## RD Sharma Class 10 Solutions Chapter 8 Circles Ex 8.2

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 8 Circles Ex 8.2

Other Exercises

Question 1.
If PT is a tangent at T to a circle whose centre is O and OP = 17 cm, OT = 8 cm. Find the length of the tangent segment PT.
Solution:
PT is the tangent to the circle with centre O, at T
Radius OT = 8 cm, OP = 17 cm

PT is the tangent segment
Now in right ∆OPT,
OP² = OT² + PT² (Pythagoras Theorem)
=> (17)² = (8)² + PT²
=> 289 = 64 + PT²
=> PT² = 289 – 64 = 225 = (15)²
PT = 15 cm

Question 2.
Find the length of a tangent drawn to a circle with radius 5 cm, from a point 13 cm from the centre of the circle.
Solution:
From a point P outside the circle with centre O, PT is the tangent to the circle and radius
OT = 5 cm, OP = 15 cm

OT ⊥ PT
Now in right ∆OPT,
OP² = OT² + PT² (Pythagoras Theorem)
(13)² = (5)² + PT²
=> 169 = 25 + PT²
=> PT² = 169 – 25 = 144 = (12)²
PT = 12 cm

Question 3.
A 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 10 cm. Find the radius of the circle.
Solution:
From a point P outside the circle of centre 0 and radius OT, PT is the tangent to the circle
OP = 26 cm, PT = 10 cm

Now in right ∆OPT
OP² = OT² + PT² (Pythagoras Theorem)
=> (26)² = r² + (10)²
=> 676 = r² + 100
=> 676 – 100 = r²
=> r² = 576 = (24)²
r = 24
Hence radius of the circle = 24 cm

Question 4.
If from any point on the common chord of two intersecting circles, tangents be drawn to the circles, prove that they are equal.
Solution:
Given : QR is the common chord of two circles intersecting each other at Q and R
P is a point on RQ when produced From PT and RS are the tangents drawn to tire circles with centres O and C respectively

To prove : PT = PS
Proof: PT is the tangent and PQR is the secant to the circle with centre O
PT² = PQ x PR ….(i)
Similarly PS is the tangent and PQR is the secant to the circle with centre C
PS² = PQ x PR ….(ii)
From (i) and (ii)
PT² = PS²
PT = PS
Hence proved.

Question 5.
If the sides of a quadrilateral touch a circle, prove that the sum of a pair of opposite sides is equal to the sum of the other pair.
Solution:
Given : The sides of a quadrilateral ABCD touch the circle at P, Q, R and S respectively

To prove : AB + CD = AP + BC
Proof : AP and AS are the tangents to the circle from A
AP = AS ….(i)
Similarly BP = BQ ……(ii)
CR = CQ ….(iii)
and DR = DS ….(iv)
AP + BP + CR + DR = AS + BQ + CQ + DS
=> (AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)
=> AB + CD = AD + BC
Hence proved.

Question 6.
Out of the two concentric circles, the radius of the outer circle is 5 cm and the chord AC of length 8 cm is a tangent to the inner circle. Find the radius of the inner circle. [NCERT Exemplar]
Solution:
Let C1 and C2 be the two circles having same centre O. AC is a chord which touches the C1 at point D

Join OD.
Also, OD ⊥ AC
AD = DC = 4 cm
[perpendicular line OD bisects the chord]
In right angled ∆AOD,
[by Pythagoras theorem, i.e.,
(hypotenuse)² = (base)² + (perpendicular)²]
=> DO² = 5² – 4² = 25 – 16 = 9
=> DO = 3 cm
Radius of the inner circle OD = 3 cm

Question 7.
A chord PQ of a circle is parallel to the tangent drawn at a point R of the circle. Prove that R bisects the arc PRQ. [NCERT Exemplar]
Solution:
Given : Chord PQ is parallel tangent at R.
To prove : R bisects the arc PRQ.

Proof: ∠1 = ∠2 [alternate interior angles]
∠1 = ∠3
[angle between tangent and chord is equal to angle made by chord in alternate segment]
∠2 = ∠3
=> PR = QR
[sides opposite to equal angles are equal]
=> PR = QR
So, R bisects PQ.

Question 8.
Prove that a diameter AB of a circle bisects all those chords which are parallel to the tangent at the point A. [NCERT Exemplar]
Solution:
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.
∠MAO = 90°
[Tangent at any point of a circle is perpendicular to the radius through the point of contact]
∠CEO = ∠MAO [corresponding angles]
∠CEO = 90°
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.

Question 9.
If AB, AC, PQ are the tangents in the figure, and AB = 5 cm, find the perimeter of ∆APQ.

Solution:
Given : AB, AC and PQ are the tangents to the circle as shown in the figure above and AB = 5 cm
To find : The perimeter of ∆APQ
Proof: PB and PX are the tangents to the circle
PB = PX
Similarly QC and QX are the tangents from
QC = QX
and AB and AC are the tangents from A
AB = AC
Now perimeter of ∆APQ
= AP + PQ + AQ
= AP + PX + QX + AQ
= AP + PB + QC + AQ { PB = PX and QC = QX}
= AB + AC
= AB + AB (AB=AC)
= 2 AB = 2 x 5 = 10 cm

Question 10.
Prove that the intercept of a tangent between two parallel tangents to a circle subtends a right angle at the centre.
Solution:
Given : PQ and RS are parallel tangents of a circle
RMP is the intercept of the tangent between PQ and RS
RO and PQ are joined

Question 11.
In the figure, PQ is tangent at a point R of the circle with centre O. If ∠TRQ = 30°, find m ∠PRS
Solution:
In the figure,
PRQ is tangent to the circle with centre O at R

RT and RS are joined such that ∠TRQ = 30°
Let ∠PRS = x°
Now ∠SRX = 90° (angle in a semicircle)
But ∠TRQ + ∠SRT + ∠PRS = 180° (Angles of a line)
=> 30° + 90° + x° = 180°
=> 120° + x° = 180°
=> x° = 180° – 120° = 60°
∠PRS = 60°

Question 12.
If PA and PB are tangents from an outside point P, such that PA = 10 cm and ∠APB = 60°. Find the length of chord AB.
Solution:
PA and PB are the tangents from a point PQ outside the circle with centre O
PA = 10 cm and ∠APB = 60°

Tangents drawn from a point outside the circle are equal
PA = PB = 10 cm ∠PAB = ∠PBA
(Angles opposite to equal sides)
But in ∆APB,
∠APB + ∠PAB + ∠PBA = 180° (Angles of a triangle)
=> 60° + ∠PAB + ∠PAB = 180°
=> 2 ∠PAB = 180° – 60° = 120°
∠PAB = 60°
∠PBA = ∠PAB = 60°
PA = PB = AB = 10 cm
Hence length of chord AB = 10 cm

Question 13.
In a right triangle ABC in which ∠B = 90°, a circle is drawn with AB as diameter intersecting the hypotenuse AC at P. Prove that the tangent to the circle at P bisects BC. [NCERT Exemplar]
Solution:
Let O be the centre of the given circle. Suppose, the tangent at P meets BC at Q.
Join BP.

To prove : BQ = QC
[angles in alternate segment]
Proof : ∠ABC = 90°
[tangent at any point of circle is perpendicular to radius through the point of contact]
In ∆ABC, ∠1 + ∠5 = 90°
[angle sum property, ∠ABC = 90°]
∠3 = ∠1
[angle between tangent and the chord equals angle made by the chord in alternate segment]
∠3 + ∠5 = 90° ……..(i)
Also, ∠APB = 90° [angle in semi-circle]
∠3 + ∠4 = 90° …….(ii)
[∠APB + ∠BPC = 180°, linear pair]
From Eqs. (i) and (ii), we get
∠3 + ∠5 = ∠3 + ∠4
∠5 = ∠4
=> PQ = QC
[sides opposite to equal angles are equal]
Also, QP = QB
[tangents drawn from an internal point to a circle are equal]
=> QB = QC
Hence proved.

Question 14.
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 = 14 cm, find the perimeter of ∆PCD.
Solution:
PA and PB are the tangents drawn from a point P out side the circle with centre O
CD is another tangents to the circle at point E which intersects PA and PB at C and D respectively

PA = 14 cm
PA and PB are the tangents to the circle from P
PA = PB = 14 cm
Now CA and CE are the tangents from C
CA = CE ….(i)
Similarly DB and DE are the tangents from D
DB = DE ….(ii)
Now perimeter of ∆PCD
= PC + PD + CD
= PC + PD + CE + DE
= PC + CE + PD + DE
= PC + CA + PD = DB {From (i) and (ii)}
= PA + PB
= 14 + 14
= 28 cm

Question 15.
In the figure, ABC is a right triangle right-angled at B such that BC = 6 cm and AB = 8 cm. Find the radius of its incircle. [CBSE 2002]

Solution:
In right ∆ABC, ∠B = 90°
BC = 6 cm, AB = 8 cm
Let r be the radius of incircle whose centre is O and touches the sides A B, BC and CA at P, Q and R respectively
AP and AR are the tangents to the circle AP = AR
Similarly CR = CQ and BQ = BP
OP and OQ are radii of the circle
OP ⊥ AB and OQ ⊥ BC and ∠B = 90° (given)
BPOQ is a square
BP = BQ = r
AR = AP = AB – BD = 8 – r
and CR = CQ = BC – BQ = 6 – r
But AC² = AB² + BC² (Pythagoras Theorem)
= (8)² + (6)² = 64 + 36 = 100 = (10)²
AC = 10 cm
=> AR + CR = 10
=> 8 – r + 6 – r = 10
=> 14 – 2r = 10
=> 2r = 14 – 10 = 4
=> r = 2
Radius of the incircle = 2 cm

Question 16.
Prove that the tangent drawn at the mid-point of an arc of a circle is parallel to the chord joining the end points of the arc. [NCERT Exemplar]
Solution:
Let mid-point of an arc AMB be M and TMT’ be the tangent to the circle.
Join AB, AM and MB.

Since, arc AM = arc MB
=> Chord AM = Chord MB
In ∆AMB, AM = MB
=> ∠MAB = ∠MBA ……(i)
[equal sides corresponding to the equal angle]
Since, TMT’ is a tangent line.
∠AMT = ∠MBA
[angle in alternate segment are equal]
∠AMT = ∠MAB [from Eq. (i)]
But ∠AMT and ∠MAB are alternate angles, which is possible only when AB || TMT’
Hence, the tangent drawn at the mid-point of an arc of a circle is parallel to the chord joining the end points of the arc.
Hence proved

Question 17.
From a point P, two tangents PA and PB are drawn to a circle with centre O. If OP = diameter of the circle show that ∆APB is equilateral.
Solution:
Given : From a point P outside the circle with centre O, PA and PB are the tangerts to the circle such that OP is diameter.
AB is joined.

To prove: APB is an equilateral triangle
Const : Join OP, AQ, OA
Proof : OP = 2r
=> OQ + QP = 2r
=> OQ = QP = r (OQ = r)
Now in right ∆OAP,
OP is its hypotenuse and Q is its mid point
OA = AQ = OQ
(mid-point of hypotenuse of a right triangle is equidistances from its vertices)
∆OAQ is equilateral triangle ∠AOQ = 60°
Now in right ∆OAP,
∠APO = 90° – 60° = 30°
=> ∠APB = 2 ∠APO = 2 x 30° = 60°
But PA = PB (Tangents from P to the circle)
=> ∠PAB = ∠PBA = 60°
Hence ∆APB is an equilateral triangle.

Question 18.
Two tangents segments PA and PB are drawn to a circle with centre O such that ∠APB = 120°. Prove that OP = 2 AP. [CBSE 2014]
Solution:
Given : From a point P. Out side the circle with centre O, PA and PB are tangerts drawn and ∠APB = 120°
OP is joined To prove : OP = 2 AP
Const: Take mid point M of OP and join AM, join also OA and OB.

Proof : In right ∆OAP,
∠OPA = $$\frac { 1 }{ 2 }$$ ∠APB = $$\frac { 1 }{ 2 }$$ x 120° = 60°
∠AOP = 90° – 60° = 30°
M is mid point of hypotenuse OP of ∆OAP
MO = MA = MP
∠OAM = ∠AOM = 30° and ∠PAM = 90° – 30° = 60°
∆AMP is an equilateral triangle
MA = MP = AP
But M is mid point of OP
OP = 2 MP = 2 AP
Hence proved.

Question 19.
If ∆ABC is isosceles with AB = AC and C (0, r) is the incircle of the ∆ABC touching BC at L. Prove that L bisects BC.
Solution:
Given: In ∆ABC, AB = AC and a circle with centre O and radius r touches the side BC of ∆ABC at L.

To prove : L is mid point of BC.
Proof : AM and AN are the tangents to the circle from A
AM = AN
But AB = AC (given)
AB – AN = AC – AM
BN = CM
Now BL and BN are the tangents from B
BL = BN
Similarly CL and CM are tangents
CL = CM
But BM = CM (proved)
BL = CL
L is mid point of BC.

Question 20.
AB is a diameter and AC is a chord of a circle with centre O such that ∠BAC = 30°. The tangent at C intersects AB at a point D. Prove that BC = BD. [NCERT Exemplar]
Solution:
To prove, BC = BD
Join BC and OC.

Given, ∠BAC = 30°
=> ∠BCD = 30°
[angle between tangent and chord is equal to angle made by chord in the alternate segment]
∠ACD = ∠ACO + ∠OCD
∠ACD = 30° + 90° = 120°
[OC ⊥ CD and OA = OC = radius => ∠OAC = ∠OCA = 30°]
In ∆ACD,
[since, sum of all interior angles of a triangle is 180°]
=> 30° + 120° + ∠ADC = 180°
=> ∠ADC = 180° – (30° + 120°) = 30°
Now, in ∆BCD,
∠BCD = ∠BDC = 30°
=> BC = BD
[since, sides opposite to equal angles are equal]

Question 21.
In the figure, a circle touches all the four sides of a quadrilateral ABCD with AB = 6 cm, BC = 7 cm, and CD = 4 cm. Find AD. [CBSE 2002]
Solution:
A circle touches the sides AB, BC, CD and DA of a quadrilateral ABCD at P, Q, R and S respectively.
AB = 6 cm, BC = 7 cm, CD = 4cm

AP and AS are the tangents to the circle
AP = AS
Similarly,
BP = BQ
CQ = CR
and OR = DS
AB + CD = AD + BC
=> 6 + 4 = 7 + x
=> 10 = 7 + x
=>x = 10 – 7 = 3

Question 22.
Prove that the perpendicular at the point contact to the tangent to a circle passes through the centre of the circle.
Solution:
Given : TS is a tangent to the circle with centre O at P, OP is joined

To prove : OP is perpendicular to TS which passes through the centre of the circle
Construction : Draw a line OR which intersect the circle at Q and meets the tangent TS at R
Proof: OP = OQ
(radii of the same circle) and OQ < OR => OP < OR
Similarly we can prove that OP is less than all lines which can be drawn from O to TS
OP is the shortest
OP is perpendicular to TS
Perpendicular through P, will pass through the centre of the circle
Hence proved.

Question 23.
Two circles touch externally at a point P. From a point T on the tangent at P, tangents TQ and TR are drawn to the circles with points of contact Q and R respectively. Prove that TQ = TR.

Solution:
Given : Two circles with centres O and C touch each other externally at P. PT is its common tangent
From a point T on PT, TR and TQ are the tangents drawn to the circles

To prove : TQ = TR
Proof : From T, TR and TP are two tangents to the circle with centre O
TR = TP ….(i)
Similarly, from T,
TQ and TP are two tangents to the circle with centre C
TQ = TP ….(ii)
From (i) and (ii)
TQ = TR
Hence proved.

Question 24.
A is a point at a distance 13 cm from the centre O of a circle of radius 5 cm. AP and AQ are the tangents to the circle at P and Q. If a tangent BC is drawn at a point R lying on the minor arc PQ to intersect AP at B and AQ at C, find the perimeter of the ∆ABC. [NCERT Exemplar]
Solution:
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 5 cm.

To find : Perimeter of ∆ABC.
Proof : ∠OPA = 90°
[Tangent at any point of a circle is perpendicular to the radius through the point of contact]
OA² = OP² + PA² [by Pythagoras Theorem]
(13)² = 5² + PA²
=> PA² = 144 = 12²
=> PA = 12 cm
Now, perimeter of ∆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 x (12) = 24 cm
[AP = AQ tangent from internal point to a circle are equal]
Hence, the perimeter of ∆ABC = 24 cm.

Question 25.
In the figure, a circle is inscribed in a quadrilateral ABCD in which ∠B = 90°. If AD = 23 cm, AB = 29 cm and DS = 5 cm, find the radius r of the circle.

Solution:
In the figure, O is the centre of the circle inscribed in a quadrilateral ABCD and ∠B = 90°
AD = 23 cm, AB = 29 cm, DS = 5 cm

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
∠OPB = ∠OQB = 90°
PBQO is a square
DS and DR are tangents to the circle
DR = DS = 5 cm
AR = AD – DR = 23 – 5 = 18 cm
AR and AQ are the tangents to the circle
AQ = AR = 18 cm But AB = 29 cm
BQ = AB – AQ = 29 – 18 = 11 cm
Side of square PBQO is 11 cm
OP = 11 cm
Hence radius of the circle = 11 cm

Question 26.
In the figure, there are two concentric circles with centre O of radii 5 cm and 3 cm. From an external point P, tangents PA and PB are drawn to these circles. If AP =12 cm, find the length of BP. [CBSE 2010]

Solution:
Two concentric circles with centre O with radii 5 cm and 3 cm respectively from a
point P, PA and BP are tangents drawn to there circles

AP = 12 cm
To find BP
In right ∆OAP,
OP² = OA² + AP² (Pythagoras Theorem)
= (5)² + (12)² = 25 + 144
= 169 = (13)²
OP = 13 cm
Now in right ∆OBP,
OP² = OB² + BP²
=> (13)² = (3)² + BP²
=> 169 = 9 + BP²
=> BP² = 169 – 9 = 160 = 16 x 10
BP = √(16 x 10) = 4√10 cm

Question 27.
In the figure, AB is a chord of length 16 cm of a circle of radius 10 cm. The tangents at A and B intersect at a point P. Find the length of PA. [CBSE 2010]

Solution:
In the figure, AB is the chord of the circle with centre O and radius 10 cm.
Two tangents from P are drawn to the circle touching it at A and B respectively
AB is joined with intersects OP at L

Question 28.
In the 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. [CBSE 2010]

Solution:
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.

Question 29.
In the figure, BDC is a tangent to the given circle at point D such that BD = 30 to the circle and meet when produced at A making BAC a right angle triangle. Calculate (i) AF (ii) radius of the circle.

Solution:
In the figure, BDC is a tangent to the given circle with centre O and D is a point such that
BD = 30 cm and CD = 7 cm

BE and CF are other two tangents drawn from B and C respectively which meet at A on producing this and ∆BAC is a right angle so formed
To find : (i) AF and (ii) radius of the circle
Join OE and OF
OE = OF radii of the circle
OE ⊥ AB and OF ⊥ AC
OEAF is a square
BD and BE are the tangents from B
BE = BD = 30 cm and similarly
CF = CD = 7 cm
Let r be the radius of the circle
OF = AF = AE = r
AB = 30 + r and AC = 7 + r and BC = 30 + 7 = 37 cm
Now in right ∆ABC

Question 30.
If d1, d2 (d2 > d1) 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 $${ d }_{ 2 }^{ 2 }={ c }^{ 2 }+{ d }_{ 1 }^{ 2 }$$. [NCERT Exemplar]
Solution:
Let AB be a chord of a circle which touches the other circle at C. Then ∆OCB is right triangle.

Question 31.
In the given figure, tangents PQ and PR are drawn from an external point P to a circle with centre O, such that ∠RPQ = 30°. A chord RS is drawn parallel to the tangent PQ. Find ∠RQS. [CBSE 2015, NCERT Exemplar]

Solution:
In the given figure,
PQ and PR are tangents to the circle with centre O drawn from P
∠RPQ = 30°
Chord RS || PQ is drawn
To find ∠RQS
PQ = PR (tangents to the circle)
∠PRQ = ∠PQR But ∠RPQ = 30°

Question 32.
From an external point P, tangents PA = PB are drawn to a circle with centre O. If ∠PAB = 50°, then find ∠AOB. [CBSE 2016]
Solution:

PA = PB [tangents drawn from external point are equal]
∠PBA = ∠PAB = 50° [angles equal to opposite sides]
∠APB = 180° – 50° – 50° = 80° [angle-sum property of a A]
∠AOB + ∠APB = 180° [sum of opposite angles of a cyclic quadrilateral is 180°]
∠AOB + 80° = 180°
∠AOB = 180°- 80° = 100°

Question 33.
In the figure, two tangents AB and AC are drawn to a circle with centre O such that ∠BAC = 120°. Prove that OA = 2AB.

Solution:
Given : In the figure, O is the centre of the circle.
AB and AC are the tangents to the circle from A such that
∠BAC = 120° .
To prove : OA = 2AB
Proof : In ∆OAB and ∆OAC
∠OBA = ∠OCA – 90° (OB and OC are radii)
OA = OA (common)
OB = OC (radii of the circle)
∆OAB ~ ∆OAC
∠OAB = ∠OAC = 60°
Now in right ∆OAB,

Question 34.
The lengths of three concecutive sides of a quadrilateral circumscribing a circle are 4 cm, 5 cm, and 7 cm respectively. Determine the length of the fourth side.
Solution:
In quadrilateral ABCD which is circumerscribing it
BC = 4 cm, CD = 5 cm and DA = 7 cm

We know that if a quad, is circumscribed in a circle, then
AB + CD = AD + BC
=> AB + 5 = 4 + 7
=> AB + 5 = 11
AB = 11 – 5 = 6
AB = 6 cm

Question 35.
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. [NCERT Exemplar]
Solution:
Joint AO, OC and O’D, O’B
Now, in ∆EO’D and ∆EO’B
O’E = O’E [common side]
ED = EB

Question 36.
In the figure, common tangents PQ and RS to two circles intersect at A. Prove that PQ = RS.

Solution:
Given : Two common tangents PQ and RS intersect each other at A.
To prove : PQ = RS
Proof: From A, AQ and AR are two tangents are drawn to the circle with centre O.

AP = AR ….(i)
Similarly AQ and AS are the tangents to the circle with centre C
AQ = AS ….(ii)
AP + AQ = AR + AS
=> PQ = RS
Hence proved.

Question 37.
Two concentric circles are of diameters 30 cm and 18 cm. Find the length of the chord of the larger circle which touches the smaller circle. [CBSE 2014]
Solution:
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

Question 38.
AB and CD are common tangents to two circles of equal radii. Prove that AB = CD. [NCERT Exemplar]
Solution:
Given : AB and CD are tangents to two circles of equal radii.
To prove :

Construction : Join OA, OC, O’B and O’D
Proof: Now, ∠OAB = 90°
[tangent at any point of a circle is perpendicular to radius through the point of contact]
Thus, AC is a straight line.
Also, ∠OAB + ∠OCD = 180°
AB || CD
Similarly, BD is a straight line and ∠O’BA = ∠O’DC = 90°
Also, AC = BD
∠A = ∠B = ∠C = ∠D = 90°
andAC = BD
ABCD is a rectangle
Hence, AB = CD
[opposite sides of rectangle are equal]

Question 39.
A triangle PQR is drawn to circumscribe a circle of radius 8 cm such that the segments QT and TR, into which QR is divided by the point of contact T, are of lengths 14 cm and 16 cm respectively. If area of ∆PQR is 336 cm², find the sides PQ and PR. [CBSE 2014]
Solution:
∆PQR is circumscribed by a circle with centre O and radius 8 cm
T is point of contact which divides the line segment OT into two parts such that
QT = 14 cm and TR = 16 cm

Area of ∆PQR = 336 cm²
Let PS = x cm
QT and QS are tangents to the circle from Q
QS = QT = 14 cm
Similarly RU and RT are tangents to the circle
RT = RU = 16 cm
Similarly PS and PU are tangents from P
PS = PU = x
Now PQ = x + 14 and PR = x + 16 and QR = 14 + 16 = 30 cm
Now area of ∆PQR = Area of ∆POQ + area of ∆QOR + area of ∆POR

Question 40.
In the figure, the tangent at a point C of a circle and a diameter AB when extended itersect at P. If ∠PCA = 110°, find ∠CBA. [NCERT Exemplar]

Solution:
Here, AB is a diameter of the circle from point C and a tangent is drawn which meets at a point P.
Join OC. Here, OC is radius.

Since, tangent at any point of a circle is perpendicular to the radius through point of contact circle.
OC ⊥ PC
Now, ∠PCA = 110° [given]
=> ∠PCO + ∠OCA = 110°
=> 90° + ∠OCA = 110°
=> ∠OCA = 20°
OC = OA = Radius of circle
∠OCA = ∠OAC = 20°
[since, two sides are equal, then their opposite angles are equal]
Since, PC is a tangent, so
∠BCP = ∠CAB = 20°
[angles in a alternate segment are equal]
In ∆PBC, ∠P + ∠C + ∠A= 180°
∠P = 180° – (∠C + ∠A)
∠P = 180° – (110° + 20°)
∠P = 180° – 130° = 50°
In ∆PBC,
∠BPC + ∠PCB + ∠PBC = 180°
[sum of all interior angles of any triangle is 180°]
=> 50° + 20° + ∠PBC = 180°
=> ∠PBC = 180° – 70°
∠PBC = 110°
Since, ∆PB is a straight line.
∠PBC + ∠CBA = 180°
∠CBA = 180° – 110° = 70°

Question 41.
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 ∠BAT = ∠ACB. [NCERT Exemplar]

Solution:
Since, AC is a diameter line, so angle in semicircle makes an angle 90°.
∠ABC = 90° [by property]
In ∆ABC,
∠CAB + ∠ABC + ∠ACB = 180°
[ sum of all interior angles of any triangle is 180°]
=> ∠CAB + ∠ACB = 180° – 90° = 90° ……….(i)
Since, diameter of a circle is perpendicular to the tangent.
i.e. CA ⊥ AT
∠CAT = 90°
=> ∠CAB + ∠BAT = 90° …….(ii)
From Eqs. (i) and (ii),
∠CAB + ∠ACB = ∠CAB + ∠BAT
=> ∠ACB = ∠BAT
Hence proved.

Question 42.
In the given figure, a ∆ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC are of lengths 8 cm and 6 cm respectively. Find the lengths of sides AB and AC, when area of ∆ABC is 84 cm². [CBSE 2015]

Solution:
In the given figure,
In ∆ABC is circle is inscribed touching it at D, E and F respectively.
Radius of the circle (r) = 4cm
OD ⊥ BC, then
OD = 4 cm, BD = 8 cm, DC = 6 cm
Join OE and OF

Question 43.
In the given figure, AB is a diameter of a circle with centre O and AT is a tangent. If ∠AOQ = 58°, find ∠ATQ. [CBSE 2015]

Solution:
In the given figure,
AB is the diameter, AT is the tangent
and ∠AOQ = 58°
To find ∠ATQ
Arc AQ subtends ∠AOQ at the centre and ∠ABQ at the remaining part of the circle
∠ABQ = $$\frac { 1 }{ 2 }$$ ∠AOQ = $$\frac { 1 }{ 2 }$$ x 58° = 29°
Now in ∆ABT,
∠BAT = 90° ( OA ⊥ AT)
∠ABT + ∠ATB = 90°
=> ∠ABT + ∠ATQ = 90°
=> 29° + ∠ATQ = 90°
=> ∠ATQ = 90°- 29° = 61°

Question 44.
In the figure, OQ : PQ = 3:4 and perimeter of ∆POQ = 60 cm. Determine PQ, QR and OP.

Solution:
In the figure, OQ : PQ = 3:4

Perimeter of ∆POQ = 60 cm
To find PQ, QR and OP
OQ : PQ = 3 : 4
Let OQ = 3x and PQ = 4x
Now in right ∆OPQ,
OP² = OQ² + PQ² = (3x)² + (4x)² = 9x² + 16x² = 25x² = (5x)²
OP = 5x
But OQ + QP + OP = 60 cm
3x + 4x + 5x = 60
=> 12x = 60
x = 5
PQ = 4x = 4 x 5 = 20 cm
QR = 2 OQ = 2 x 3x = 6 x 5 = 30 cm
OP = 5x = 5 x 5 = 25 cm

Question 45.
Equal circles with centre O and O’ touch each other at X. OO’ produced to meet a circle with centre O’, at A. AC is a tangent to the circle whose centre is O. O’ D is perpendicular to AC. Find the value of $$\frac { DO’ }{ CO }$$

Solution:
Two equal circles with centre O and O’ touch each other externally at X
OO’ produced to meet at A

AC is the tangent of circle with centre O,
O’D ⊥ AC is drawn OC is joined
AC is tangent and OC is the radius
OC ⊥ AC
O’D ⊥ AC
OC || O’D
Now O’A = $$\frac { 1 }{ 2 }$$ A x or $$\frac { 1 }{ 2 }$$ AO
∠A = ∠A (common)
∠AO’D = ∠AOC (corresponding angles)

Question 46.
In the figure, BC is a tangent to the circle with centre O. OE bisects AP. Prove that ∆AEO ~ ∆ABC.

Solution:
Given : In the figure, BC is a tangent to the circle with centre O at B.
AB is diameter AC is joined which intersects the circle at P
OE bisects AP

To prove : ∆AEO ~ ∆ABC
Proof: In ∆OAE and ∆OPE
OE = OE (common)
OA = OP ‘ (radii of the same circle)
EA = EP (given)
∆OAE = ∆OPE (SSS axiom)
∠OEA = ∠OEP
But ∠OEA + ∠OEP = 180°
∠OEA = 90°
Now in ∆AEO and ∆ABC
∠OEA = ∠ABC (each 90°)
∠A = ∠A (common)
∆AEO ~ ∆ABC (AA axiom)
Hence proved.

Question 47.
In the figure, PO ⊥ QO. The tangents to the circle at P and Q intersect at a point T. Prove that PQ and OT are right bisectors of each other.

Solution:
Given : In the figure, O is the centre of the circle
PO ⊥ QO
They tangents at P and Q intersect each other at T

To prove : PQ and OT are right bisector of each other
Proof : PT and QT are tangents to the circle
PT = QT
OP and OQ are radii of the circle and ∠POQ = 90° ( PO ⊥ QO)
OQTP is a square Where PQ and OT are diagonals
Diagonals of a square bisect each other at right angles
PQ and OT bisect each other at right angles
Hence PQ and QT are right bisectors of each other.

Question 48.
In the figure, O is the centre of the circle and BCD is tangent to it at C. Prove that ∠BAC + ∠ACD = 90°.

Solution:
Given : In the figure, O is the centre of the circle BCD is a tangent, CP is a chord

prove : ∠BAC + ∠ACD = 90°
Proof: ∠ACD = ∠CPA (Angles in the alternate segment)
But in ∆ACP,
∠ACP = 90° (Angle in a semicircle)
∠PAC + ∠CPA = 90°
=> ∠BAC + ∠ACD = 90°
(∠ACD = ∠CPA proved)
Hence proved.

Question 49.
Prove that the centre of a circle touching two intersecting lines lies on the angle bisector of the lines. [NCERT Exemplar]
Solution:
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 lines lies on the angle bisector of the lines.

Construction : Join OR, and OQ.
In ∆POR and ∆POQ
∠PRO = ∠PQO = 90°
[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.
∆PRO = ∆PQO [RHS]
Hence, ∠RPO = ∠QPO [by CPCT]
Thus, O lies on angle bisector of PR and PQ.
Hence proved.

Question 50.
In the figure, there are two concentric circles with centre O. PRT and PQS are tangents to the inner circle from a point P lying on the outer circle. If PR = 5 cm, find the lengths of PS. [CBSE 2017]

Solution:
Construction : Join OS and OP.
Consider ∆POS
We have,
PO = OS

∆POS is an isosceles triangle.
We know that in an isosceles triangle, if a line drawn perpendicular to the base of the triangle from the common vertex of the equal sides, then that line will bisect the base (unequal side).
And PQ = PR = 5 cm
[PRT and PQS are tangents to the inner circle to the inner circle from a point P lying on the outer circle]
We have, PQ = QS
It is given that, PQ = 5 cm
QS = 5 cm
From the figure, we have
PS = PQ + QS
=> PS = 5 + 5
=> PS = 10 cm

Question 51.
In the figure, PQ is a tangent from an external point P to a circle with centre O and OP cuts the circle at T and QOP is a diameter. If ∠POR = 130° and S is a point on the circle, find ∠1 + ∠2. [CBSE 2017]

Solution:
Construction : Join RT.

Given, ∠POR = 130°
∠POQ = 180°- (∠POR) = 180° – 130° = 50°
Since, PQ is a tangent
∠PQO = 90°
Now, In ∆POQ,
∠POQ + ∠PQO + ∠QPO = 180°
=> 50° + 90° + ∠1 = 180°
=> ∠1 = 180° – 140°
=> ∠1 = 40°
Now, In ∆RST
∠RST = $$\frac { 1 }{ 2 }$$ ∠ROT
[Angle which is subtended by on arc at the centre of a circle is double the size of the angle subtended at any point on the circumference]
=> ∠2 = $$\frac { 1 }{ 2 }$$ x 130° = 65°
Now ∠1 + ∠2 = 40° + 65° = 105°

Question 52.
In the figure, PA and PB are tangents to the circle from an external point P. CD is another tangent touching the circle at Q. If PA = 12 cm, QC = QD = 3 cm, then find PC = PD. [CBSE 2017]

Solution:
PA = PB = 12 cm …(i)
QC = AC = 3cm …(ii)
QD = BD = 3 cm …(iii)
[Tangents drawn from an external point are equal]
To find : PC + PD
= (PA – AC) + (PB – BD)
= (12 – 3) + (12 – 3) [From (i), (ii), and (iii)]
= 9 + 9 = 18 cm

Hope given RD Sharma Class 10 Solutions Chapter 8 Circles Ex 8.2 are helpful to complete your math homework.

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## RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.3

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.3

Other Exercises

Question 1.
In a ∆ABC, AD is the bisector of ∠A, meeting side BC at D.
(i) If BD = 2.5 cm, AB = 5 cm and AC = 4.2 cm, find DC. (C.B.S.E. 1996)
(ii) If BD = 2 cm, AB = 5 cm and DC = 3 cm, find AC. (C.B.S.E. 1992)
(iii) If AB = 3.5 cm, AC = 4.2 cm and DC = 2.8 cm, find BD. (C.B.S.E. 1992)
(iv) If AB = 10 cm, AC = 14 cm and BC = 6 cm, find BD and DC.
(v) If AC = 4.2 cm, DC = 6 cm and BC = 10 cm, find AB. (C.B.S.E. 1997C)
(vi) If AB = 5.6 cm, AC = 6 cm and DC = 3 cm, find BC. (C.B.S.E. 2001C)
(vii) If AD = 5.6 cm, BC = 6 cm and BD = 3.2 cm, find AC. (C.B.S.E. 2001C)
(viii) If AB = 10 cm, AC = 6 cm and BC = 12 cm, find BD and DC. (C.B.S.E. 2001)
Solution:
In ∆ABC, AD is the angle bisector of ∠A which meet BC at D
(i) BD = 2.5 cm, AB = 5 cm and AC = 4.2 cm

=> 6x = 10 (12 – x) = 120 – 10x
=> 6x + 10x = 120
=> 16x = 120
x = 7.5
BD = 7.5 cm and DC = 12 – 7.5 = 4.5 cm

Question 2.
In the figure, AE is the bisector of the exterior ∠CAD meeting BC produced in E. If AB = 10 cm, AC = 6 cm and BC = 12 cm, find CE.

Solution:
In ∆ABC, AE is the bisector of exterior ∠A which meets BC produced at E.
AB = 10 cm, AC = 6 cm, BC = 12 cm Let CE = x, then BE = BC + CE = (12 + x)

Question 3.
In the figure, ∆ABC is a triangle such that $$\frac { AB }{ AC }$$ = $$\frac { BD }{ DC }$$ , ∠B = 70°, ∠C = 50°. Find ∠BAD.

Solution:

Question 4.
In the figure, check whether AD is the bisector of ∠A of ∆ABC in each of the following :

(i) AB = 5 cm, AC = 10 cm, BD = 1.5 cm and CD = 3.5 cm
(ii) AB = 4 cm, AC = 6 cm, BD = 1.6 cm and CD = 2.4 cm
(iii) AB = 8 cm, AC = 24 cm, BD = 6 cm and BC = 24 cm
(iv) AB = 6 cm, AC = 8 cm, BD = 1.5 cm and CD = 2 cm
(v) AB = 5 cm, AC = 12 cm, BD = 2.5 cm and BC = 9 cm
Solution:
(i) AB = 5 cm, AC = 10 cm, BD = 1.5 cm, CD = 3.5 cm

Question 5.
In figure, AD bisects ∠A, AB = 12 cm AC = 20 cm, and BD = 5 cm. Determine CD.

Solution:

Question 6.
In the figure, In ∆ABC, if ∠1 = ∠2, prove that $$\frac { AB }{ AC }$$ = $$\frac { BD }{ DC }$$.

Solution:
Given : In ∆ABC,
AD is a line drawn from A meeting BC in D Such that ∠1 = ∠2

Question 7.
D, E and F are the points on sides BC, CA and AB respectively of ∆ABC such that AD bisects ∠A, BE bisects ∠B and CF bisects ∠C. If AB = 5 cm, BC = 8 cm and CA = 4 cm, determine AF, CE and BD.
Solution:
In ∆ABC, AD, BE and CF are the bisector of ∠A, ∠B and ∠C respectively
AB = 5 cm, BC = 8 cm and CA = 4 cm

Hope given RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.3 are helpful to complete your math homework.

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## RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.2

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.2

Other Exercises

Question 1.
In a ∆ABC, D and E are points on the sides AB and AC respectively such that DE || BC.
(i) If AD = 6 cm, DB = 9 cm and AE = 8 cm, find AC. (C.B.S.E. 1995)
(ii) If $$\frac { AD }{ DB }$$ = $$\frac { 3 }{ 4 }$$ and AC = 15 cm, find AE. (C.B.S.E. 1994)
(iii) If $$\frac { AD }{ DB }$$ = $$\frac { 2 }{ 3 }$$ and AC = 18 cm, find AE. (C.B.S.E. 1994C)
(iv) If AD = 4, AE = 8, DB = x – 4, and EC = 3x – 19, find x. (C.B.S.E. 1992C)
(v) If AD = 8 cm, AB = 12 cm and AE = 12 cm, find CE. (C.B.S.E. 1992C)
(vi) If AD = 4 cm, DB = 4.5 cm and AE = 8 cm, find AC. (C.B.S.E. 1992C)
(vii) If AD = 1 cm, AB = 6 cm and AC = 9 cm, findAE. (C.B.S.E. 1992C)
(viii) If $$\frac { AD }{ DB }$$ = $$\frac { 4 }{ 5 }$$ and EC = 2.5 cm, find AE.
(ix) If AD = x, DB = x – 2, AE = x + 2 and EC = x – 1, find the value of x. (C.B.S.E. 1993C)
(x) If AD = 8x – 7, DB = 5x – 3, AE = 4x – 3 and EC = (3x – 1), find the value of x.
(xi) If AD = 4x – 3, AE = 8x – 7, BD = 3x – 1 and CE = 5x – 3, find the value of x. (C.B.S.E. 2002)
(xii) If AD = 2.5 cm, BD = 3.0 cm and AE = 3.75 cm, find the length of AC. (C.B.S.E. 2006C)
Solution:

Question 2.
In a ∆ABC, D and E are points on the sides AB and AC respectively. For each of the following cases show that DE || BC:
(i) AB = 12 cm, AD = 8 cm, AE = 12 cm and AC = 18 cm. (C.B.S.E. 1991)
(ii) AB = 5.6 cm, AD = 1.4 cm, AC = 7.2 cm and AE = 1.8 cm. (C.B.S.E. 1990)
(ii) AB = 10.8 cm, BD = 4.5 cm, AC = 4.8 cm and AE = 2.8 cm.
(iv) AD = 5.7 cm, BD = 9.5 cm, AE = 3.3 cm and EC = 5.5 cm.
Solution:
In ∆ABC, D and E are points on the sides AB and AC respectively
(i) AB = 12 cm, AD = 8 cm, AE = 12 cm and AC = 18 cm
DB = AB – AD = 12 – 8 = 4 cm and EC = AC – AE = 18 – 12 = 6 cm

Question 3.
In a ∆ABC, P and Q are points on sides AB and AC respectively, such that PQ || BC. If AP = 2.4 cm, AQ = 2 cm, QC = 3 cm and BC = 6 cm, find AB and PQ.
Solution:
In ∆ABC,
P and Q are points on AB and AC respectively such that PQ || BC
AP = 2.4 cm, AQ = 2 cm, QC = 3 cm and BC = 6 cm

Question 4.
In a ∆ABC, D and E are points on AB and AC respectively such that DE || BC. If AD = 2.4 cm, AE = 3.2 cm, DE = 2 cm and BC = 5 cm, find BD and CE. (C.B.S.E. 2001C)
Solution:
In the ∆ABC, DE || BC
AD = 2.4 cm, AE = 3.2 cm, DE = 2 cm and BC = 5 cm

Question 5.
In the figure, state if PQ || EF.

Solution:
In ∆DEF
PQ intersects DE and DF at P and Q respectively
Such that DP = 3.9 cm, PE = 3 cm DQ = 3.6 cm, QF = 2.4 cm

Question 6.
M and N are points on the sides PQ and PR respectively of a ∆PQR. For each of the following cases, state whether MN || QR.
(i) PM = 4 cm, QM = 4.5 cm, PN = 4 cm, NR = 4.5 cm
(ii) PQ = 1.28 cm, PR = 2.56 cm, PM = 0.16 cm, PN = 0.32 cm
Solution:
(i) In the ∆PQR
M and N are points on PQ and PR respectively
PM = 4 cm, QM = 4.5 cm, PN = 4 cm, RN = 4.5 cm

Question 7.
In three line segments OA, OB, and OC, points L, M, N respectively are so chosen that LM || AB and MN || BC but neither of L, M, N nor of A, B, C are collinear. Show that LN || AC.
Solution:
Given : On OA, OB and OC, points are L, M, and N respectively
Such that LM || AB, MN || BC

Question 8.
If D and E are points on sides AB and AC respectively of a ∆ABC such that DE || BC and BD = CE. Prove that ∆ABC is isosceles. (C.B.S.E. 2007)
Solution:
Given : In ∆ABC, D and E are points on the sides AB and AC such that BD = CD

Hope given RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.2 are helpful to complete your math homework.

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## RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.1

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.1

Other Exercises

Question 1.
Fill in the blanks using the correct word given in brackets :
(i) All circles are …………… (congruent, similar).
(ii) All squares are ………… (similar, congruent).
(iii) All …………….. triangles are similar (isosceles, equilaterals) :
(iv) Two triangles are similar, if their corres-ponding angles are ……… (proportional, equal)
(v) Two triangles are similar, if their corres-ponding sides are ……………. (proportional, equal)
(vi) Two polygons of the same number of sides are similar, if (a) their corresponding angles are …………. and (b) their corres-ponding sides are …………….. (equal, proportional).
Solution:
(i) All circles are similar.
(ii) All squares are similar.
(iii) All equilaterals triangles are similar.
(iv) Two triangles are similar, if their corresponding angles are equal.
(v) Two triangles are similar, if their corresponding sides are proportional.
(vi) Two polygons of the same number of sides are similar, if (a) their corresponding angles are equal and (b) their corresponding sides are proportional.

Question 2.
Write the truth value (T/F) of each of the following statements :
(i) Any two similar figures are congruent.
(ii) Any two congruent figures are similar.
(iii) Two polygons are similar, if their corresponding sides are proportional.
(iv) Two polygons are similar if their corresponding angles are proportional.
(v) Two triangles are similar if their corresponding sides are proportional.
(vi) Two triangles are similar if their corresponding angles are proportional.
Solution:
(i) False : In some cases, the similar polygons can be congruent.
(ii) True.
(iii) False : Its corresponding angles must be equal also.
(iv) False : Angle are equal not proportional.
(v) True.
(vi) False : Sides should be proportional and corresponding angles should be equal.

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## RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS

Other Exercises

Question 1.
If the circumference and the area of a circle are numerically equal, then diameter of the circle is
(a) π/2
(b) 2π
(c) 2
(d) 4
Solution:
Let r be the radius of the circle, then Circumference = 2πr
and area = πr²
But 2πr= πr²
∴ 2r = r²
⇒ r = 2
Diameter = 2r = 2 x 2 = 4 (d)

Question 2.
If the difference between the circum-ference and radius of a circle is 37 cm., then using π =  $$\frac { 22 }{ 7 }$$  the circumference (in cm) of the circle is
(a) 154
(b) 44
(c) 14
(d) 7 [CBSE 2013]
Solution:
Difference between circumference and radius of a circle = 37 cm
∴  2πr-r = 37

Question 3.
A wire can be bent in the form of a circle of radius 56 cm. If it is bent in the form of a square, then its area will be
(a) 3520 cm²
(b) 6400 cm²
(c) 7744 cm²
(d) 8800 cm²
Solution:
Radius of a circular wire (r) = 56 cm
Circumference = 2πr = 2 x $$\frac { 22 }{ 7 }$$ x 56 cm = 352 cm
Now perimeter of square = 352 cm

Question 4.
If a wire is bent into the shape of a square, then the area of the square is 81 cm². When wire is bent into a semicircular shape, then the area of the semi-circle will be
(a) 22 cm²
(b) 44 cm²
(c) 77 cm²
(d) 154 cm²
Solution:
Area of a square wire = 81 cm²
∴ Side of square = $$\sqrt { Area }$$  = $$\sqrt { 81 }$$  cm = 9 cm ard per in eret of square =4a = 4 x 9 = 36cm
Perimeter of semicircular wire whose bent = 36 cm
Let r be the radius, then

Question 5.
A circular park has a path of uniform width around it. The difference between the outer and inner circumferences of the circular path is 132 m. Its width is
(a) 20 m
(b) 21 m
(c) 22 m
(d) 24 m
Solution:
Let R and r be the radii of the outer and inner circles of the park, then 2πR – 2πr = 132

Question 6.
The radius of a wheel is 0.25 m. The number of revolutions it will make to travel a distance of 11 km will be
(a) 2800
(b) 4000
(c) 5500
(d) 7000
Solution:
Radius of the wheel (r) = 0.25 m = 25 cm
Circumference of the wheel

Question 7.
The ratio of the outer and inner perimeters of a circular path is 23 : 22. If the path is 5 metres wide, the diameter of the inner circle is
(a) 55 m
(b) 110 m
(c) 220 m
(d) 230 m
Solution:
Ratio in the outer and inner perimeter of a circular path = 23 : 22
Width of path = 5 m
Let R and r be the radii of outer and inner path then R- r = 5 m ….(i)

Question 8.
The circumference of a circle is 100 cm. The side of a square inscribed in the circle is

Solution:
Circumference of a circle (c) = 100 cm
Diagonal of square which is inscribed in the circle

Question 9.
The area of the incircle of an equilateral triangle of side 42 cm is :
(a) 22 73 cm²
(b) 231 cm²
(c) 462 cm²
(d) 924 cm²
Solution:
Side of an equilateral triangle (a) = 42 cm
Radius of inscribed circle = $$\frac { 1 }{ 3 }$$ x altitude

Question 10.
The area of incircle of an equilateral triangle is 154 cm2. The perimeter of the triangle is
(a) 71.5 cm
(b) 71.7 cm
(c) 72.3 cm
(d) 72.7 cm
Solution:
Area of incircle of an equilateral triangle = 154 cm²

Question 11.
The area of the largest triangle that can be inscribed in a semi-circle of radius r. is
(a) r²
(b) 2 r²
(c) r³
(d) 2r³
Solution:
The largest triangle inscribed in a semi-circle of radius r, can be ΔABC as shown in the figure, whose base = AB = 2r

Question 12.
The perimeter of a triangle is 30 cm and the circumference of its incircle is 88 cm. The area of the triangle is
(a) 70 cm²
(b) 140 cm²
(c) 210 cm²
(d) 420 cm²
Solution:
The perimeter of a triangle = 30 cm
and circumference of its incircle = 88 cm

Question 13.
The area of a circle is 220 cm², the area of a square inscribed in it is
(a) 49 cm²
(b) 70 cm²
(c) 140 cm²
(d) 150 cm²
Solution:
Area of a circle = 220 cm²

Question 14.
If the circumference of a circle increases from 4π to 8π, then its area is
(a) halved
(b) doubled
(c) tripled
Solution:
In first case circumference of a circle = 4π

Question 15.
If the radius of a circle is diminished by 10%, then its area is diminished by
(a) 10%
(b) 19%
(c) 20%
(d) 36%
Solution:
Let in first case radius of a circle = r
Then area = πr²

Question 16.
If the area of a square is same as the area of a circle, then the ratio of their perimeter, in terms of 7t, is

Solution:
Let side of square = a
Perimeter = 4 a
Then area = a²
∴ Area of circle = a²

Question 17.
The area of the largest triangle that can be inscribed in a semi-circle of radius r is
(a) 2r
(b) r²
(c) r
(d) $$\sqrt { r }$$
Solution:

Question 18.
The ratio of the areas of a circle and an equilateral triangle whose diameter and a side are respectively equal, is
(a) π : $$\sqrt { 2 }$$
(b) π : $$\sqrt { 3 }$$
(c) $$\sqrt { 3 }$$ :π
(d) $$\sqrt { 2 }$$ : π
Solution:
Let side of equilateral triangle = a
Then area = $$\frac { \sqrt { 3 } }{ 4 }$$ a²
Diameter of circle = a

Question 19.
If the sum of the areas of two circles with radii r1 and r2 is equal to the area of a circle of radius r, then r + r
(a) >r²
(b) = r²
(c) < r²
(d) None of these
Solution:
Sum of area of two circles with radii r1  and r2

Question 20.
If the perimeter of a semi-circular protractor is 36 cm, then its diameter is
(a) 10 cm
(b) 12 cm
(c) 14 cm
(d) 16 cm
Solution:
Perimeter of a semicircle = 36 cm
Let d be its diameter, then

Question 21.
The perimeter of the sector OAB shown in the fiugre, is

Solution:
Radius of sector of 60° = 7 cm
∴ Perimeter = arc AB + 2 r

Question 22.
If the perimeter of a sector of a circle of radius 6.5 cm is 29 cm, then its area is
(a) 58 cm²
(b) 52 cm²
(c) 25 cm²
(d) 56 cm²
Solution:
Radius of a sector = 6.5 cm
and perimeter = 29 cm

Question 23.
If the area of a sector of a circle bounded by an arc of length 5K cm is equal to 20K cm², then its radius is
(a) 12 cm
(b) 16 cm
(c) 8 cm
(d) 10 cm
Solution:
Let r be the radius, then
Length of the arc of sector of θ angle = 5π

Question 24.
The area of the circle that can be inscribed in a square of side 10 cm is
(a) 40π cm²
(b) 30π cm²
(c) 100π cm²
(d) 25π cm²
Solution:
Side of square = 10 cm
∴ Diameter of the inscribed circle = 10 cm

Question 25.
If the difference between the circumference
(a) 154 cm²
(b) 160 cm²
(c) 200 cm²
(d) 150 cm²
Solution:
Let r be the radius of a circle then circum-ference = 2πr
∴  2πr-r = 37

Question 26.
The area of a circular path of uniform width h surrounding a circular region of radius r is
(a) π (2r + h) r
(b) π (2r + h) h
(c) π (h + r)r
(d) π (h + r) A
Solution:
Let r be the radius of inner circle h is the width of circular path

Question 27.
If AB is a chord of length 5$$\sqrt { 3 }$$  cm of a circle with centre O and radius 5 cm, then area of sector OAB is

Solution:
Radius of the circle (r) = 5 cm
AB is a chord = 5$$\sqrt { 3 }$$
Draw OM ⊥ AB which bisects the chord AB at M

Question 28.
The area of a circle whose area and circumference are numerically equal, is
(a) 2π sq. units
(b) 4π sq. units
(c) 6π sq. units
(d) 8π sq. units
Solution:
Let radius of the circle = r
∴ Area = πr²
and circumference = 2πr

Question 29.
If diameter of a circle is increased by 40%, then its area increases by
(a) 96%
(b) 40%
(c) 80%
(d) 48%
Solution:
Let the diameter of a circle in first case = 2r
Area = πr²

Question 30.
In the figure, the shaded area is
(a) 50 (π – 2) cm²
(b) 25 (π – 2) cm²
(c) 25 (π + 2) cm²
(d) 5 (π – 2) cm²

Solution:
In the figure, ∠AOB = 90°
and radius of the circle = 10 cm

Question 31.
In the figure, the area of the segment PAQ is

Solution:
a is the radius of the circle arc PAQ subtends angle 90° at the centre
∴ Area of segment PAQ

Question 32.
In the figure, the area of segment ACB is

Solution:
r is the radius of the circle and arc ACB subtends angle of 120° at the centre
Area of segment ACB = r

Question 33.
If the area of a sector of a circle bounded by an arc of length 5π cm is equal to 20rc cm², then the radius of the circle is
(a) 12 cm
(b) 16 cm
(c) 8 cm
(d) 10 cm
Solution:
Length of arc = 5π cm
area of sector = 20π cm²
Let the angle at the centre be θ

Question 34.
In the figure, the ratio of the areas of two sectors S1 and S2 is
(a) 5 : 2
(b) 3 : 5
(c) 5 : 3
(d) 4 : 5

Solution:
Let r be the radius of the circle

Question 35.
If the area of a sector of a circle is $$\frac { 5 }{ 18 }$$ of the area of the circle, then the sector angle is equal to
(a) 60°
(b) 90°
(c) 100°
(d) 120°
Solution:
Area of sector of a circle = $$\frac { 5 }{ 18 }$$ x area of circle
Let θ be its angle at the centre and r be radius

Question 36.
If the area of a sector of a circle is $$\frac { 7 }{ 20 }$$ of the area of the circle, then the sector angel is equal to
(a) 110°
(b) 130°
(c) 100°
(d) 126°
Solution:
Area of sector of a circle = $$\frac { 7 }{ 20 }$$ of the area of the circle
Let r be the radius and θ be its angle at the centre

Question 37.
In the figure, if ABC is an equilateral triangle, then shaded area is equal to?

Solution:
ΔABC is an equilateral triangle inscribed in a circle with centre O and radius r
BO and CO are joined

Question 38.
In the figure, the ara of the shaded region is
(a) 3π cm²
(b) 6π cm²
(c) 9π cm²
(d) 7π cm²

Solution:
In the figure, ∠B = ∠C = 90°, ∠D = 60?

∴ ∠A= 360° – (90° + 90° + 60°) = 360° – 240° = 120°
Radius of the sector = 3 cm

Question 39.
If the perimeter of a circle is equal to that of a square, then the ratio of their areas is
(a) 13 : 22
(b) 14 : 11
(c) 22 : 13
(d) 11 :14
Solution:
Let side of square = a units
∴ Area = a² sq. units
and perimeter = 4a units
Now perimeter of circle = 4a units

Question 40.
The radius of a circle is 20 cm. It is divided into four parts of equal area by drawing three concentric circles inside it. Then, the radius of the largest of three concentric circles drawn is

Solution:
Radius of circle (R) = 20 cm

Question 41.
The area of a sector whose perimeter is four times its radius r units, is

Solution:
Perimeter = 4r
and length of arc = 4r – 2r = 2r

Question 42.
If a chord of a circle of radius 28 cm makes an angle of 90° at the centre, then the area of the major segment is
(a) 392 cm²
(b) 1456 cm²
(c) 1848 cm²
(d) 2240 cm²
Solution:
A chord AB makes an angle of 90° at the centre
Radius of the circle = 28 cm

Question 43.
If the area of a circle inscribed in an equilateral triangle is 48π square units, then perimeter of the trianlge is
(a) 17$$\sqrt { 3 }$$ units
(b) 36 units
(c) 72 units
(d) 48$$\sqrt { 3 }$$  units
Solution:
Area of a circle inscribed in an equilateral triangle = 48π sq. units

Question 44.
The hour hand of a clock is 6 cm long. The area swept by it between 11.20 am and 11.55 am is
(a) 2.75 cm²
(b) 5.5 cm²
(c) 11 cm²
(d) 10 cm²
Solution:
Length of hour hand of a clock (r) = 6 cm
Time 11.20 am to 11.55 am = 35 minutes

Question 45.
ABCD is a square of side 4 cm. If ? is a point in the interior of the square such that ΔCED is equilateral, then area of ΔACEis
(a) 2 ($$\sqrt { 3 }$$ – 1) cm²
(b) 4 ($$\sqrt { 3 }$$ -1) cm²
(c) 6($$\sqrt { 3 }$$-1)cm²
(d) 8($$\sqrt { 3 }$$-1)cm²
Solution:
Side of square ABCD = 4 cm
and side of equilateral ΔCED = 4 cm

Question 46.
If the area of a circle is equal to the sum of the areas of two circles of diameters 10 cm and 24 cm, then diameter of the larger circle (in cm) is
(a) 34
(b) 26

(c) 17
(d) 14

Solution:

Question 47.
If π is taken as 22/7, the-distance (in metres) covered by a wheel of diameter 35 cm, in one revolution, is (a) 2.2
(b) 1.1
(b) 9.625
(d) 96.25   [CBSE 2013]
Solution:
Diameter of a wheel = 35 cm = $$\frac { 35 }{ 100 }$$ m
Circumference of the wheel = πd

Question 48.
ABCD is a rectangle whose three vertices are B (4, 0), C (4, 3) and D (0, 3). The length of one of its diagonals is
(a) 5
(b) 4

(c) 3
(d) 25 [CBSE 2014]
Solution:
Three vertices of a rectangle ABCD are B (4,0), C (4, 3) and D (0, 3) length of one of its diagonals

Question 49.
Area of the largest triangle that can be inscribed in a semi-circle of radius r units is

Solution:
Take a point C on the circumference of the semi-circle and join it by the end points of diameter A and B.

Question 50.
If the sum of the areas of two circles with radii r1 and r2 is equal to the area of a circle of radius r, then

Solution:
According to the given condition,
Area of circle = Area of first circle + Area of second circle.

Question 51.
If the sum of the circumference of two circles with radii r, and r2 is equal to the circumference of a circle of radius r, then

Solution:
According to the given condition, Circumference of circle = Circumference of first circle + Circumference of second circle

Question 52.
If the circumference of a circle and the perimeter of a square are equal, then
(a) Area of the circle = Area of the square
(b) Area of the circle < Area of the square
(c) Area of the circle > Area of the square

(d) Nothing definite can be said
Solution:
According to the given condition, Circumference of a circle = Perimeter of square 2 πr = 4 a
[where, r and a are radius of circle and side of square respectively]

Question 53.
If the perimeter of a circle is equal to that of a square, then the ratio of their areas is
(a) 22 : 7
(b) 14 : 11
(c) 7 : 22
(d) 11 : 14
Solution:
Let radius of circle be r and side of a square be a
According to the given condition,
Perimeter of a circle = Perimeter of a square

Hope given RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS are helpful to complete your math homework.

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## RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.3

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.3

Other Exercises

Question 1.
For the following arithmetic progressions write the first term a and the common difference d :
(i) -5, -1, 3, 7, …………
(ii) $$\frac { 1 }{ 5 }$$ , $$\frac { 3 }{ 5 }$$ , $$\frac { 5 }{ 5 }$$ , $$\frac { 7 }{ 5 }$$ , ……
(iii) 0.3, 0.55, 0.80, 1.05, …………
(iv) -1.1, -3.1, -5.1, -7.1, …………..
Solution:
(i) -5, -1, 3, 7, …………

Question 2.
Write the arithmetic progression when first term a and common difference d are as follows:
(i) a = 4, d = -3
(ii) a = -1, d= $$\frac { 1 }{ 2 }$$
(iii) a = -1.5, d = -0.5
Solution:
(i) First term (a) = 4
and common difference (d) = -3
Second term = a + d = 4 – 3 = 1
Third term = a + 2d = 4 + 2 x (-3) = 4 – 6 = -2
Fourth term = a + 3d = 4 + 3 (-3) = 4 – 9 = -5
Fifth term = a + 4d = 4 + 4 (-3) = 4 – 12 = -8
AP will be 4, 1, -2, -5, -8, ……….

Question 3.
In which of the following situations, the sequence of numbers formed will form an A.P?
(i) The cost of digging a well for the first metre is ₹ 150 and rises by ₹ 20 for each succeeding metre.
(ii) The amount of air present in the cylinder when a vacuum pump removes each time $$\frac { 1 }{ 4 }$$ of the remaining in the cylinder.
(iii) Divya deposited ₹ 1000 at compound interest at the rate of 10% per annum. The amount at the end of first year, second year, third year, …, and so on. [NCERT Exemplar]
Solution:
(i) Cost of digging a well for the first metre = ₹ 150
Cost for the second metre = ₹ 150 + ₹ 20 = ₹ 170
Cost for the third metre = ₹ 170 + ₹ 20 = ₹ 190
Cost for the fourth metre = ₹ 190 + ₹ 20 = ₹ 210
The sequence will be (In rupees)
150, 170, 190, 210, ………..
Which is an A.P.
Whose = 150 and d = 20
(ii) Let air present in the cylinder = 1

(iii) Amount at the end of the 1st year = ₹ 1100
Amount at the end of the 2nd year = ₹ 1210
Amount at the end of 3rd year = ₹ 1331 and so on.
So, the amount (in ₹) at the end of 1st year, 2nd year, 3rd year, … are
1100, 1210, 1331, …….
Here, a2 – a1 = 110
a3 – a2 = 121
As, a2 – a1 ≠ a3 – a2, it does not form an AP

Question 4.
Find the common difference and write the next four terms of each of the following arithmetic progressions :
(i) 1, -2, -5, -8, ……..
(ii) 0, -3, -6, -9, ……
(iii) -1, $$\frac { 1 }{ 4 }$$ , $$\frac { 3 }{ 2 }$$ , ……..
(iv) -1, – $$\frac { 5 }{ 6 }$$ , – $$\frac { 2 }{ 3 }$$ , ………..
Solution:

Question 5.
Prove that no matter what the real numbers a and b are, the sequence with nth term a + nb is always an A.P. What is the common difference ?
Solution:
an = a + nb
Let n= 1, 2, 3, 4, 5, ……….
a1 = a + b
a2 = a + 2b
a3 = a + 3b
a4 = a + 4b
a5 = a + 5b
We see that it is an A.P. whose common difference is b and a for any real value of a and b
as a2 – a1 = a + 2b – a – b = b
a3 – a2 = a + 3b – a – 2b = b
a4 – a3 = a + 4b – a – 3b = b
and a5 – a4 = a + 5b – a – 4b = b

Question 6.
Find out which of the following sequences are arithmetic progressions. For those which are arithmetic progressions, find out the common difference.

Solution:

Question 7.
Find the common difference of the A.P. and write the next two terms :
(i) 51, 59, 67, 75, …….
(ii) 75, 67, 59, 51, ………
(iii) 1.8, 2.0, 2.2, 2.4, …….
(iv) 0, $$\frac { 1 }{ 4 }$$ , $$\frac { 1 }{ 2 }$$ , $$\frac { 3 }{ 4 }$$ , ………..
(v) 119, 136, 153, 170, ………..
Solution:

Hope given RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.3 are helpful to complete your math homework.

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## RD Sharma Class 10 Solutions Chapter 2 Polynomials Ex 2.3

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 2 Polynomials Ex 2.3

Other Exercises

Question 1.
Apply division algorithm to find the quotient q(x) and remainder r(x) on dividing f(x) by g(x) in each of the following :

Solution:

Question 2.
Check whether the first polynomial is a factor of the second polynomial by applying the division algorithm.

Solution:

Question 3.
Obtain all zeros of the polynomial f(x) = 2x4 + x3 – 14x2 – 19x – 6, if two of its zeros are -2 and -1.
Solution:

Question 4.
Obtain all zeros of f(x) = x3 + 13x2 + 32x + 20, if one of its zeros is -2.
Solution:
f(x) = x3 + 13x2 + 32x + 20
One zero = -2 or x = -2
x + 2 is a factor of f (x)
Now dividing f(x) by x + 2, we get

x + 1 = 0 => x = -1
and x + 10 = 0
=> x = -10
-1 and -10
Hence zeros are -10, -1, -2

Question 5.
Obtain all zeros of the polynomial f(x) = x4 – 3x3 – x2 + 9x – 6, if two of its zeros are – √3 and √3
Solution:

Question 6.
Find all zeros of the polynomial f(x) = 2x4 – 2x3 – 7x2 + 3x + 6, if its two zeros are $$\surd \frac { 3 }{ 2 }$$ and – $$\surd \frac { 3 }{ 2 }$$
Solution:

Question 7.
Find all the zeros of the polynomial x4 + x3 – 34x2 -4x+ 120, if two of its zeros are 2 and -2. [CBSE 2008]
Solution:

Either x + 6 = 0, then x = -6
or x – 5 = 0, then x = 5
Hence other two zeros are -6, 5
and all zeros are 2, -2, -6, 5

Question 8.
Find all zeros of the polynomial 2x4 + 7x3 – 19x2 – 14x + 30, if two of its zeros are √2 and -√2
Solution:

Question 9.
Find all the zeros of the polynomial 2x3 + x2 – 6x – 3, if two of its zeros are – √3 and √3. (CBSE 2009)
Solution:
Let f(x) = 2x3 + x2 – 6x – 3
and two zeros of f(x) are – √3 and √3

Question 10.
Find all the zeros of the polynomial x3 + 3x2 – 2x – 6, if two of its zeros are – √2 and √2 (CBSE2009)
Solution:

Question 11.
What must be added to the polynomial f(x) = x4 + 2x3 – 2x2 + x – 1 so that the resulting polynomial is exactly divisible by x2 + 2x – 3 ?
Solution:

Question 12.
What must be subtracted from the polynomial f(x) = x4 + 2x3 – 13x2 – 12x + 21 so that the resulting polynomial is exactly divisible by x3 – 4x + 3 ?
Solution:

The resulting polynomial is exactly divisible by x2 – 4x + 3
Remainder = 0
=> 2x -3 – k = 0
=> k = 2x – 3
(2x – 3) must be subtracted

Question 13.
Given that √2 is a zero of the cubic polynomial 16x3 + √2x2 – 10x – 4√2 , find its other two zeroes. [NCERT Exemplar]
Solution:

Question 14.
Given that x – √5 is a factor of the cubic polynomial x3 – 3 √5 x2 + 13x – 3 √5 , find all the zeroes of the polynomial. [NCERT Exemplar|
Solution:

Hope given RD Sharma Class 10 Solutions Chapter 2 Polynomials Ex 2.3 are helpful to complete your math homework.

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## RD Sharma Class 10 Solutions Chapter 1 Real Numbers Ex 1.6

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 1 Real Numbers Ex 1.6

Other Exercises

Question 1.
Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion.

Solution:

Since, the denominator is of the form 2m x 5n, the rational number has a terminating decimal expansion.

Question 2.
Write down the decimal expansions of the following rational numbers by writing their denominators in the form 2m x 5n, where m and n are non-negative integers:

Solution:

Question 3.
Write the denominator of the rational number $$\frac { 257 }{ 5000 }$$ in the form 2m x 5n , where m, n are non-negative integers. Hence, write the decimal expansion, without actual division.
Solution:

Question 4.
What can you say about the prime factorisations of the denominators of the following rationals :
Solution:
(i) 43.123456789
This decimal fraction is terminating Its denominator will be factorised in the form of 2m x 5n where m and n are non-negative integers.

This decimal fraction is non-terminating repeating decimals.
The denominator of their fraction will be not in the form of 2m x 5n where m and n are non-negative integers.

This decimal fraction is terminating
Its denominator will be factorised in the form of 2m x 5n where m and n are non-negative integers
(iv) 0.120120012000120000 ………..
This decimal fraction in non-terminating non recurring
Its denominator will not be factorised in the form of 2m x 5n where m and n are non negative integers

Question 5.
A rational number in its decimal expansion is 327.7081. What can you say about the prime factors of q, when this number is expressed in the form $$\frac { p }{ q }$$ ? Give reasons. [NCERT Exemplar]
Solution:
327.7081 is terminating decimal number. So, it represents a rational number and also its denominator must have the form 2m x 5n.

Hence, the prime factors of q is 2 and 5.

Hope given RD Sharma Class 10 Solutions Chapter 1 Real Numbers Ex 1.6 are helpful to complete your math homework.

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## RD Sharma Class 10 Solutions Chapter 8 Circles MCQS

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 8 Circles MCQS

Other Exercises

Mark the correct alternative in each of the following :
Question 1.
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
(d) √119 cm
Solution:
(d) Radius of a circle OP = 5 cm OQ = 12 cm, PQ is tangent

OP ⊥ PQ
In right ∆OPQ,
OQ² = OP² + PQ² (Pythagoras Theorem)
=> (12)² = (5)2 + PQ²
=> 144 = 25 + PQ²
PQ² = 144 – 25 = 119
PQ = √119

Question 2.
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
(a) 7 cm
(b) 12 cm
(c) 15 cm
(d) 24.5 cm
Solution:
(a) Let PQ be the tangent from Q to the circle with O as centre
PQ = 24 cm
OQ = 25 cm

OQ ⊥ PQ
Now in right ∆OPQ,
OQ² = OP² + PQ² (Pythagoras Theorem)
=> (25)² = r² + (24)²
=> 625 = r² + 576
=> r² = 625 – 576 = 49 = (7)²
r = 7
Radius of the circle = 7 cm

Question 3.
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) √7 cm
(b) 7 cm
(c) 5 cm
(d) 25 cm
Solution:
(c) Let AB be the tangent from A to the circle of centre O, then
OB = 3 cm
BA = 4 cm

OB ⊥ BA
In right ∆OBA,
OA² = OB² + BA² (Pythagoras Theorem) = (3)² + (4)² = 9 + 16 = 25 = (5)²
OA = 5
Distance of A from the centre O = 5 cm

Question 4.
If tangents PA and PB from a point P to a circle with centre O are inclined to each other at an angle of 80° then ∠POA is equal to
(a) 50°
(b) 60°
(c) 70°
(d) 80°
Solution:
(a) PA and PB are the tangents to the circle from P and ∠APB = 80°

∠AOB = 180° – ∠APB = 180°- 80° = 100°
But OP is the bisector of ∠AOB
∠POA = ∠POB = $$\frac { 1 }{ 2 }$$ ∠AOB
=> ∠POA = $$\frac { 1 }{ 2 }$$ x 100° = 50°

Question 5.
If TP and TQ are two tangents to a circle with centre O so that ∠POQ = 110°, then, ∠PTQ is equal to
(a) 60°
(b) 70°
(c) 80°
(d) 90°
Solution:
(b) TP and TQ are the tangents from T to the circle with centre O and OP, OQ are joined and ∠POQ = 110°

But ∠POQ + ∠PTQ = 180°
=> 110° + ∠PTQ = 180°
=> ∠PTQ = 180° – 110° = 70°

Question 6.
PQ is a tangent to a circle with centre O at the point P. If ∆OPQ is an isosceles triangle, then ∠OQP is equal to
(a) 30°
(b) 45°
(c) 60°
(d) 90°
Solution:
(b) In a circle with centre O, PQ is a tangent to the circle at P and ∆OPQ is an isosceles triangle such that OP = PQ

OP is radius of the circle
OP ⊥ PQ
OP = PQ
∠POQ = ∠OQP
But ∠POQ = ∠PQO = 90° (OP ⊥ PQ)
∠OQP = ∠POQ = 45°

Question 7.
Two equal circles touch each other externally at C and AB is a common tangent to the circles. Then, ∠ACB =
(a) 60°
(b) 45°
(c) 30°
(d) 90°
Solution:
(d) Two circles with centres O and O’ touch each other at C externally
A common tangent is drawn which touches the circles at A and B respectively.
Join OA, O’B and O’O which passes through C

AO = BO’ (radii of the equal circle)
AB || OO’
=> AOO’B is a rectangle
Draw another common tangent through C which intersects AB at D, then DA = DC = DB
AC and BC are the diagonals of equal square
AC = BC
∠DAC = ∠DBC = 45°
∠ACB = 90°

Question 8.
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 ∆ABC. The radius of the circle is
(a) 1 cm
(b) 2 cm
(c) 3 cm
(d) 4 cm
Solution:
(b) In a right ∆ABC, ∠B = 90°
BC = 6 cm, AB = 8 cm

AC² = AB² + BC² (Pythagoras Theorem) = (8)² + (6)² = 64 + 36 = 100 = (10)²
AC = 10 cm
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 and 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
=> 10 = 8 – r + 6 – r
10 = 14 – 2r
=> 2r = 14 – 10 = 4
=> r = 2
Radius of the incircle = 2 cm

Question 9.
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 ∠POR = 120°, then ∠OPQ is
(a) 60°
(b) 45°
(c) 30°
(d) 90°
Solution:
(c) PQ is a tangent to the circle with centre O, from P, QOR is the diameter and ∠POR = 120°

OQ is radius and PQ is tangent to the circle
OQ ⊥ QP or ∠OQP = 90°
But ∠QOP + ∠POR = 180° (Linear pair)
=> ∠QOP + 120° = 180°
∠QOP = 180° – 120° = 60°
Now in ∆POQ
∠QOP + ∠OQP + ∠OPQ = 180° (Angles of a triangle)
=> 60° + 90° + ∠OPQ = 180°
=> 150° + ∠OPQ = 180°
=> ∠OPQ = 180° – 150° = 30°

Question 10.
If four sides of a quadrilateral ABCD are tangential to a circle, then
(a) AC + AD = BD + CD
(b) AB + CD = BC + AD
(c) AB + CD = AC + BC
(d) AC + AD = BC + DB
Solution:
(b) 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

Question 11.
The length of the tangent drawn from a point 8 cm away from the centre of a circle of radius 6 cm is
(a) √7 cm
(b) 2√7cm
(c) 10 cm
(d) 5 cm
Solution:
(b) Radius of the circle = 6 cm
and distance of the external point from the centre = 8 cm
Length of tangent = √{(8)² – (6)²}
= √(64 – 36) = √28
= √(4 x 7) = 2√7 cm

Question 12.
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
(c) 8 cm
(d) 12 cm
Solution:
(c) AB and CD are two common tangents to the two circles which touch each other externally at C and intersect AB in D

CD = 4 cm
DA and DC are tangents to the first circle from D
CD = AD = 4 cm
Similarly DC and DB are tangents to the second circle from D
CD = DB = 4 cm
AB = AD + DB = 4 + 4 = 8 cm

Question 13.
In the adjoining figure, if AD, AE and BC are tangents to the circle at D, E and F respectively. Then,

(a) AD = AB + BC + CA
(b) 2AD = AB + BC + CA
(c) 3AD = AB + BC + CA
(d) 4AD = AB + BC + CA
Solution:
(b) AD, AE and BC are the tangents to the circle at D, E and F respectively
D and AE are tangents to the circle from A
Similarly, CD = CF and BE = BF ….(ii)
Now AB + AC + BC = AE – BE + AD – CD + CF + BF
= AD – BE + AD – CD + BE + BE
= 2AD [From (i) and (ii)]
or 2 AD = AB + BC + CA

Question 14.
In the 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
(d) 5 cm
Solution:
(d) 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

OQ = $$\frac { 1 }{ 2 }$$ QS = $$\frac { 1 }{ 2 }$$ x 6 = 3 cm
OQ ⊥ QR
Now in right ∆OQR
OR² = QR² + QO² = (3)² + (4)² = 9 + 16 = 25 = (5)²
OR = 5 cm

Question 15.
In the figure, the perimeter of ∆ABC is
(a) 30 cm
(b) 60 cm
(c) 45 cm
(d) 15 cm

Solution:
(a) ∆ABC is circumscribed of circle with centre O
AQ = 4 cm, CP = 5 cm and BR = 6 cm
AQ and AR the tangents to the circle AQ = AR = 4 cm
Similarly BP and BR are tangents,
BP = BR = 6 cm
and CP and CQ are the tangents
CQ = CP = 5 cm

AB = AR + BR = 4 + 6 = 10 cm
BC = BP + CP = 6 + 5 = 11 cm
AC = AQ + CQ = 4 + 5 = 9 cm
Perimeter of ∆ABC = AB + BC + AC = 10 + 11 + 9 = 30 cm

Question 16.
In the figure, AP is a tangent to the circle with centre O such that OP = 4 cm and ∠OPA = 30°. Then, AP =
(a) 2√2 cm
(b) 2 cm
(c) 2√3 cm
(d) 3√2 cm

Solution:
(c) In the figure, AP is the tangent to the circle with centre O such that
OP = 4 cm, ∠OPA = 30°
Join OA, let AP = x

Question 17.
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
(c) 24 cm
(d) 36 cm
Solution:
(c) OP is radius, PA is the tangent
OP ⊥ AP

Now in right ∆OAP,
OA² = OP² + AP²
(15)² = (9)² + AP²
225 = 81 + AP²
=> AP² = 225 – 81 = 144 = (12)²
AP = 12 cm
But AP = AQ = 12 cm (tangents from A to the circle)
AP + AQ = 12+ 12 = 24 cm

Question 18.
At one end of a diameter PQ of a circle of radius 5 cm, tangent XPY is drawn to the circle. The length of chord AB parallel to XY and at a distance of 8 cm from P is
(a) 5 cm
(b) 6 cm
(c) 7 cm
(d) 8 cm
Solution:
(d) 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 ∆OAR,
OA² = OR² + AR²
(5)² = (3)² + AR²
25 = 9 + AR²
=> AR² = 25 – 9 = 16 = (4)²
AR = 4 cm
But R is mid-point of AB
AB = 2 AR = 2 x 4 = 8 cm

Question 19.
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°
(b) 60°
(c) 90°
(d) 180°
Solution:
(c) 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 ∆OPT
∠OTP = 90°
∠OPT + ∠POT = 180° – 90° = 90°

Question 20.
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
(d) 7 cm

Solution:
(d) In the figure, ∆ABC is the circumscribed a circle
AB = 12 cm, BC = 8 cm and AC = 10 cm
Let AD = a, DB = b and EC = c, then
AF = a, BE = b and FC = c

But AB + BC + AC = 12 + 8 + 10 = 30
a + b + b + c + c + a = 30
=> 2 (a + b + c) = 30
a + b + c = 15
Subtracting BC or b + c from this a = 15 – 8 = 7

Question 21.
In the figure, if AP = PB, then
(a) AC = AB
(b) AC = BC
(c) AQ = QC
(d) AB = BC

Solution:
(b) 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
AQ + CQ = BR + CR
AC = BC

Question 22.
In the figure, if AP = 10 cm, then BP =
(a) √91 cm
(b) √127 cm
(c) √119 cm
(d) √109 cm

Solution:
(b) In the figure,
OA = 6 cm, OB = 3 cm and AP = 10 cm

OA is radius and AP is the tangent
OA ⊥ AP
Now in right ∆OAP
OP² = AP² + OA² = (10)² + (6)² = 100 + 36 = 136
Similarly BP is tangent and OB is radius
OP² = OB² + BP²
136 = (3)² + BP2
=> 136 = 9 + BP²
=> BP² = 136 – 9 = 127
BP = √127 cm

Question 23.
In the figure, if PR is tangent to the circle at P and Q is the centre of the circle, then ∠POQ =
(a) 110°
(b) 100°
(c) 120°
(d) 90°

Solution:
(c) In the figure, PR is the tangent to the circle at P.
O is the centre of the circle ∠QPR = 60°

OP is the radius and PR is the tangent OPR = 90°
=> ∠OPQ + ∠QPR = 90°
=> ∠OPQ + 60° = 90°
=> ∠OPQ = 90° – 60° = 30°
OP = OQ (radii of the circle)
∠OQP = 30°
In ∆OPQ,
∠OPQ + ∠OQP + ∠POQ = 180°
=> 30° + 30° + ∠PQR = 180°
=> 60° + ∠POQ = 180°
∠POQ = 180° – 60° = 120°

Question 24.
In the figure, if quadrilateral PQRS circumscribes a circle, then PD + QB =
(a) PQ
(b) QR
(c) PR
(d) PS

Solution:
(a) In the figure, quadrilateral PQRS is circumscribed a circle

PD = PA (tangents from P to the circle)
Similarly QA = QB
PD + QB = PA + QA = PQ

Question 25.
In the figure, two equal circles touch each other at T, if QP = 4.5 cm, then QR =
(a) 9 cm
(b) 18 cm
(c) 15 cm
(d) 13.5 cm

Solution:
(a) 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

Question 26.
In the figure, APB is a tangent to a circle with centre O at point P. If ∠QPB = 50°, then the measure of ∠POQ is
(a) 100°
(b) 120°
(c) 140°
(d) 150°

Solution:
(a) In the figure, APB is a tangent to the circle with centre O

∠QPB = 50°
OP is radius and APB is a tangent
OP ⊥ AB
=> ∠OPB = 90°
=> ∠OPQ + ∠QPB = 90°
∠OPQ + 50° = 90°
=> ∠OPQ = 90° – 50° = 40°
But OP = OQ
∠OPQ = OQP = 40°
∠POQ = 180°- (40° + 40°) = 180° – 80° = 100°

Question 27.
In the figure, if tangents PA and PB are drawn to a circle such that ∠APB = 30° and chord AC is drawn parallel to the tangent PB, then ∠ABC =
(a) 60°
(b) 90°
(c) 30°
(d) None of these

Solution:
(c) In the figure, PA and PB are the tangents to the circle with centre O

∠APB = 30°
Chord AC || BP,
AB is joined
PA = PB
∠PAB = ∠PBA
But ∠PAB + ∠PBA = 180° – 30° = 150°
=> ∠BPA + ∠PBA = 150°
=> 2 ∠PBA = 150°
=> ∠PBA = 75°
AC || BC
∠BAC = ∠PBA = 75°
But ∠PBA = ∠ACB = 75° (Angles in the alternate segment)
∠ABC = 180° – (75° + 75°) = 180° – 150° = 30°

Question 28.
In the figure, PR =
(a) 20 cm
(b) 26 cm
(c) 24 cm
(d) 28 cm

Solution:
(b) 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 ∆OQP
OP² = (OQ)² + PQ² = (3)² + (4)² = 9 + 16 = 25 = (5)²
OP = 5 cm
Similarly in right ∆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

Question 29.
Two circles of same radii r and centres O and O’ touch each other at P as shown in figure. If OO’ 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’ =
(a) $$\frac { 3 }{ 2 }$$
(b) 2
(c) 3
(d) $$\frac { 1 }{ 4 }$$

Solution:
(c) Two circles of equal radii touch each other externally at P. OO’ produced meets at A

From A, AT is the tangent to the circle (O, r)
O’Q ⊥ AT
Now AO : AO’ = 3r : r
= 3 : 1 = $$\frac { 3 }{ 1 }$$

Question 30.
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
(c) 8 cm
(d) 10 cm

Solution:
(c) In the figure, two concentric circles of radii 3 cm and 5 cm with centre O
Chord BC touches the inner circle at P
Draw a tangent AB to the inner circle
Join OQ and OA

OQ is radius and AQB is the tangent
OQ ⊥ AB and OQ bisects AB
AQ = QB
Similarly, BP = PC or P is mid-point of BC
But BQ and BP are tangents from B
QB = BP = AQ
In right ∆OAQ,
OA² = AQ² + OQ²
(5)² = AQ² + (3)²
=> AQ² = (5)² – (3)²
=> AQ² = 25 – 9 = 16 = (4)²
AQ = 4 cm
BC = 2 BP = 2 BQ = 2 AQ = 2 x 4 = 8 cm

Question 31.
In the 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
(c) 15 cm
(d) 18 cm

Solution:
(c) In the figure, two concentric circles with centre O
From a point P on the outer circle,
PRT and PQS are the tangents are drawn to the inner circle at R and Q respectively
PR = 7.5 cm
Join OR and OQ

PT is chord and OR is radius
R is mid-point of PT
Similarly Q is mid-point of PS
But PR = PQ (tangents from P)
PT = 2 PR and PS = 2 PQ
PS = 2 PQ = 2 PR = 2 x 7.5 = 15 cm

Question 32.
In the figure, if AB = 8 cm and PE = 3 cm, then AE =
(a) 11 cm
(b) 7 cm
(c) 5 cm
(d) 3 cm

Solution:
(c) In the figure, AB and AC are the tangents to the circle from A
DE is another tangent drawn from P
AB = 8 cm, PE = 3 cm

AB = AC (tangents drawn from A to the circle)
Similarly PE = EC and DP = DB
Now AE = AC – CE = AB – PE = 8 – 3 = 5 cm

Question 33.
In the figure, PQ and PR are tangents drawn from P to a circle with centre O. If ∠OPQ = 35°, then
(a) a = 30°, b = 60°
(b) a = 35°, b = 55°
(c) a = 40°, b = 50°
(d) a = 45°, b = 45°

Solution:
(b) In the figure, PQ and PR are the tangents drawn from P to the circle with centre O
∠OPQ = 35°
PO is joined

PQ = PR (tangents from P to the circle)
∠OPQ = ∠OPR
=> 35° = a
=> a = 35°
OQ is radius and PQ is tangent
OQ ⊥ PQ
=> ∠OQP = 90°
In ∆OQP,
∠POQ + ∠QPO = 90°
=> b + 35° = 90°
=> b = 90° – 35° = 55°
a = 35°, b = 55°

Question 34.
In the figure, if TP and TQ are tangents drawn from an external point T to a circle with centre O such that ∠TQP = 60°, then
(a) 25°
(b) 30°
(c) 40°
(d) 60°

Solution:
(b) In the figure, TP and TQ are the tangents drawn from T to the circle with centre O
OP, OQ and PQ are joined
∠TQP = 60°
TP = TQ (Tangents from T to the circle)
∠TQP = ∠TPQ = 60°
∠PTQ = 180° – (60° + 60°) = 180° – 120° = 60°
and ∠POQ = 180° – ∠PTQ = 180° – 60° = 120°
But OP = OQ (radii of the same circle)
∠OPQ = ∠OQP
But ∠OPQ + ∠OQP = 180° – 120° = 60°
But ∠OPQ = 30°

Question 35.
In the 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 [CBSE 2012]

(a) 11 cm
(b) 10 cm
(c) 14 cm
(d) 15 cm
Solution:
(b) In the figure,
PA = 4 cm, BP = 3 cm, AC = 11 cm
AP and AR are the tangents from A to the circle
AP = AR
=> AR = 4 cm
Similarly BP and BQ are tangents
BQ = BP = 3 cm
AC =11 cm
AR + CR = 11 cm
4 + CR =11 cm
CR = 11 – 4 = 7 cm
CQ and CR are tangents to the circle
CQ = CR = 7 cm
Now, BC = BQ + CQ = 3 + 7 = 10 cm

Question 36.
In the figure, a circle touches the side DF of AEDF at H and touches ED and EF produced at K and M respectively. If EK = 9 cm, then the perimeter of ∆EDF is [CBSE 2012]
(a) 18 cm
(b) 13.5 cm
(c) 12 cm
(d) 9 cm

Solution:
(a) In ∆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 ∆DEF
= ED + DF + EF
= ED + DH + FH + EF
= ED + DK + EM + EF
= EK + EM
= 9 + 9
= 18 cm

Question 37.
In the 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 [CBSE 2013]

(a) 3 cm
(b) 5 cm
(c) 4 cm
(d) 6 cm
Solution:
(b) If figure, DE and DF are tangents to the circle drawn from D.
A is the centre of the circle.
DE = 5 cm and DE ⊥ DF
Join AE, AF

DE is the tangent and AE is radius
AE ⊥ DE
Similarly, AF ⊥ DF
But ∠D = 90° (given)
AFDE is a square
AE = DE (side of square)
But DE = 5 cm
AE = 5 cm
Radius of circle is 5 cm

Question 38.
In the 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 = 29 cm, AD = 23 cm, ∠B = 90° and DS = 5 cm, then the radius of the circle (in cm) is [CBSE 2013]

(a) 11
(b) 18
(c) 6
(d) 15
Solution:
(a) In the figure, a circle touches the sides of a quadrilateral ABCD
∠B = 90°, OP = OQ = r
AB = 29 cm, AD = 23 cm, DS = 5 cm
∠B = 90°
BA is tangent and OQ is radius
∠OQB = 90°
Similarly OP is radius and BC is tangents
∠OPB = 90°
But ∠B = 90° (given)
PBQO is a square
DS = 5 cm
But DS and DR are tangents to the circles
DR = 5 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
OPBQ is a square
OQ = BQ = 11 cm
Radius of the circle = 11 cm

Question 39.
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
(a) 4
(b) 3
(c) 2
(d) 1
Solution:
(c)

Question 40.
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 ∠APB is
(a) 30°
(b) 45°
(c) 60°
(d) 90°
Solution:
(d) We have, AT = TP and TB = TP (Lengths of the tangents from ext. point T to the circles)
∠TAP = ∠TPA = x (say)
and ∠TBP = ∠TPB = y (say)
Also, in triangle APB,
x + x + x + y + y = 180°
=> 2x + 2y = 180°
=> x + y = 90°
=> ∠APB = 90°

Question 41.
In the figure, PQ and PR are two tangents to a circle with centre O. If ∠QPR= 46, then ∠QOR equals
(a) 67°
(b) 134°
(c) 44°
(d) 46°

Solution:
(b) ∠OQP = 90°
[Tangent is ⊥ to the radius through the point of contact]
∠ORP = 90°
∠OQP + ∠QPR + ∠ORP + ∠QOR = 360° [Angle sum property of a quad.]
90° + 46° + 90° + ∠QOR = 360°
∠QOR = 360° – 90° – 46° – 90° = 134°

Question 42.
In the 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 = 3.8 cm, then the length of QR (in cm) is [CBSE2014]
(a) 3.8
(b) 7.6
(c) 5.7
(d) 1.9

Solution:
(b) In the figure, QR is common tangent to the two circles touching each other externally at T
Tangent at T meets QR at P
PT = 3.8 cm
PT and PQ are tangents from P
PT = PQ = 3.8 cm
Similarly PT and PR are tangents
PT = PR = 3.8 cm
QR = 3.8 + 3.8 = 7.6 cm

Question 43.
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 =
(a) 10
(b) 9
(c) 8
(d) 7 (CBSE 2014)

Solution:
(b) In 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

Question 44.
If angle between two radii of a circle is 130°, the angle between the tangent at the ends of radii is (NCERT Exemplar)
(a) 90°
(b) 50°
(c) 70°
(d) 40°
Solution:
(b) O is the centre of the circle.
Given, ∠POQ = 130°
PT and QT are tangents drawn from external point T to the circle.

∠OPT = ∠OQT = 90° [Radius is perpendicular to the tangent at point of contact]
∠PTQ + ∠OPT + ∠OQT + ∠POQ = 360°
=> ∠PTQ + 90° + 90° + 130° = 360°
=> ∠PTQ = 360° – 310° = 50°
Thus, the angle between the tangents is 50°.

Question 45.
If two tangents inclined at a angle of 60° are drawn to a circle of radius 3 cm, then length of each tangent is equal to [NCERT Exemplar]
(a) $$\frac { 3\surd 3 }{ 2 }$$ cm
(b) 6 cm
(c) 3 cm
(d) 3√3 cm
Solution:
(d) 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 ∠APC
∠APO = ∠CPO = 30°
Also, OA ⊥ AP
Tangent at any point of a circle is perpendicular to the radius through the point of contact.

Hence, the length of each tangent is 3√3 cm

Question 46.
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 [NCERT Exemplar]
(a) 3 cm
(b) 6 cm
(c) 9 cm
(d) 1 cm
Solution:
(b) Let O be the centre of two concentric circles C1 and C2, whose radii are r1 = 4 cm and r2 = 5 cm.
Now, we draw a chord AC of circle C2, which touches the circle C1 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 ∆OBC, by using Pythagoras theorem,
OC² = BC² + BO² [(hypotenuse)² = (base)² + (perpendicular)²]
=> 5² = BC² + 4²
=> BC² = 25 – 16 = 9
=> BC = 3 cm
Length of chord AC = 2 BC = 2 x 3 = 6 cm

Question 47.
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 [NCERT Exemplar]
(a) 4 cm
(b) 5 cm
(b) 6 cm
(d) 8 cm
Solution:
(d) First, draw a circle of radius 5 cm having centre O.
A tangent XY is drawn at point A.

A chord CD is drawn which is parallel to XY and at a distance of 8 cm from A.
Now, ∠OAY = 90°
[Tangent and any point of a circle is perpendicular to the radius through the point of contact]
∠OAY + ∠OED = 180°
[sum of cointerior is 180°]
=> ∆OED = 180°
Also, AE = 8 cm, Join OC
Now, in right angled ∆OBC
OC² = OE² + EC²
=> EC² = OC² – OE² [by Pythagoras theorem]
EC² = 5² – 3² [OC = radius = 5 cm, OE = AE – AO = 8 – 5 = 3 cm]
EC² = 25 – 9 = 16
=> EC = 4 cm
Hence, length of chord CD = 2 CE = 2 x 4 = 8 cm
[Since, perpendicular from centre to the chord bisects the chord]

Question 48.
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 [NCERT Exemplar]
(a) 60 cm²
(b) 65 cm²
(c) 30 cm²
(d) 32.5 cm²
Solution:
(a) 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.
OQ ⊥ QP [since, AP is a tangent line]
In right angled ∆PQO,
OP² = OQ² + QP²
=> 13² = 5² + QP²
=> QP² = 169 – 25 = 144 = 12²
=> QP = 12 cm

Now, area of ∆OQP = $$\frac { 1 }{ 2 }$$ x QP x QO = $$\frac { 1 }{ 2 }$$ x 12 x 5 = 30 cm²
Area of quadrilateral QORP = 2 ∆OQP = 2 x 30 = 60 cm²

Question 49.
If PA and PB are tangents to the circle with centre O such that ∠APB = 50°, then ∠OAB is equal to
(a) 25°
(b) 30°
(c) 40°
(d) 50°
Solution:
(a) Given, PA and PB are tangent lines.
PA = PB [Since, the length of tangents drawn from an ∠PBA = ∠PAB = θ [say]

In ∆PAB,
∠P + ∠A + ∠B = 180°
[since, sum of angles of a triangle = 180°
50°+ θ + θ = 180°
2θ = 180° – 50° = 130°
θ = 65°
Also, OA ⊥ PA
[Since, tangent at any point of a circle is perpendicular to the radius through the point of contact]
∠PAO = 90°
=> ∠PAB + ∠BAO = 90°
=> 65° + ∠BAO = 90°
=> ∠BAO = 90° – 65° = 25°

Question 50.
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 [NCERT Exemplar]
(a) 10 cm
(b) 7.5 cm
(c) 5 cm
(d) 2.5 cm
Solution:
(c)

Question 51.
In the figure, if ∠AOB = 125°, then ∠COD is equal to [NCERT Exemplar]

(a) 45°
(b) 35°
(c) 55°
(d) 62$$\frac { 1 }{ 2 }$$°
Solution:
(c) We know that, the opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
∠AOB + ∠COD = 180°
=> ∠COD = 180° – ∠AOB = 180° – 125° = 55°

Question 52.
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 ∠BQR = 70°, then ∠AQB is equal to [NCERT Exemplar]

(a) 20°
(b) 40°
(c) 35°
(d) 45°
Solution:
(b) Given, AB || PR
∠ABQ = ∠BQR = 70° [alternate angles]
Also QD is perpendicular to AB and QD bisects AB.
In ∆QDA and ∆QDB
∠QDA = ∠QDB [each 90°]
QD = QD [common side]
∆ADQ = ∆BDQ [by SAS similarity criterion]
Then, ∠QAD = ∠QBD …(i) [c.p,c.t.]
Also, ∠ABQ = ∠BQR [alternate interior angle]
∠ABQ = 70° [∠BQR = 70°]
Hence, ∠QAB = 70° [from Eq. (i)]
Now, in ∆ABQ,
∠A + ∠B + ∠Q = 180°
=> ∠Q = 180° – (70° + 70°) = 40°

Hope given RD Sharma Class 10 Solutions Chapter 8 Circles MCQS are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

## RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS

Other Exercises

Mark the correct alternative in each of the following :

Question 1.
If 6 θ is an acute angle such that cos θ =

Solution:

Question 2.

Solution:

Question 3.

Solution:

Question 4.

Solution:

Question 5.

Solution:

Question 6.

Solution:

Question 7.

Solution:

Question 8.
If 6 is an acute angle such that tan2 6 = 8 $$\frac { 8 }{ 7 }$$, then the value of

Solution:

Question 9.
If 3 cos θ = 5 sin 6, then the value of

Solution:

Question 10.
If tan2 45° – cos2 30° = x sin 45° cos 45°, then x =

Solution:

Question 11.
The value of cos217° – sin2 73° is

Solution:
cos2 17° – sin2 73° = cos2 (90° – 73°) – sin2 73°
= sin2 73° – sin2 73° = 0 (c)

Question 12.

Solution:

Question 13.

Solution:

Question 14.
If A and B are complementary angles then
(a) A – sin B
(b) cos A = cos B
(c) A = tan B
(d) sec A = cosec B
Solution:
∵ A and B are complementary angles
∴ A + B = 90°
⇒ A – 90° – B
sec A = sec (90° – B) = cosec B (d)

Question 15.
If x sin (90° – θ) cot (90° – 6) – cos (90° – θ), then x =
(a) 0
(b) 1

(c) -1
(d) 2,

Solution:

Question 16.
If x tan 45° cos 60° = sin 60° cot 60°, then x is equal to

Solution:

Question 17.
If angles A, B, C of a AABC form an increasing AP, then sin B =

Solution:

Question 18.
If 6 is an acute angle such that sec2 θ =

Solution:

Question 19.
The value of tan 1° tan 2° tan 3s tan 89° is
(a) 1
(b) -1

(c) 0
(d) None of these

Solution:
tan 1° tan 2° tan 3° tan 44° tan 45° tan 46° tan 89°
= tan 1° tan 2° tan 3° tan 44° tan 45° tan
(90° – 44°) tan (90° – 43°) tan (90° – 1°)’
= tan 1° tan 2° tan 3° tan 44° tan 45° cot
44° cot 43°….. cot 1°

Question 20.
The value of cos 1″ cos 2° cos 3° cos 180° is
(a) 1
(b) 0
(c) -1
(d) None of these
Solution:
None of these, because we deal here only an angle O<θ ≤ 90° (d)

Question 21.
The value of tan 10° tan 15° tan 75° tan 80° is
(a) -1
(b) 0
(c) 1
(d) None of these
Solution:
tan 10° tan 15° tan 75° tan 80°
= tan 10° tan 15° tan (90° – 15°) tan (90° – 10°) .
= tan 10° tan 15° cot 15° cot 10°
= tan 10° cot 10° tan 15° cot 15°
{∵ tan θ cot θ = 1}
= 1×1 = 1 (C)

Question 22.
The value of

Solution:

Question 23.
If 6 and 20 – 45° arc acute angles such that sin 0 = cos (20 – 45°), then tan 0 is equal to

Solution:

Question 24.
If 5θ and 4θ are acute angles satisfying sin 5θ = cos 4θ, then 2sin 3θ-√3 tan 3θ is equal to
(a) 1
(b) 0
(c) -1
(d) 1 + √3
Solution:

Question 25.
If A + B = 90°, then
Solution:

Question 26.

Solution:

Question 27.

Solution:

Question 28.
sin 2A = 2 sin A is true when A =
(a) 0°
(b) 30°

(c) 45°
(d) 60°

Solution:

Question 29.

Solution:

Question 30.
If A, B and C are interior angles of a

Solution:

Question 31.
If cos θ = $$\frac { 2 }{ 3 }$$, then 2 sec2 θ + 2 tan2 θ-7 is equal to
(a) 1
(b) 0
(c) 3
(d) 4
Solution:

Question 32.
tan 5° x tan 30° x 4 tan 85° is equal to

Solution:

Question 33.

(a)-2
(b) 2
(c) 1
(d) 0
Solution:

Question 34.
In the figure, the value of cos Φ is

Solution:

Question 35.
In the figure, AD = 4 cm, BD = 3 cm and CB = 12 cm, find cot θ.

Solution:

Hope given RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

## RD Sharma Class 10 Solutions Chapter 9 Constructions Ex 9.3

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 9 Constructions Ex 9.3

Other Exercises

Question 1.
Draw a circle of radius 6 cm. From a point 10 cm away from its centre, construct the pair of tangents to the circle and measure their lengths.
Solution:
Steps of construction :
(i)
Draw a circle with O centre and 6 cm radius.
(ii) Take a point P, 10 cm away from the centre O.
(iii) Join PO and bisect it at M.
(iv) With centre M and diameter PO, draw a circle intersecting the given circle at T and S.
(v) Join PT and PS.
Then PT and PS are the required tangents.

Question 2.
Draw a circle of radius 3 cm. Take two points P and Q on one of its extended diameter each at a distance of 7 cm from its centre. Draw tangents to the circle from these two points P and Q.
Solution:
Steps of construction :
(i)
Draw a circle with centre O and radius 3 cm.
(ii) Draw a diameter and produce it to both sides.
(iii) Take two points P and Q on this diameter with a distance of 7 cm each from the centre O.
(iv) Bisect PO at M and QO at N
(v) With centres M and N, draw circle on PO and QO as diameter which intersect the given circle at S, T and S’, T’ respectively.
(vi) Join PS, PT, QS’ and QT’.
Then PS, PT, QS’ and QT’ are the required tangents to the given circle.

Question 3.
Draw a line segment AB of length 8 cm. Taking A as centre, draw a circle of radius 4 cm and taking B as centre, draw another circle of radius 3 cm. Construct tangents to each circle from the centre of the other circle. [CBSE 2013]
Solution:
Steps of construction :
(i)
Draw a line segment AB = 8 cm.
(ii) With centre A and radius 4 cm and with centre B and radius 3 cm, circles are drawn.
(iii) Bisect AB at M.
(iv) With centre M and diameter AB, draw a circle which intersects the two circles at S’, T’ and S, T respectively.
(v) Join AS, AT, BS’and BT’.
Then AS, AT, BS’ and BT’ are the required tangent.

Question 4.
Draw two tangents to a circle of raidus 3.5 cm from a point P at a distance of 6.2 cm from its centre.
Solution:
Steps of construction :
(i) Draw a circle with centre O and radius 3.5 cm
(ii) Take a point P which is 6.2 cm from O.
(iii) Bisect PO at M and draw a circle with centre M and diameter OP which intersects the given circle at T and S respectively.
(iv) Join PT and PS.
PT and PS are the required tangents to circle.

Question 5.
Draw a pair of tangents to a circle of radius 4.5 cm, which are inclined to each other at an angle of 45°.            [CBSE 2013]
Solution:
Steps of construction :
Angle at the centre 180° – 45° = 135°
(i) Draw a circle with centre O and radius 4.5 cm.

(ii) At O, draw an angle ∠TOS = 135°
(iii) At T and S draw perpendicular which meet each other at P.
PT and PS are the tangents which inclined each other 45°.

Question 6.
Draw a right triangle ABC in which AB = 6 cm, BC = 8 cm and ∠B = 90°. Draw BD perpendicular from B on AC and draw a circle passing through the points B, C and D. Construct tangents from A to this circle.
Solution:
Steps of Construction :
Draw a line segment BC = 8 cm
From B draw an angle of 90°
Draw an arc $$\breve { BA }$$  = 6cm cutting the angle at A.
Join AC.
ΔABC is the required A.
Draw ⊥ bisector of BC cutting BC at M.
Take M as centre and BM as radius, draw a circle.
Take A as centre and AB as radius draw an arc cutting the circle at E. Join AE.
AB and AE are the required tangents.
Justification :
∠ABC = 90°                                            (Given)
Since, OB is a radius of the circle.
∴ AB is a tangent to the circle.
Also AE is a tangent to the circle.

Question 7.
Draw two concentric circles of radii 3 cm and 5 cm. Construct a tangent to the smaller circle from a point on the larger circle. Also, measure its length.                      [CBSE 2016]
Solution:
Given, two concentric circles of radii 3 cm and 5 cm with centre O. We have to draw pair of tangents from point P on outer circle to the other.

Steps of construction :
(i) Draw two concentric circles with centre O and radii 3 cm and 5 cm.
(ii) Taking any point P on outer circle. Join OP.
(iii) Bisect OP, let M’ be the mid-point of OP.
Taking M’ as centre and OM’ as radius draw a circle dotted which cuts the inner circle as M and P’.
(iv) Join PM and PP’. Thus, PM and PP’ are the required tangents.
(v) On measuring PM and PP’, we find that PM = PP’ = 4 cm.
Actual calculation:
In right angle ΔOMP, ∠PMO = 90°
∴ PM2 = OP2 – OM2
[by Pythagoras theorem i.e. (hypotenuse)2 = (base)2 + (perpendicular)2]
⇒ PM2 = (5)2 – (3)2 = 25 – 9 = 16
⇒ PM = 4 cm
Hence, the length of both tangents is 4 cm.

Hope given RD Sharma Class 10 Solutions Chapter 9 Constructions Ex 9.3 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.