are tangent from point P
there, CA = CE
DB = ED
Now $\text{perimeter}\ \text{of}\ \triangle\text{PCD}= \text{PC + CD + PD}$
$\Rightarrow\text{perimeter}\ \text{of}\ \triangle\text{PCD}= \text{PC + CA + DB + PD}$
$\Rightarrow\text{perimeter}\ \text{of}\ \triangle\text{PCD}= \text{PA + PB}$
$\Rightarrow\text{perimeter}\ \text{of}\ \triangle\text{PCD}= 2 \text{PA}$
$\Rightarrow\text{perimeter}\ \text{of}\ \triangle\text{PCD}= \text{PC} + 2 × 14$
Hence, $\Rightarrow\text{perimeter}\ \text{of}\ \triangle\text{PCD}$ is 28cm
View full question & answer→
Question 293 Marks
In the given figure, O is the centre of the circle and BCD is tangent to it at C. Prove that $\angle\text{BAC}+\angle\text{ACD}=90^{\circ}.$
Answer
In the given figure, let us join D an A.

Consider$\triangle\text{OCA}$ . We have,
OC = OA (Radii of the same circle)
We know that angles opposite to equal sides of a triangle will be equal.
It is clear from the figure that,
$\angle\text{DCA}+\angle\text{OCA}=\angle\text{OCD}$
Now from (1)
$\angle\text{DCA}+\angle\text{OAC}=\angle\text{OCD}$
Now as BD is tangent therefore $\angle\text{OCD}=90^{\circ}$
Therefore $\angle\text{OCA}=\angle\text{OAC}\ \angle\text{DCA}+\angle\text{OAC}=90^{\circ}$
From the figure we can see that $\angle\text{OAC}=\angle\text{BAC}$
$\angle\text{DAC}+\angle\text{BAC}=90^{\circ}$
Thus we have proved.
View full question & answer→
Generate Paper FreeAll Circles topics