{"id":624561,"date":"2023-09-01T11:56:53","date_gmt":"2023-09-01T11:56:53","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=624561"},"modified":"2023-09-01T11:57:14","modified_gmt":"2023-09-01T11:57:14","slug":"kc-sinha-exercise-11-1-mathematics-solution-class-10-chapter-11-circles","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-11-1-mathematics-solution-class-10-chapter-11-circles\/","title":{"rendered":"KC Sinha: Exercise 11.1 – Mathematics Solution Class 10 Chapter 11 Circles"},"content":{"rendered":"\n\n\n\n\n

Question 1<\/h4>\n\n\n\n

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.<\/strong><\/p>\n\n\n\n

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 \u2220ABO = 90\u00b0
Consider \u0394ABO
Using Pythagoras theorem
\u21d2 AB2<\/sup> + OB2<\/sup> = AO2<\/sup>
\u21d2 42<\/sup> + OB2<\/sup> = 52<\/sup>
\u21d2 16 + OB2<\/sup> = 25
\u21d2 OB2<\/sup> = 25 \u2013 16
\u21d2 OB2<\/sup> = 9
\u21d2 OB = \u00b13
As length cannot be negative
\u21d2 OB = 3 cm
Hence length of radius is 3 cm<\/p>\n\n\n\n

\n\n\n\n

Question 2<\/h4>\n\n\n\n

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.<\/strong><\/p>\n\n\n\n

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 \u2220ABO = 90\u00b0
Consider \u0394ABO
Using Pythagoras theorem
\u21d2 AB2<\/sup> + OB2<\/sup> = AO2<\/sup>
\u21d2 AB2<\/sup> + 202<\/sup> = 292<\/sup>
\u21d2 AB2<\/sup> = 292<\/sup> – 202<\/sup>
Using a2<\/sup> \u2013 b2<\/sup> = (a + b)(a \u2013 b)
\u21d2 AB2<\/sup> = (29 \u2013 20)(29 + 20)
\u21d2 AB2<\/sup> = 9 \u00d7 49
\u21d2 AB = \u221a(9 \u00d7 49)
\u21d2 AB = \u00b1 (3 \u00d7 7)
\u21d2 AB = \u00b121
As length cannot be negative
\u21d2 AB = 21 m
Hence Rajesh has to cover 21 m to reach the flower bed along the tangential path.<\/p>\n\n\n\n

\n\n\n\n

Question 3<\/h4>\n\n\n\n

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.<\/strong><\/p>\n\n\n\n

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 \u2220ABO = 90\u00b0
Consider \u0394ABO
Using Pythagoras theorem
\u21d2 AB2<\/sup> + OB2<\/sup> = AO2<\/sup>
\u21d2 AB2<\/sup> + 32<\/sup> = 52<\/sup>
\u21d2 AB2<\/sup> + 9 = 25
\u21d2 AB2<\/sup> = 25 \u2013 9
\u21d2 AB2<\/sup> = 16
\u21d2 AB = \u00b14
As length cannot be negative
\u21d2 AB = 4 cm
Hence length of tangent is 4 cm<\/p>\n\n\n\n

\n\n\n\n

Question 4<\/h4>\n\n\n\n

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.<\/strong><\/p>\n\n\n\n

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 \u2220PBO = 90\u00b0
Consider \u0394PBO
Using Pythagoras theorem
\u21d2 PB2<\/sup> + OB2<\/sup> = PO2<\/sup>
\u21d2 122<\/sup> + OB2<\/sup> = 132<\/sup>
\u21d2 OB2<\/sup> = 132<\/sup> – 122<\/sup>
\u21d2 OB2<\/sup> = 169 \u2013 144
\u21d2 OB2<\/sup> = 25
\u21d2 OB = \u00b15
As length cannot be negative
\u21d2 OB = 5 cm
Hence length of radius is 5 cm<\/p>\n\n\n\n

\n\n\n\n

Question 5<\/h4>\n\n\n\n

If d1<\/sub>, d2<\/sub> (d2<\/sub><d1<\/sub>) are the diameters of two concentric circles and chord of one circle of length C is tangent to other circle, then prove that d2<\/sub>2<\/sup> = C2<\/sup> + d1<\/sub>2<\/sup>.<\/strong><\/p>\n\n\n\n

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<\/sub> and d2<\/sub> hence the radius will be d12 and d22
In \u0394OAB
\u21d2 OA = OB \u2026radius of the outer circle
Hence \u0394OAB 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<\/p>\n\n\n\n

Hence  AD=DB=C2
Consider \u0394ODB
\u21d2 \u2220ODB = 90\u00b0 \u2026radius perpendicular to tangent
Using Pythagoras theorem
\u21d2 OD2<\/sup> + BD2<\/sup> = OB2<\/sup>
\u21d2d2222+C222=d2122
Multiply the whole by 22<\/sup>
\u21d2 d2<\/sub>2<\/sup> + C2<\/sup> = d1<\/sub>2<\/sup>
Hence proved<\/p>\n\n\n\n

\n\n\n\n

Question 6<\/h4>\n\n\n\n

Prove that the line segment joining the point of contact of two parallel tangents to a circle is a diameter of the circle.<\/strong><\/p>\n\n\n\n

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
\u21d2 OA perpendicular to PA and RB \u2026(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
\u21d2 OB perpendicular to PA and RB \u2026(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.<\/p>\n\n\n\n

\n\n\n\n

Question 7<\/h4>\n\n\n\n

Prove that the perpendicular at the point of contact of the tangent to a circle passes through the centre.<\/strong><\/p>\n\n\n\n

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
\u21d2 \u2220ABR = 90\u00b0 \u2026(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
\u21d2 \u2220OBR = 90\u00b0 \u2026(ii)
Equation (i) and (ii) implies that
\u21d2 \u2220ABR = \u2220OBR
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.<\/p>\n\n\n\n

Hence, the perpendicular at the point of contact of the tangent to a circle passes through the centre<\/p>\n\n\n\n

\n\n\n\n

Question 8<\/h4>\n\n\n\n

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.<\/strong><\/p>\n\n\n\n

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 \u0394OAB
\u21d2 OA = OB \u2026radius of outer circle
Hence \u0394OAB 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 \u0394ODB
\u21d2 \u2220ODB = 90\u00b0 \u2026radius perpendicular to tangent
Using Pythagoras theorem
\u21d2 OD2<\/sup> + BD2<\/sup> = OB2<\/sup>
\u21d2 62<\/sup> + BD2<\/sup> = 102<\/sup>
\u21d2 36 + BD2<\/sup> = 100
\u21d2 BD2<\/sup> = 100 \u2013 36
\u21d2 BD2<\/sup> = 64
\u21d2 BD = \u00b18
As length cannot be negative
\u21d2 BD = 8 cm
\u21d2 AB = 2 \u00d7 8 \u2026since AB = 2BD
\u21d2 AB = 16 cm<\/p>\n\n\n\n

\n\n\n\n

Question 9<\/h4>\n\n\n\n

(i) A circle is inscribed in a \u0394ABC having sides BC, CA and AB 16 cm, 20 cm and 24 cm respectively as shown in the figure Find AD, BE and CF.<\/strong><\/p>\n\n\n\n

(ii) If AF=4cm, BE=3cm, AC=11cm, then find BC.<\/strong>
<\/strong>
Sol :
i) Tangents drawn from external point are equal
AD and AF are tangents from point A
\u21d2 AD = AF = a
BF and BE are tangents from point B
\u21d2 BD = BE = b
CD and CE are tangents from point C
\u21d2 CF = CE = c

From figure
We have AC = AF + FC
\u21d2 20 = a + c \u2026(i)
Also, AB = AD + DB
\u21d2 24 = a + b \u2026(ii)
And CB = CE + EB
\u21d2 16 = c + b \u2026(iii)
Add (i), (ii) and (iii)
\u21d2 20 + 24 + 16 = a + c + a + b + c + b
\u21d2 60 = 2(a + b + c)
\u21d2 a + b + c = 30 \u2026(iv)
Substitute (i) in (iv)
\u21d2 20 + b = 30
\u21d2 b = 10
Substitute (ii) in (iv)
\u21d2 24 + c = 30
\u21d2 c = 6
Substitute (iii) in (iv)
\u21d2 16 + a = 30
\u21d2 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
\u21d2 CF = CE = c
From figure
AC = AF + FC
\u21d2 11 = 4 + c
\u21d2 c = 7 cm
Hence EC = c = 7 cm
We have BC = BE + EC
\u21d2 BC = 3 + 7
\u21d2 BC = 10 cm
Hence BC is 10 cm<\/p>\n\n\n\n

\n\n\n\n

Question 10<\/p>\n\n\n\n

In the given figure, ABCD is a quadrilateral in which \u2220D=90\u00b0. 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.<\/strong>
<\/strong><\/p>\n\n\n\n

Sol :
r is the radius which is OR = r
Consider quadrilateral DROS
\u21d2 \u2220RDS = 90\u00b0 \u2026given
\u21d2 \u2220DRO = 90\u00b0 \u2026radius is perpendicular to the tangent
\u21d2 DR = DS \u2026tangents drawn from the same point are equal<\/p>\n\n\n\n

As the adjacent angles are 90\u00b0 and adjacent sides are same hence DROS is a square
Hence OR = DR = r \u2026(i)
As tangents drawn from the same point are equal
BQ and BP are tangents drawn from B
\u21d2 BQ = BP
\u21d2 BQ = 27 cm \u2026BP is 27 cm given
From figure
\u21d2 BC = BQ + QC
\u21d2 38 = 27 + QC \u2026BC is 38 cm given
\u21d2 QC = 11 cm
CQ and CR are tangents drawn from C
\u21d2 CQ = CR \u2026tangents from same point
\u21d2 CR = 11 cm
Again from figure
\u21d2 CD = CR + RD
\u21d2 25 = 11 + r \u2026CD is 25 given and RD = r from (i)
\u21d2 r = 14 cm
Hence r radius is 14 cm<\/p>\n\n\n\n

\n\n\n\n

Question 11<\/h4>\n\n\n\n

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.<\/strong>
<\/strong><\/p>\n\n\n\n

Sol :
Consider \u0394POA
OA = 6 cm \u2026radius of the outer circle
PA = 10 cm \u2026given
\u2220OAP = 90\u00b0 \u2026radius is perpendicular to the tangent
Hence \u0394POA is right-angled triangle
Using Pythagoras
\u21d2 OA2<\/sup> + AP2<\/sup> = OP2<\/sup>
\u21d2 62<\/sup> + 102<\/sup> = OP2<\/sup>
\u21d2 36 + 100 = OP2<\/sup>
\u21d2 OP2<\/sup> = 136 \u2026(i)
Consider \u0394PBO
OB = 4 cm \u2026radius of inner circle
\u2220OBP = 90\u00b0 \u2026radius is perpendicular to the tangent
Hence \u0394POB is right-angled triangle
Using Pythagoras
\u21d2 OB2<\/sup> + BP2<\/sup> = OP2<\/sup>
Using (i)
\u21d2 42<\/sup> + BP2<\/sup> = 136
\u21d2 16 + BP2<\/sup> = 136
\u21d2 BP2<\/sup> = 120
\u21d2 BP = 10.9 cm
Hence length of PB is 10.9 cm<\/p>\n\n\n\n

\n\n\n\n

Question 12<\/h4>\n\n\n\n

Show that the tangents at the extremities of any chord of a circle make equal angles with the chord.<\/strong><\/p>\n\n\n\n

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 \u2220APQ = \u2220AQP
Consider \u0394OPQ
\u21d2 OP = OQ \u2026radius
Hence \u0394OPQ is an isosceles triangle
\u21d2 \u2220OPQ = \u2220OQP \u2026base angles of isosceles triangle \u2026(a)
As radius OP is perpendicular to tangent AP at point of contact P
\u21d2 \u2220APO = 90\u00b0
From figure \u2220APO = \u2220APQ + \u2220OPQ
\u21d2 90\u00b0 = \u2220APQ + \u2220OPQ
\u21d2 \u2220APQ = 90\u00b0 – \u2220OPQ \u2026(i)
As radius OQ is perpendicular to tangent AQ at point of contact Q
\u21d2 \u2220AQO = 90\u00b0
From figure \u2220AQO = \u2220APQ + \u2220OPQ
\u21d2 90\u00b0 = \u2220AQP + \u2220OQP
\u21d2 \u2220AQP = 90\u00b0 – \u2220OQP
Using (a)
\u21d2 \u2220AQP = 90\u00b0 – \u2220OPQ \u2026(ii)
Using (i) and (ii), we can say that
\u21d2 \u2220APQ = \u2220AQP
Hence proved
Hence, the tangents at the extremities of any chord of a circle make equal angles with the chord<\/p>\n\n\n\n

\n\n\n\n

Question 13<\/h4>\n\n\n\n

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.<\/strong>
<\/strong><\/p>\n\n\n\n

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
\u21d2 BC = BS + SC
\u21d2 7 = b + c
\u21d2 b = 7 \u2013 c \u2026 BC is 7 cm given \u2026(i)
Also,
\u21d2 DC = DP + PC
\u21d2 4 = d + c
\u21d2 c = 4 \u2013 d \u2026 DC is 4 cm given \u2026(ii)
And
\u21d2 AB = AR + RB
\u21d2 6 = a + b \u2026 AB is 6 cm given
\u21d2 a = 6 \u2013 b
Using (i)
\u21d2 a = 6 \u2013 (7 \u2013 c)
Using (ii)
\u21d2 a = 6 \u2013 (7 \u2013 (4 \u2013 d))
\u21d2 a = 6 \u2013 (7 \u2013 4 + d)
\u21d2 a = 6 \u2013 7 + 4 \u2013 d
\u21d2 a + d = 3
\u21d2 AQ + QD = 3 \u2026since AQ = a and QD = d
From figure AQ + QD = AD
\u21d2 AD = 3 cm
Hence AD is 3 cm<\/p>\n\n\n\n

\n\n\n\n

Question 14<\/h4>\n\n\n\n

(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.<\/strong><\/p>\n\n\n\n

(ii) If PA = 11cm, PD = 7 cm, then DE = ?<\/strong>
<\/strong>
Sol :
i) From P we have tangents PA and PB
Hence PA = PB \u2026tangents from same point are equal \u2026(a)
Point C is on PA
From C we have tangents CA and CE
\u21d2 CA = CE \u2026tangents from same point are equal \u2026(i)
Point D is on PB
From D we have two tangents DE and DB
\u21d2 DE = DB \u2026 tangents from same point are equal \u2026(ii)
Consider \u0394PCD
\u21d2 perimeter of \u0394PCD = PC + CD + PD
From figure CD = CE + ED
\u21d2 perimeter of \u0394PCD = PC + CE + ED + PD
Using (i) and (ii)
\u21d2 perimeter of \u0394PCD = PC + CA + DB + PD
From figure we have
PC + CA = PA and DB + PD = PB
\u21d2 perimeter of \u0394PCD = PA + PB
Using (a)
\u21d2 perimeter of \u0394PCD = PA + PA
\u21d2 perimeter of \u0394PCD = 2(PA)
PA is 14 cm given
\u21d2 perimeter of \u0394PCD = 2 \u00d7 14
\u21d2 perimeter of \u0394PCD = 28 cm
ii) PA = 11 cm \u2026given
using (a)
PB = 11 cm
From figure
\u21d2 PB = PD + DB
Using (ii)
\u21d2 PB = PD + DE
\u21d2 11 = 7 + DE \u2026PD is 7 cm given
\u21d2 DE = 5 cm
Hence DE = 5 cm<\/p>\n\n\n\n

\n\n\n\n

Question 15<\/h4>\n\n\n\n

In two concentric circles, prove that all chords of the outer circle which touch the inner arc of equal length.<\/strong><\/p>\n\n\n\n

Sol :
Let O be the centre of concentric circles with radius \u2018r\u2019 and \u2018R\u2019 (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 \u0394OAB
OA = OB \u2026radius
Hence \u0394OAB 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
\u21d2 AD = BD \u2026(a)
Now consider \u0394ODB
\u21d2 \u2220ODB = 90\u00b0 \u2026radius is perpendicular to tangent
Using Pythagoras
\u21d2 OD2<\/sup> + BD2<\/sup> = OB2<\/sup>
The radius are OB = R and OD = r
\u21d2 r2<\/sup> + BD2<\/sup> = R2<\/sup>
\u21d2 BD2<\/sup> = R2<\/sup> – r2<\/sup>
\u21d2 BD = \u221a(R2<\/sup> – r2<\/sup>) \u2026(i)
From figure
\u21d2 AB = AD + BD
Using (a)
\u21d2 AB = BD + BD
\u21d2 AB = 2BD
Using (i)
\u21d2 AB = 2\u221a(R2<\/sup> – r2<\/sup>)
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\u2019t 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.<\/p>\n\n\n\n

\n
KC Sinha Solutions<\/a><\/div>\n\n\n\n
KC Sinha Class 10 Solutions<\/a><\/div>\n<\/div>\n","protected":false},"excerpt":{"rendered":"

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 cmTangent is AB whose length is 4 cmOB is radius as shownNow we […]<\/p>\n","protected":false},"author":302,"featured_media":624584,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"newspack_featured_image_position":"","newspack_post_subtitle":"","newspack_article_summary_title":"Overview:","newspack_article_summary":"","newspack_hide_updated_date":false,"newspack_show_updated_date":false,"footnotes":""},"categories":[24],"tags":[],"boards":[],"class_list":["post-624561","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-class-10","entry"],"yoast_head":"\nKC Sinha: Exercise 11.1 - Mathematics Solution Class 10 Chapter 11 Circles - IndCareer Schools<\/title>\n<meta name=\"description\" content=\"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. 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