Maharashtra Board Solutions for Class 10-Maths (Part 2): Chapter 3- Circle

Class 10: Maths Chapter 3 solutions. Complete Class 10 Maths Chapter 3 Notes.

Maharashtra Board Solutions for Class 10-Maths (Part 2): Chapter 3- Circle

Maharashtra Board 10th Maths Chapter 3, Class 10 Maths Chapter 3 solutions

Practice Set 3.1

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.
i. What is the measure of ∠CAB? 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 ∠ABC? Why?

Solution:
i. line AB is the tangent to the circle with centre C and radius AC. [Given]
∴ ∠CAB = 90° (i) [Tangent theorem]
ii. seg CA ⊥ line AB [From (i)]
radius = l(AC) = 6 cm
∴ The distance of point C from line AB is 6 cm.
iii. In ∆CAB, ∠CAB = 90° [From (i)]
∴ BC² = AB² + AC² . [Pythagoras theorem]

= 6² + 6²
= 2 × 6²
∴ BC = 2×62−−−−−√ [Taking square root of both sides]
= 6 2–√ cm
∴ d(B, C) = 6 cm
iv. In ∆ABC,
AC = AB = 6cm
∴ ∠ABC = ∠ACB [Isosceles triangle theorem]
Let ∠ABC = ∠ACB =x
In ∆ABC,
∠CAB + ∠ABC + ∠ACB = 180° [Sum of the measures of angles of a triangle is 180°]
∴ 90° + x + x = 180°
∴ 90 + 2x = 180°
∴ 2x = 180°- 90°
∴ x = 90∘2
∴ x = 45°
∴ ∠ABC = 45°

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
i. What is the length of each tangent segment?
ii. What is the measure of ∠MRO?
iii. What is the measure of ∠MRN?

Solution:
seg RM and seg RN are tangents to the circle with centre O. [Given]
∴ ∠OMR = ∠ONR = 90° [Tangent theorem]

i. In ∆OMR, ∠OMR = 90°
∴ OR² = OM² + RM² [Pythagoras theorem]
∴ 10² = 5² + RM²
∴ 100 = 25 + RM²
∴ RM² = 75
∴ RM = 75−−√ [Taking square root of both sides]
∴ RM = RN [Tangent segment theorem]
Length of each tangent segment is 5 3–√ cm.
ii. In ∆RMO,
∠OMR = 90° [Tangent theorem]
OM = 5 cm and OR = 10 cm
∴ OM = 12 OR
∴ ∠MRO = 30° (i) [Converse of 30° – 60° – 90° theorem]
Similarly, ∠NRO = 30°
iii. But, ∠MRN = ∠MRO + ∠NRO [Angle addition property]
= 30° + 30° [From (i)]
∴ ∠MRN = 60°

Question 3.
Seg RM and seg RN are tangent segments of a circle with centre O. Prove that seg OR bisects ∠MRN as well as ∠MON.

Solution:
Proof:
In ∆OMR and ∆ONR,
seg RM ≅ seg RN [Tangent segment theorem]
seg OM ≅ seg ON [Radii of the same circle]
seg OR ≅ seg OR [Common side]
∴ ∆OMR ≅ ∆ONR [SSS test of congruency]
{∴ ∠MRO ≅ ∠NRO
∠MOR ≅ ∠NOR } [c.a.c.t.]
∴ seg OR bisects ∠MRN and ∠MON.

Question 4.
What is the distance between two parallel tangents of a circle having radius 4.5 cm? Justify your answer.
Solution:
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.

OM = ON = 4.5 [Given]
line AB || line RS [Construction]
line PQ || line RS [Given]
∴ line AB || line PQ || line RS
Now, ∠OMP = ∠ONR = 90° (i) [Tangent theorem]
For line PQ || line AB,
∠OMP = ∠AON = 90° (ii) [Corresponding angles and from (i)]
For line RS || line AB,
∠ONR = ∠AOM = 90° (iii) [Corresponding angles and from (i)]
∠AON + ∠AOM = 90° + 90° [From (ii) and (iii)]
∴ ∠AON + ∠AOM = 180°
∴ ∠AON and ∠AOM form a linear pair.
∴ ray OM and ray ON are opposite rays.
∴ Points M, O, N are collinear. (iv)
∴ MN = OM + ON [M – O – N, From (iv)]
∴ MN = 4.5 + 4.5
∴ MN = 9 cm
∴ Distance between two parallel tangents PQ and RS is 9 cm.

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)

Solution:
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]

12 × 24 = 12 units
Also, seg OP ⊥ chord QR [The segment joining centre of a circle and midpoint of a chord is perpendicular to the chord]
In ∆OPQ, ∠OPQ = 90°
∴ OQ² = OP² + QP² [Pythagoras theorem]
= 10² + 12²
= 100 + 144
= 244
∴ OQ = 244−−−√ = 261−−√ units.
∴ The radius of the circle is 261−−√ units.

Question 2.
In the adjoining figure, M is the centre of the circle and seg AB is a diameter, seg MS ⊥ chord AD, seg MT ⊥ chord AC, ∠DAB ≅ ∠CAB.
i. Prove that: chord AD ≅ 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.

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:
Proof:
i. ∠DAB ≅ ∠CAB [Given]
∴ ∠SAM ≅ ∠TAM (i) [A – S – D, A – M – B, A -T – C]
In ∆SAM and ∆TAM,
∠SAM ≅ ∠TAM [From (i)]
∠ASM ≅ ∠ATM [Each angle is of measure 90°]
seg AM ≅ seg AM [Common side]
∴ ∆SAM ≅ ∆TAM [AAS [SAA] test of congruency]
∴ side MS ≅ side MT [c.s.c.t]
But, seg MS ⊥ chord AD [Given]
seg MT ⊥chord AC
∴ chord AD ≅ 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.

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:

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.
∴ PA = PB … [Perpendicular bisector theorem]
∴ PA = PB = radius
∴ The circle with centre P and radius PA passes through point B.

ii. The circle with any other point Q and radius QA is drawn.
QA = QB = radius … [Perpendicular bisector theorem]
∴ The circle with centre Q and radius QA passes through point B.

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)

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:

i. Let points A, B, C be any three non collinear points.
Draw the perpendicular bisector of seg AB (line l).
∴ Points A and B are equidistant from any point of line l ….(i)[Perpendicular bisector theorem]
Draw the perpendicular bisector of seg BC (line m) to intersect line l at point P.
∴ Points B and C are equidistant from any point of line m ….(ii) [Perpendicular bisector theorem]
∴ PA = PB …[From (i)]
PB = PC … [From (ii)]
∴ PA = PB = PC = radius
∴ 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.

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:

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.
∴ 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.

Question 6.
Which theorem do we use in proving that hypotenuse is the longest side of a right angled triangle? (Textbook pg. no. 52)

Solution:
In ∆ABC,
∠ABC = 90°
∴ ∠BAC < 90° and ∠ACB < 90° [Given]
∴ ∠ABC > ∠BAC and ∠ABC > ∠ACB
∴ AC > BC and AC > AB [Side opposite to greater angle is greater]
∴ 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.

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 ≅ seg DQ
Construction: Draw seg AP and seg AQ.

Solution:
Proof:
In ∆PAD and ∆QAD,
seg PA ≅ [segQA] [Radii of the same circle]
seg AD ≅ seg AD [Common side]

∠APD = ∠AQD = 90° [Tangent theorem]
∴ ∆PAD = ∆QAD [By Hypotenuse side test]
∴ seg DP = seg DQ [c.s.c.t]

Practice Set 3.2

Question 1.
Two circles having radii 3.5 cm and 4.8 cm touch each other internally. Find the distance between their centres.
Solution:
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

The two circles touch each other internally.
∴ By theorem of touching circles,
P – Q – R
PQ = PR – QR
= 4.8 – 3.5
= 1.3 cm
[The distance between the centres of circles touching internally is equal to the difference in their radii]

Question 2.
Two circles of radii 5.5 cm and 4.2 cm touch each other externally. Find the distance between their centres.
Solution:
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

The two circles touch each other externally.
∴ By theorem of touching circles,
P – Q – 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]

Question 3.
If radii of two circles are 4 cm and 2.8 cm. Draw figure of these circles touching each other
i. externally
ii. internally.
Solution:
i. Circles touching externally:

ii. Circles touching internally:

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 –

i. seg AP || seg BQ,
ii. ∆APR ~ ∆RQB, and
iii. Find ∠RQB if ∠PAR = 35°.
Solution:
The circles with centres P and Q touch each other at R.
∴ By theorem of touching circles,
P – R – Q
i. In ∆PAR,
seg PA = seg PR [Radii of the same circle]
∴ ∠PRA ≅ ∠PAR (i) [Isosceles triangle theorem]
Similarly, in ∆QBR,
seg QR = seg QB [Radii of the same circle]
∴ ∠RBQ ≅ ∠QRB (ii) [Isosceles triangle theorem]
But, ∠PRA ≅ ∠QRB (iii) [Vertically opposite angles]
∴ ∠PAR ≅ ∠RBQ (iv) [From (i) and (ii)]
But, they are a pair of alternate angles formed by transversal AB on seg AP and seg BQ.
∴ seg AP || seg BQ [Alternate angles test]
ii. In ∆APR and ∆RQB,
∠PAR ≅ ∠QRB [From (i) and (iii)]
∠APR ≅ ∠RQB [Alternate angles]
∴ ∆APR – ∆RQB [AA test of similarity]
iii. ∠PAR = 35° [Given]
∴ ∠RBQ = ∠PAR= 35° [From (iv)]
In ∆RQB,
∠RQB + ∠RBQ + ∠QRB = 180° [Sum of the measures of angles of a triangle is 180°]
∴ ∠RQB + ∠RBQ + ∠RBQ = 180° [From (ii)]
∴ ∠RQB + 2 ∠RBQ = 180°
∴ ∠RQB + 2 × 35° = 180°
∴ ∠RQB + 70° = 180°
∴ ∠RQB = 110°

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.

Construction : Draw seg AF ⊥ seg BD.
Solution:
i. The circles with centres A and B touch each other at E. [Given]
∴ By theorem of touching circles,
A – E – B

∴ ∠ACD = ∠BDC = 90° [Tangent theorem]
∠AFD = 90° [Construction]
∴ ∠CAF = 90° [Remaining angle of ꠸AFDC]
∴ ꠸AFDC is a rectangle. [Each angle is of measure 900]
∴ AC = DF = 4 cm [Opposite sides of a rectangle]
Now, BD = BF + DF [B – F – C]
∴ 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]

ii. Now, in ∆AFB, ∠AFB = 90° [Construction]
∴ AB² = AF² + BF² [Pythagoras theorem]
∴ 10² = AF² + 2²
∴ 100 = AF² + 4
∴ AF² = 96
∴ AF = 96−−√ [Taking square root of both sides]
= 16×6−−−−−√
= 4 6–√ cm
But, CD = AF [Opposite sides of a rectangle]
∴ CD = 4 6–√ cm

Question 1.
Take three collinear points X – Y – 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:

Line l is a common tangent of the two circles.

Question 2.
Take points Y – X – 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:

Line l is a common tangent of the two circles.

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.

The point common to the circles and the line is called their common point of contact.

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

2. Circles touching internally:
For circles touching internally, the distance between their centres is equal to difference of their radii,
i. e. AB = AC – BC

Question 3.
The circles shown in the given figure are called externally touching circles. Why? (iexthook pg. no. 57)

Answer:
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 – T – S].
Hence the given circles are externally touching circles.

Question 4.
The circles shown in the given figure are called internally touching circles, why? (Textbook pg. no. 57)

Answer:
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 – N – M].
Hence, the given circles are internally touching circles.

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)

Solution:
i. Here, circle with centres A and B touch each other externally at point C.
∴ d(A, B) = d(A, C) + d(B ,C)
= 3 + 4
∴ 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.
∴ d(A, B) = d(A, C) – d(B, C)
= 4 – 3
∴ d(A,B) = 1 cm
[The distance between the centres of circles touching internally is equal to the difference in their radii]

Practice Set 3.3

Question 1.
In the adjoining figure, points G, D, E, F are concyclic points of a circle with centre C.
∠ECF = 70°, m(arc DGF) = 200°. Find m(arc DE) and m(arc DEF).

Solution:
m(arc EF) = m∠ECF [Definition of measure of minor arc]
∴ m(arc EF) = 70°
i. m(arc DE) + m(arc DGF)
+ m(arc EF) = 360° [Measure of a circle is 360°]
∴ m(arc DE) = 360° – m(arc DGF) – m(arc EF)
= 360° – 200° – 70°
∴ m(arc DE) = 90°
ii. m(arc DEF) = m(arc DE) + m(arc EF) [Arc addition property]
= 90° + 70°
∴ m(arc DEF) = 160°

Question 2.
In the adjoining figure, AQRS is an equilateral triangle. Prove that,
i. arc RS ≅ arc QS ≅ arc QR
ii. m(arc QRS) = 240°.

Solution:
Proof:
i. ∆QRS is an equilateral triangle, [Given]
∴ seg RS ≅ seg QS ≅ seg QR [Sides of an equilateral triangle]
∴ arc RS ≅ arc QS ≅ 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° [Measure of a circle is 360°, arc addition property]
∴ x + x + x = 360°
∴ 3x = 360°
∴ x = 360∘3=120∘
∴ m(arc RS) = m(arc QS) = m(arc QR) = 120° (i)
Now, m(arc QRS) = m(arc QR) + m(arc RS) [Arc addition property]
= 120° + 120° [From (i)]
∴ m(arc QRS) = 240°

Question 3.
In the adjoining figure, chord AB ≅ chord CD. Prove that, arc AC = arc BD.

Solution:
Proof:
chord AB ≅ chord CD [Given]
∴ arc AB ≅ arc CD [Corresponding arcs of congruents chords of a circle are congruent]
∴ m(arc AB) = m(arc CD)
∴ m(arc AC) + m(arc BC) = m(arc BC) + m(arc BD) [Arc addition property]
∴ m(arc AC) = m(arc BD)
∴ arc AC ≅ arc BD

Question 1.
Theorem : The chords corresponding to congruent arcs of a circle (or congruent circles) are congruent. (Textbook pg. no. 61)
Given: B is the centre of circle.
arc APC ≅ arc DQE
To prove: chord AC ≅ chord DE

Proof:
[m(arc APC) = ∠ABC (i) [Definition of measure of
m(arc DQE) = ∠DBE] (ii) minor arc]
arc APC ≅ arc ∠DQE (iii) [Given]
∴ ∠ABC ≅ ∠DBE [From (i), (ii) and (iii)]
In ∆ABC and ∆DBE,
side AB ≅ side DB [Radil of the same circle]
side [CB] side [EB] [Radii of the same circle]
∠ABC ≅∠DBE [From (iii), Measures of congruent arcs]
∴ ∆ABC ≅ ∆DBE [SAS test of congruency]
∴ chord AC ≅ chord DE [c.s.c.t]

Question 2.
Theorem: Corresponding arcs of congruent chords of a circle (or congruent circles) are congruent (Textbook pg. no. 61)
Given: O is the centre of circle, chord PQ = chord RS
To prove: arc PMQ = arc RNS

Proof:
In ∆POQ and ∆ROS,
[side PO ≅ side RO
side OQ ≅ side OS] [Radii of the same circle]
chord PQ ≅ chord RS [Given]
∴ ∆POQ ≅ ∆ROS [SSS test of congruency]
∴ ∠POQ ≅ ∠ROS (i) [c.a.c.t.]
m(arc PMQ) = ∠POQ (ii)
m(arc RNS) = ∠ROS (iii) [Definition of measure of minor arc]
∴ arc PMQ ≅ arc RNS [From (i), (ii) and (iii)]

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 ≅ arc DQE
To prove: chord AC ≅ chord DE

Proof:
[m(arc APC) = ∠ABC (i)
m(arc DQE) = ∠DRE] (ii) [Definition of measure of minor arc]
arc APC ≅ arc DQE (iii) [Given]
∴ ∠ABC = ∠DRE (iv) [From (i), (ii) and (iii)]
In ∆ABC and ∆DRE,
[side AB ≅ side DR [Radii of congruent circles]
side CB ≅ side ER] [From (iv)]
∠ABC ≅ ∠DRE
∴ ∆ABC ≅ ∆DRE [SAS test of congruency]

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)
Statement:
The chords corresponding to congruent major arcs of a circle are congruent.
Given: B is the centre of circle.
arc AXC ≅ arc DXE
To prove: chord AC ≅ chord DE

Proof:
m(major arc) = 360° – m(minor arc)
∴ m(arc AXC) = 360° – m(arc APC) (i)
m(arc DXE) = 360° – m(arc DQE) (ii)
m(arc AXC) = m(arc DXE) (iii) [Given]
∴ 360° – m(arc APC) = 360°- m(arc DQE) [From (i), (ii) and (iii)]
∴ m(arc APC) = m(arc DQE) (iv)
∴ m(arc APC) = ∠ABC (v) [Definition of measure of minor arc]
m(arc DQE) = ∠DBE (vi)
∴ ∠ABC = ∠DBE (vii) [From (iv), (v) and (vi)]
In ∆ABC and ∆DBE,
[side AB ≅ side DB
Side CB ≅ side EB] [Radii of the same circle]
∠ABC ≅ ∠DBE [From (vii)]
∴ ∆ABC ≅ ∆DBE [SAS test of congruency]
∴ chord AC ≅ chord DE [c.s.c.t.]

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:
i. Yes, the major arcs corresponding to congruent chords are congruent.

Proof:
In ∆POQ and ∆ROS,
seg OP ≅ seg OR [Radii of the same circle]
seg OQ ≅ seg OS [Radii of the same circle]
seg PQ ≅ seg RS [Given]
∴ ∆POQ ≅ ∆ROS [SSS test of congruence]
∴ ∠POQ ≅ ∠SOR (i) [c.a.c.t]
[ m(arc PMQ) = ∠POQ (ii)
m(arc RNS) = ∠SOR ] (iii) [Definition of measure of minor arc]
∴ m(arc PMQ) = m(arc RNS)
m(minor arc) = 360° – m(major arc) (iv) [From (i), (ii) and (iii)]
m(arc PMQ) = 360° – m(arc PXQ) (v)
and m(arc RNS) = 360° – m(arc RXS) (vi)
∴ 360°- m(arc PXQ) = 360°- m(arc RXS) [From (iv), (v) and (vi)]
∴ m(arc PXQ) = m(arc RXS)

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 ≅ arc RYS

Proof:
seg PQ and seg RS are the diameters of the same circle. [Given]
∴ arc PYQ and arc RYS are semicircular arcs.
∴ m(arc PYQ) = m(arc RYS) = 180° [Measure of a semicircular arc is 180°]
∴ arc PYQ ≅ arc RYS

Practice Set 3.4

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.
i. ∠AOB
ii. ∠ACB
iii. arc AB
iv. arc ACB.

Solution:
i. seg OA = seg OB = radius…… (i) [Radii of the same circle]
seg AB = radius…… (ii) [Given]
∴ seg OA = seg OB = seg AB [From (i) and (ii)]
∴ ∆OAB is an equilateral triangle.
∴ m∠AOB = 60° [Angle of an equilateral triangle]
ii. m ∠ACB = 12 m ∠AOB [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 × 60°
∴ m ∠ACB = 30°
iii. m(arc AB) = m ∠AOB [Definition of measure of minor arc]
∴ m(arc AB) = 60°
iv. m(arc ACB) + m(arc AB) = 360° [Measure of a circle is 360°]
∴ m(arc ACB) = 360° – m(arc AB)
= 360° – 60°
∴ m(arc ACB) = 300°

Question 2.
In the adjoining figure, ꠸PQRS is cyclic, side PQ ≅ side RQ, ∠PSR = 110°. Find
i. measure of ∠PQR
ii. m (arc PQR)
iii. m (arc QR)
iv. measure of ∠PRQ

Solution:
i. ꠸PQRS is a cyclic quadrilateral. [Given]
∴ ∠PSR + ∠PQR = 180° [Opposite angles of a cyclic quadrilateral are supplementary]
∴ 110° + ∠PQR = 180°
∴ ∠PQR = 180° – 110°
∴ m ∠PQR = 70°
ii. ∠PSR= 12 m (arcPQR) [Inscribed angle theorem]
110°= 12 m (arcPQR)
∴ m(arc PQR) = 220°
iii. In ∆PQR,
side PQ ≅ side RQ [Given]
∴ ∠PRQ = ∠QPR [Isosceles triangle theorem]
Let ∠PRQ = ∠QPR = x
Now, ∠PQR + ∠QPR + ∠PRQ = 180° [Sum of the measures of angles of a triangle is 180°]
∴ ∠PQR + x + x= 180°
∴ 70° + 2x = 180°
∴ 2x = 180° – 70°
∴ 2x = 110°
∴ x=110∘2=55∘
∴ ∠PRQ = ∠QPR = 55°….. (i)
But, ∠QPR = 12 m(arc QR) [Inscribed angle theorem]
∴ 55° = 12 m(arc QR)
∴ m(arc QR) = 110°
iv. ∠PRQ = ∠QPR =55° [From (i)]
∴ m ∠PRQ = 55°

Question 3.
□ MRPN is cyclic, ∠R = (5x -13)°, ∠N = (Ax + 4)°. Find measures of ∠R and ∠N.
Solution:
□ MRPN is a cyclic quadrilateral. [Given]
∴ ∠R + ∠N = 180° [Opposite angles of a cyclic quadrilateral are supplementary]

∴ 5x – 13 + 4x + 4 = 180
∴ 9x – 9 = 180
∴ 9x = 189
∴ x = 1899
∴ x = 21
∴ ∠R = 5x – 13
= 5 × 21 – 13
= 105 – 13
= 92°
∠N = 4x + 4
= 4 × 21 +4
= 84 +4
= 88°
∴ m∠R = 92° and m ∠N = 88°

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 ∠RTS is an acute angle.
Given: O is the centre of the circle, seg RS is the diameter of the circle.
To prove: ∠RTS is an acute angle.
Construction: Let seg RT intersect the circle at point P. Join PS and PT.

Proof:
seg RS is the diameter. [Given]
∴ ∠RPS = 90° [Angle inscribed in a semicircle]
Now, ∠RPS is the exterior angle of ∆PTS.
∴ ∠RPS > ∠PTS [Exterior angle is greater than the remote interior angles]

∴ 90° > ∠PTS
i.e. ∠PTS < 90°
i.e, ∠RTS < 90° [R – P -T]
∠RTS is an acute angle.

Question 5.
Prove that, any rectangle is a cyclic quadrilateral.
Given: ꠸ABCD is a rectangle.
To prove: ꠸ABCD is a cyclic quadrilateral.

Proof:
꠸XBCD is a rectangle. [Given]
∴ ∠A = ∠B = ∠C = ∠D = 90° [Angles of a rectangle]
Now, ∠A + ∠C = 90° + 90°
∴ ∠A + ∠C = 180°
∴ ꠸ABCD is a cyclic quadrilateral. [Converse of cyclic quadrilateral theorem]

Question 6.
In the adjoining figure, altitudes YZ and XT of ∆WXY intersect at P. Prove that,
i. □ WZPT is cyclic.
ii. Points X, Z, T, Y are concyclic.
Given: seg YZ ⊥ side XW
seg XT ⊥ side WY
To prove: i. □WZPT is cyclic.
ii. Points X, Z, T, Y are concyclic.

Proof:
i. segYZ ⊥ side XW [Given]
∴∠PZW = 90°…… (i)
seg XT I side WY [Given]
∴ ∠PTW = 90° ……(ii)
∠PZW + ∠PTW = 90° + 90° [Adding (i) and (ii)]
∴∠PZW + ∠PTW = 180°
∴□WZPT Ls a cyclic quadrilateral. [Converse of cyclic quadrilateral theorem]
ii. ∠XZY = ∠YTX = 90° [Given]
∴ Points X and Y on line XY subtend equal angles on the same side of line XY.
∴ 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]

Question 7.
In the adjoining figure, m (arc NS) = 125°, m(arc EF) = 37°, find the measure of ∠NMS.
Solution:

Chords EN and FS intersect externally at point M.
m∠NMS = 12 [m (arc NS) – m(arc EF)]
= 12 (125° – 37°) = 12 × 88°
∴ m∠NMS = 44°

Question 8.
In the adjoining figure, chords AC and DE intersect at B. If ∠ABE = 108°, m(arc AE) = 95°, find m (arc DC).

Solution:
Chords AC and DE intersect internally at point B.
∴ ∠ABE = 12 [m(arc AE) + m(arc DC)]
∴ 108° = 12 [95° + m(arc DC)]
∴ 108° × 2 = 95° + m(arc DC)
∴ 95° + m(arc DC) = 216°
∴ m(arc DC) = 216° – 95°
∴ m(arc DC) = 121°

Question 1.
Draw a sufficiently large circle of any radius as shown in the figure below. Draw a chord AB and central ∠ACB. Take any point D on the major arc and point E on the minor arc.

i. Measure ∠ADB and ∠ACB and compare the measures.
ii. Measure ∠ADB and ∠AEB. Add the measures.
iii. Take points F, G, H on the arc ADB. Measure ∠AFB, ∠AGB, ∠AHB. Compare these measures with each other as well as with measure of ∠ADB.
iv. Take any point I on the arc AEB. Measure ∠AIB and compare it with ∠AEB. (Textbook pg, no. 64)
Answer:
i. ∠ACB = 2 ∠ADB.
ii. ∠ADB + ∠AEB = 180°.
iii. ∠AHB = ∠ADB = ∠AFB = ∠AGB
iv. ∠AEB = ∠AIB

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 ∠PRQ, ∠PSQ, ∠PTQ. What do you observe? (Textbook pg. no.65)

Answer:
∠PRQ = ∠PSQ = ∠PTQ = 90°
[Student should draw and verily the above answers.]

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)
Given: Chord AB and chord CD intersect at E in the exterior of the circle.
To prove: ∠AEC = 12 [m(arc AC) – m(arc BD)]
Construction: Draw seg AD.

Proof:
∠ADC is the exterior angle of ∆ADE.
∴ ∠ADC = ∠DAE + ∠AED [Remote interior angle theorem]
∴ ∠ADC = ∠DAE + ∠AEC [C – D – E]
∴ ∠AEC = ∠ADC – ∠DAE ……(i)
∠ADC = 12 m(arc AC) (ii) [Inscribed angle theorem]
∠DAE = 12 m(arc BD) (iii) [A – B – E, Inscribed angle theorem]
∴ ∠AEC = 12 m(arc AC) – 12 m (arc BD) [From (i), (ii) and (iii)]
∴ ∠AEC = 12 m(arc AC) – m (arc BD)

Question 4.
Angles inscribed in the same arc are congruent.
Write ‘given’ and ‘to prove’ with the help of the given figure.
Think of the answers of the following questions and write the proof.
i. Which arc is intercepted by ∠PQR ?
ii. Which arc is intercepted by ∠PSR ?
iii. What is the relation between an inscribed angle and the arc intercepted by it? (Textbook: pg. no. 68)

Given: C is the centre of circle. ∠PQR and ∠PSR are inscribed in same arc PTR.
To prove: ∠PQR ≅ ∠PSR
Proof:
i. arc PTR is intercepted by ∠PQR.
ii. arc PTR is intercepted by ∠PSR.
iii. ∠PQR = 12 m(arc PTR), and (i) [inscribed angle theorem]
∠PSR = 12 m(arcPTR) (ii) [Inscribed angle theorem]
∴ ∠PQR ≅ ∠PSR [From (i) and (ii)]

Question 5.
Angle inscribed in a semicircle is a right angle. With the help of given figure write ‘given’, ‘to prove’ and ‘the proof. (Textbook pg. no. 68)

Given: M is the centre of circle. ∠ABC is inscribed in arc ABC.
Arcs ABC and AXC are semicircles.
To prove: ∠ABC = 90°
Proof:
∠ABC = 12 m(arc AXC) (i) [Inscribed angle theorem]
arc AXC is a semicircle.
∴ m(arc AXC) = 180° (ii) [Measure of semicircular arc is 1800]
∴ ∠ABC = 12 × 180°
∴ ∠ABC = 90° [From (i) and (ii)]

Question 6.
Theorem: Opposite angles of a cyclic quadrilateral are supplementry.
Fill in the blanks and complete the following proof. (Textbook pg. no. 68)
Given: □ ABCD is cyclic.
To prove: ∠B + ∠D = 180°
∠A + ∠C = 180°

Proof:
arc ABC is intercepted by the inscribed angle ∠ADC.
∴ ∠ADC m(arcABC) (i) [Inscribed angle theorem]
Similarly, ∠ABC is an inscribed angle. It intercepts arc ADC.
∴ ABC = 12 m(arc ADC) (ii) [Inscribed angle theorem]
∴ ∠ADC + ∠ABC
= 12 m(arcABC) + 12 m(arc ADC) [Adding (i) and (ii)]
∴ ∠D + ∠B = 12 m(areABC) + m(arc ADC)]
∴ ∠B + ∠D = 12 × 360° [arc ABC and arc ADC constitute a complete circle]
= 180°
∴ ∠B + ∠D = 180°
Similarly we can prove,
∠A + ∠C = 180°

Question 7.
In the above theorem, after proving ∠B + ∠D = 180°, can you use another way to prove ∠A + ∠C = 180°? (Textbook pg. no. 69)
Proof:
Yes, we can prove ∠A + ∠C = 180° by another way.
∠B + ∠D = 180°
In ꠸ABCD,
∠A + ∠B + ∠C + ∠D = 360° [Sum of the measures of all angles of a quadrilateral is 360°.]
∴ ∠A + ∠C + 180° = 360°
∴ ∠A + ∠C = 360° – 180°
∴ ∠A + ∠C = 180°

Question 8.
An exterior angle of a cyclic quadrilateral is congruent to the angle opposite to its adjacent interior angle. (Textbook pg. no. 69)
Given: ꠸ABCD is a cyclic quadrilateral.
∠BCE is the exterior angle of ꠸ABCD.
To prove: ∠BCE ≅ ∠BAD

Proof:
∠BCE + ∠BCD = 180°…… (i) [Angles in a linear pair]
꠸ABCD is a cyclic quadrilateral. [Given]
∠BAD + ∠BCD = 180°………. (ii) [Opposite angles of a cyclic quadrilateral are supplementary]
∴ ∠BCE + ∠BCD = ∠BAD + ∠BCD [From (i) and (ii)]
∴ ∠BCE = ∠BAD

Question 9.
Theorem : If a pair of opposite angles of a quadrilateral is supplementary, then the quadrilateral is cyclic. (Textbook pg. no. 69)
Given: In ꠸ABCD, ∠A + ∠C = 180°
To prove: ꠸ABCD is a cyclic quadrilateral.
Proof:
(Indirect method)
Suppose ꠸ABCD is not a cyclic quadrilateral.
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 D – E – C.
∴ ꠸ABED is a cyclic quadrilateral.
∠DAB + ∠DEB = 180° (i) [Opposite angles of a cyclic quadrilateral are supplementary]
∠DAB + ∠DCB = 180° (ii) [Given]
∴ ∠DAB + ∠DEB = ∠DAB + ∠DCB [From (i) and (ii)]
∴ ∠DEB = ∠DCB

But, ∠DEB ≠ ∠DCB as ∠DEB is the exterior angle of ∆BEC.
∴ Our supposition is wrong.
∴ ꠸ABCD is a cyclic quadrilateral.
Case II: Point C lies inside the circle.
Then, take point E on the circle such that
D – C – E
∴ □ABED is a cyclic quadrilateral.
∠DAB + ∠DEB = 180° (iii) [Opposite angles of a cyclic quadrilateral are supplementary]
∠DAB + ∠DCB = 180° (iv) [Given]
∴ ∠DAB + ∠DEB = ∠DAB + ∠DCB [From (iii) and (iv)]
∴ ∠DEB = ∠DCB
But ∠DEB ≠ ∠DCB as ∠DCB is the exterior angle of ∆BCE.
∴ Our supposition is wrong.
∴ □ABCD is a cyclic quadrilateral.

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)
Given: Points B and C lie on the same side of the line AD.
∠ABD = ∠ACD
To prove: Points A, B, C, D are concyclic. i.e., □ABCD is a cyclic quadrilateral.

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 – E – C.
∠ABD ≅ ∠AED (i) [Angles inscribed in the same arc]
∠ABD ≅ ∠ACD (ii) [Given]
∴ ∠AED ≅ ∠ACD [From (i) and (ii)]
∴ ∠AED ≅ ∠ECD [A – E – C]

But, ∠AED ≅ ∠ECD as ∠AED is the exterior angle of ∆ECD.
∴ Our supposition is wrong.
∴ 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 – C – E.
∠ABD ≅ ∠AED (iii) [Angles inscribed in the same arc]
∠ABD ≅ ∠ACD (iv) [Given]
∴ ∠AED ≅ ∠ACD [From (iii) and (iv)]
∴ ∠CED ≅ ∠ACD [A – C – E]

But, ∠CED ≅ ∠ACD as ∠ACD is the exterior angle of ∆ECD.
∴ Our supposition is wrong.
∴ Points A, B, C, D are concyclic points.

Question 11.
The above theorem is converse of a certain theorem. State it. (Textbook pg. no. 70)
Answer:
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.

Practice Set 3.5

Question 1.
In the adjoining figure, ray PQ touches the circle at point Q. PQ = 12, PR = 8, find PS and RS.
Solution:
i. Ray PQ is a tangent to the circle at point Q and seg PS is the secant. [Given]
∴ PR × PS = PQ² [Tangent secant segments theorem]

∴ 8 × PS = 12²
∴ 8 × PS = 144
∴ PS = 1448
∴ PS = 18 units
ii. Now, PS = PR + RS [P – R – S]
∴ 18 = 8 + RS
∴ RS = 18 – 8
∴ RS = 10 units

Question 2.
In the adjoining figure, chord MN and chord RS intersect at point D.
i. If RD = 15, DS = 4, MD = 8 find DN
ii. If RS = 18, MD = 9, DN = 8 find DS

Solution:
i. Chords MN and RS intersect internally at point D. [Given]
∴ MD × DN = RD × DS [Theorem of internal division of chords]
∴ 8 × DN = 15 × 4
∴ DN = 15×48
∴ DN = 7.5 units
ii. Let the value of RD be x.
RS = RD + DS [R – D – S]
∴ 18 = x + DS
∴ DS = 18 – x
Now, MD × DN = RD × DS [Theorem of internal division of chords]
∴ 9 × 8 = x(18 – x)
∴ 72 = 18x – x²
∴ x2 – 18x + 72 = 0
∴ x2 – 12x – 6x + 72 = 0
∴ x (x – 12) – 6 (x – 12) = 0
∴ (x – 12) (x – 6) = 0
∴ x – 12 = 0 or x – 6 = 0
∴ x = 12 or x = 6
∴ DS = 18 – 12 or DS = 18 – 6
∴ DS = 6 units or DS = 12 units

Question 3.
In the adjoining figure, O is the centre of the circle and B is a point of contact. Seg OE ⊥ seg AD, AB = 12, AC = 8, find
i. AD
ii. DC
iii. DE.

Solution:
i. Line AB is the tangent at point B and seg AD is the secant. [Given]
∴ AC × AD = AB² [Tangent secant segments theorem]
∴ 8 × AD = 122
∴ 8 × AD = 144
∴ AD = 1448
∴ AD = 18 units
ii. AD = AC + DC [A – C – D]
∴ 18 = 8 + DC
∴ DC = 18 – 8
∴ DC = 10 units
iii. seg OE ⊥ seg AD [Given]
i.e. seg OE ⊥ seg CD [A – C – D]
∴ DE = 12 DC [Perpendicular drawn from the centre of the circle to the chord bisects the chord]
= 12 × 10
∴ DE = 5 units

Question 4.
In the adjoining figure, if PQ = 6, QR = 10, PS = 8, find TS.
Solution:
PR = PQ + QR [P-Q-R]
∴ PR = 6 + 10 = 16 units

Chords TS and RQ intersect externally at point P.
PQ × PR = PS × PT [Theorem of external division of chords]
∴ 6 × 16 = 8 × PT
∴ PT = 6×168 = 12 units
But, PT = PS + TS [P – S – T]
∴ 12 = 8 + TS
∴ TS = 12 – 8
∴ TS = 4 units

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 × GE = 4r².
Given: seg EF is the diameter.
seg DF is a tangent to the circle,
radius = r

To prove: DE × GE = 4r²
Construction: Join seg GF.
Proof:
seg EF is the diameter. [Given]
∴ ∠EGF = 90° (i) [Angle inscribed in a semicircle]
seg DF is a tangent to the circle at F. [Given]

∴ ∠EFD = 90° (ii) [Tangent theorem]
In ∆DFE,
∠EFD = 90 ° [From (ii)]
seg FG ⊥ side DE [From (i)]
∴ ∆EFD ~ ∆EGF [Similarity of right angled triangles]
∴ EFGE = DEEF [Corresponding sides of similar triangles]
∴ DE × GE = EF²
∴ DE × GE = (2r)² [diameter = 2r]
∴ DE × GE = 4r²

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)

Given: ∠ABC 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 ∠ABC.
To prove: ∠ABC = 12 m(arc ADB)
Proof:
Case I: Centre M lies on arm BA of ∠ABC.
∠MBC = 90° [Trangnet theorem]

i.e. ∠ABC 90° (i) [A – M – B]
arc ADB is a semicircular arc.
∴ m(arc ADB) = 180° (ii) [Measure ofa semicircle is 180°]
∴ ∠ABC = 12 m(arc ADB) [(From (i) and (ii)]

Case II: Centre M lies in the exterior of ∠ABC.
Draw radii MA and MB.
∴ ∠MBA = ∠MAB [Isosceles triangle theorem]

Let, ∠MHA = ∠MAB =x, ∠ABC = y In ∆ABM,
∠AMB + ∠MBA + ∠MAB = 180° [Sum of the measures of all the angles of a triangle is 1800]
∴ ∠AMB + x + x = 180°
∴ ∠AMB = 180° – 2x …… (i)
Now, ∠MBC = ∠MBA + ∠ABC [Angle addition property]
∴ 90° = x + y [Tangent theorem]
∴ x = 90° – y ……(ii)
∠AMB = 180° – 2 (90° – y) [From (i) and (ii)]
∴ ∠AMB = 180° – 180° + 2y
∴ 2y = ∠AMB
∴ y = 12 ∠AMB
∴ ∠ABC = 12 ∠AMB
∴ ∠ABC = 12 m(arc ADB) [Definition of measure of minor arc]

Case III: Centre M lies in the interior of ∠ABC.
Ray BE is the opposite ray of ray BC.
Now, ∠ABE = 12 m (arc AFB) (i) [Proved in case II]

∠ABC + ∠ABE = 180° [Angles in a linear pair]
∴ 180 – ∠ABC = ∠ABE
∴ 180 – ∠ABC = 12 m(arc AFB) [From (i)]
= 12 [360 – m (arc ADB)]
∴ 180 – ∠ABC = 180 – 12 m(arc ADB)
∴ -∠ABC = – 12 m(arc ADB)
∴ ∠ABC = 12 m(arc ADB)

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:
Yes, the theorem 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 × EB = CE × ED
Construction: Draw seg AD and seg CB.

Proof:
In ∆CEB and ∆AED,
∠CEB = ∠DEA [Vertically opposite angles]
∠CBE = ∠ADE [Angles inscribed in the same arc]
∴ ∆CEB ~ ∆AED [by AA test of similarity]
∴ CEAE = EBED [Corresponding sides of similar triangles]
∴ AE × EB = CE × ED

Question 3.
In figure, seg PQ is a diameter of a circle with centre O. R is any point on the circle, seg RS ⊥ seg PQ. Prove that, SR is the geometric mean of PS and SQ. [That is, SR² = PS × SQ] (Textbook pg. no. 81)
Given: seg PQ is the diameter.
seg RS ⊥ seg PQ
To prove: SR² = PS × SQ
Construction: Extend ray RS, let it intersect the circle at point T.

Proof:
seg PQ ⊥ seg RS [Given]
∴ seg OS ⊥ chord RT [R – S – T, P – S – O]
∴ 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.
∴ SR × TS = PS × SQ [Theorem of internal division of chords]
∴ SR × SR = PS × SQ [From (i)]
∴ SR² = PS × SQ

Question 4.
Theorem: If secants containing chords AB and CD of a circle intersect outside the circle in point E, then
AE × EB = CE × ED. (Textbook pg. no. 78)
Given: Chords AB and CD of a circle intersect outside the circle in point E.
To prove: AE × EB = CE × ED
Construction: Draw seg AD and seg BC.

Proof:
In ∆ADE and ∆CBE,
∠AED = ∠CEB [Common angle]
∠DAE ≅ ∠BCE [Angles inscribed in the same arc]
∴ ∆ADE ~ ∆CBE [AA testof similaritv]
∴ AECE = EDEB [Corresponding sides of similar triangles]
∴ AE × EB = CE × ED

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 × EB = ET².
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 × EB = ET²
Construction: Draw seg TA and seg TB.

Proof:
In ∆EAT and ∆ETB,
∠AET ≅ ∠TEB [Common angle]
∠ETA ≅ ∠EBT [Theorem of angle between tangent and secant, E – A – B]
∴ ∆EAT ~ ∆ETB [AA test of similarity]
∴ EAET = ETEB [Corresponding sides of similar triangles]
∴ EA × EB = ET²

Question 6.
In the figure in the above example, if seg PR and seg RQ are drawn, what is the nature of ∆PRQ. (Textbook pg. no, 81)
Answer:
seg PQ is the diameter of the circle.
∴ ∠PRQ = 90°

∴ ∆PRQ is a right angled triangle. [Angle inscribed in a semicircle]

Question 7.
Have you previously proved the property proved in the above example? (Textbook pg. no. 81)
Answer:
Yes. It is the theorem of geometric mean.
∆PSR ~ ∆RSQ [Similarity of right angled triangles]

∴ PSSR = SRSQ [Corresponding sides of similar triangles]
∴ SR² = PS × SQ

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Maharashtra Board Solutions for Class 10-Maths (Part 2): Chapter 3- Circle

Download PDF: Maharashtra Board Solutions for Class 10-Maths (Part 2): Chapter 3- Circle PDF

Chapterwise Maharashtra Board Solutions Class 10 Maths (Part 2) :

• Chapter 1- Similarity
• Chapter 2- Pythagoras Theorem
• Chapter 3- Circle
• Chapter 4- Geometric Constructions
• Chapter 5- Co-ordinate Geometry
• Chapter 6- Trigonometry
• Chapter 7- Mensuration

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About Maharashtra State Board (MSBSHSE)

The Maharashtra State Board of Secondary and Higher Secondary Education or MSBSHSE (Marathi: महाराष्ट्र राज्य माध्यमिक आणि उच्च माध्यमिक शिक्षण मंडळ), is an autonomous and statutory body established in 1965. The board was amended in the year 1977 under the provisions of the Maharashtra Act No. 41 of 1965.

The Maharashtra State Board of Secondary & Higher Secondary Education (MSBSHSE), Pune is an independent body of the Maharashtra Government. There are more than 1.4 million students that appear in the examination every year. The Maha State Board conducts the board examination twice a year. This board conducts the examination for SSC and HSC. 

The Maharashtra government established the Maharashtra State Bureau of Textbook Production and Curriculum Research, also commonly referred to as Ebalbharati, in 1967 to take up the responsibility of providing quality textbooks to students from all classes studying under the Maharashtra State Board. MSBHSE prepares and updates the curriculum to provide holistic development for students. It is designed to tackle the difficulty in understanding the concepts with simple language with simple illustrations. Every year around 10 lakh students are enrolled in schools that are affiliated with the Maharashtra State Board.

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• NCERT Solutions for 9th Class Maths : Chapter 7 Triangles
• NCERT Solutions for Class 10th Mathematics: Chapter 6 – Triangles
• Maharashtra Board Solutions for Class 10-Maths (Part 2): Chapter 1- Similarity
• Maharashtra Board Solutions for Class 10-Maths (Part 2): Chapter 2- Pythagoras Theorem

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