{"id":565923,"date":"2021-12-29T06:51:09","date_gmt":"2021-12-29T06:51:09","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=565923"},"modified":"2021-12-29T11:52:33","modified_gmt":"2021-12-29T11:52:33","slug":"maharashtra-board-solutions-for-class-10-maths-part-2-chapter-3-circle","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/maharashtra-board-solutions-for-class-10-maths-part-2-chapter-3-circle\/","title":{"rendered":"Maharashtra Board Solutions for Class 10-Maths (Part 2): Chapter 3- Circle"},"content":{"rendered":"\n

Class 10: Maths Chapter 3 solutions. Complete Class 10 Maths Chapter 3 Notes.<\/p>\n\n\n\n

Maharashtra Board Solutions for Class 10-Maths (Part 2): Chapter 3- Circle<\/h2>\n\n\n\n

Maharashtra Board 10th Maths Chapter 3, Class 10 Maths Chapter 3 solutions<\/p>\n\n\n\n

Practice Set 3.1<\/h4>\n\n\n\n

Question 1.
In the adjoining figure, the radius of a circle with centre C is 6 cm, line AB is a tangent at A. Answer the following questions.<\/strong>
i. What is the measure of \u2220CAB? Why?
ii. What is the distance of point C from line AB? Why?
iii. d(A, B) = 6 cm, find d(B, C).
iv. What is the measure of \u2220ABC? Why?
\"Maharashtra
Solution:
<\/strong>i. line AB is the tangent to the circle with centre C and radius AC. [Given]
\u2234 \u2220CAB = 90\u00b0 (i) [Tangent theorem]
ii. seg CA \u22a5 line AB [From (i)]
radius = l(AC) = 6 cm
\u2234 The distance of point C from line AB is 6 cm.
iii. In \u2206CAB, \u2220CAB = 90\u00b0 [From (i)]
\u2234 BC2<\/sup> = AB2<\/sup> + AC2<\/sup> . [Pythagoras theorem]
\"Maharashtra
= 62<\/sup> + 62<\/sup>
= 2 \u00d7 62<\/sup>
\u2234 BC = 2\u00d762\u2212\u2212\u2212\u2212\u2212\u221a [Taking square root of both sides]
= 6 2\u2013\u221a cm
\u2234 d(B, C) = 6 cm
iv. In \u2206ABC,
AC = AB = 6cm
\u2234 \u2220ABC = \u2220ACB [Isosceles triangle theorem]
Let \u2220ABC = \u2220ACB =x
In \u2206ABC,
\u2220CAB + \u2220ABC + \u2220ACB = 180\u00b0 [Sum of the measures of angles of a triangle is 180\u00b0]
\u2234 90\u00b0 + x + x = 180\u00b0
\u2234 90 + 2x = 180\u00b0
\u2234 2x = 180\u00b0- 90\u00b0
\u2234 x = 90\u22182
\u2234 x = 45\u00b0
\u2234 \u2220ABC = 45\u00b0<\/p>\n\n\n\n

Question 2.
In the adjoining figure, O is the centre of the circle. From point R, seg RM and seg RN are tangent segments touching the circle at M and N. If (OR) = 10 cm and radius of the circle = 5 cm, then<\/strong>
i. What is the length of each tangent segment?
ii. What is the measure of \u2220MRO?
iii. What is the measure of \u2220MRN?
\"Maharashtra
Solution:
seg RM and seg RN are tangents to the circle with centre O. [Given]
\u2234 \u2220OMR = \u2220ONR = 90\u00b0 [Tangent theorem]
\"Maharashtra
i. In \u2206OMR, \u2220OMR = 90\u00b0
\u2234 OR2<\/sup> = OM2<\/sup> + RM2<\/sup> [Pythagoras theorem]
\u2234 102<\/sup> = 52<\/sup> + RM2<\/sup>
\u2234 100 = 25 + RM2<\/sup>
\u2234 RM2<\/sup> = 75
\u2234 RM = 75\u2212\u2212\u221a [Taking square root of both sides]
\u2234 RM = RN [Tangent segment theorem]
Length of each tangent segment is 5 3\u2013\u221a cm.
ii. In \u2206RMO,
\u2220OMR = 90\u00b0 [Tangent theorem]
OM = 5 cm and OR = 10 cm
\u2234 OM = 12 OR
\u2234 \u2220MRO = 30\u00b0 (i) [Converse of 30\u00b0 \u2013 60\u00b0 \u2013 90\u00b0 theorem]
Similarly, \u2220NRO = 30\u00b0
iii. But, \u2220MRN = \u2220MRO + \u2220NRO [Angle addition property]
= 30\u00b0 + 30\u00b0 [From (i)]
\u2234 \u2220MRN = 60\u00b0<\/p>\n\n\n\n

Question 3.
Seg RM and seg RN are tangent segments of a circle with centre O. Prove that seg OR bisects \u2220MRN as well as \u2220MON.
\"Maharashtra
Solution:<\/strong>
Proof:
In \u2206OMR and \u2206ONR,
seg RM \u2245 seg RN [Tangent segment theorem]
seg OM \u2245 seg ON [Radii of the same circle]
seg OR \u2245 seg OR [Common side]
\u2234 \u2206OMR \u2245 \u2206ONR [SSS test of congruency]
{\u2234 \u2220MRO \u2245 \u2220NRO
\u2220MOR \u2245 \u2220NOR } [c.a.c.t.]
\u2234 seg OR bisects \u2220MRN and \u2220MON.<\/p>\n\n\n\n

Question 4.
What is the distance between two parallel tangents of a circle having radius 4.5 cm? Justify your answer.
Solution:<\/strong>
Let the lines PQ and RS be the two parallel tangents to circle at M and N respectively. Through centre O, draw line AB || line RS.
\"Maharashtra
OM = ON = 4.5 [Given]
line AB || line RS [Construction]
line PQ || line RS [Given]
\u2234 line AB || line PQ || line RS
Now, \u2220OMP = \u2220ONR = 90\u00b0 (i) [Tangent theorem]
For line PQ || line AB,
\u2220OMP = \u2220AON = 90\u00b0 (ii) [Corresponding angles and from (i)]
For line RS || line AB,
\u2220ONR = \u2220AOM = 90\u00b0 (iii) [Corresponding angles and from (i)]
\u2220AON + \u2220AOM = 90\u00b0 + 90\u00b0 [From (ii) and (iii)]
\u2234 \u2220AON + \u2220AOM = 180\u00b0
\u2234 \u2220AON and \u2220AOM form a linear pair.
\u2234 ray OM and ray ON are opposite rays.
\u2234 Points M, O, N are collinear. (iv)
\u2234 MN = OM + ON [M \u2013 O \u2013 N, From (iv)]
\u2234 MN = 4.5 + 4.5
\u2234 MN = 9 cm
\u2234 Distance between two parallel tangents PQ and RS is 9 cm.<\/p>\n\n\n\n

Question 1.
In the adjoining figure, seg QR is a chord of the circle with centre O. P is the midpoint of the chord QR. If QR = 24, OP = 10, find radius of the circle. To find solution of the problem, write the theorems that are useful. Using them, solve the problem. (Textbook pg. no. 48)
\"Maharashtra
Solution:<\/strong>
Theorems which are useful to find solution:
i. The segment joining the centre of a circle and the midpoint of a chord is perpendicular to the chord.
ii. In a right angled triangle, sum of the squares of the perpendicular sides is equal to square of its hypotenuse.
QP = 12 (QR) [P is the midpoint of chord QR]
\"Maharashtra
12 \u00d7 24 = 12 units
Also, seg OP \u22a5 chord QR [The segment joining centre of a circle and midpoint of a chord is perpendicular to the chord]
In \u2206OPQ, \u2220OPQ = 90\u00b0
\u2234 OQ2<\/sup> = OP2<\/sup> + QP2<\/sup> [Pythagoras theorem]
= 102<\/sup> + 122<\/sup>
= 100 + 144
= 244
\u2234 OQ = 244\u2212\u2212\u2212\u221a = 261\u2212\u2212\u221a units.
\u2234 The radius of the circle is 261\u2212\u2212\u221a units.<\/p>\n\n\n\n

Question 2.
In the adjoining figure, M is the centre of the circle and seg AB is a diameter, seg MS \u22a5 chord AD, seg MT \u22a5 chord AC, \u2220DAB \u2245 \u2220CAB.<\/strong>
i. Prove that: chord AD \u2245 chord AC.
ii. To solve this problem which theorems will you use?
a. The chords which are equidistant from the centre are equal in length.
b. Congruent chords of a circle are equidistant from the centre.
\"Maharashtra
iii. Which of the following tests of congruence of triangles will be useful?
a. SAS b. ASA c. SSS d. AAS e. Hypotenuse-side test.
Using appropriate test and theorem write the proof of the above example. (Textbook pg. no, 48)
Solution:
<\/strong>Proof:
i. \u2220DAB \u2245 \u2220CAB [Given]
\u2234 \u2220SAM \u2245 \u2220TAM (i) [A \u2013 S \u2013 D, A \u2013 M \u2013 B, A -T \u2013 C]
In \u2206SAM and \u2206TAM,
\u2220SAM \u2245 \u2220TAM [From (i)]
\u2220ASM \u2245 \u2220ATM [Each angle is of measure 90\u00b0]
seg AM \u2245 seg AM [Common side]
\u2234 \u2206SAM \u2245 \u2206TAM [AAS [SAA] test of congruency]
\u2234 side MS \u2245 side MT [c.s.c.t]
But, seg MS \u22a5 chord AD [Given]
seg MT \u22a5chord AC
\u2234 chord AD \u2245 chord AC [Chords of a circle equidistant from the centre are congruent]
ii. Theorem used for solving the problem:
The chords which are equidistant from the centre are equal in length.
iii. Test of congruency useful in solving the above problem is AAS ISAAI test of congruency.<\/p>\n\n\n\n

Question 3.
i. Draw segment AB. Draw perpendicular bisector l of the segment AB. Take point P on the line l as centre,
PA as radius and draw a circle. Observe that the circle passes through point B also. Find the reason.
ii. Taking any other point Q on the line l, if a circle is drawn with centre Q and radius QA, will it pass through B? Think.
iii. How many such circles can be drawn, passing through A and B? Where will their centres lie? (Textbook pg. no. 49)
Solution:<\/strong>
\"Maharashtra
i. Draw the circle with centre P and radius PA.
line l is the perpendicular bisector of seg AB.
Every point on the perpendicular bisector of a segment is equidistant from the endpoints of the segment.
\u2234 PA = PB \u2026 [Perpendicular bisector theorem]
\u2234 PA = PB = radius
\u2234 The circle with centre P and radius PA passes through point B.<\/p>\n\n\n\n

ii. The circle with any other point Q and radius QA is drawn.
QA = QB = radius \u2026 [Perpendicular bisector theorem]
\u2234 The circle with centre Q and radius QA passes through point B.<\/p>\n\n\n\n

iii. We can draw infinite number of circles passing through A and B.
All their centres will lie on the perpendicular bisector of AB (i.e., line l)<\/p>\n\n\n\n

Question 4.
i. Take any three non-collinear points. What should be done to draw a circle passing through all these points? Draw a circle through these points.
ii. Is it possible to draw one more circle passing through these three points? Think of it. (Textbook pg. no. 49)
Solution:<\/strong>
\"Maharashtra
i. Let points A, B, C be any three non collinear points.
Draw the perpendicular bisector of seg AB (line l).
\u2234 Points A and B are equidistant from any point of line l \u2026.(i)[Perpendicular bisector theorem]
Draw the perpendicular bisector of seg BC (line m) to intersect line l at point P.
\u2234 Points B and C are equidistant from any point of line m \u2026.(ii) [Perpendicular bisector theorem]
\u2234 PA = PB \u2026[From (i)]
PB = PC \u2026 [From (ii)]
\u2234 PA = PB = PC = radius
\u2234 With PA as radius the required circle is drawn through points A, B, C.
ii. It is not possible to draw more than one circle passing through these three points.<\/p>\n\n\n\n

Question 5.
Take 3 collinear points D, E, F. Try to draw a circle passing through these points. If you are not able to draw a circle, think of the reason. (Textbook pg. no. 49)
Solution:<\/strong>
\"Maharashtra
Let D, E, F be the collinear points.
The perpendicular bisector of DE and EF drawn (i.e., line l and line m) do not intersect at a common point.
\u2234 There is no single common point (centre of circle) from which a circle can be drawn passing through points D, E and F.
Hence, we cannot draw a circle passing through points D, E and F.<\/p>\n\n\n\n

Question 6.
Which theorem do we use in proving that hypotenuse is the longest side of a right angled triangle? (Textbook pg. no. 52)
\"Maharashtra
Solution:<\/strong>
In \u2206ABC,
\u2220ABC = 90\u00b0
\u2234 \u2220BAC < 90\u00b0 and \u2220ACB < 90\u00b0 [Given]
\u2234 \u2220ABC > \u2220BAC and \u2220ABC > \u2220ACB
\u2234 AC > BC and AC > AB [Side opposite to greater angle is greater]
\u2234 Hypotenuse is the longest side in right angled triangle.
We use theorem, If two angles of a triangle are not equal, then the side opposite to the greater angle is greater than the side opposite to the smaller angle.<\/p>\n\n\n\n

Question 7.
Theorem: Tangent segments drawn from an external point to a circle are congruent
Draw radius AP and radius AQ and complete the following proof of the theorem.
Given: A is the centre of the circle.
Tangents through external point D touch the circle at the points P and Q.
To prove: seg DP \u2245 seg DQ
Construction: Draw seg AP and seg AQ.
\"Maharashtra
Solution:<\/strong>
Proof:
In \u2206PAD and \u2206QAD,
seg PA \u2245 [segQA] [Radii of the same circle]
seg AD \u2245 seg AD [Common side]
\"Maharashtra
\u2220APD = \u2220AQD = 90\u00b0 [Tangent theorem]
\u2234 \u2206PAD = \u2206QAD [By Hypotenuse side test]
\u2234 seg DP = seg DQ [c.s.c.t]<\/p>\n\n\n\n

Practice Set 3.2<\/h4>\n\n\n\n

Question 1.
Two circles having radii 3.5 cm and 4.8 cm touch each other internally. Find the distance between their centres.
Solution:<\/strong>
Let the two circles having centres P and Q touch each other internally at point R.
Here, QR = 3.5 cm, PR = 4.8 cm
\"Maharashtra
The two circles touch each other internally.
\u2234 By theorem of touching circles,
P \u2013 Q \u2013 R
PQ = PR \u2013 QR
= 4.8 \u2013 3.5
= 1.3 cm
[The distance between the centres of circles touching internally is equal to the difference in their radii]<\/p>\n\n\n\n

Question 2.
Two circles of radii 5.5 cm and 4.2 cm touch each other externally. Find the distance between their centres.
Solution:<\/strong>
Let the two circles having centres P and R touch each other externally at point Q.
Here, PQ = 5.5 cm, QR = 4.2 cm
\"Maharashtra
The two circles touch each other externally.
\u2234 By theorem of touching circles,
P \u2013 Q \u2013 R
PR = PQ + QR
= 5.5 + 4.2
= 9.7 cm
[The distance between the centres of the circles touching externally is equal to the sum of their radii]<\/p>\n\n\n\n

Question 3.
If radii of two circles are 4 cm and 2.8 cm. Draw figure of these circles touching each other<\/strong>
i. externally
ii. internally.
Solution:
<\/strong>i. Circles touching externally:
\"Maharashtra
ii. Circles touching internally:
\"Maharashtra<\/p>\n\n\n\n

Question 4.
In the adjoining figure, the circles with centres P and Q touch each other at R A line passing through R meets the circles at A and B respectively. Prove that \u2013<\/strong>
\"Maharashtra
i. seg AP || seg BQ,
ii. \u2206APR ~ \u2206RQB, and
iii. Find \u2220RQB if \u2220PAR = 35\u00b0.
Solution:
<\/strong>The circles with centres P and Q touch each other at R.
\u2234 By theorem of touching circles,
P \u2013 R \u2013 Q
i. In \u2206PAR,
seg PA = seg PR [Radii of the same circle]
\u2234 \u2220PRA \u2245 \u2220PAR (i) [Isosceles triangle theorem]
Similarly, in \u2206QBR,
seg QR = seg QB [Radii of the same circle]
\u2234 \u2220RBQ \u2245 \u2220QRB (ii) [Isosceles triangle theorem]
But, \u2220PRA \u2245 \u2220QRB (iii) [Vertically opposite angles]
\u2234 \u2220PAR \u2245 \u2220RBQ (iv) [From (i) and (ii)]
But, they are a pair of alternate angles formed by transversal AB on seg AP and seg BQ.
\u2234 seg AP || seg BQ [Alternate angles test]
ii. In \u2206APR and \u2206RQB,
\u2220PAR \u2245 \u2220QRB [From (i) and (iii)]
\u2220APR \u2245 \u2220RQB [Alternate angles]
\u2234 \u2206APR \u2013 \u2206RQB [AA test of similarity]
iii. \u2220PAR = 35\u00b0 [Given]
\u2234 \u2220RBQ = \u2220PAR= 35\u00b0 [From (iv)]
In \u2206RQB,
\u2220RQB + \u2220RBQ + \u2220QRB = 180\u00b0 [Sum of the measures of angles of a triangle is 180\u00b0]
\u2234 \u2220RQB + \u2220RBQ + \u2220RBQ = 180\u00b0 [From (ii)]
\u2234 \u2220RQB + 2 \u2220RBQ = 180\u00b0
\u2234 \u2220RQB + 2 \u00d7 35\u00b0 = 180\u00b0
\u2234 \u2220RQB + 70\u00b0 = 180\u00b0
\u2234 \u2220RQB = 110\u00b0<\/p>\n\n\n\n

Question 5.
In the adjoining figure, the circles with centres A and B touch each other at E. Line l is a common tangent which touches the circles at C and D respectively. Find the length of seg CD if the radii of the circles are 4 cm, 6 cm.
\"Maharashtra
Construction : Draw seg AF \u22a5 seg BD.
Solution:<\/strong>
i. The circles with centres A and B touch each other at E. [Given]
\u2234 By theorem of touching circles,
A \u2013 E \u2013 B
\"Maharashtra
\u2234 \u2220ACD = \u2220BDC = 90\u00b0 [Tangent theorem]
\u2220AFD = 90\u00b0 [Construction]
\u2234 \u2220CAF = 90\u00b0 [Remaining angle of \ua838AFDC]
\u2234 \ua838AFDC is a rectangle. [Each angle is of measure 900]
\u2234 AC = DF = 4 cm [Opposite sides of a rectangle]
Now, BD = BF + DF [B \u2013 F \u2013 C]
\u2234 6 = BF + 4 BF = 2 cm
Also, AB = AE + EB
= 4 + 6 = 10 cm
[The distance between the centres of circles touching externally is equal to the sum of their radii]<\/p>\n\n\n\n

ii. Now, in \u2206AFB, \u2220AFB = 90\u00b0 [Construction]
\u2234 AB2<\/sup> = AF2<\/sup> + BF2<\/sup> [Pythagoras theorem]
\u2234 102<\/sup> = AF2<\/sup> + 22<\/sup>
\u2234 100 = AF2<\/sup> + 4
\u2234 AF2<\/sup> = 96
\u2234 AF = 96\u2212\u2212\u221a [Taking square root of both sides]
= 16\u00d76\u2212\u2212\u2212\u2212\u2212\u221a
= 4 6\u2013\u221a cm
But, CD = AF [Opposite sides of a rectangle]
\u2234 CD = 4 6\u2013\u221a cm<\/p>\n\n\n\n

Question 1.
Take three collinear points X \u2013 Y \u2013 Z as shown in figure.
Draw a circle with centre X and radius XY. Draw another circle with centre Z and radius YZ.
Note that both the circles intersect each other at the single point Y. Draw a line \/ through point Y and perpendicular to seg XZ. What is line l (Textbook pg. no. 56)
Solution:<\/strong>
\"Maharashtra
Line l is a common tangent of the two circles.<\/p>\n\n\n\n

Question 2.
Take points Y \u2013 X \u2013 Z as shown in the figure. Draw a circle with centre Z and radius ZY.
Also draw a circle with centre X and radius XY. Note that both the circles intersect each other at the point Y.
Draw a line l perpendicular to seg YZ through point Y. What is line l? (Textbook pg. no. 56)
Solution:<\/strong>
\"Maharashtra
Line l is a common tangent of the two circles.<\/p>\n\n\n\n

If two circles in the same plane intersect with a line in the plane in only one point, they are said to be touching circles and the line is their common tangent.<\/p>\n\n\n\n

The point common to the circles and the line is called their common point of contact.<\/p>\n\n\n\n

1. Circles touching externally:
For circles touching externally, the distance between their centres is equal to sum of their radii, i.e. AB = AC + BC
\"Maharashtra<\/p>\n\n\n\n

2. Circles touching internally:
For circles touching internally, the distance between their centres is equal to difference of their radii,
i. e. AB = AC \u2013 BC
\"Maharashtra<\/p>\n\n\n\n

Question 3.
The circles shown in the given figure are called externally touching circles. Why? (iexthook pg. no. 57)
\"Maharashtra
Answer:<\/strong>
Circles with centres R and S lie in the same plane and intersect with a line l in the plane in one and only one point T [R \u2013 T \u2013 S].
Hence the given circles are externally touching circles.<\/p>\n\n\n\n

Question 4.
The circles shown in the given figure are called internally touching circles, why? (Textbook pg. no. 57)
\"Maharashtra
Answer:<\/strong>
Circles with centres N and M lie in the same plane and intersect with a line p in the plane in one and only one point T [K \u2013 N \u2013 M].
Hence, the given circles are internally touching circles.<\/p>\n\n\n\n

Question 5.
In the given figure, the radii of the circles with centres A and B are 3 cm and 4 cm respectively. Find
i. d(A,B) in figure (a)
ii. d(A,B) in figure (b) (Textbook pg. no. 57)
\"Maharashtra
Solution:<\/strong>
i. Here, circle with centres A and B touch each other externally at point C.
\u2234 d(A, B) = d(A, C) + d(B ,C)
= 3 + 4
\u2234 d(A,B) = 7 cm
[The distance between the centres of circles touching externally is equal to the sum of their radii]
ii. Here, circle with centres A and 13 touch each other internally at point C.
\u2234 d(A, B) = d(A, C) \u2013 d(B, C)
= 4 \u2013 3
\u2234 d(A,B) = 1 cm
[The distance between the centres of circles touching internally is equal to the difference in their radii]<\/p>\n\n\n\n

Practice Set 3.3<\/h4>\n\n\n\n

Question 1.
In the adjoining figure, points G, D, E, F are concyclic points of a circle with centre C.
\u2220ECF = 70\u00b0, m(arc DGF) = 200\u00b0. Find m(arc DE) and m(arc DEF).
\"Maharashtra
Solution:<\/strong>
m(arc EF) = m\u2220ECF [Definition of measure of minor arc]
\u2234 m(arc EF) = 70\u00b0
i. m(arc DE) + m(arc DGF)
+ m(arc EF) = 360\u00b0 [Measure of a circle is 360\u00b0]
\u2234 m(arc DE) = 360\u00b0 \u2013 m(arc DGF) \u2013 m(arc EF)
= 360\u00b0 \u2013 200\u00b0 \u2013 70\u00b0
\u2234 m(arc DE) = 90\u00b0
ii. m(arc DEF) = m(arc DE) + m(arc EF) [Arc addition property]
= 90\u00b0 + 70\u00b0
\u2234 m(arc DEF) = 160\u00b0<\/p>\n\n\n\n

Question 2.
In the adjoining figure, AQRS is an equilateral triangle. Prove that,
i. arc RS \u2245 arc QS \u2245 arc QR
ii. m(arc QRS) = 240\u00b0.
\"Maharashtra
Solution:<\/strong>
Proof:
i. \u2206QRS is an equilateral triangle, [Given]
\u2234 seg RS \u2245 seg QS \u2245 seg QR [Sides of an equilateral triangle]
\u2234 arc RS \u2245 arc QS \u2245 arc QR [Corresponding arcs of congruents chords of a circle are congruent]
ii. Let m(arc RS) = m(arc QS)= m(arc QR) = x
m(arc RS) + m(arc QS) + m(arc QR) = 360\u00b0 [Measure of a circle is 360\u00b0, arc addition property]
\u2234 x + x + x = 360\u00b0
\u2234 3x = 360\u00b0
\u2234 x = 360\u22183=120\u2218
\u2234 m(arc RS) = m(arc QS) = m(arc QR) = 120\u00b0 (i)
Now, m(arc QRS) = m(arc QR) + m(arc RS) [Arc addition property]
= 120\u00b0 + 120\u00b0 [From (i)]
\u2234 m(arc QRS) = 240\u00b0<\/p>\n\n\n\n

Question 3.
In the adjoining figure, chord AB \u2245 chord CD. Prove that, arc AC = arc BD.
\"Maharashtra
Solution:<\/strong>
Proof:
chord AB \u2245 chord CD [Given]
\u2234 arc AB \u2245 arc CD [Corresponding arcs of congruents chords of a circle are congruent]
\u2234 m(arc AB) = m(arc CD)
\u2234 m(arc AC) + m(arc BC) = m(arc BC) + m(arc BD) [Arc addition property]
\u2234 m(arc AC) = m(arc BD)
\u2234 arc AC \u2245 arc BD<\/p>\n\n\n\n

Question 1.
Theorem : The chords corresponding to congruent arcs of a circle (or congruent circles) are congruent. (Textbook pg. no. 61)<\/strong>
Given: B is the centre of circle.
arc APC \u2245 arc DQE
To prove: chord AC \u2245 chord DE
\"Maharashtra
Proof:
[m(arc APC) = \u2220ABC (i) [Definition of measure of
m(arc DQE) = \u2220DBE] (ii) minor arc]
arc APC \u2245 arc \u2220DQE (iii) [Given]
\u2234 \u2220ABC \u2245 \u2220DBE [From (i), (ii) and (iii)]
In \u2206ABC and \u2206DBE,
side AB \u2245 side DB [Radil of the same circle]
side [CB] side [EB] [Radii of the same circle]
\u2220ABC \u2245\u2220DBE [From (iii), Measures of congruent arcs]
\u2234 \u2206ABC \u2245 \u2206DBE [SAS test of congruency]
\u2234 chord AC \u2245 chord DE [c.s.c.t]<\/p>\n\n\n\n

Question 2.
Theorem: Corresponding arcs of congruent chords of a circle (or congruent circles) are congruent (Textbook pg. no. 61)<\/strong>
Given: O is the centre of circle, chord PQ = chord RS
To prove: arc PMQ = arc RNS
\"Maharashtra
Proof:
In \u2206POQ and \u2206ROS,
[side PO \u2245 side RO
side OQ \u2245 side OS] [Radii of the same circle]
chord PQ \u2245 chord RS [Given]
\u2234 \u2206POQ \u2245 \u2206ROS [SSS test of congruency]
\u2234 \u2220POQ \u2245 \u2220ROS (i) [c.a.c.t.]
m(arc PMQ) = \u2220POQ (ii)
m(arc RNS) = \u2220ROS (iii) [Definition of measure of minor arc]
\u2234 arc PMQ \u2245 arc RNS [From (i), (ii) and (iii)]<\/p>\n\n\n\n

Question 3.
Prove the two theorems on textbook pg.no.61 for congruent circles. (Textbook pg. no. 62)
Theorem : The chords corresponding to congruent arcs of congruent circles are congruent
Given: In congruent circles with centres B and R,
arc APC \u2245 arc DQE<\/strong>
To prove: chord AC \u2245 chord DE
\"Maharashtra
Proof:
[m(arc APC) = \u2220ABC (i)
m(arc DQE) = \u2220DRE] (ii) [Definition of measure of minor arc]
arc APC \u2245 arc DQE (iii) [Given]
\u2234 \u2220ABC = \u2220DRE (iv) [From (i), (ii) and (iii)]
In \u2206ABC and \u2206DRE,
[side AB \u2245 side DR [Radii of congruent circles]
side CB \u2245 side ER] [From (iv)]
\u2220ABC \u2245 \u2220DRE
\u2234 \u2206ABC \u2245 \u2206DRE [SAS test of congruency]<\/p>\n\n\n\n

Question 4.
While proving the first theorem of the two, we assume that the minor arc APC and minor arc DQE are congruent. Can you prove the same theorem by assuming that corresponding major arcs congruent? (Textbook pg. no. 62)<\/strong>
Statement:
The chords corresponding to congruent major arcs of a circle are congruent.
Given: B is the centre of circle.
arc AXC \u2245 arc DXE
To prove: chord AC \u2245 chord DE
\"Maharashtra
Proof:
m(major arc) = 360\u00b0 \u2013 m(minor arc)
\u2234 m(arc AXC) = 360\u00b0 \u2013 m(arc APC) (i)
m(arc DXE) = 360\u00b0 \u2013 m(arc DQE) (ii)
m(arc AXC) = m(arc DXE) (iii) [Given]
\u2234 360\u00b0 \u2013 m(arc APC) = 360\u00b0- m(arc DQE) [From (i), (ii) and (iii)]
\u2234 m(arc APC) = m(arc DQE) (iv)
\u2234 m(arc APC) = \u2220ABC (v) [Definition of measure of minor arc]
m(arc DQE) = \u2220DBE (vi)
\u2234 \u2220ABC = \u2220DBE (vii) [From (iv), (v) and (vi)]
In \u2206ABC and \u2206DBE,
[side AB \u2245 side DB
Side CB \u2245 side EB] [Radii of the same circle]
\u2220ABC \u2245 \u2220DBE [From (vii)]
\u2234 \u2206ABC \u2245 \u2206DBE [SAS test of congruency]
\u2234 chord AC \u2245 chord DE [c.s.c.t.]<\/p>\n\n\n\n

Question 5.
i. In the second theorem, are the major arcs corresponding to congruent chords congruent?
ii. Is the theorem true, when the chord PQ and chord RS are diameters of the circle? (Textbook pg. no. 62)
Solution:<\/strong>
i. Yes, the major arcs corresponding to congruent chords are congruent.
\"Maharashtra
Proof:
In \u2206POQ and \u2206ROS,
seg OP \u2245 seg OR [Radii of the same circle]
seg OQ \u2245 seg OS [Radii of the same circle]
seg PQ \u2245 seg RS [Given]
\u2234 \u2206POQ \u2245 \u2206ROS [SSS test of congruence]
\u2234 \u2220POQ \u2245 \u2220SOR (i) [c.a.c.t]
[ m(arc PMQ) = \u2220POQ (ii)
m(arc RNS) = \u2220SOR ] (iii) [Definition of measure of minor arc]
\u2234 m(arc PMQ) = m(arc RNS)
m(minor arc) = 360\u00b0 \u2013 m(major arc) (iv) [From (i), (ii) and (iii)]
m(arc PMQ) = 360\u00b0 \u2013 m(arc PXQ) (v)
and m(arc RNS) = 360\u00b0 \u2013 m(arc RXS) (vi)
\u2234 360\u00b0- m(arc PXQ) = 360\u00b0- m(arc RXS) [From (iv), (v) and (vi)]
\u2234 m(arc PXQ) = m(arc RXS)<\/p>\n\n\n\n

ii. Yes, the major arcs corresponding to congruent chords (diameters) are congruent.
Given: O is the centre of circle.
seg PQ and seg RS are the diameters.
To prove: arc PYQ \u2245 arc RYS
\"Maharashtra
Proof:
seg PQ and seg RS are the diameters of the same circle. [Given]
\u2234 arc PYQ and arc RYS are semicircular arcs.
\u2234 m(arc PYQ) = m(arc RYS) = 180\u00b0 [Measure of a semicircular arc is 180\u00b0]
\u2234 arc PYQ \u2245 arc RYS<\/p>\n\n\n\n

Practice Set 3.4<\/h4>\n\n\n\n

Question 1.
In the adjoining figure, in a circle with centre O, length of chord AB is equal to the radius of the circle. Find measure of each of the following.<\/strong>
i. \u2220AOB
ii. \u2220ACB
iii. arc AB
iv. arc ACB.
\"Maharashtra
Solution:
<\/strong>i. seg OA = seg OB = radius\u2026\u2026 (i) [Radii of the same circle]
seg AB = radius\u2026\u2026 (ii) [Given]
\u2234 seg OA = seg OB = seg AB [From (i) and (ii)]
\u2234 \u2206OAB is an equilateral triangle.
\u2234 m\u2220AOB = 60\u00b0 [Angle of an equilateral triangle]
ii. m \u2220ACB = 12 m \u2220AOB [Measure of an angle subtended by an arc at a point on the circle is half of the measure of the angle subtended by the arc at the centre]
= 12 \u00d7 60\u00b0
\u2234 m \u2220ACB = 30\u00b0
iii. m(arc AB) = m \u2220AOB [Definition of measure of minor arc]
\u2234 m(arc AB) = 60\u00b0
iv. m(arc ACB) + m(arc AB) = 360\u00b0 [Measure of a circle is 360\u00b0]
\u2234 m(arc ACB) = 360\u00b0 \u2013 m(arc AB)
= 360\u00b0 \u2013 60\u00b0
\u2234 m(arc ACB) = 300\u00b0<\/p>\n\n\n\n

Question 2.
In the adjoining figure, \ua838PQRS is cyclic, side PQ \u2245 side RQ, \u2220PSR = 110\u00b0. Find<\/strong>
i. measure of \u2220PQR
ii. m (arc PQR)
iii. m (arc QR)
iv. measure of \u2220PRQ
\"Maharashtra
Solution:
<\/strong>i. \ua838PQRS is a cyclic quadrilateral. [Given]
\u2234 \u2220PSR + \u2220PQR = 180\u00b0 [Opposite angles of a cyclic quadrilateral are supplementary]
\u2234 110\u00b0 + \u2220PQR = 180\u00b0
\u2234 \u2220PQR = 180\u00b0 \u2013 110\u00b0
\u2234 m \u2220PQR = 70\u00b0
ii. \u2220PSR= 12 m (arcPQR) [Inscribed angle theorem]
110\u00b0= 12 m (arcPQR)
\u2234 m(arc PQR) = 220\u00b0
iii. In \u2206PQR,
side PQ \u2245 side RQ [Given]
\u2234 \u2220PRQ = \u2220QPR [Isosceles triangle theorem]
Let \u2220PRQ = \u2220QPR = x
Now, \u2220PQR + \u2220QPR + \u2220PRQ = 180\u00b0 [Sum of the measures of angles of a triangle is 180\u00b0]
\u2234 \u2220PQR + x + x= 180\u00b0
\u2234 70\u00b0 + 2x = 180\u00b0
\u2234 2x = 180\u00b0 \u2013 70\u00b0
\u2234 2x = 110\u00b0
\u2234 x=110\u22182=55\u2218
\u2234 \u2220PRQ = \u2220QPR = 55\u00b0\u2026.. (i)
But, \u2220QPR = 12 m(arc QR) [Inscribed angle theorem]
\u2234 55\u00b0 = 12 m(arc QR)
\u2234 m(arc QR) = 110\u00b0
iv. \u2220PRQ = \u2220QPR =55\u00b0 [From (i)]
\u2234 m \u2220PRQ = 55\u00b0<\/p>\n\n\n\n

Question 3.
\u25a1 MRPN is cyclic, \u2220R = (5x -13)\u00b0, \u2220N = (Ax + 4)\u00b0. Find measures of \u2220R and \u2220N.
Solution:<\/strong>
\u25a1 MRPN is a cyclic quadrilateral. [Given]
\u2234 \u2220R + \u2220N = 180\u00b0 [Opposite angles of a cyclic quadrilateral are supplementary]
\"Maharashtra
\u2234 5x \u2013 13 + 4x + 4 = 180
\u2234 9x \u2013 9 = 180
\u2234 9x = 189
\u2234 x = 1899
\u2234 x = 21
\u2234 \u2220R = 5x \u2013 13
= 5 \u00d7 21 \u2013 13
= 105 \u2013 13
= 92\u00b0
\u2220N = 4x + 4
= 4 \u00d7 21 +4
= 84 +4
= 88\u00b0
\u2234 m\u2220R = 92\u00b0 and m \u2220N = 88\u00b0<\/p>\n\n\n\n

Question 4.
In the adjoining figure, seg RS is a diameter of the circle with centre O. Point T lies in the exterior of the circle. Prove that \u2220RTS is an acute angle.<\/strong>
Given: O is the centre of the circle, seg RS is the diameter of the circle.
To prove: \u2220RTS is an acute angle.
Construction: Let seg RT intersect the circle at point P. Join PS and PT.
\"Maharashtra
Proof:
seg RS is the diameter. [Given]
\u2234 \u2220RPS = 90\u00b0 [Angle inscribed in a semicircle]
Now, \u2220RPS is the exterior angle of \u2206PTS.
\u2234 \u2220RPS > \u2220PTS [Exterior angle is greater than the remote interior angles]
\"Maharashtra
\u2234 90\u00b0 > \u2220PTS
i.e. \u2220PTS < 90\u00b0
i.e, \u2220RTS < 90\u00b0 [R \u2013 P -T]
\u2220RTS is an acute angle.<\/p>\n\n\n\n

Question 5.
Prove that, any rectangle is a cyclic quadrilateral.
Given: \ua838ABCD is a rectangle.
To prove: \ua838ABCD is a cyclic quadrilateral.
\"Maharashtra
Proof:
\ua838XBCD is a rectangle. [Given]
\u2234 \u2220A = \u2220B = \u2220C = \u2220D = 90\u00b0 [Angles of a rectangle]
Now, \u2220A + \u2220C = 90\u00b0 + 90\u00b0
\u2234 \u2220A + \u2220C = 180\u00b0
\u2234 \ua838ABCD is a cyclic quadrilateral. [Converse of cyclic quadrilateral theorem]<\/p>\n\n\n\n

Question 6.
In the adjoining figure, altitudes YZ and XT of \u2206WXY intersect at P. Prove that,<\/strong>
i. \u25a1 WZPT is cyclic.
ii. Points X, Z, T, Y are concyclic.
Given: seg YZ \u22a5 side XW
seg XT \u22a5 side WY
To prove: i. \u25a1WZPT is cyclic.
ii. Points X, Z, T, Y are concyclic.
\"Maharashtra
Proof:
i. segYZ \u22a5 side XW [Given]
\u2234\u2220PZW = 90\u00b0\u2026\u2026 (i)
seg XT I side WY [Given]
\u2234 \u2220PTW = 90\u00b0 \u2026\u2026(ii)
\u2220PZW + \u2220PTW = 90\u00b0 + 90\u00b0 [Adding (i) and (ii)]
\u2234\u2220PZW + \u2220PTW = 180\u00b0
\u2234\u25a1WZPT Ls a cyclic quadrilateral. [Converse of cyclic quadrilateral theorem]
ii. \u2220XZY = \u2220YTX = 90\u00b0 [Given]
\u2234 Points X and Y on line XY subtend equal angles on the same side of line XY.
\u2234 Points X, Z, T and Y are concydic. [If two points on a given line subtend equal angles at two distinct points which lie on the same side of the line, then the four points are concyclic]<\/p>\n\n\n\n

Question 7.
In the adjoining figure, m (arc NS) = 125\u00b0, m(arc EF) = 37\u00b0, find the measure of \u2220NMS.
Solution:<\/strong>
\"Maharashtra
Chords EN and FS intersect externally at point M.
m\u2220NMS = 12 [m (arc NS) \u2013 m(arc EF)]
= 12 (125\u00b0 \u2013 37\u00b0) = 12 \u00d7 88\u00b0
\u2234 m\u2220NMS = 44\u00b0<\/p>\n\n\n\n

Question 8.
In the adjoining figure, chords AC and DE intersect at B. If \u2220ABE = 108\u00b0, m(arc AE) = 95\u00b0, find m (arc DC).<\/strong>
\"Maharashtra
Solution:
<\/strong>Chords AC and DE intersect internally at point B.
\u2234 \u2220ABE = 12 [m(arc AE) + m(arc DC)]
\u2234 108\u00b0 = 12 [95\u00b0 + m(arc DC)]
\u2234 108\u00b0 \u00d7 2 = 95\u00b0 + m(arc DC)
\u2234 95\u00b0 + m(arc DC) = 216\u00b0
\u2234 m(arc DC) = 216\u00b0 \u2013 95\u00b0
\u2234 m(arc DC) = 121\u00b0<\/p>\n\n\n\n

Question 1.
Draw a sufficiently large circle of any radius as shown in the figure below. Draw a chord AB and central \u2220ACB. Take any point D on the major arc and point E on the minor arc.<\/strong>
\"Maharashtra
i. Measure \u2220ADB and \u2220ACB and compare the measures.
ii. Measure \u2220ADB and \u2220AEB. Add the measures.
iii. Take points F, G, H on the arc ADB. Measure \u2220AFB, \u2220AGB, \u2220AHB. Compare these measures with each other as well as with measure of \u2220ADB.
iv. Take any point I on the arc AEB. Measure \u2220AIB and compare it with \u2220AEB. (Textbook pg, no. 64)
Answer:
<\/strong>i. \u2220ACB = 2 \u2220ADB.
ii. \u2220ADB + \u2220AEB = 180\u00b0.
iii. \u2220AHB = \u2220ADB = \u2220AFB = \u2220AGB
iv. \u2220AEB = \u2220AIB<\/p>\n\n\n\n

Question 2.
Draw a sufficiently large circle with centre C as shown in the figure. Draw any diameter PQ. Now take points R, S, T on both the semicircles. Measure \u2220PRQ, \u2220PSQ, \u2220PTQ. What do you observe? (Textbook pg. no.65)<\/strong>
\"Maharashtra
Answer:
<\/strong>\u2220PRQ = \u2220PSQ = \u2220PTQ = 90\u00b0
[Student should draw and verily the above answers.]<\/p>\n\n\n\n

Question 3.
Prove that, if two lines containing chords of a circle intersect each other outside the circle, then the measure of angle between them is half the difference in measures of the arcs intercepted by the angle. (Textbook pg. no. 72)<\/strong>
Given: Chord AB and chord CD intersect at E in the exterior of the circle.
To prove: \u2220AEC = 12 [m(arc AC) \u2013 m(arc BD)]
Construction: Draw seg AD.
\"Maharashtra
Proof:
\u2220ADC is the exterior angle of \u2206ADE.
\u2234 \u2220ADC = \u2220DAE + \u2220AED [Remote interior angle theorem]
\u2234 \u2220ADC = \u2220DAE + \u2220AEC [C \u2013 D \u2013 E]
\u2234 \u2220AEC = \u2220ADC \u2013 \u2220DAE \u2026\u2026(i)
\u2220ADC = 12 m(arc AC) (ii) [Inscribed angle theorem]
\u2220DAE = 12 m(arc BD) (iii) [A \u2013 B \u2013 E, Inscribed angle theorem]
\u2234 \u2220AEC = 12 m(arc AC) \u2013 12 m (arc BD) [From (i), (ii) and (iii)]
\u2234 \u2220AEC = 12 m(arc AC) \u2013 m (arc BD)<\/p>\n\n\n\n

Question 4.
Angles inscribed in the same arc are congruent.
Write \u2018given\u2019 and \u2018to prove\u2019 with the help of the given figure.
Think of the answers of the following questions and write the proof.<\/strong>
i. Which arc is intercepted by \u2220PQR ?
ii. Which arc is intercepted by \u2220PSR ?
iii. What is the relation between an inscribed angle and the arc intercepted by it? (Textbook: pg. no. 68)
\"Maharashtra
Given: C is the centre of circle. \u2220PQR and \u2220PSR are inscribed in same arc PTR.
To prove: \u2220PQR \u2245 \u2220PSR
Proof:
i. arc PTR is intercepted by \u2220PQR.
ii. arc PTR is intercepted by \u2220PSR.
iii. \u2220PQR = 12 m(arc PTR), and (i) [inscribed angle theorem]
\u2220PSR = 12 m(arcPTR) (ii) [Inscribed angle theorem]
\u2234 \u2220PQR \u2245 \u2220PSR [From (i) and (ii)]<\/p>\n\n\n\n

Question 5.
Angle inscribed in a semicircle is a right angle. With the help of given figure write \u2018given\u2019, \u2018to prove\u2019 and \u2018the proof. (Textbook pg. no. 68)<\/strong>
\"Maharashtra
Given: M is the centre of circle. \u2220ABC is inscribed in arc ABC.
Arcs ABC and AXC are semicircles.
To prove: \u2220ABC = 90\u00b0
Proof:
\u2220ABC = 12 m(arc AXC) (i) [Inscribed angle theorem]
arc AXC is a semicircle.
\u2234 m(arc AXC) = 180\u00b0 (ii) [Measure of semicircular arc is 1800]
\u2234 \u2220ABC = 12 \u00d7 180\u00b0
\u2234 \u2220ABC = 90\u00b0 [From (i) and (ii)]<\/p>\n\n\n\n

Question 6.
Theorem: Opposite angles of a cyclic quadrilateral are supplementry.
Fill in the blanks and complete the following proof. (Textbook pg. no. 68)<\/strong>
Given: \u25a1 ABCD is cyclic.
To prove: \u2220B + \u2220D = 180\u00b0
\u2220A + \u2220C = 180\u00b0
\"Maharashtra
Proof:
arc ABC is intercepted by the inscribed angle \u2220ADC.
\u2234 \u2220ADC m(arcABC) (i) [Inscribed angle theorem]
Similarly, \u2220ABC is an inscribed angle. It intercepts arc ADC.
\u2234 ABC = 12 m(arc ADC) (ii) [Inscribed angle theorem]
\u2234 \u2220ADC + \u2220ABC
= 12 m(arcABC) + 12 m(arc ADC) [Adding (i) and (ii)]
\u2234 \u2220D + \u2220B = 12 m(areABC) + m(arc ADC)]
\u2234 \u2220B + \u2220D = 12 \u00d7 360\u00b0 [arc ABC and arc ADC constitute a complete circle]
= 180\u00b0
\u2234 \u2220B + \u2220D = 180\u00b0
Similarly we can prove,
\u2220A + \u2220C = 180\u00b0<\/p>\n\n\n\n

Question 7.
In the above theorem, after proving \u2220B + \u2220D = 180\u00b0, can you use another way to prove \u2220A + \u2220C = 180\u00b0? (Textbook pg. no. 69)<\/strong>
Proof:
Yes, we can prove \u2220A + \u2220C = 180\u00b0 by another way.
\u2220B + \u2220D = 180\u00b0
In \ua838ABCD,
\u2220A + \u2220B + \u2220C + \u2220D = 360\u00b0 [Sum of the measures of all angles of a quadrilateral is 360\u00b0.]
\u2234 \u2220A + \u2220C + 180\u00b0 = 360\u00b0
\u2234 \u2220A + \u2220C = 360\u00b0 \u2013 180\u00b0
\u2234 \u2220A + \u2220C = 180\u00b0<\/p>\n\n\n\n

Question 8.
An exterior angle of a cyclic quadrilateral is congruent to the angle opposite to its adjacent interior angle. (Textbook pg. no. 69)<\/strong>
Given: \ua838ABCD is a cyclic quadrilateral.
\u2220BCE is the exterior angle of \ua838ABCD.
To prove: \u2220BCE \u2245 \u2220BAD
\"Maharashtra
Proof:
\u2220BCE + \u2220BCD = 180\u00b0\u2026\u2026 (i) [Angles in a linear pair]
\ua838ABCD is a cyclic quadrilateral. [Given]
\u2220BAD + \u2220BCD = 180\u00b0\u2026\u2026\u2026. (ii) [Opposite angles of a cyclic quadrilateral are supplementary]
\u2234 \u2220BCE + \u2220BCD = \u2220BAD + \u2220BCD [From (i) and (ii)]
\u2234 \u2220BCE = \u2220BAD<\/p>\n\n\n\n

Question 9.
Theorem : If a pair of opposite angles of a quadrilateral is supplementary, then the quadrilateral is cyclic. (Textbook pg. no. 69)<\/strong>
Given: In \ua838ABCD, \u2220A + \u2220C = 180\u00b0
To prove: \ua838ABCD is a cyclic quadrilateral.
Proof:
(Indirect method)
Suppose \ua838ABCD is not a cyclic quadrilateral.
We can still draw a circle passing through three non collinear points A, B, D.
\"Maharashtra
Case I: Point C lies outside the circle.
Then, take point E on the circle
such that D \u2013 E \u2013 C.
\u2234 \ua838ABED is a cyclic quadrilateral.
\u2220DAB + \u2220DEB = 180\u00b0 (i) [Opposite angles of a cyclic quadrilateral are supplementary]
\u2220DAB + \u2220DCB = 180\u00b0 (ii) [Given]
\u2234 \u2220DAB + \u2220DEB = \u2220DAB + \u2220DCB [From (i) and (ii)]
\u2234 \u2220DEB = \u2220DCB
\"Maharashtra
But, \u2220DEB \u2260 \u2220DCB as \u2220DEB is the exterior angle of \u2206BEC.
\u2234 Our supposition is wrong.
\u2234 \ua838ABCD is a cyclic quadrilateral.
Case II: Point C lies inside the circle.
Then, take point E on the circle such that
D \u2013 C \u2013 E
\u2234 \u25a1ABED is a cyclic quadrilateral.
\u2220DAB + \u2220DEB = 180\u00b0 (iii) [Opposite angles of a cyclic quadrilateral are supplementary]
\u2220DAB + \u2220DCB = 180\u00b0 (iv) [Given]
\u2234 \u2220DAB + \u2220DEB = \u2220DAB + \u2220DCB [From (iii) and (iv)]
\u2234 \u2220DEB = \u2220DCB
But \u2220DEB \u2260 \u2220DCB as \u2220DCB is the exterior angle of \u2206BCE.
\u2234 Our supposition is wrong.
\u2234 \u25a1ABCD is a cyclic quadrilateral.<\/p>\n\n\n\n

Question 10.
Theorem: If two points on a given line subtend equal angles at two distinct points which lie on the same side of the line, then the four points are concyclic. (Textbook pg. no. 70)<\/strong>
Given: Points B and C lie on the same side of the line AD.
\u2220ABD = \u2220ACD
To prove: Points A, B, C, D are concyclic. i.e., \u25a1ABCD is a cyclic quadrilateral.
\"Maharashtra
Proof:
Suppose points A, B, C, D are not concyclic points.
We can still draw a circle passing through three non collinear points A, B, D.
Case I: Point C lies outside the circle.
Then, take point E on the circle such that
A \u2013 E \u2013 C.
\u2220ABD \u2245 \u2220AED (i) [Angles inscribed in the same arc]
\u2220ABD \u2245 \u2220ACD (ii) [Given]
\u2234 \u2220AED \u2245 \u2220ACD [From (i) and (ii)]
\u2234 \u2220AED \u2245 \u2220ECD [A \u2013 E \u2013 C]
\"Maharashtra
But, \u2220AED \u2245 \u2220ECD as \u2220AED is the exterior angle of \u2206ECD.
\u2234 Our supposition is wrong.
\u2234 Points A, B, C, D are concyclic points.
Case II: Point C lies inside the circle. Then, take point E on the circle such that A \u2013 C \u2013 E.
\u2220ABD \u2245 \u2220AED (iii) [Angles inscribed in the same arc]
\u2220ABD \u2245 \u2220ACD (iv) [Given]
\u2234 \u2220AED \u2245 \u2220ACD [From (iii) and (iv)]
\u2234 \u2220CED \u2245 \u2220ACD [A \u2013 C \u2013 E]
\"Maharashtra
But, \u2220CED \u2245 \u2220ACD as \u2220ACD is the exterior angle of \u2206ECD.
\u2234 Our supposition is wrong.
\u2234 Points A, B, C, D are concyclic points.<\/p>\n\n\n\n

Question 11.
The above theorem is converse of a certain theorem. State it. (Textbook pg. no. 70)
Answer:<\/strong>
If four points are concyclic, then the line joining any two points subtend equal angles at the other two points which are on the same side of that line.<\/p>\n\n\n\n

Practice Set 3.5<\/h4>\n\n\n\n

Question 1.
In the adjoining figure, ray PQ touches the circle at point Q. PQ = 12, PR = 8, find PS and RS.
Solution:<\/strong>
i. Ray PQ is a tangent to the circle at point Q and seg PS is the secant. [Given]
\u2234 PR \u00d7 PS = PQ2<\/sup> [Tangent secant segments theorem]
\"Maharashtra
\u2234 8 \u00d7 PS = 122<\/sup>
\u2234 8 \u00d7 PS = 144
\u2234 PS = 1448
\u2234 PS = 18 units
ii. Now, PS = PR + RS [P \u2013 R \u2013 S]
\u2234 18 = 8 + RS
\u2234 RS = 18 \u2013 8
\u2234 RS = 10 units<\/p>\n\n\n\n

Question 2.
In the adjoining figure, chord MN and chord RS intersect at point D.<\/strong>
i. If RD = 15, DS = 4, MD = 8 find DN
ii. If RS = 18, MD = 9, DN = 8 find DS
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Solution:
<\/strong>i. Chords MN and RS intersect internally at point D. [Given]
\u2234 MD \u00d7 DN = RD \u00d7 DS [Theorem of internal division of chords]
\u2234 8 \u00d7 DN = 15 \u00d7 4
\u2234 DN = 15\u00d748
\u2234 DN = 7.5 units
ii. Let the value of RD be x.
RS = RD + DS [R \u2013 D \u2013 S]
\u2234 18 = x + DS
\u2234 DS = 18 \u2013 x
Now, MD \u00d7 DN = RD \u00d7 DS [Theorem of internal division of chords]
\u2234 9 \u00d7 8 = x(18 \u2013 x)
\u2234 72 = 18x \u2013 x2<\/sup>
\u2234 x2 \u2013 18x + 72 = 0
\u2234 x2 \u2013 12x \u2013 6x + 72 = 0
\u2234 x (x \u2013 12) \u2013 6 (x \u2013 12) = 0
\u2234 (x \u2013 12) (x \u2013 6) = 0
\u2234 x \u2013 12 = 0 or x \u2013 6 = 0
\u2234 x = 12 or x = 6
\u2234 DS = 18 \u2013 12 or DS = 18 \u2013 6
\u2234 DS = 6 units or DS = 12 units<\/p>\n\n\n\n

Question 3.
In the adjoining figure, O is the centre of the circle and B is a point of contact. Seg OE \u22a5 seg AD, AB = 12, AC = 8, find<\/strong>
i. AD
ii. DC
iii. DE.
\"Maharashtra
Solution:
<\/strong>i. Line AB is the tangent at point B and seg AD is the secant. [Given]
\u2234 AC \u00d7 AD = AB2<\/sup> [Tangent secant segments theorem]
\u2234 8 \u00d7 AD = 122
\u2234 8 \u00d7 AD = 144
\u2234 AD = 1448
\u2234 AD = 18 units
ii. AD = AC + DC [A \u2013 C \u2013 D]
\u2234 18 = 8 + DC
\u2234 DC = 18 \u2013 8
\u2234 DC = 10 units
iii. seg OE \u22a5 seg AD [Given]
i.e. seg OE \u22a5 seg CD [A \u2013 C \u2013 D]
\u2234 DE = 12 DC [Perpendicular drawn from the centre of the circle to the chord bisects the chord]
= 12 \u00d7 10
\u2234 DE = 5 units<\/p>\n\n\n\n

Question 4.
In the adjoining figure, if PQ = 6, QR = 10, PS = 8, find TS.
Solution:<\/strong>
PR = PQ + QR [P-Q-R]
\u2234 PR = 6 + 10 = 16 units
\"Maharashtra
Chords TS and RQ intersect externally at point P.
PQ \u00d7 PR = PS \u00d7 PT [Theorem of external division of chords]
\u2234 6 \u00d7 16 = 8 \u00d7 PT
\u2234 PT = 6\u00d7168 = 12 units
But, PT = PS + TS [P \u2013 S \u2013 T]
\u2234 12 = 8 + TS
\u2234 TS = 12 \u2013 8
\u2234 TS = 4 units<\/p>\n\n\n\n

Question 5.
In the adjoining figure, seg EF is a diameter and seg DF is a tangent segment. The radius of the circle is r. Prove that, DE \u00d7 GE = 4r2<\/sup>.<\/strong>
Given: seg EF is the diameter.
seg DF is a tangent to the circle,
radius = r
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To prove: DE \u00d7 GE = 4r2<\/sup>
Construction: Join seg GF.
Proof:
seg EF is the diameter. [Given]
\u2234 \u2220EGF = 90\u00b0 (i) [Angle inscribed in a semicircle]
seg DF is a tangent to the circle at F. [Given]
\"Maharashtra
\u2234 \u2220EFD = 90\u00b0 (ii) [Tangent theorem]
In \u2206DFE,
\u2220EFD = 90 \u00b0 [From (ii)]
seg FG \u22a5 side DE [From (i)]
\u2234 \u2206EFD ~ \u2206EGF [Similarity of right angled triangles]
\u2234 EFGE = DEEF [Corresponding sides of similar triangles]
\u2234 DE \u00d7 GE = EF2<\/sup>
\u2234 DE \u00d7 GE = (2r)2<\/sup> [diameter = 2r]
\u2234 DE \u00d7 GE = 4r2<\/sup><\/p>\n\n\n\n

Question 1.
Theorem: If an angle has its vertex on the circle, its one side touches the circle and the other intersects the circle in one more point, then the measure of the angle is half the measure of its intercepted arc. (Textbook pg.no. 75 and 76)<\/strong>
\"Maharashtra
Given: \u2220ABC is any angle, whose vertex B lies on the circle with centre M.
Line BC is tangent at B and line BA is secant intersecting the circle at point A.
Arc ADB is intercepted by \u2220ABC.
To prove: \u2220ABC = 12 m(arc ADB)
Proof:
Case I: Centre M lies on arm BA of \u2220ABC.
\u2220MBC = 90\u00b0 [Trangnet theorem]
\"Maharashtra
i.e. \u2220ABC 90\u00b0 (i) [A \u2013 M \u2013 B]
arc ADB is a semicircular arc.
\u2234 m(arc ADB) = 180\u00b0 (ii) [Measure ofa semicircle is 180\u00b0]
\u2234 \u2220ABC = 12 m(arc ADB) [(From (i) and (ii)]<\/p>\n\n\n\n

Case II: Centre M lies in the exterior of \u2220ABC.
Draw radii MA and MB.
\u2234 \u2220MBA = \u2220MAB [Isosceles triangle theorem]
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Let, \u2220MHA = \u2220MAB =x, \u2220ABC = y In \u2206ABM,
\u2220AMB + \u2220MBA + \u2220MAB = 180\u00b0 [Sum of the measures of all the angles of a triangle is 1800]
\u2234 \u2220AMB + x + x = 180\u00b0
\u2234 \u2220AMB = 180\u00b0 \u2013 2x \u2026\u2026 (i)
Now, \u2220MBC = \u2220MBA + \u2220ABC [Angle addition property]
\u2234 90\u00b0 = x + y [Tangent theorem]
\u2234 x = 90\u00b0 \u2013 y \u2026\u2026(ii)
\u2220AMB = 180\u00b0 \u2013 2 (90\u00b0 \u2013 y) [From (i) and (ii)]
\u2234 \u2220AMB = 180\u00b0 \u2013 180\u00b0 + 2y
\u2234 2y = \u2220AMB
\u2234 y = 12 \u2220AMB
\u2234 \u2220ABC = 12 \u2220AMB
\u2234 \u2220ABC = 12 m(arc ADB) [Definition of measure of minor arc]<\/p>\n\n\n\n

Case III: Centre M lies in the interior of \u2220ABC.
Ray BE is the opposite ray of ray BC.
Now, \u2220ABE = 12 m (arc AFB) (i) [Proved in case II]
\"Maharashtra
\u2220ABC + \u2220ABE = 180\u00b0 [Angles in a linear pair]
\u2234 180 \u2013 \u2220ABC = \u2220ABE
\u2234 180 \u2013 \u2220ABC = 12 m(arc AFB) [From (i)]
= 12 [360 \u2013 m (arc ADB)]
\u2234 180 \u2013 \u2220ABC = 180 \u2013 12 m(arc ADB)
\u2234 -\u2220ABC = \u2013 12 m(arc ADB)
\u2234 \u2220ABC = 12 m(arc ADB)<\/p>\n\n\n\n

Question 2.
We have proved the above theorem by drawing seg AC and seg DB. Can the theorem be proved by drawing seg AD and seg CB, instead of seg AC and seg DB? (Textbook pg. no. 77)
Solution:
<\/strong>Yes, the theor<\/strong>em can be proved by drawing seg AD and seg CB.
Given: P is the centre of circle, chords AB and CD intersect internally at point E.
To prove: AE \u00d7 EB = CE \u00d7 ED
Construction: Draw seg AD and seg CB.
\"Maharashtra
Proof:
In \u2206CEB and \u2206AED,
\u2220CEB = \u2220DEA [Vertically opposite angles]
\u2220CBE = \u2220ADE [Angles inscribed in the same arc]
\u2234 \u2206CEB ~ \u2206AED [by AA test of similarity]
\u2234 CEAE = EBED [Corresponding sides of similar triangles]
\u2234 AE \u00d7 EB = CE \u00d7 ED<\/p>\n\n\n\n

Question 3.
In figure, seg PQ is a diameter of a circle with centre O. R is any point on the circle, seg RS \u22a5 seg PQ. Prove that, SR is the geometric mean of PS and SQ. [That is, SR2<\/sup> = PS \u00d7 SQ] (Textbook pg. no. 81)<\/strong>
Given: seg PQ is the diameter.
seg RS \u22a5 seg PQ
To prove: SR2<\/sup> = PS \u00d7 SQ
Construction: Extend ray RS, let it intersect the circle at point T.
\"Maharashtra
Proof:
seg PQ \u22a5 seg RS [Given]
\u2234 seg OS \u22a5 chord RT [R \u2013 S \u2013 T, P \u2013 S \u2013 O]
\u2234 segSR = segTS (i) [Perpendicular drawn from the centre of the circle to the chord bisects the chord]
Chords PQ and RT intersect internally at point S.
\u2234 SR \u00d7 TS = PS \u00d7 SQ [Theorem of internal division of chords]
\u2234 SR \u00d7 SR = PS \u00d7 SQ [From (i)]
\u2234 SR2<\/sup> = PS \u00d7 SQ<\/p>\n\n\n\n

Question 4.
Theorem: If secants containing chords AB and CD of a circle intersect outside the circle in point E, then
AE \u00d7 EB = CE \u00d7 ED. (Textbook pg. no. 78)<\/strong>
Given: Chords AB and CD of a circle intersect outside the circle in point E.
To prove: AE \u00d7 EB = CE \u00d7 ED
Construction: Draw seg AD and seg BC.
\"Maharashtra
Proof:
In \u2206ADE and \u2206CBE,
\u2220AED = \u2220CEB [Common angle]
\u2220DAE \u2245 \u2220BCE [Angles inscribed in the same arc]
\u2234 \u2206ADE ~ \u2206CBE [AA testof similaritv]
\u2234 AECE = EDEB [Corresponding sides of similar triangles]
\u2234 AE \u00d7 EB = CE \u00d7 ED<\/p>\n\n\n\n

Question 5.
Theorem: Point E is in the exterior of a circle. A secant through E intersects the circle at points A and B, and a tangent through E touches the circle at point T, then EA \u00d7 EB = ET2<\/sup>.<\/strong>
Given: Secant through point E intersects the circle in points A and B.
Tangent drawn through point E touches the circle in point T.
To prove: EA \u00d7 EB = ET2<\/sup>
Construction: Draw seg TA and seg TB.
\"Maharashtra
Proof:
In \u2206EAT and \u2206ETB,
\u2220AET \u2245 \u2220TEB [Common angle]
\u2220ETA \u2245 \u2220EBT [Theorem of angle between tangent and secant, E \u2013 A \u2013 B]
\u2234 \u2206EAT ~ \u2206ETB [AA test of similarity]
\u2234 EAET = ETEB [Corresponding sides of similar triangles]
\u2234 EA \u00d7 EB = ET2<\/sup><\/p>\n\n\n\n

Question 6.
In the figure in the above example, if seg PR and seg RQ are drawn, what is the nature of \u2206PRQ. (Textbook pg. no, 81)
Answer:<\/strong>
seg PQ is the diameter of the circle.
\u2234 \u2220PRQ = 90\u00b0
\"Maharashtra
\u2234 \u2206PRQ is a right angled triangle. [Angle inscribed in a semicircle]<\/p>\n\n\n\n

Question 7.
Have you previously proved the property proved in the above example? (Textbook pg. no. 81)
Answer:<\/strong>
Yes. It is the theorem of geometric mean.
\u2206PSR ~ \u2206RSQ [Similarity of right angled triangles]
\"Maharashtra
\u2234 PSSR = SRSQ [Corresponding sides of similar triangles]
\u2234 SR2<\/sup> = PS \u00d7 SQ<\/p>\n\n\n\n

Download PDF<\/strong><\/h2>\n\n\n\n

Maharashtra Board Solutions for Class 10-Maths (Part 2): Chapter 3- Circle<\/p>\n\n\n\n

Download PDF<\/strong>: Maharashtra Board Solutions for Class 10-Maths (Part 2): Chapter 3- Circle PDF<\/a><\/p>\n\n\n\n

Chapterwise Maharashtra Board Solutions Class 10 Maths (Part 2) :<\/strong><\/h2>\n\n\n\n