Maharashtra Board Class 10 Maths Chapter 3 Circle Set 3.3 Solutions

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).
ℹ️ चित्र व्याख्या (Diagram Explanation): एक वृत्त जिसका केंद्र C है, पर चार बिंदु G, D, E, F हैं। ये बिंदु वृत्त की परिधि पर स्थित हैं, और वृत्त के केंद्र C से इन बिंदुओं तक रेखाखंड खींचे गए हैं, जो केंद्र में कोण बनाते हैं।
Answer: 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°
In simple words: Using the definition of a minor arc and the property that the total measure of a circle is 360°, we subtracted the given arc measures from 360° to find m(arc DE). Then, using the arc addition property, we added m(arc DE) and m(arc EF) to find m(arc DEF).

🎯 Exam Tip: Remember the basic properties of circles, such as the measure of a minor arc being equal to its central angle and the sum of measures of all arcs in a circle being 360°. These are crucial for solving arc-related problems.

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°.
ℹ️ चित्र व्याख्या (Diagram Explanation): एक वृत्त के अंदर एक समबाहु त्रिभुज AQRS दर्शाया गया है। त्रिभुज के शीर्ष Q, R, S वृत्त की परिधि पर स्थित हैं, और केंद्र A (जो वृत्त का केंद्र भी प्रतीत होता है) से इन शीर्षों तक रेखाएँ खींची गई हैं।
Answer: Proof:
i. AQRS 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 = \( \frac{360°}{3} \) = 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°
In simple words: We first established that since AQRS is an equilateral triangle, its sides (chords) are congruent, leading to congruent arcs. Then, by letting each arc measure be 'x' and using the property that a full circle measures 360°, we found 'x'. Finally, m(arc QRS) was found by adding the measures of its constituent arcs.

🎯 Exam Tip: When dealing with equilateral triangles inscribed in a circle, remember that all sides are equal, and consequently, the arcs subtended by these chords are also equal. This relationship is key to solving problems involving arc measures.

Question 3. In the adjoining figure, chord AB ≅ chord CD. Prove that, arc AC ≅ arc BD.
ℹ️ चित्र व्याख्या (Diagram Explanation): एक वृत्त जिसके केंद्र का उल्लेख नहीं है, उसके अंदर दो जीवाएँ AB और CD एक-दूसरे को काटती हुई दर्शाई गई हैं। जीवाओं के अंतबिंदु A, B, C, D वृत्त की परिधि पर स्थित हैं।
Answer: 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
In simple words: By using the property that congruent chords subtend congruent arcs, we established the equality of measures for arc AB and arc CD. Then, applying the arc addition property and subtracting the common arc BC from both sides, we proved that arc AC is congruent to arc BD.

🎯 Exam Tip: The key principle here is that equal chords in a circle subtend equal arcs. Also, the arc addition property allows for manipulation of arc measures to derive desired relationships. Clearly state the properties used in each step of the proof.

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
ℹ️ चित्र व्याख्या (Diagram Explanation): एक वृत्त जिसका केंद्र B है, के अंदर दो जीवाएँ AC और DE दर्शाई गई हैं। बिंदु P और Q चापों पर स्थित हैं, जिससे चाप APC और DQE बनते हैं। वृत्त पर बिंदु A, C, D, E भी स्थित हैं।
Answer: 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 [Radii of the same circle]
side CB ≅ side EB]
∠ABC ≅ ∠DBE [From (iv)]
∴ ∆ABC ≅ ∆DBE [SAS test of congruency]
∴ chord AC ≅ chord DE [c.s.c.t]
In simple words: To prove that chords corresponding to congruent arcs are congruent, we used the definition of minor arc measure to relate arc congruence to central angle congruence. By proving the triangles formed by the chords and radii (ΔABC and ΔDBE) congruent using the SAS test, we concluded that the chords AC and DE are congruent.

🎯 Exam Tip: For theorem proofs, ensure each step is justified with a valid geometrical reason. The congruence criteria (SAS, SSS, ASA, AAS) are fundamental for proving triangle congruence, which often leads to proving chord congruence.

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
ℹ️ चित्र व्याख्या (Diagram Explanation): एक वृत्त जिसका केंद्र O है, के अंदर दो जीवाएँ PQ और RS दर्शाई गई हैं। बिंदु M और N चापों पर स्थित हैं, जिससे चाप PMQ और RNS बनते हैं। केंद्र O से सभी बिंदुओं P, Q, R, S तक रेखाएँ खींची गई हैं।
Answer: 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)]
In simple words: To prove that corresponding arcs of congruent chords are congruent, we first proved the triangles formed by the chords and radii (∆POQ and ∆ROS) congruent using the SSS test. This congruence implies that the central angles (∠POQ and ∠ROS) are equal, which in turn means their corresponding minor arcs (arc PMQ and arc RNS) are congruent.

🎯 Exam Tip: The SSS congruence criterion is often used when dealing with chords and radii. Remember that congruent central angles subtend congruent arcs, which is a direct consequence of the definition of the measure of a minor arc.

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
ℹ️ चित्र व्याख्या (Diagram Explanation): दो अलग-अलग सर्वांगसम वृत्त दर्शाए गए हैं, जिनके केंद्र क्रमशः B और R हैं। पहले वृत्त में जीवा AC है और दूसरे वृत्त में जीवा DE है। इन जीवाओं द्वारा निर्मित चाप APC और DQE भी दर्शाए गए हैं।
Answer: 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]
∠ABC ≅ ∠DRE [From (iv)]
∴ ∆ABC ≅ ∆DRE [SAS test of congruency]
In simple words: This proof establishes that if arcs in congruent circles are congruent, then their corresponding chords are also congruent. We achieve this by demonstrating that the central angles subtended by these arcs are equal, and then using the SAS congruence criterion to show that the triangles formed by the chords and radii are congruent, ultimately proving the chord congruence.

🎯 Exam Tip: When dealing with congruent circles, remember that their radii are equal. This is a critical point for applying congruence tests (like SAS) to prove relationships between chords and arcs in different circles.

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.
ℹ️ चित्र व्याख्या (Diagram Explanation): एक वृत्त जिसका केंद्र B है, के अंदर दो जीवाएँ AC और DE दर्शाई गई हैं। वृत्त पर बिंदु X और P चापों AXC और APC बनाते हैं, जबकि बिंदु Q और E चापों DQE और DXE बनाते हैं। केंद्र B से सभी बिंदुओं A, C, D, E, X, Q तक रेखाएँ खींची गई हैं।
Answer: 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.]
In simple words: This proof demonstrates that if major arcs are congruent, their corresponding minor arcs are also congruent. By using this, we revert to the proof method for congruent minor arcs, ultimately showing that the central angles and thus the triangles formed by the chords are congruent, leading to the congruence of the chords themselves.

🎯 Exam Tip: Understanding the relationship between major and minor arcs (summing to 360°) is crucial. If major arcs are congruent, it directly implies that their corresponding minor arcs are also congruent, simplifying the proof back to a known theorem.

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)
Answer:
i. Yes, the major arcs corresponding to congruent chords are congruent.
Proof:
ℹ️ चित्र व्याख्या (Diagram Explanation): एक वृत्त जिसका केंद्र O है, के अंदर दो जीवाएँ PQ और RS दर्शाई गई हैं। बिंदु X और M चापों PXQ और PMQ बनाते हैं, जबकि बिंदु N और S चापों RNS और RXS बनाते हैं। केंद्र O से सभी बिंदुओं P, Q, R, S, X, N तक रेखाएँ खींची गई हैं।
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) [From (i), (ii) and (iii)]
m(minor arc) = 360° - m(major arc) (iv)
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)
In simple words: Yes, major arcs corresponding to congruent chords are congruent. We first prove that if chords are congruent, their subtended central angles are equal, implying minor arcs are congruent. Using the relationship that a major arc is 360° minus its corresponding minor arc, we show that the major arcs must also be congruent.

🎯 Exam Tip: This question tests a deeper understanding of arc relationships. If minor arcs are congruent, their complements (major arcs) must also be congruent. Be precise with the application of arc addition and subtraction properties.

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
ℹ️ चित्र व्याख्या (Diagram Explanation): एक वृत्त जिसका केंद्र O है, के अंदर दो व्यास PQ और RS दर्शाए गए हैं। बिंदु Y चाप PYQ पर स्थित है, जबकि बिंदु X चाप PQX पर स्थित है। ये व्यास वृत्त को अर्धवृत्तों में विभाजित करते हैं, और केंद्र O से होकर गुजरते हैं।
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
In simple words: When chords are diameters, they are congruent and divide the circle into two semicircles. Since each semicircle measures 180 degrees, the major arcs (which are essentially semicircles in this case) corresponding to these diameters are congruent.

🎯 Exam Tip: Diameters are the longest chords in a circle and always bisect the circle into two semicircles. This implies that any two diameters in the same circle are congruent, and the arcs they subtend are always 180 degrees, making them congruent as well.