{"id":549644,"date":"2021-10-20T11:54:39","date_gmt":"2021-10-20T11:54:39","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=549644"},"modified":"2021-10-28T10:31:15","modified_gmt":"2021-10-28T10:31:15","slug":"rs-aggarwal-solutions-for-class-9-maths-chapter-11-circle","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/rs-aggarwal-solutions-for-class-9-maths-chapter-11-circle\/","title":{"rendered":"RS Aggarwal Solutions for Class 9 Maths Chapter 11\u2013Circle"},"content":{"rendered":"\n

Class 9: Maths Chapter 11 solutions. Complete Class 9 Maths Chapter 11 Notes.<\/p>\n\n\n\n

RS Aggarwal Solutions for Class 9 Maths Chapter 11\u2013Circle<\/h2>\n\n\n\n

RS Aggarwal 9th Maths Chapter 11, Class 9 Maths Chapter 11 solutions<\/p>\n\n\n\n

Ex 11A Solutions<\/h4>\n\n\n\n

Question 1.<\/strong>
Solution:<\/strong>
Let AB be a chord of a circle with centre O. OC\u22a5AB and OA be the radius of the circle, then
AB = 16cm, OA = 10cm .
<\/p>\n\n\n\n

\"RS<\/figure>\n\n\n\n


OC \u22a5 AB.
OC bisects AB at C
AC = 12 AB = 12 x 16 = 8cm
<\/p>\n\n\n\n

\"RS<\/figure>\n\n\n\n

Question 2.<\/strong>
Solution:<\/strong>
Let AB be the chord of the circle with centre O and OC \u22a5 AB, OA be the radius of the circle,
then OC = 3cm, OA = 5cm
<\/p>\n\n\n\n

\"RS<\/figure>\n\n\n\n


Now in right \u2206 OAC,
OA\u00b2 = AC\u00b2 = OC\u00b2 (Pythagoras Theorem)
<\/p>\n\n\n\n

\"RS<\/figure>\n\n\n\n

Question 3.<\/strong>
Solution:<\/strong>
Let AB be the chord, OA be the radius of
the circle OC \u22a5 AB
then AB = 30 cm, OC = 8cm
<\/p>\n\n\n\n

\"RS<\/figure>\n\n\n\n

Question 4.<\/strong>
Solution:<\/strong>
AB and CD are parallel chords of a circle with centre O.
<\/p>\n\n\n\n

\"RS<\/figure>\n\n\n\n
\"RS<\/figure>\n\n\n\n
\"RS<\/figure>\n\n\n\n

Question 5.<\/strong>
Solution:<\/strong>
Let AB and CD be two chords of a circle with centre O.
OA and OC are the radii of the circle. OL\u22a5AB and OM\u22a5CD.
<\/p>\n\n\n\n

\"RS<\/figure>\n\n\n\n\n\n

Question 6.<\/strong>
Solution:<\/strong>
In the figure, a circle with centre O, CD is its diameter AB is a chord such that CD\u22a5AB.
AB = 12cm, CE = 3cm.
Join OA.
\u2235 COD\u22a5AB which intersects AB at E
<\/p>\n\n\n\n

\"RS<\/figure>\n\n\n\n\n\n

Question 7.<\/strong>
Solution:<\/strong>
A circle with centre O, AB is diameter which bisects chord CD at E
i.e. CE = ED = 8cm and EB = 4cm
Join OC.
Let radius of the circle = r
<\/p>\n\n\n\n

\"RS<\/figure>\n\n\n\n
\"RS<\/figure>\n\n\n\n

Question 8.<\/strong>
Solution:<\/strong>
Given : O is the centre of a circle AB is a chord and BOC is the diameter. OD\u22a5AB
To prove : AC || OD and AC = 20D
Proof : OD\u22a5AB
\u2235 D is midpoint of AB
<\/p>\n\n\n\n

\"RS<\/figure>\n\n\n\n

Question 9.<\/strong>
Solution:<\/strong>
Given : O is the centre of the circle two
chords AB and CD intersect each other at P inside the circle. PO bisects \u2220BPD.
To prove : AB = CD.
<\/p>\n\n\n\n

\"RS<\/figure>\n\n\n\n
\"RS<\/figure>\n\n\n\n

Question 10.<\/strong>
Solution:<\/strong>
Given : PQ is the diameter of the circle with centre O which is perpendicular to one chord AB and chord AB || CD.
PQ intersects AB and CD at E and F respectively
To prove : PQ\u22a5CD and PQ bisects CD.
<\/p>\n\n\n\n

\"RS<\/figure>\n\n\n\n

Question 11.<\/strong>
Solution:<\/strong>
Two circles with centre O and O\u2019 intersect each other.
<\/p>\n\n\n\n

\"RS<\/figure>\n\n\n\n


To prove : The two circles cannot intersect each other at more than two points.
Proof : Let if opposite, the two circles intersect each other at three points P, Q and R.
Then these three points are non-collinear. But, we know that through three non- collinear points, one and only one circle can be drawn.
\u2235 Our supposition is wrong
Hence two circle can not intersect each other at not more than two points.
Hence proved<\/p>\n\n\n\n

Question 12.<\/strong>
Solution:<\/strong>
Given : Two circles with centres O and O\u2019 intersect each other at A and B. AB is a common chord. OO\u2019 is joined.
AO and AO is joined.
<\/p>\n\n\n\n

\"RS<\/figure>\n\n\n\n
\"RS<\/figure>\n\n\n\n

Question 13.<\/strong>
Solution:<\/strong>
Given : Two equal circles intersect each other at P and Q.
A straight line drawn through
P, is drawn which meets the circles at A and B respectively
To prove : QA = QB
<\/p>\n\n\n\n

\"RS<\/figure>\n\n\n\n

Question 14.<\/strong>
Solution:<\/strong>
Given : A circle with centre 0. AB and CD are two chords and diameter PQ bisects AB and CD at L and M
To Prove : AB || CD.
<\/p>\n\n\n\n

\"RS<\/figure>\n\n\n\n

Question 15.<\/strong>
Solution:<\/strong>
Given : Two circles with centres A and B touch each other at C internally. A, B arc joined. PQ is the perpendicular bisector of AB intersecting it at L and meeting the bigger circle at P and Q respectively and radii of the circles are 5cm and 3cm. i.e. AC = 5cm,BC = 3cm
Required : To find the lenght of PQ
<\/p>\n\n\n\n

\"RS<\/figure>\n\n\n\n
\"RS<\/figure>\n\n\n\n

Question 16.<\/strong>
Solution:<\/strong>
Given : AB is a chord of a circle with centre O. AB is produced to C such that BC = OB, CO is joined and produced to meet the circle at D.
\u2220 ACD = y\u00b0, \u2220 AOD = x\u00b0
To prove : x = 3y
<\/p>\n\n\n\n

\"RS<\/figure>\n\n\n\n
\"RS<\/figure>\n\n\n\n

Question 17.<\/strong>
Solution:<\/strong>
Given : O is the centre of a circle AB and AC are two chords such that AB = AC
OP\u22a5AB and OQ\u22a5AC.
which intersect AB and AC at M and N
respectively. PB and QC are joined.
To prove : PB = QC.
<\/p>\n\n\n\n

\"RS<\/figure>\n\n\n\n
\"RS<\/figure>\n\n\n\n

Question 18.<\/strong>
Solution:<\/strong>
Given : In a circle with centre O, BC is its diameter. AB and CD are two chords such that AB || CD.
To prove : AB = CD
Const. Draw OL\u22a5AB
OM\u22a5CD.
Proof : In \u2206 OLB and \u2206 OCM,
OB = OC (radii of the same circle)
\u2220 OLB = \u2220 OMC (each 90\u00b0)
\u2220 OBL = \u2220 OCM (alternate angles)
<\/p>\n\n\n\n

\"RS<\/figure>\n\n\n\n

Question 19.<\/strong>
Solution:<\/strong>
Equilateral \u2206 ABC in inscribed in a circle in which
AB = BC = CA = 9cm.
<\/p>\n\n\n\n

\"RS<\/figure>\n\n\n\n
\"RS<\/figure>\n\n\n\n
\"RS<\/figure>\n\n\n\n
\"RS<\/figure>\n\n\n\n

Question 20.<\/strong>
<\/p>\n\n\n\n

\"RS<\/figure>\n\n\n\n


Solution:<\/strong>
Given : AB and AC are two equal chords of a circle with centre O
To Prove : O lies on the bisector of \u2220 BAC
<\/p>\n\n\n\n

\"RS<\/figure>\n\n\n\n

Question 21.<\/strong>
Solution:<\/strong>
Given : OPQR is a square with centre O, a circle is drawn which intersects the square at X and Y.
To Prove : Q = QY
Const. Join OX and OY
<\/p>\n\n\n\n

\"RS<\/figure>\n\n\n\n

Ex 11B<\/h4>\n\n\n\n

Question 1.<\/strong>
Solution:<\/strong>
(i) O is the centre of the circle
\u2220OAB = 40\u00b0, \u2220OCB = 30\u00b0
Join OB.
<\/p>\n\n\n\n

\"RS<\/figure>\n\n\n\n
\"RS<\/figure>\n\n\n\n
\"RS<\/figure>\n\n\n\n
\"RS<\/figure>\n\n\n\n

Question 2.<\/strong>
Solution:<\/strong>
O is the centre of the cirlce and \u2220AOB = 70\u00b0
\u2235 Arc AB subtends \u2220AOB at the centre and \u2220ACB at the remaining part of the circle.
\u2235 \u2220ACB = 12 \u2220AOB = 12 x 70\u00b0
=> \u2220ACB = 35\u00b0
or \u2220OCA = 35\u00b0
In \u2206OAC,
OA = OC (radii of the same circle)
\u2234 \u2220OAC = \u2220OCA = 35\u00b0 Ans.<\/p>\n\n\n\n

Question 3.<\/strong>
Solution:<\/strong>
In the figure, O is the centre of the circle. \u2220PBC = 25\u00b0, \u2220APB =110\u00b0
\u2220 APB + \u2220 BPC = 180\u00b0 (Linear pair)
=> 110\u00b0 + \u2220 BPC = 180\u00b0
<\/p>\n\n\n\n

\"RS<\/figure>\n\n\n\n
\"RS<\/figure>\n\n\n\n

Question 4.<\/strong>
Solution:<\/strong>
O is the centre of the circle
\u2220ABD = 35\u00b0 and \u2220B AC = 70\u00b0
BOD is the diameter of the circle
\u2220BAD = 90\u00b0 (Angle in a semi circle)
But \u2220ADB + \u2220ABD + \u2220BAD = 180\u00b0 (Angles of a triangle)
=> \u2220ADB + 35\u00b0 + 90\u00b0 = 180\u00b0
=> \u2220ADB + 125\u00b0 = 180\u00b0
=> \u2220ADB = 180\u00b0 \u2013 125\u00b0 = 55\u00b0
But \u2220ACB = \u2220ADB (Angles in the same segment of the circle)
\u2220ACB = 55\u00b0 Ans.<\/p>\n\n\n\n

Question 5.<\/strong>
Solution:<\/strong>
O is the centre of a circle and \u2220ACB = 50\u00b0
\u2234 arc AB subtends \u2220 AOB at the centre and \u2220 ACB at the remaining part of the circle.
\u2234 \u2220 AOB = 2 \u2220 ACB
= 2 x 50\u00b0 = 100
\u2234 OA = OB (radii of the same circle)
\u2234 \u2220 OAB = \u2220 OBA (Angles opposite to equal sides)
Now in \u2206 OAB,
\u2220 OAB + \u2220 OBA + \u2220 AOB = 180\u00b0
=> \u2220 OAB + \u2220 OAB + \u2220 AOB = 180\u00b0 (\u2220OAB = \u2220OBA)
=> 2 \u2220 OAB + 100\u00b0= 180\u00b0
=> 2 \u2220 OAB = 180\u00b0 \u2013 100\u00b0 = 80\u00b0
=> \u2220OAB = 80o2 = 40\u00b0
Hence, OAB = 40\u00b0 Ans.<\/p>\n\n\n\n

Question 6.<\/strong>
Solution:<\/strong>
(i) In the figure,
\u2220ABD = 54\u00b0 and \u2220BCD = 43\u00b0
\u2220BAD = \u2220BCD (Angles in the same segment of a circle)
\u2220BAD = 43\u00b0
<\/p>\n\n\n\n

\"RS<\/figure>\n\n\n\n
\"RS<\/figure>\n\n\n\n

Question 7.<\/strong>
Solution:<\/strong>
Chord DE || diameter AC of the circle with centre O.
\u2220CBD = 60\u00b0
\u2220CBD = \u2220 CAD
(Angles in the same segment of a circle)
\u2220CAD = 60\u00b0
Now in \u2206 ADC,
<\/p>\n\n\n\n

\"RS<\/figure>\n\n\n\n

Question 8.<\/strong>
Solution:<\/strong>
In the figure,
chord CD || diameter AB of the circle with centre O.
\u2220 ABC = 25\u00b0
Join CD and DO.
AB || CD
\u2220 ABC = \u2220 BCD (alternate angles)
<\/p>\n\n\n\n

\"RS<\/figure>\n\n\n\n
\"RS<\/figure>\n\n\n\n

Question 9.<\/strong>
Solution:<\/strong>
AB and CD are two straight lines passing through O, the centre of the circle and \u2220AOC = 80\u00b0, \u2220CDE = 40\u00b0
\u2220 CED = 90\u00b0 (Angle in a semi circle)
<\/p>\n\n\n\n

\"RS<\/figure>\n\n\n\n
\"RS<\/figure>\n\n\n\n

Question 10.<\/strong>
Solution:<\/strong>
O is the centre of the circle and \u2220AOB = 40\u00b0, \u2220BDC = 100\u00b0
Arc AB subtends \u2220AOB at the centre and \u2220 ACB at the remaining part of the circle
\u2220 AOB = 2 \u2220 ACB
<\/p>\n\n\n\n

\"RS<\/figure>\n\n\n\n
\"RS<\/figure>\n\n\n\n

Question 11.<\/strong>
Solution:<\/strong>
Chords AC and BD of a circle with centre O, intersect each other at E at right angles.
\u2220 OAB = 25\u00b0. Join OB.
In \u2206 OAB,
OA = OB (radii of the same circle)
<\/p>\n\n\n\n

\"RS<\/figure>\n\n\n\n
\"RS<\/figure>\n\n\n\n

Question 12.<\/strong>
Solution:<\/strong>
In the figure, O is the centre of a circle \u2220 OAB = 20\u00b0 and \u2220 OCB = 55\u00b0 .
In \u2206 OAB,
<\/p>\n\n\n\n

\"RS<\/figure>\n\n\n\n
\"RS<\/figure>\n\n\n\n

Question 13.<\/strong>
Solution:<\/strong>
Given : A \u2206 ABC is inscribed in a circle with centre O and \u2220 BAC = 30\u00b0
To Prove : BC = radius of the circle
Const. Join OB and OC
<\/p>\n\n\n\n

\"RS<\/figure>\n\n\n\n
\"RS<\/figure>\n\n\n\n

Question 14.<\/strong>
Solution:<\/strong>
In a circle with centre O and PQ is its diameter. \u2220PQR = 65\u00b0, \u2220SPR = 40\u00b0 and \u2220PQM = 50\u00b0
(i) \u2220PRQ = 90\u00b0 (Angle in a semicircle) and \u2220PQR + \u2220RPQ + \u2220PQR = 180\u00b0 (Angles of a triangle)
<\/p>\n\n\n\n

\"RS<\/figure>\n\n\n\n
\"RS<\/figure>\n\n\n\n
\"RS<\/figure>\n\n\n\n

Ex 11C<\/h4>\n\n\n\n

Question 1.<\/strong>
Solution:<\/strong>
In cyclic quad. ABCD, \u2220 DBC = 60\u00b0 and \u2220BAC = 40\u00b0
\u2234\u2220 CAD and \u2220 CBD are in the same segment of the circle.
\u2234\u2220 CAD = \u2220 CBD or \u2220 DBC
=> \u2220 CAD = 60\u00b0
\u2234\u2220BAD = \u2220BAC + \u2220CAD
= 40\u00b0 + 60\u00b0 = 100\u00b0
But in cyclic quad. ABCD,
\u2220BAD + \u2220BCD = 180\u00b0
(Sum of opposite angles)
=> 100\u00b0 + \u2220BCD = 180\u00b0
=> \u2220BCD = 180\u00b0 \u2013 100\u00b0
\u2234 \u2220 BCD = 80\u00b0
Hence (i) \u2220BCD = 80\u00b0 and
(ii) \u2220CAD = 60\u00b0 Ans.<\/p>\n\n\n\n

Question 2.<\/strong>
Solution:<\/strong>
In the figure, POQ is diameter, PQRS is a cyclic quad, and \u2220 PSR =150\u00b0 In cyclic quad. PQRS.
\u2220 PSR + \u2220PQR = 180\u00b0
(Sum of opposite angles)
=> 150\u00b0 + \u2220PQR = 180\u00b0
=> \u2220PQR = 180\u00b0- 150\u00b0 = 30\u00b0
=> \u2220PQR =180\u00b0 \u2013 150\u00b0 = 30\u00b0
Now in \u2206 PQR,
\u2234\u2220 PRQ = 90\u00b0 (Angle in a semicircle)
\u2234\u2220 RPQ + \u2220PQR = 90\u00b0
=> \u2220RPQ + 30\u00b0 = 90\u00b0
=> \u2220RPQ = 90\u00b0 \u2013 30\u00b0 = 60\u00b0 Ans.<\/p>\n\n\n\n

Question 3.<\/strong>
Solution:<\/strong>
In cyclic quad. ABCD,
AB || DC and \u2220BAD = 100\u00b0
\u2220 ADC = \u2220BAD =180\u00b0
(co-interior angles)
<\/p>\n\n\n\n\n\n


=> \u2220 ADC + 100\u00b0 = 180\u00b0
=> \u2220ADC = 180\u00b0 \u2013 100\u00b0 = 80\u00b0
\u2234 ABCD is a cyclic quadrilateral.
\u2234 \u2220BAD + \u2220BCD = 180\u00b0
=> 100\u00b0 + \u2220 BCD = 180\u00b0
=> \u2220BCD = 180\u00b0 \u2013 100\u00b0
=> \u2220BCD = 80\u00b0
Similarly \u2220ABC + \u2220ADC = 180\u00b0
=> \u2220ABC + 80\u00b0 = 180\u00b0
=> \u2220ABC = 180\u00b0 \u2013 80\u00b0 = 100\u00b0
Hence (i) \u2220BCD = 80\u00b0 (ii) \u2220ADC = 80\u00b0 and (iii) \u2220ABC = 100\u00b0 Ans.<\/p>\n\n\n\n

Question 4.<\/strong>
Solution:<\/strong>
O is the centre of the circle and arc ABC subtends an angle of 130\u00b0 at the centre i.e. \u2220AOC = 130\u00b0. AB is produced to P
Reflex \u2220AOC = 360\u00b0 \u2013 130\u00b0 = 230\u00b0
Now, arc AC subtends reflex \u2220 AOC at the centre and \u2220 ABC at the remaining out of the circle.
<\/p>\n\n\n\n

\"RS<\/figure>\n\n\n\n

Question 5.<\/strong>
Solution:<\/strong>
In the figure, ABCD is a cyclic quadrilateral in which BA is produced to F and AE is drawn parallel to CD.
\u2220ABC = 92\u00b0 and \u2220FAE = 20\u00b0
ABCD is a cyclic quadrilateral.
<\/p>\n\n\n\n

\"RS<\/figure>\n\n\n\n
\"RS<\/figure>\n\n\n\n

Question 6.<\/strong>
Solution:<\/strong>
In the figure, BD = DC and \u2220CBD = 30\u00b0
In \u2206 BCD,
BD = DC (given)
\u2220 BCD = \u2220 CBD
(Angles opposite to equal sides)
= 30\u00b0
But \u2220BCD + \u2220CBD + \u2220BDC = 180\u00b0 (Angles of a triangle)
=> 30\u00b0+ 30\u00b0+ \u2220BDC = 180\u00b0
=> 60\u00b0+ \u2220BDC = 180\u00b0
=> \u2220 BDC =180\u00b0 \u2013 60\u00b0 = 120\u00b0
But ABDC is a cyclic quadrilateral
\u2220BAC + \u2220BDC = 180\u00b0
=> \u2220BAC + 120\u00b0= 180\u00b0
=> \u2220 BAC = 180\u00b0 \u2013 120\u00b0 = 60\u00b0
Hence \u2220 BAC = 60\u00b0 Ans.<\/p>\n\n\n\n

Question 7.<\/strong>
Solution:<\/strong>
(i) Arc ABC subtends \u2220 AOC at the centre , and \u2220 ADC at the remaining part of the circle.
\u2220 AOC = 2 \u2220 ADC
<\/p>\n\n\n\n

\"RS<\/figure>\n\n\n\n
\"RS<\/figure>\n\n\n\n

Question 8.<\/strong>
Solution:<\/strong>
In the figure, ABC is an equilateral triangle inscribed is a circle
Each angle is of 60\u00b0.
\u2220 BAC = \u2220 BDC
(Angles in the same segment)
\u2220BDC = 60\u00b0
BECD is a cyclic quadrilateral.
\u2220BDC + \u2220BEC = 180\u00b0
(opposite angles of cyclic quad.)
=> 60\u00b0+ \u2220BEC = 180\u00b0
=> \u2220BEC = 180\u00b0 \u2013 60\u00b0= 120\u00b0
Hence \u2220BDC = 60\u00b0 and \u2220BEC = 120\u00b0 Ans.<\/p>\n\n\n\n

Question 9.<\/strong>
Solution:<\/strong>
ABCD is a cyclic quadrilateral.
\u2220BCD + \u2220BAD = 180\u00b0
(opposite angles of a cyclic quad.)
=> 100\u00b0+ \u2220BAD = 180\u00b0
so \u2220BAD = 180\u00b0 \u2013 100\u00b0 = 80\u00b0
Now in \u2206 ABD,
\u2220BAD + \u2220ABD + \u2220ADB = 180\u00b0 (Angles of a triangle)
<\/p>\n\n\n\n

\"RS<\/figure>\n\n\n\n


=> 80\u00b0 + 50\u00b0 + \u2220ADB = 180\u00b0
=> 130\u00b0+ \u2220ADB = 180\u00b0
=> \u2220ADB = 180\u00b0 \u2013 130\u00b0 = 50\u00b0
Hence, \u2220ADB = 50\u00b0 Ans.<\/p>\n\n\n\n

Question 10.<\/strong>
Solution:<\/strong>
Arc BAD subtends \u2220 BOD at the centre and \u2220BCD at the remaining part of the circle.
<\/p>\n\n\n\n

\"RS<\/figure>\n\n\n\n

Question 11.<\/strong>
Solution:<\/strong>
In \u2206 OAB,
OA = OB (radii of the same circle)
\u2220OAB = \u2220OBA = 50\u00b0
and Ext \u2220BOD = \u2220OAB + \u2220OBA
=>x\u00b0 = 50\u00b0 + 50\u00b0 \u2013 100\u00b0
ABCD is a cyclic quadrilateral
\u2220BAD + \u2220BCD = 180\u00b0
(opposite angles of a cyclic quad.)
=> 50\u00b0+ y\u00b0 = 180\u00b0
=> y\u00b0 = 180\u00b0 \u2013 50\u00b0 = 130\u00b0
Hence x = 100\u00b0 and y = 130\u00b0 Ans.<\/p>\n\n\n\n

Question 12.<\/strong>
Solution:<\/strong>
Sides AD and AB of cyclic quadrilateral ABCD are produced to E and F respectively.
\u2220CBF = 130\u00b0, \u2220CDE = x.
\u2220CBF + \u2220CBA = 180\u00b0 (Linear pair)
=> 130\u00b0+ \u2220CBA = 180\u00b0
=> \u2220CBA = 180\u00b0 \u2013 130\u00b0 = 50\u00b0
But Ext. \u2220 CDE = Interior opp. \u2220 CBA (In cyclic quad. ABCD)
=> x = 50\u00b0 Ans.<\/p>\n\n\n\n

Question 13.<\/strong>
Solution:<\/strong>
In a circle with centre O AB is its diameter and DO || CB is drawn. \u2220BCD = 120\u00b0
To Find : (i) \u2220BAD (ii) ABD
(iii) \u2220CBD (iv) \u2220ADC
(v) Show that \u2206 AOD is an equilateral triangle.
(i) ABCD is a cyclic quadrilateral.
\u2220BCD + \u2220BAD = 180\u00b0
120\u00b0 + \u2220BAD = 180\u00b0
<\/p>\n\n\n\n

\"RS<\/figure>\n\n\n\n
\"RS<\/figure>\n\n\n\n

Question 14.<\/strong>
Solution:<\/strong>
AB = 6cm, BP = 2cm, DP = 2.5cm
Let CD = xcm
Two chords AB and CD
<\/p>\n\n\n\n

\"RS<\/figure>\n\n\n\n

Question 15.<\/strong>
Solution:<\/strong>
O is the centre of the circle
\u2220 AOD = 140\u00b0 and \u2220CAB = 50\u00b0
BD is joined.
(i) ABDC is a cyclic quadrilateral.
<\/p>\n\n\n\n

\"RS<\/figure>\n\n\n\n
\"RS<\/figure>\n\n\n\n

Question 16.<\/strong>
Solution:<\/strong>
Given : ABCD is a cyclic quadrilateral whose sides AB and DC are produced to meet each other at E.
To Prove : \u2206 EBC ~ \u2206 EDA
Proof : In \u2206 EBC and \u2206 EDA
\u2220 E = \u2220 E (common)
\u2220ECB = \u2220EAD
{Exterior angle of a cyclic quad, is equal to its interior opposite angle}
and \u2220 EBC = \u2220EDA
\u2206 EBC ~ \u2206 EDA (AAS axiom)
Hence proved<\/p>\n\n\n\n

Question 17.<\/strong>
Solution:<\/strong>
Solution Given : In an isosceles \u2206 ABC, AB = AC
A circle is drawn x in such a way that it passes through B and C and intersects AB and AC at D and E respectively.
DE is joined.
To Prove : DE || BC
Proof : In \u2206 ABC,
AB = AC (given)
\u2220 B = \u2220 C (angles opposite to equal sides)
But \u2220 ADE = \u2220 C (Ext. angle of a cyclic quad, is equal E to its interior opposite angle)
\u2220ADE = \u2220B
But, these are corresponding angles
DE || BC.
Hence proved.<\/p>\n\n\n\n

Question 18.<\/strong>
Solution:<\/strong>
Given : \u2206 ABC is an isosceles triangle in which AB = AC.
D and E are midpoints of AB and AC respectively.
DE is joined.
To Prove : D, B, C, E are concyclic.
Proof: D and E are midpoints of sides AB and AC respectively.
DE || BC
In \u2206 ABC, AB = AC
\u2220B = \u2220C
But \u2220 ADE = \u2220 B (alternate angles)
\u2220ADE =\u2220C
But \u2220ADE is exterior angle of quad. DBCE which is equal to its interior opposite angle C.
DBCE is a cyclic quadrilateral.
Hence D, B, C, E are con cyclic.
Hence proved.<\/p>\n\n\n\n

Question 19.<\/strong>
Solution:<\/strong>
Given : ABCD is a cyclic quadrilateral whose perpendicular bisectors l, m, n, p of the side are drawn
<\/p>\n\n\n\n\n\n


To prove : l, m, n and p are concurrent.
Proof : The sides AB, BC, CD and DA are the chords of the circle passing through the vertices\u2019s of quad. A, B, C and D. and perpendicular bisectors of a chord always passes through the centre of the circle.
l,m, n and p which are the perpendicular bisectors of the sides of cyclic quadrilateral will pass through O, the same point Hence, l, m, n and p are concurrent.
Hence proved.<\/p>\n\n\n\n

Question 20.<\/strong>
Solution:<\/strong>
Given : ABCD is a rhombus and four circles are drawn on the sides AB, BC, CD and DA as diameters. Diagonal AC and BD intersect each other at O.
<\/p>\n\n\n\n

\"RS<\/figure>\n\n\n\n\n\n

Question 21.<\/strong><\/p>\n\n\n\n

Solution:<\/strong>
Given: ABCD is a rectangle whose diagonals AC and BD intersect each other at O.
To prove : O is the centre of the circle passing through A, B, C and D
<\/p>\n\n\n\n

\"RS<\/figure>\n\n\n\n

Question 22.<\/strong>
Solution:<\/strong>
Construction.
(i) Let A, B and C are three points
(ii) With A as centre and BC as radius draw an arc
(iii) With centre C, and radius AB, draw another arc which intersects the first arc at D.
D is the required point.
Join BD and CD, AC and BA and CB
<\/p>\n\n\n\n

\"RS<\/figure>\n\n\n\n

BC = BC (common)
AC = BD (const.)
AB = DC
\u2234 \u2206 ABC \u2245 \u2206 DBC (SSS axiom)
\u2234 \u2220BAC \u2245 \u2220BDC (c.p.c.t.)
But these are angles on the same sides of BC
Hence these are angles in the same segment of a circle
A, B, C, D are concyclic Hence D lies on the circle passing througtvA, B and C.
Hence proved.<\/p>\n\n\n\n

Question 23.<\/strong>
Solution:<\/strong>
Given : ABCD is a cylic quadrilateral (\u2220B \u2013 \u2220D) = 60\u00b0
To prove : The small angle of the quad, is 60\u00b0
<\/p>\n\n\n\n

\"RS<\/figure>\n\n\n\n
\"RS<\/figure>\n\n\n\n

Question 24.<\/strong>
<\/p>\n\n\n\n

\"RS<\/figure>\n\n\n\n


Solution:<\/strong>
Given : ABCD is a quadrilateral in which AD = BC and \u2220 ADC = \u2220BCD
To prove : A, B, C and D lie on a circle
<\/p>\n\n\n\n

\"RS<\/figure>\n\n\n\n
\"RS<\/figure>\n\n\n\n

Question 25.<\/strong>
Solution:<\/strong>
Given : In the figure, two circles intersect each other at D and C
\u2220BAD = 75\u00b0, \u2220DCF = x\u00b0 and \u2220DEF = y\u00b0
<\/p>\n\n\n\n

\"RS<\/figure>\n\n\n\n

Question 26.<\/strong>
Solution:<\/strong>
Given : ABCD is a cyclic quadrilateral whose diagonals AC and BD intersect at O at right angle.
<\/p>\n\n\n\n

\"RS<\/figure>\n\n\n\n
\"RS<\/figure>\n\n\n\n
\"RS<\/figure>\n\n\n\n

Question 27.<\/strong>
Solution:<\/strong>
In a circle, two chords AB and CD intersect each other at E when produced.
AD and BC are joined.
<\/p>\n\n\n\n

\"RS<\/figure>\n\n\n\n

Question 28.<\/strong>
Solution:<\/strong>
Given : Two parallel chords AB and CD of a circle BD and AC are joined and produced to meet at E.
<\/p>\n\n\n\n

\"RS<\/figure>\n\n\n\n

Question 29.<\/strong>
Solution:<\/strong>
Given : In a circle with centre O, AB is its diameter. ADE and CBE are lines meeting at E such that \u2220BAD = 35\u00b0 and \u2220BED = 25\u00b0.
To Find : (i) \u2220DBC (ii) \u2220DCB (iii) \u2220BDC
Solution. Join BD and AC,
<\/p>\n\n\n\n

\"RS<\/figure>\n\n\n\n
\"RS<\/figure>\n\n\n\n

RS Aggarwal Solutions for Class 9 Maths Chapter 11: Download PDF<\/strong><\/h2>\n\n\n\n

RS Aggarwal Solutions for Class 9 Maths Chapter 11\u2013Circle<\/p>\n\n\n\n

Download PDF<\/strong>: RS Aggarwal Solutions for Class 9 Maths Chapter 11\u2013Circle PDF<\/a><\/p>\n\n\n\n

Chapterwise RS Aggarwal Solutions for Class 9 Maths :<\/strong><\/h2>\n\n\n\n