NCERT 9th Maths Chapter 10, class 9 Maths Chapter 10 solutions<\/p>\n\n\n\n
Page No: 171<\/p>\n\n\n\n
Exercise 10.1<\/strong><\/p>\n\n\n\n 1. Fill in the blanks:<\/strong> Answer<\/strong><\/p>\n\n\n\n (i) The centre of a circle lies in interior<\/p>\n\n\n\n of the circle. (exterior\/interior) of the circle. (exterior\/interior) of the circle. when its ends are the ends of a diameter. of the circle. parts.<\/p>\n\n\n\n 2. Write True or False: Give reasons for your answers.<\/strong> Answer<\/strong><\/p>\n\n\n\n (i) True. Exercise 10.2<\/strong><\/p>\n\n\n\n NCERT 9th Maths Chapter 10, class 9 Maths Chapter 10 solutions<\/p>\n\n\n\n 1. Recall that two circles are congruent if they have the same radii. Prove that equal chords of congruent circles subtend equal angles at their centres.<\/strong><\/p>\n\n\n\n Answer<\/strong><\/p>\n\n\n\n A circle is a collection of points whose every every point is equidistant from the centre. Thus, two circles can only be congruent when they the distance of every point of the both circle is equal from the centre.<\/p>\n\n\n\n Given, 2. Prove that if chords of congruent circles subtend equal angles at their centres, then the chords are equal.<\/strong><\/p>\n\n\n\n Answer<\/strong> Page No: 176<\/p>\n\n\n\n NCERT 9th Maths Chapter 10, class 9 Maths Chapter 10 solutions<\/p>\n\n\n\n Exercise 10.3<\/strong><\/p>\n\n\n\n 1. Draw different pairs of circles. How many points does each pair have in common? What is the maximum number of common points?<\/strong><\/p>\n\n\n\n Answer<\/strong><\/p>\n\n\n\n No point is common.<\/p>\n\n\n\n One point P is common.<\/p>\n\n\n\n One point P is common.<\/p>\n\n\n\n Two points P and Q are common.<\/p>\n\n\n\n No point is common.<\/p>\n\n\n\n 2. Suppose you are given a circle. Give a construction to find its centre.<\/strong><\/p>\n\n\n\n Answer<\/strong><\/p>\n\n\n\n Steps of construction: 3. If two circles intersect at two points, prove that their centres lie on the perpendicular bisector of the common chord.<\/strong><\/p>\n\n\n\n Answer<\/strong><\/p>\n\n\n\n Given, Page No: 179<\/p>\n\n\n\n NCERT 9th Maths Chapter 10, class 9 Maths Chapter 10 solutions<\/p>\n\n\n\n Exercise 10.4<\/strong><\/p>\n\n\n\n 1. Two circles of radii 5 cm and 3 cm intersect at two points and the distance between their centres is 4 cm. Find the length of the common chord.<\/strong><\/p>\n\n\n\n Answer<\/strong><\/p>\n\n\n\n OP = 5cm, PS = 3cm and OS = 4cm. In \u0394POR, In \u0394PRS, 2. If two equal chords of a circle intersect within the circle, prove that the segments of one chord are equal to corresponding segments of the other chord.<\/strong><\/p>\n\n\n\n Answer<\/strong><\/p>\n\n\n\n Given, Proof, 3. If two equal chords of a circle intersect within the circle, prove that the line joining the point of intersection to the centre makes equal angles with the chords.<\/strong><\/p>\n\n\n\n Answer<\/strong><\/p>\n\n\n\n Given, In \u0394OEM and \u0394OEN, Answer<\/strong><\/p>\n\n\n\n OM \u22a5 AD is drawn from O. 5. Three girls Reshma, Salma and Mandip are playing a game by standing on a circle of radius 5m drawn in a park. Reshma throws a ball to Salma, Salma to Mandip, Mandip to Reshma. If the distance between Reshma and Salma and between Salma and Mandip is 6m each, what is the distance between Reshma and Mandip?<\/strong><\/p>\n\n\n\n Answer<\/strong><\/p>\n\n\n\n Let A, B and C represent the positions of Reshma, Salma and Mandip respectively. 6. A circular park of radius 20m is situated in a colony. Three boys Ankur, Syed and David are sitting at equal distance on its boundary each having a toy telephone in his hands to talk each other. Find the length of the string of each phone.<\/strong><\/p>\n\n\n\n Answer<\/strong><\/p>\n\n\n\n Let A, B and C represent the positions of Ankur, Syed and David respectively. All three boys at equal distances thus ABC is an equilateral triangle. Page No: 184<\/p>\n\n\n\n NCERT 9th Maths Chapter 10, class 9 Maths Chapter 10 solutions<\/p>\n\n\n\n Exercise 10.5<\/strong><\/p>\n\n\n\n 1. In Fig. 10.36, A,B and C are three points on a circle with centre O such that \u2220BOC = 30\u00b0 and \u2220AOB = 60\u00b0. If D is a point on the circle other than the arc ABC, find \u2220ADC.<\/strong><\/p>\n\n\n\n Answer<\/strong><\/p>\n\n\n\n Here, \u2220BOC Page No: 185<\/p>\n\n\n\n 2. A chord of a circle is equal to the radius of the circle. Find the angle subtended by the chord at a point on the minor arc and also at a point on the major arc.<\/strong><\/p>\n\n\n\n Answer<\/strong> \u2220ADB = 180\u00b0 (Opposite angles of cyclic quadrilateral) Thus, angle subtend by the chord at a point on the minor arc and also at a point on the major arc are 150\u00b0 and 30\u00b0 respectively.<\/p>\n\n\n\n 3. In Fig. 10.37, \u2220PQR = 100\u00b0, where P, Q and R are points on a circle with centre O. Find \u2220OPR. Reflex \u2220POR = 2 \u00d7 \u2220PQR = 2 \u00d7 100\u00b0 = 200\u00b0 \u2220ORP +<\/p>\n\n\n\n \u2220POR = 180\u00b0 (Sum of the angles in a triangle) \u2220OPR + 160\u00b0 = 180\u00b0 4. In Fig. 10.38, \u2220ABC = 69\u00b0, \u2220 ACB = 31\u00b0, find \u2220BDC.<\/strong><\/p>\n\n\n\n Answer<\/strong><\/p>\n\n\n\n \u2220BAC = \u2220BDC (Angles in the segment of the circle) \u2220ABC +<\/p>\n\n\n\n \u2220ACB = 180\u00b0 (Sum of the angles in a triangle) 5. In Fig. 10.39, A, B, C and D are four points on a circle. AC and BD intersect at a point E such that \u2220BEC = 130\u00b0 and \u2220ECD = 20\u00b0. Find \u2220 BAC.<\/strong><\/p>\n\n\n\n Answer<\/strong><\/p>\n\n\n\n \u2220BAC = \u2220CDE (Angles in the segment of the circle) 6. ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If \u2220DBC = 70\u00b0, \u2220BAC is 30\u00b0, find \u2220BCD. Further, if AB = BC, find \u2220ECD.<\/strong><\/p>\n\n\n\n Answer<\/strong> 7. If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle.<\/strong><\/p>\n\n\n\n Answer<\/strong><\/p>\n\n\n\n Let ABCD be a cyclic quadrilateral and its diagonal AC and BD are the diameters of the circle through the vertices of the quadrilateral.<\/p>\n\n\n\n \u2220ABC = \u2220BCD = \u2220CDA = \u2220DAB = 90\u00b0 (Angles in the semi-circle) NCERT 9th Maths Chapter 10, class 9 Maths Chapter 10 solutions<\/p>\n\n\n\n 8. If the non-parallel sides of a trapezium are equal, prove that it is cyclic.<\/strong><\/p>\n\n\n\n Answer<\/strong><\/p>\n\n\n\n Given, Proof: \u2220C = 180\u00b0 (sum of the co-interior angles) Page No: 186<\/p>\n\n\n\n 9. Two circles intersect at two points B and C. Through B, two line segments ABD and PBQ are drawn to intersect the circles at A, D and P, Q respectively (see Fig. 10.40). Prove that \u2220ACP = \u2220QCD.<\/strong><\/p>\n\n\n\n Answer<\/p>\n\n\n\n Chords AP and DQ are joined. NCERT 9th Maths Chapter 10, class 9 Maths Chapter 10 solutions<\/p>\n\n\n\n 10. If circles are drawn taking two sides of a triangle as diameters, prove that the point of intersection of these circles lie on the third side.<\/strong><\/p>\n\n\n\n Answer<\/strong><\/p>\n\n\n\n Given,<\/p>\n\n\n\n Two circles are drawn on the sides AB and AC of the triangle \u0394ABC as diameters. The circles intersected at D.<\/p>\n\n\n\n Construction,<\/p>\n\n\n\n AD is joined.<\/p>\n\n\n\n To prove,<\/p>\n\n\n\n D lies on BC. We have to prove that BDC is a straight line.<\/p>\n\n\n\n Proof: 11. ABC and ADC are two right triangles with common hypotenuse AC. Prove that \u2220CAD = \u2220CBD.<\/strong><\/p>\n\n\n\n Answer<\/strong><\/p>\n\n\n\n Given, 12. Prove that a cyclic parallelogram is a rectangle.<\/strong><\/p>\n\n\n\n Answer<\/strong><\/p>\n\n\n\n Given, Proof: \u22202 = 180\u00b0 (Opposite angles of a cyclic parallelogram) Chapter 10 Circles NCERT Solutions will help you in understanding about circles, other related terms and some properties of a circle. NCERT Textbooks for Class 9 are prepared in such a way that students can understand topics in an interesting way.<\/p>\n\n\n\n \u2022 Circles and Its Related Terms: A Review: The collection of all the points in a plane, which are at a fixed distance from a fixed point in the plane, is called a circle. The fixed point is called the centre of the circle and the fixed distance is called the radius of the circle. A piece of a circle between two points is called an arc. The length of the complete circle is called its circumference. The region between a chord and either of its arcs is called a segment of the circle. When two arcs are equal, that is, each is a semicircle, then both segments and both sectors become the same and each is known as a semicircular region.<\/p>\n\n\n\n \u2022 Angle Subtended by a Chord at a Point: Equal chords of a circle subtend equal angles at the centre. If the angles subtended by the chords of a circle at the centre are equal, then the chords are equal.<\/p>\n\n\n\n \u2022 Perpendicular from the Centre to a Chord: The perpendicular from the centre of a circle to a chord bisects the chord. The line drawn through the centre of a circle to bisect a chord is perpendicular to the chord.<\/p>\n\n\n\n \u2022 Circle through Three Points: There is one and only one circle passing through three given non-collinear points.<\/p>\n\n\n\n \u2022 Equal Chords and Their Distances from the Centre: The length of the perpendicular from a point to a line is the distance of the line from the point.<\/p>\n\n\n\n (i) Equal chords of a circle (or of congruent circles) are equidistant from the centre (or centres).<\/p>\n\n\n\n (ii) Chords equidistant from the centre of a circle are equal in length.<\/p>\n\n\n\n \u2022 Angle Subtended by an Arc of a Circle: If two chords of a circle are equal, then their corresponding arcs are congruent and conversely, if two arcs are congruent, then their corresponding chords are equal.<\/p>\n\n\n\n (i) The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.<\/p>\n\n\n\n (ii) Angles in the same segment of a circle are equal.<\/p>\n\n\n\n (iii) If a line segment joining two points subtends equal angles at two other points lying on the same side of the line containing the line segment, the four points lie on a circle (i.e. they are concyclic).<\/p>\n\n\n\n \u2022 Cyclic Quadrilaterals: A quadrilateral ABCD is called cyclic if all the four vertices of it lie on a circle. The sum of either pair of opposite angles of a cyclic quadrilateral is 180\u00ba. If the sum of a pair of opposite angles of a quadrilateral is 180\u00ba, the quadrilateral is cyclic.<\/p>\n\n\n\n There are total six exercises in the Chapter 10 Class 9 NCERT Solutions<\/strong> which can be used to complete homework and understand the concepts behind the questions. You can find exercisewise Class 9 Maths NCERT Solutions from the links given below.<\/p>\n\n\n\n These NCERT Solutions prepared by Indcareer Schools experts have taken every care so students can easily understand and clear their doubts. These solutions are prerequisites before going for supplementary Maths Books.<\/p>\n\n\n\n FAQ on <\/strong>Chapter <\/strong>10 Circles<\/strong> Chapter 10 Circles Class 9 NCERT Solutions for Maths contains total six exercises in which the last one if optional. You will get detailed and accurate NCERT Solutions of every question so you can always find them whenever any difficulty occur.<\/p>\n\n\n\n Since, the diameter of circle is double its radius.<\/p>\n\n\n\n \u2234 Diameter = 2 x Radius<\/p>\n\n\n\n \u21d2 (1\/2)x Diameter = Radius<\/p>\n\n\n\n \u21d2 Radius = (1\/2) x 3.8 cm<\/p>\n\n\n\n = 1\/2 x 38\/10 cm<\/p>\n\n\n\n = 19\/10 = 1.9 cm.<\/p>\n\n\n\n The length of the complete circle is called its circumference, whereas a piece of a circle between two points is called an arc. If the length of an arc is less than the semicircle, then it is a minor arc, otherwise, it is a major arc.<\/p>\n\n\n\n A chord, passing through the centre is called a diameter of the circle.<\/p>\n\n\n\n NCERT 9th Maths Chapter 10, class 9 Maths Chapter 10 solutions<\/p>\n\n\n\n NCERT Solutions for 9th Class Maths : Chapter 10 Circles<\/p>\n\n\n\n Download PDF<\/strong>: NCERT Solutions for 9th Class Maths : Chapter 10 Circles PDF<\/a><\/p>\n\n\n\n The National Council of Educational Research and Training is an autonomous organization of the Government of India which was established in 1961 as a literary, scientific, and charitable Society under the Societies Registration Act. Its headquarters are located at Sri Aurbindo Marg in New Delhi. Visit the Official NCERT website<\/a> to learn more. <\/p>\n\n\n\n
(i) The centre of a circle lies in ____________ of the circle. (exterior\/ interior)
(ii) A point, whose distance from the centre of a circle is greater than its radius lies in __________ of the circle. (exterior\/ interior)
(iii) The longest chord of a circle is a _____________ of the circle.
(iv) An arc is a ___________ when its ends are the ends of a diameter.
(v) Segment of a circle is the region between an arc and _____________ of the circle.
(vi) A circle divides the plane, on which it lies, in _____________ parts.<\/p>\n\n\n\n
(ii) A point, whose distance from the centre of a circle is greater than its radius lies in exterior<\/p>\n\n\n\n
(iii) The longest chord of a circle is a diameter<\/p>\n\n\n\n
(iv) An arc is a semicircle<\/p>\n\n\n\n
(v) Segment of a circle is the region between an arc and chord<\/p>\n\n\n\n
(vi) A circle divides the plane, on which it lies, in three<\/p>\n\n\n\n
(i) Line segment joining the centre to any point on the circle is a radius of the circle.
(ii) A circle has only finite number of equal chords.
(iii) If a circle is divided into three equal arcs, each is a major arc.
(iv) A chord of a circle, which is twice as long as its radius, is a diameter of the circle.
(v) Sector is the region between the chord and its corresponding arc.
(vi) A circle is a plane figure.<\/p>\n\n\n\n
All the line segment from the centre to the circle is of equal length.
(ii) False.
We can draw infinite numbers of equal chords.
(iii) False.
We get major and minor arcs for unequal arcs. So, for equal arcs on circle we can’t say it is major arc or minor arc.
(iv) True.
A chord which is twice as long as radius must pass through the centre of the circle and is diameter to the circle.
(v) False.
Sector is the region between the arc and the two radii of the circle.
(vi) True.
A circle can be drawn on the plane.<\/p>\n\n\n\n![\"NCERT](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/02\/class-9-maths-chapter-10-ncert-1.jpg\")
<\/figure>\n\n\n\n
AB = CD (Equal chords)
To prove,
\u2220AOB = \u2220COD
Proof,
In \u0394AOB and \u0394COD,
OA = OC (Radii)
OB = OD (Radii)
AB = CD (Given)
\u2234 \u0394AOB \u2245 \u0394COD (SSS congruence condition)
Thus, \u2220AOB = \u2220COD by CPCT.
Equal chords of congruent circles subtend equal angles at their centres.<\/p>\n\n\n\n![](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/02\/class-9-maths-chapter-10-ncert-1-1.jpg\")
Given,
\u2220AOB = \u2220COD (Equal angles)
To prove,
AB = CD
Proof,
In \u0394AOB and \u0394COD,
OA = OC (Radii)
\u2220AOB = \u2220COD (Given)
OB = OD (Radii)
\u2234 \u0394AOB \u2245 \u0394COD (SAS congruence condition)
Thus, AB = CD by CPCT.
If chords of congruent circles subtend equal angles at their centres, then the chords are equal.<\/p>\n\n\n\n![\"NCERT](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/02\/class-9-maths-chapter-10-ncert-2.jpg\")
<\/figure>\n\n\n\n![\"NCERT](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/02\/class-9-maths-chapter-10-ncert-3.jpg\")
<\/figure>\n\n\n\n![\"NCERT](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/02\/class-9-maths-chapter-10-ncert-4.jpg\")
<\/figure>\n\n\n\n![\"NCERT](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/02\/class-9-maths-chapter-10-ncert-6-1.jpg\")
<\/figure>\n\n\n\n![\"NCERT](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/02\/class-9-maths-chapter-10-ncert-5.jpg\")
<\/figure>\n\n\n\n![\"NCERT](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/02\/class-9-maths-chapter-10-ncert-7.jpg\")
<\/figure>\n\n\n\n
Step I: A circle is drawn.
Step II: Two chords AB and CD are drawn.
Step III: Perpendicular bisector of the chords AB and CD are drawn.
Step IV: Let these two perpendicular bisector meet at a point. The point of intersection of these two perpendicular bisector is the centre of the circle.<\/p>\n\n\n\n![\"NCERT](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/02\/class-9-maths-chapter-10-ncert-8.jpg\")
<\/figure>\n\n\n\n
Two circles which intersect each other at P and Q.
To prove,
OO’ is perpendicular bisector of PQ.
Proof,
In \u0394POO’ and \u0394QOO’,
OP = OQ (Radii)
OO’ = OO’ (Common)
O’P = OQ (Radii)
\u2234 \u0394POO’ \u2245 \u0394QOO’ (SSS congruence condition)
Thus,
\u2220POO’ = \u2220QOO’ — (i)
In \u0394POR and \u0394QOR,
OP = OQ (Radii)
\u2220POR = \u2220QOR (from i)
OR = OR (Common)
\u2234 \u0394POR \u2245 \u0394QOR (SAS congruence condition)
Thus,
\u2220PRO = \u2220QRO
also,
\u2220PRO + \u2220QRO = 180\u00b0
\u21d2 \u2220PRO = \u2220QRO = 180\u00b0\/2 = 90\u00b0
Hence,
OO’ is perpendicular bisector of PQ.<\/p>\n\n\n\n
also, PQ = 2PR
Let RS be x.<\/p>\n\n\n\n![\"NCERT](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/02\/class-9-maths-chapter-10-ncert-9.jpg\")
<\/figure>\n\n\n\n
OP2 <\/sup>= OR2 <\/sup>+PR2<\/sup>
\u21d2 52 <\/sup>= (4-x)2 <\/sup>+PR2<\/sup>
\u21d2 25 = 16 + x2 <\/sup>– 8x +PR2<\/sup>
\u21d2 PR2<\/sup> = 9 – x2<\/sup> + 8x — (i)<\/p>\n\n\n\n
PS2 <\/sup>= PR2 <\/sup>+RS2<\/sup>
\u21d2 32 <\/sup>= PR2<\/sup> +x2<\/sup>
\u21d2 PR2<\/sup> = 9 – x2<\/sup> — (ii)
Equating (i) and (ii),
9 – x2<\/sup> + 8x = 9 – x2<\/sup>
\u21d2 8x = 0
\u21d2 x = 0
Putting the value of x in (i) we get,
PR2<\/sup> = 9 – 02<\/sup>
\u21d2 PR = 3cm
Length of the cord PQ = 2PR = 2\u00d73 = 6cm<\/p>\n\n\n\n
AB and CD are chords intersecting at E.
AB = CD
To prove,
AE = DE and CE = BE
Construction,
OM \u22a5 AB and ON \u22a5 CD. OE is joined.<\/p>\n\n\n\n![\"NCERT](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/02\/class-9-maths-chapter-10-ncert-10.jpg\")
<\/figure>\n\n\n\n
OM bisects AB (OM \u22a5 AB)
ON bisects CD (ON \u22a5 CD)
As AB = CD thus,
AM = ND — (i)
and MB = CN — (ii)
In \u0394OME and \u0394ONE,
\u2220OME = \u2220ONE (Perpendiculars)
OE = OE (Common)
OM = ON (AB = CD and thus equidistant from the centre)
\u0394OME \u2245 \u0394ONE by RHS congruence condition.
ME = EN by CPCT — (iii)
From (i) and (ii) we get,
AM + ME = ND + EN
\u21d2 AE = ED
From (ii) and (iii) we get,
MB – ME = CN – EN
\u21d2 EB = CE<\/p>\n\n\n\n
AB and CD are chords intersecting at E.
AB = CD, PQ is the diameter.
To prove,
\u2220BEQ = \u2220CEQ
Construction,
OM \u22a5 AB and ON \u22a5 CD. OE is joined.<\/p>\n\n\n\n![\"NCERT](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/02\/class-9-maths-chapter-10-ncert-11.jpg\")
<\/figure>\n\n\n\n
OM = ON (Equal chords are equidistant from the centre)
OE = OE (Common)
\u2220OME = \u2220ONE (Perpendicular)
\u0394OEM \u2245 \u0394OEN by RHS congruence condition.
Thus,
\u2220MEO = \u2220NEO by CPCT
\u21d2 \u2220BEQ = \u2220CEQ
4. If a line intersects two concentric circles (circles with the same centre) with centre O at A, B, C and D, prove that AB = CD (see Fig. 10.25).<\/strong><\/p>\n\n\n\n![\"NCERT](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/02\/class-9-maths-chapter-10-ncert-12.jpg\")
<\/figure>\n\n\n\n
OM bisects AD as OM \u22a5 AD.
\u21d2 AM = MD — (i)
also, OM bisects BC as OM \u22a5 BC.
\u21d2 BM = MC — (ii)
From (i) and (ii),
AM – BM = MD – MC
\u21d2 AB = CD<\/p>\n\n\n\n![\"NCERT](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/02\/class-9-maths-chapter-10-ncert-13.jpg\")
<\/figure>\n\n\n\n
AB = 6cm and BC = 6cm.
Radius OA = 5cm
BM \u22a5 AC is drawn.
ABC is an isosceles triangle as AB = BC, M is mid-point of AC. BM is perpendicular bisector of AC and thus it passes through the centre of the circle.
Let AM = y and OM = x then BM = (5-x).
Applying Pythagoras theorem in \u0394OAM,
OA2 <\/sup>= OM2 <\/sup>+AM2<\/sup>
\u21d2 52 <\/sup>= x2 <\/sup>+y2 <\/sup>— (i)Applying Pythagoras theorem in \u0394AMB,
AB2 <\/sup>= BM2 <\/sup>+AM2<\/sup>
\u21d2 62 <\/sup>= (5-x)2 <\/sup>+y2<\/sup> — (ii)Subtracting (i) from (ii), we get
36 – 25 = (5-x)2 <\/sup>-x2<\/sup> –
\u21d2 11 = 25 – 10x
\u21d2 10x = 14 \u21d2 x= 7\/5
Substituting the value of x in (i), we get
y2 <\/sup>+ 49\/25 = 25
\u21d2 y2<\/sup> = 25 – 49\/25
\u21d2 y2<\/sup> = (625 – 49)\/25
\u21d2 y2<\/sup> = 576\/25
\u21d2 y = 24\/5
Thus,
AC = 2\u00d7AM = 2\u00d7y = 2\u00d7(24\/5) m = 48\/5 m = 9.6 m
Distance between Reshma and Mandip is 9.6 m.<\/p>\n\n\n\n![\"NCERT](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/02\/class-9-maths-chapter-10-ncert-14.jpg\")
<\/figure>\n\n\n\n
AD \u22a5 BC is drawn. Now, AD is median of \u0394ABC and it passes through the centre O.
Also, O is the centroid of the \u0394ABC. OA is the radius of the triangle.
OA = 2\/3 AD
Let the side of a triangle a metres then BD = a\/2 m.
Applying Pythagoras theorem in \u0394ABD,
AB2 <\/sup>= BD2 <\/sup>+AD2<\/sup>
\u21d2 AD2 <\/sup>= AB2 <\/sup>– BD2 <\/sup>
\u21d2 AD2 <\/sup>= a2 <\/sup>-(a\/2)2<\/sup>
\u21d2 AD2 <\/sup>= 3a2<\/sup>\/4
\u21d2 AD = \u221a3a\/2
OA = 2\/3 AD \u21d2 20 m = 2\/3 \u00d7 \u221a3a\/2
\u21d2 a = 20\u221a3 m
Length of the string is 20\u221a3 m.<\/p>\n\n\n\n![\"NCERT](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/02\/class-9-maths-chapter-10-ncert-15.jpg\")
<\/figure>\n\n\n\n
\u2220AOC = \u2220AOB +<\/p>\n\n\n\n
\u21d2 \u2220AOC = 60\u00b0 + 30\u00b0
\u21d2 \u2220AOC = 90\u00b0
We know that angle subtended by an arc at centre is double the angle subtended by it any point on the remaining part of the circle.
\u2220ADC = 1\/2\u2220AOC = 1\/2 \u00d7 90\u00b0 = 45\u00b0<\/p>\n\n\n\n![](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/02\/class-9-maths-chapter-10-ncert-16.jpg\")
Given,
AB is equal to the radius of the circle.
In \u0394OAB,
OA = OB = AB = radius of the circle.
Thus, \u0394OAB is an equilateral triangle.
\u2220AOC = 60\u00b0
also,
\u2220ACB = 1\/2 \u2220AOB = 1\/2 \u00d7 60\u00b0 = 30\u00b0
ACBD is a cyclic quadrilateral,
\u2220ACB +<\/p>\n\n\n\n
\u21d2 \u2220ADB = 180\u00b0 – 30\u00b0 = 150\u00b0<\/p>\n\n\n\n![](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/02\/class-9-maths-chapter-10-ncert-17.jpg\")
Answer<\/strong><\/p>\n\n\n\n
\u2234 \u2220POR = 360\u00b0 – 200\u00b0 = 160\u00b0
In \u0394OPR,
OP = OR (radii of the circle)
\u2220OPR = \u2220ORP
Now,
\u2220OPR +<\/p>\n\n\n\n
\u21d2 \u2220OPR +<\/p>\n\n\n\n
\u21d2 2\u2220OPR = 180\u00b0 – 160\u00b0
\u21d2 \u2220OPR = 10\u00b0<\/p>\n\n\n\n![\"NCERT](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/02\/class-9-maths-chapter-10-ncert-16-1.jpg\")
<\/figure>\n\n\n\n
In \u0394ABC,
\u2220BAC +<\/p>\n\n\n\n
\u21d2 \u2220BAC + 69\u00b0 + 31\u00b0 = 180\u00b0
\u21d2 \u2220BAC = 180\u00b0 – 100\u00b0
\u21d2 \u2220BAC = 80\u00b0
Thus, \u2220BDC = 80\u00b0<\/p>\n\n\n\n![\"NCERT](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/02\/class-9-maths-chapter-10-ncert-18.jpg\")
<\/figure>\n\n\n\n
In \u0394CDE,
\u2220CEB = \u2220CDE + \u2220DCE (Exterior angles of the triangle)
\u21d2 130\u00b0 = \u2220CDE + 20\u00b0
\u21d2 \u2220CDE = 110\u00b0
Thus, \u2220BAC = 110\u00b0<\/p>\n\n\n\n![](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/02\/class-9-maths-chapter-10-ncert-19.jpg\")
For chord CD,
\u2220CBD = \u2220CAD (Angles in same segment)
\u2220CAD = 70\u00b0
\u2220BAD = \u2220BAC + \u2220CAD = 30\u00b0 + 70\u00b0 = 100\u00b0
\u2220BCD + \u2220BAD = 180\u00b0 (Opposite angles of a cyclic quadrilateral)
\u21d2 \u2220BCD + 100\u00b0 = 180\u00b0
\u21d2 \u2220BCD = 80\u00b0
In \u0394ABC
AB = BC (given)
\u2220BCA = \u2220CAB (Angles opposite to equal sides of a triangle)
\u2220BCA = 30\u00b0
also, \u2220BCD = 80\u00b0
\u2220BCA + \u2220ACD = 80\u00b0
\u21d2 30\u00b0 + \u2220ACD = 80\u00b0
\u2220ACD = 50\u00b0
\u2220ECD = 50\u00b0<\/p>\n\n\n\n![\"NCERT](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/02\/class-9-maths-chapter-10-ncert-20.jpg\")
<\/figure>\n\n\n\n
Thus, ABCD is a rectangle as each internal angle is 90\u00b0.<\/p>\n\n\n\n
ABCD is a trapezium where non-parallel sides AD and BC are equal.
Construction,
DM and CN are perpendicular drawn on AB from D and C respectively.
To prove,
ABCD is cyclic trapezium.<\/p>\n\n\n\n![\"NCERT](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/02\/class-9-maths-chapter-10-ncert-21.jpg\")
<\/figure>\n\n\n\n
In \u0394DAM and \u0394CBN,
AD = BC (Given)
\u2220AMD = \u2220BNC (Right angles)
DM = CN (Distance between the parallel lines)
\u0394DAM \u2245 \u0394CBN by RHS congruence condition.
Now,
\u2220A = \u2220B by CPCT
also, \u2220B +<\/p>\n\n\n\n
\u21d2 \u2220A + \u2220C = 180\u00b0
Thus, ABCD is a cyclic quadrilateral as sum of the pair of opposite angles is 180\u00b0.<\/p>\n\n\n\n![\"NCERT](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/02\/class-9-maths-chapter-10-ncert-22.jpg\")
<\/figure>\n\n\n\n
For chord AP,
\u2220PBA = \u2220ACP (Angles in the same segment) — (i)
For chord DQ,
\u2220DBQ = \u2220QCD (Angles in same segment) — (ii)
ABD and PBQ are line segments intersecting at B.
\u2220PBA = \u2220DBQ (Vertically opposite angles) — (iii)
By the equations (i), (ii) and (iii),
\u2220ACP = \u2220QCD<\/p>\n\n\n\n![\"NCERT](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/02\/class-9-maths-chapter-10-ncert-23.jpg\")
<\/figure>\n\n\n\n
\u2220ADB = \u2220ADC = 90\u00b0 (Angle in the semi circle)
Now,
\u2220ADB + \u2220ADC = 180\u00b0
\u21d2 \u2220BDC is straight line.
Thus, D lies on the BC.<\/p>\n\n\n\n
AC is the common hypotenuse. \u2220B = \u2220D = 90\u00b0.
To prove,
\u2220CAD = \u2220CBD![](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/02\/class-9-maths-chapter-10-ncert-24.jpg\")
Proof:
Since, \u2220ABC and \u2220ADC are 90\u00b0. These angles are in the semi circle. Thus, both the triangles are lying in the semi circle and AC is the diameter of the circle.
\u21d2 Points A, B, C and D are concyclic.
Thus, CD is the chord.
\u21d2 \u2220CAD = \u2220CBD (Angles in the same segment of the circle)<\/p>\n\n\n\n
ABCD is a cyclic parallelogram.
To prove,
ABCD is rectangle.<\/p>\n\n\n\n![\"NCERT](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/02\/class-9-maths-chapter-10-ncert-25.jpg\")
<\/figure>\n\n\n\n
\u22201 +<\/p>\n\n\n\n
also, Opposite angles of a cyclic parallelogram are equal.
Thus,
\u22201 = \u22202
\u21d2 \u22201 + \u22201 = 180\u00b0
\u21d2 \u22201 = 90\u00b0
One of the interior angle of the paralleogram is right angled. Thus, ABCD is a rectangle.<\/p>\n\n\n\nNCERT Solutions for Class 9 Maths Chapters:<\/h4>\n\n\n\n
<\/p>\n\n\n\nHow many exercises in Chapter 10 Circles?<\/h4>\n\n\n\n
The diameter of circle is 3.8 cm. Find the length of its radius.<\/h4>\n\n\n\n
What is an arc?<\/h4>\n\n\n\n
What is a diameter?<\/h4>\n\n\n\n
NCERT Solutions for 9th Class Maths : Chapter 10: Download PDF<\/strong><\/h2>\n\n\n\n
Chapterwise NCERT Solutions for Class 9 Maths :<\/strong><\/h2>\n\n\n\n
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About NCERT<\/h2>\n\n\n\n