| Exercise 11.1 Exercise 11.2 |
Exercise 11.1
Question 1
The length of a tangent from a point A at a distance 5cm from the centre of a circle is 4cm. Find the radius of the circle.
Sol :Let the centre of circle be O so that AO = 5 cm
Tangent is AB whose length is 4 cm
OB is radius as shown
Now we know that radius is perpendicular to point of contact
Hence OB is perpendicular to AB
Hence ∠ABO = 90°
Consider ΔABO
Using Pythagoras theorem
⇒ AB2 + OB2 = AO2
⇒ 42 + OB2 = 52
⇒ 16 + OB2 = 25
⇒ OB2 = 25 – 16
⇒ OB2 = 9
⇒ OB = ±3
As length cannot be negative
⇒ OB = 3 cm
Hence length of radius is 3 cm
Question 2
Rajesh is 29 m away from the centre of a circular flower bed. Find the distance he has to cover to reach the flower bed along the tangential path if the radius of the flower bed is 20m.
Sol :Let the centre of circular flower bed be O and radius OB = 20 m
Let Rajesh is at point A, and he has to travel tangential path to reach flower bed which is AB as shown
Now we know that radius is perpendicular to point of contact
Hence OB is perpendicular to AB
Hence ∠ABO = 90°
Consider ΔABO
Using Pythagoras theorem
⇒ AB2 + OB2 = AO2
⇒ AB2 + 202 = 292
⇒ AB2 = 292 - 202
Using a2 – b2 = (a + b)(a – b)
⇒ AB2 = (29 – 20)(29 + 20)
⇒ AB2 = 9 × 49
⇒ AB = √(9 × 49)
⇒ AB = ± (3 × 7)
⇒ AB = ±21
As length cannot be negative
⇒ AB = 21 m
Hence Rajesh has to cover 21 m to reach the flower bed along the tangential path.
Question 3
Find the length of the tangent drawn from a point, whose distance from the centre of a circle is 5c, and the radius of the circle is 3 cm.
Sol :Let A be the point at distance of 5 cm from the centre as AO = 5 cm
AB is the tangent at point B as shown
OB is the radius which is 3 cm
Now we know that radius is perpendicular to point of contact
Hence OB is perpendicular to AB
Hence ∠ABO = 90°
Consider ΔABO
Using Pythagoras theorem
⇒ AB2 + OB2 = AO2
⇒ AB2 + 32 = 52
⇒ AB2 + 9 = 25
⇒ AB2 = 25 – 9
⇒ AB2 = 16
⇒ AB = ±4
As length cannot be negative
⇒ AB = 4 cm
Hence length of tangent is 4 cm
Question 4
A point P is 13 cm from the centre of the circle. The length of the tangent drawn from P to the circle is 12cm. Find the radius of the circle.
Sol :Let the centre of circle be O so that PO = 13 cm
Tangent is PB whose length is 12 cm
OB is radius as shown
Now we know that radius is perpendicular to point of contact
Hence OB is perpendicular to PB
Hence ∠PBO = 90°
Consider ΔPBO
Using Pythagoras theorem
⇒ PB2 + OB2 = PO2
⇒ 122 + OB2 = 132
⇒ OB2 = 132 - 122
⇒ OB2 = 169 – 144
⇒ OB2 = 25
⇒ OB = ±5
As length cannot be negative
⇒ OB = 5 cm
Hence length of radius is 5 cm
Question 5
If d1, d2 (d2<d1) are the diameters of two concentric circles and chord of one circle of length C is tangent to other circle, then prove that d22 = C2 + d12.
Sol :Let the two concentric circles have the centre O and let AB be the chord of an outer circle whose length is D and which is also tangent to the inner circle at point D as shown
The diameters are given as d1 and d2 hence the radius will be $\frac{\mathrm{d}_{1}}{2}$ and $\frac{\mathrm{d}_{2}}{2}$
In ΔOAB
⇒ OA = OB …radius of the outer circle
Hence ΔOAB is an isosceles triangle
As radius is perpendicular to tangent OC is perpendicular to AB
OC is altitude from the apex, and in an isosceles triangle, the altitude is also the median
Hence $\mathrm{AD}=\mathrm{DB}=\frac{\mathrm{C}}{2}$
Consider ΔODB
⇒ ∠ODB = 90° …radius perpendicular to tangent
Using Pythagoras theorem
⇒ OD2 + BD2 = OB2
$\Rightarrow \frac{\mathrm{d}_{2}^{2}}{2^{2}}+\frac{\mathrm{C}^{2}}{2^{2}}=\frac{\mathrm{d}_{1}^{2}}{2^{2}}$
Multiply the whole by 22
⇒ d22 + C2 = d12
Hence proved
Let lines AP and BR are parallel tangents to circle having centre O
We have to prove that AB is the diameter
To prove AB as diameter, we have to prove that AB passes through O which means that points A, O and B are on the same line or collinear
OA is perpendicular to PA at A because the line from the centre is perpendicular to the tangent at the point of contact
PA || RB
Hence OA is also perpendicular to RB
⇒ OA perpendicular to PA and RB …(i)
Similarly, OB is perpendicular to RB at B because the line from the centre is perpendicular to the tangent at the point of contact
PA || RB
Hence OB is also perpendicular to PA
⇒ OB perpendicular to PA and RB …(ii)
From (i) and (ii) we can say that OA and OB can be same line or parallel lines, but we have a common point O which implies that OA and OB are same lines
Hence A, O, B lies on the same line, i.e. A, O and B are collinear
Thus AB passes through O
Hence AB is the diameter
Hence, the line segment joining the point of contact of two parallel tangents to a circle is a diameter of the circle.
Let there be a circle with centre O and BR as tangent with the point of contact as B
Let AB be the line perpendicular to BR
⇒ ∠ABR = 90° …(i)
As OB is the radius of the circle and we know that radius is perpendicular to the tangent at the point of contact
OB is perpendicular to BR
⇒ ∠OBR = 90° …(ii)
Equation (i) and (ii) implies that
⇒ ∠ABR = ∠OBR
This is only possible iff A and O lie on the same line or A and O are the same points
Case 1: Suppose A and O are on the same line
If A and O are on the same line, then the perpendicular AB to tangent BR has passed through the centre
Case 2: suppose A and O are the same points
As O itself is the centre of the circle, and A and O are the same points hence the perpendicular to the tangent at the point of contact passes through the circle
In any scenario, the line has to pass through the centre.
Hence, the perpendicular at the point of contact of the tangent to a circle passes through the centre
Let the two concentric circles have the centre O and let AB be the chord of an outer circle whose length is D and which will also be tangent to the inner circle at point D because it is given that the chord touches the inner circle.
The radius of inner circle OD = 6 cm and the radius of outer circle OB = 10 cm
In ΔOAB
⇒ OA = OB …radius of outer circle
Hence ΔOAB is isosceles triangle
As radius is perpendicular to tangent OC is perpendicular to AB
OC is altitude from apex and in isosceles triangle the altitude is also the median
Hence AD = DB
Hence AB = 2DB
Consider ΔODB
⇒ ∠ODB = 90° …radius perpendicular to tangent
Using Pythagoras theorem
⇒ OD2 + BD2 = OB2
⇒ 62 + BD2 = 102
⇒ 36 + BD2 = 100
⇒ BD2 = 100 – 36
⇒ BD2 = 64
⇒ BD = ±8
As length cannot be negative
⇒ BD = 8 cm
⇒ AB = 2 × 8 …since AB = 2BD
⇒ AB = 16 cm
Sol :
i) Tangents drawn from external point are equal
AD and AF are tangents from point A
⇒ AD = AF = a
BF and BE are tangents from point B
⇒ BD = BE = b
CD and CE are tangents from point C
⇒ CF = CE = c
From figure
We have AC = AF + FC
⇒ 20 = a + c …(i)
Also, AB = AD + DB
⇒ 24 = a + b …(ii)
And CB = CE + EB
⇒ 16 = c + b …(iii)
Add (i), (ii) and (iii)
⇒ 20 + 24 + 16 = a + c + a + b + c + b
⇒ 60 = 2(a + b + c)
⇒ a + b + c = 30 …(iv)
Substitute (i) in (iv)
⇒ 20 + b = 30
⇒ b = 10
Substitute (ii) in (iv)
⇒ 24 + c = 30
⇒ c = 6
Substitute (iii) in (iv)
⇒ 16 + a = 30
⇒ a = 14
Hence AD = a = 14 cm, BE = b = 10 cm and CF = c = 6 cm
ii)
Tangents drawn from external point are equal
CF and CE are tangents from point C
⇒ CF = CE = c
From figure
AC = AF + FC
⇒ 11 = 4 + c
⇒ c = 7 cm
Hence EC = c = 7 cm
We have BC = BE + EC
⇒ BC = 3 + 7
⇒ BC = 10 cm
Hence BC is 10 cm
Question 10
r is the radius which is OR = r
Consider quadrilateral DROS
⇒ ∠RDS = 90° …given
⇒ ∠DRO = 90° …radius is perpendicular to the tangent
⇒ DR = DS …tangents drawn from the same point are equal
As the adjacent angles are 90° and adjacent sides are same hence DROS is a square
Hence OR = DR = r …(i)
As tangents drawn from the same point are equal
BQ and BP are tangents drawn from B
⇒ BQ = BP
⇒ BQ = 27 cm …BP is 27 cm given
From figure
⇒ BC = BQ + QC
⇒ 38 = 27 + QC …BC is 38 cm given
⇒ QC = 11 cm
CQ and CR are tangents drawn from C
⇒ CQ = CR …tangents from same point
⇒ CR = 11 cm
Again from figure
⇒ CD = CR + RD
⇒ 25 = 11 + r …CD is 25 given and RD = r from (i)
⇒ r = 14 cm
Hence r radius is 14 cm
Consider ΔPOA
OA = 6 cm …radius of the outer circle
PA = 10 cm …given
∠OAP = 90° …radius is perpendicular to the tangent
Hence ΔPOA is right-angled triangle
Using Pythagoras
⇒ OA2 + AP2 = OP2
⇒ 62 + 102 = OP2
⇒ 36 + 100 = OP2
⇒ OP2 = 136 …(i)
Consider ΔPBO
OB = 4 cm …radius of inner circle
∠OBP = 90° …radius is perpendicular to the tangent
Hence ΔPOB is right-angled triangle
Using Pythagoras
⇒ OB2 + BP2 = OP2
Using (i)
⇒ 42 + BP2 = 136
⇒ 16 + BP2 = 136
⇒ BP2 = 120
⇒ BP = 10.9 cm
Hence length of PB is 10.9 cm
Let the circle with centre O and chord PQ with tangents from point A as AP and AQ as shown
We have to prove that ∠APQ = ∠AQP
Consider ΔOPQ
⇒ OP = OQ …radius
Hence ΔOPQ is an isosceles triangle
⇒ ∠OPQ = ∠OQP …base angles of isosceles triangle …(a)
As radius OP is perpendicular to tangent AP at point of contact P
⇒ ∠APO = 90°
From figure ∠APO = ∠APQ + ∠OPQ
⇒ 90° = ∠APQ + ∠OPQ
⇒ ∠APQ = 90° - ∠OPQ …(i)
As radius OQ is perpendicular to tangent AQ at point of contact Q
⇒ ∠AQO = 90°
From figure ∠AQO = ∠APQ + ∠OPQ
⇒ 90° = ∠AQP + ∠OQP
⇒ ∠AQP = 90° - ∠OQP
Using (a)
⇒ ∠AQP = 90° - ∠OPQ …(ii)
Using (i) and (ii), we can say that
⇒ ∠APQ = ∠AQP
Hence proved
Hence, the tangents at the extremities of any chord of a circle make equal angles with the chord
Mark the touching points as P, Q, R and S as shown
As tangents from a point are of equal length we have
AQ = AR = a
BR = BS = b
CP = CS = c
DP = DQ = d
From figure
⇒ BC = BS + SC
⇒ 7 = b + c
⇒ b = 7 – c … BC is 7 cm given …(i)
Also,
⇒ DC = DP + PC
⇒ 4 = d + c
⇒ c = 4 – d … DC is 4 cm given …(ii)
And
⇒ AB = AR + RB
⇒ 6 = a + b … AB is 6 cm given
⇒ a = 6 – b
Using (i)
⇒ a = 6 – (7 – c)
Using (ii)
⇒ a = 6 – (7 – (4 – d))
⇒ a = 6 – (7 – 4 + d)
⇒ a = 6 – 7 + 4 – d
⇒ a + d = 3
⇒ AQ + QD = 3 …since AQ = a and QD = d
From figure AQ + QD = AD
⇒ AD = 3 cm
Hence AD is 3 cm
Sol :
i) From P we have tangents PA and PB
Hence PA = PB …tangents from same point are equal …(a)
Point C is on PA
From C we have tangents CA and CE
⇒ CA = CE …tangents from same point are equal …(i)
Point D is on PB
From D we have two tangents DE and DB
⇒ DE = DB … tangents from same point are equal …(ii)
Consider ΔPCD
⇒ perimeter of ΔPCD = PC + CD + PD
From figure CD = CE + ED
⇒ perimeter of ΔPCD = PC + CE + ED + PD
Using (i) and (ii)
⇒ perimeter of ΔPCD = PC + CA + DB + PD
From figure we have
PC + CA = PA and DB + PD = PB
⇒ perimeter of ΔPCD = PA + PB
Using (a)
⇒ perimeter of ΔPCD = PA + PA
⇒ perimeter of ΔPCD = 2(PA)
PA is 14 cm given
⇒ perimeter of ΔPCD = 2 × 14
⇒ perimeter of ΔPCD = 28 cm
ii) PA = 11 cm …given
using (a)
PB = 11 cm
From figure
⇒ PB = PD + DB
Using (ii)
⇒ PB = PD + DE
⇒ 11 = 7 + DE …PD is 7 cm given
⇒ DE = 5 cm
Hence DE = 5 cm
Let O be the centre of concentric circles with radius ‘r’ and ‘R’ (R>r) and AB be the chord which touches the inner circle at point D
We have to prove that AB has a fixed length
Consider ΔOAB
OA = OB …radius
Hence ΔOAB is an isosceles triangle
Radius OD is perpendicular to tangent AB at the point of contact D
Hence OD is the altitude, and we know that the altitude from the apex of the isosceles triangle is also the median
⇒ AD = BD …(a)
Now consider ΔODB
⇒ ∠ODB = 90° …radius is perpendicular to tangent
Using Pythagoras
⇒ OD2 + BD2 = OB2
The radius are OB = R and OD = r
⇒ r2 + BD2 = R2
⇒ BD2 = R2 - r2
⇒ BD = √(R2 - r2) …(i)
From figure
⇒ AB = AD + BD
Using (a)
⇒ AB = BD + BD
⇒ AB = 2BD
Using (i)
⇒ AB = 2√(R2 - r2)
Here observe that AB only depends on R and r which are fixed radius of inner circle and outer circle.
And as the radius of both the circle will not change however one may draw the chord the radius will always be fixed.
And hence AB won’t change AB is fixed length
Hence proved
Hence, in two concentric circles, all chords of the outer circle which touch the inner circle are of equal length.
Consider ΔODB
⇒ ∠ODB = 90° …radius perpendicular to tangent
Using Pythagoras theorem
⇒ OD2 + BD2 = OB2
$\Rightarrow \frac{\mathrm{d}_{2}^{2}}{2^{2}}+\frac{\mathrm{C}^{2}}{2^{2}}=\frac{\mathrm{d}_{1}^{2}}{2^{2}}$
Multiply the whole by 22
⇒ d22 + C2 = d12
Hence proved
Question 6
Prove that the line segment joining the point of contact of two parallel tangents to a circle is a diameter of the circle.
Sol :Let lines AP and BR are parallel tangents to circle having centre O
We have to prove that AB is the diameter
To prove AB as diameter, we have to prove that AB passes through O which means that points A, O and B are on the same line or collinear
OA is perpendicular to PA at A because the line from the centre is perpendicular to the tangent at the point of contact
PA || RB
Hence OA is also perpendicular to RB
⇒ OA perpendicular to PA and RB …(i)
Similarly, OB is perpendicular to RB at B because the line from the centre is perpendicular to the tangent at the point of contact
PA || RB
Hence OB is also perpendicular to PA
⇒ OB perpendicular to PA and RB …(ii)
From (i) and (ii) we can say that OA and OB can be same line or parallel lines, but we have a common point O which implies that OA and OB are same lines
Hence A, O, B lies on the same line, i.e. A, O and B are collinear
Thus AB passes through O
Hence AB is the diameter
Hence, the line segment joining the point of contact of two parallel tangents to a circle is a diameter of the circle.
Question 7
Prove that the perpendicular at the point of contact of the tangent to a circle passes through the centre.
Sol :Let there be a circle with centre O and BR as tangent with the point of contact as B
Let AB be the line perpendicular to BR
⇒ ∠ABR = 90° …(i)
As OB is the radius of the circle and we know that radius is perpendicular to the tangent at the point of contact
OB is perpendicular to BR
⇒ ∠OBR = 90° …(ii)
Equation (i) and (ii) implies that
⇒ ∠ABR = ∠OBR
This is only possible iff A and O lie on the same line or A and O are the same points
Case 1: Suppose A and O are on the same line
If A and O are on the same line, then the perpendicular AB to tangent BR has passed through the centre
Case 2: suppose A and O are the same points
As O itself is the centre of the circle, and A and O are the same points hence the perpendicular to the tangent at the point of contact passes through the circle
In any scenario, the line has to pass through the centre.
Hence, the perpendicular at the point of contact of the tangent to a circle passes through the centre
Question 8
Two concentric circles are of radii 10 cm, and 6 cm Find the length of the chord of the larger circle which touches the smaller circle.
Sol :Let the two concentric circles have the centre O and let AB be the chord of an outer circle whose length is D and which will also be tangent to the inner circle at point D because it is given that the chord touches the inner circle.
The radius of inner circle OD = 6 cm and the radius of outer circle OB = 10 cm
In ΔOAB
⇒ OA = OB …radius of outer circle
Hence ΔOAB is isosceles triangle
As radius is perpendicular to tangent OC is perpendicular to AB
OC is altitude from apex and in isosceles triangle the altitude is also the median
Hence AD = DB
Hence AB = 2DB
Consider ΔODB
⇒ ∠ODB = 90° …radius perpendicular to tangent
Using Pythagoras theorem
⇒ OD2 + BD2 = OB2
⇒ 62 + BD2 = 102
⇒ 36 + BD2 = 100
⇒ BD2 = 100 – 36
⇒ BD2 = 64
⇒ BD = ±8
As length cannot be negative
⇒ BD = 8 cm
⇒ AB = 2 × 8 …since AB = 2BD
⇒ AB = 16 cm
Question 9
(i) A circle is inscribed in a ΔABC having sides BC, CA and AB 16 cm, 20 cm and 24 cm respectively as shown in the figure Find AD, BE and CF.
(ii) If AF=4cm, BE=3cm, AC=11cm, then find BC.Sol :
i) Tangents drawn from external point are equal
AD and AF are tangents from point A
⇒ AD = AF = a
BF and BE are tangents from point B
⇒ BD = BE = b
CD and CE are tangents from point C
⇒ CF = CE = c
From figure
We have AC = AF + FC
⇒ 20 = a + c …(i)
Also, AB = AD + DB
⇒ 24 = a + b …(ii)
And CB = CE + EB
⇒ 16 = c + b …(iii)
Add (i), (ii) and (iii)
⇒ 20 + 24 + 16 = a + c + a + b + c + b
⇒ 60 = 2(a + b + c)
⇒ a + b + c = 30 …(iv)
Substitute (i) in (iv)
⇒ 20 + b = 30
⇒ b = 10
Substitute (ii) in (iv)
⇒ 24 + c = 30
⇒ c = 6
Substitute (iii) in (iv)
⇒ 16 + a = 30
⇒ a = 14
Hence AD = a = 14 cm, BE = b = 10 cm and CF = c = 6 cm
ii)
Tangents drawn from external point are equal
CF and CE are tangents from point C
⇒ CF = CE = c
From figure
AC = AF + FC
⇒ 11 = 4 + c
⇒ c = 7 cm
Hence EC = c = 7 cm
We have BC = BE + EC
⇒ BC = 3 + 7
⇒ BC = 10 cm
Hence BC is 10 cm
Question 10
In the given figure, ABCD is a quadrilateral in which ∠D=90°. A circle C (O,r) touches the sides AB, BC, CD and DA at P,Q,R,S respectively, If BC =38 cm, CD=25 cm and BP=27 cm, find the value of r.
Sol :r is the radius which is OR = r
Consider quadrilateral DROS
⇒ ∠RDS = 90° …given
⇒ ∠DRO = 90° …radius is perpendicular to the tangent
⇒ DR = DS …tangents drawn from the same point are equal
As the adjacent angles are 90° and adjacent sides are same hence DROS is a square
Hence OR = DR = r …(i)
As tangents drawn from the same point are equal
BQ and BP are tangents drawn from B
⇒ BQ = BP
⇒ BQ = 27 cm …BP is 27 cm given
From figure
⇒ BC = BQ + QC
⇒ 38 = 27 + QC …BC is 38 cm given
⇒ QC = 11 cm
CQ and CR are tangents drawn from C
⇒ CQ = CR …tangents from same point
⇒ CR = 11 cm
Again from figure
⇒ CD = CR + RD
⇒ 25 = 11 + r …CD is 25 given and RD = r from (i)
⇒ r = 14 cm
Hence r radius is 14 cm
Question 11
In the given figure, O is the centre of two concentric circles of radii 4 cm and 6cm respectively. PA and PB are tangents to the outer and inner circle respectively. If PA=10cm, find the length of PB up to one place of decimal.
Sol :Consider ΔPOA
OA = 6 cm …radius of the outer circle
PA = 10 cm …given
∠OAP = 90° …radius is perpendicular to the tangent
Hence ΔPOA is right-angled triangle
Using Pythagoras
⇒ OA2 + AP2 = OP2
⇒ 62 + 102 = OP2
⇒ 36 + 100 = OP2
⇒ OP2 = 136 …(i)
Consider ΔPBO
OB = 4 cm …radius of inner circle
∠OBP = 90° …radius is perpendicular to the tangent
Hence ΔPOB is right-angled triangle
Using Pythagoras
⇒ OB2 + BP2 = OP2
Using (i)
⇒ 42 + BP2 = 136
⇒ 16 + BP2 = 136
⇒ BP2 = 120
⇒ BP = 10.9 cm
Hence length of PB is 10.9 cm
Question 12
Show that the tangents at the extremities of any chord of a circle make equal angles with the chord.
Sol :Let the circle with centre O and chord PQ with tangents from point A as AP and AQ as shown
We have to prove that ∠APQ = ∠AQP
Consider ΔOPQ
⇒ OP = OQ …radius
Hence ΔOPQ is an isosceles triangle
⇒ ∠OPQ = ∠OQP …base angles of isosceles triangle …(a)
As radius OP is perpendicular to tangent AP at point of contact P
⇒ ∠APO = 90°
From figure ∠APO = ∠APQ + ∠OPQ
⇒ 90° = ∠APQ + ∠OPQ
⇒ ∠APQ = 90° - ∠OPQ …(i)
As radius OQ is perpendicular to tangent AQ at point of contact Q
⇒ ∠AQO = 90°
From figure ∠AQO = ∠APQ + ∠OPQ
⇒ 90° = ∠AQP + ∠OQP
⇒ ∠AQP = 90° - ∠OQP
Using (a)
⇒ ∠AQP = 90° - ∠OPQ …(ii)
Using (i) and (ii), we can say that
⇒ ∠APQ = ∠AQP
Hence proved
Hence, the tangents at the extremities of any chord of a circle make equal angles with the chord
Question 13
In the given figure, a circle touches all the four sides of a quadrilateral ABCD whose three sides are AB = 6cm, BC = 7cm, and CD = 4cm. Find AD.
Sol :Mark the touching points as P, Q, R and S as shown
As tangents from a point are of equal length we have
AQ = AR = a
BR = BS = b
CP = CS = c
DP = DQ = d
From figure
⇒ BC = BS + SC
⇒ 7 = b + c
⇒ b = 7 – c … BC is 7 cm given …(i)
Also,
⇒ DC = DP + PC
⇒ 4 = d + c
⇒ c = 4 – d … DC is 4 cm given …(ii)
And
⇒ AB = AR + RB
⇒ 6 = a + b … AB is 6 cm given
⇒ a = 6 – b
Using (i)
⇒ a = 6 – (7 – c)
Using (ii)
⇒ a = 6 – (7 – (4 – d))
⇒ a = 6 – (7 – 4 + d)
⇒ a = 6 – 7 + 4 – d
⇒ a + d = 3
⇒ AQ + QD = 3 …since AQ = a and QD = d
From figure AQ + QD = AD
⇒ AD = 3 cm
Hence AD is 3 cm
Question 14
(i) 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 the point E and PA = 14 cm, find the perimeter of 
PCD.
(ii) If PA = 11cm, PD = 7 cm, then DE = ?PCD.Sol :
i) From P we have tangents PA and PB
Hence PA = PB …tangents from same point are equal …(a)
Point C is on PA
From C we have tangents CA and CE
⇒ CA = CE …tangents from same point are equal …(i)
Point D is on PB
From D we have two tangents DE and DB
⇒ DE = DB … tangents from same point are equal …(ii)
Consider ΔPCD
⇒ perimeter of ΔPCD = PC + CD + PD
From figure CD = CE + ED
⇒ perimeter of ΔPCD = PC + CE + ED + PD
Using (i) and (ii)
⇒ perimeter of ΔPCD = PC + CA + DB + PD
From figure we have
PC + CA = PA and DB + PD = PB
⇒ perimeter of ΔPCD = PA + PB
Using (a)
⇒ perimeter of ΔPCD = PA + PA
⇒ perimeter of ΔPCD = 2(PA)
PA is 14 cm given
⇒ perimeter of ΔPCD = 2 × 14
⇒ perimeter of ΔPCD = 28 cm
ii) PA = 11 cm …given
using (a)
PB = 11 cm
From figure
⇒ PB = PD + DB
Using (ii)
⇒ PB = PD + DE
⇒ 11 = 7 + DE …PD is 7 cm given
⇒ DE = 5 cm
Hence DE = 5 cm
Question 15
In two concentric circles, prove that all chords of the outer circle which touch the inner arc of equal length.
Sol :Let O be the centre of concentric circles with radius ‘r’ and ‘R’ (R>r) and AB be the chord which touches the inner circle at point D
We have to prove that AB has a fixed length
Consider ΔOAB
OA = OB …radius
Hence ΔOAB is an isosceles triangle
Radius OD is perpendicular to tangent AB at the point of contact D
Hence OD is the altitude, and we know that the altitude from the apex of the isosceles triangle is also the median
⇒ AD = BD …(a)
Now consider ΔODB
⇒ ∠ODB = 90° …radius is perpendicular to tangent
Using Pythagoras
⇒ OD2 + BD2 = OB2
The radius are OB = R and OD = r
⇒ r2 + BD2 = R2
⇒ BD2 = R2 - r2
⇒ BD = √(R2 - r2) …(i)
From figure
⇒ AB = AD + BD
Using (a)
⇒ AB = BD + BD
⇒ AB = 2BD
Using (i)
⇒ AB = 2√(R2 - r2)
Here observe that AB only depends on R and r which are fixed radius of inner circle and outer circle.
And as the radius of both the circle will not change however one may draw the chord the radius will always be fixed.
And hence AB won’t change AB is fixed length
Hence proved
Hence, in two concentric circles, all chords of the outer circle which touch the inner circle are of equal length.
| S.no | Chapters | Links |
|---|---|---|
| 1 | Real numbers | Exercise 1.1 Exercise 1.2 Exercise 1.3 Exercise 1.4 |
| 2 | Polynomials | Exercise 2.1 Exercise 2.2 Exercise 2.3 |
| 3 | Pairs of Linear Equations in Two Variables | Exercise 3.1 Exercise 3.2 Exercise 3.3 Exercise 3.4 Exercise 3.5 |
| 4 | Trigonometric Ratios and Identities | Exercise 4.1 Exercise 4.2 Exercise 4.3 Exercise 4.4 |
| 5 | Triangles | Exercise 5.1 Exercise 5.2 Exercise 5.3 Exercise 5.4 Exercise 5.5 |
| 6 | Statistics | Exercise 6.1 Exercise 6.2 Exercise 6.3 Exercise 6.4 |
| 7 | Quadratic Equations | Exercise 7.1 Exercise 7.2 Exercise 7.3 Exercise 7.4 Exercise 7.5 |
| 8 | Arithmetic Progressions (AP) | Exercise 8.1 Exercise 8.2 Exercise 8.3 Exercise 8.4 |
| 9 | Some Applications of Trigonometry: Height and Distances | Exercise 9.1 |
| 10 | Coordinates Geometry | Exercise 10.1 Exercise 10.2 Exercise 10.3 Exercise 10.4 |
| 11 | Circles | Exercise 11.1 Exercise 11.2 |
| 12 | Constructions | Exercise 12.1 |
| 13 | Area related to Circles | Exercise 13.1 |
| 14 | Surface Area and Volumes | Exercise 14.1 Exercise 14.2 Exercise 14.3 Exercise 14.4 |
| 15 | Probability | Exercise 15.1 |
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