Maharashtra Board Class 8 Maths Chapter 17 Circle Chord and Arc Set 17.2 Solutions

Get the most accurate MSBSHSE Solutions for Class 8 Maths Chapter 17 Circle Chord and Arc Set 17.2 here. Updated for the 2026-27 academic session, these solutions are based on the latest MSBSHSE textbooks for Class 8 Maths. Our expert-created answers for Class 8 Maths are available for free download in PDF format.

Detailed Chapter 17 Circle Chord and Arc Set 17.2 MSBSHSE Solutions for Class 8 Maths

For Class 8 students, solving MSBSHSE textbook questions is the most effective way to build a strong conceptual foundation. Our Class 8 Maths solutions follow a detailed, step-by-step approach to ensure you understand the logic behind every answer. Practicing these Chapter 17 Circle Chord and Arc Set 17.2 solutions will improve your exam performance.

Class 8 Maths Chapter 17 Circle Chord and Arc Set 17.2 MSBSHSE Solutions PDF

Question 1. The diameters PQ and RS of the circle with centre C are perpendicular to each other at C. State, why arc PS and arc SQ are congruent. Write the other arcs which are congruent to arc PS.
ℹ️ चित्र व्याख्या (Diagram Explanation): एक वृत्त केंद्र C के साथ दिखाया गया है। PQ और RS वृत्त के दो व्यास हैं जो केंद्र C पर एक दूसरे के लंबवत हैं। बिंदु P, R, S, Q वृत्त की परिधि पर स्थित हैं।
Answer: diameter PQ \( \perp \) diameter RS ... [Given]
\( \therefore \) m\( \angle \)PCS = m\( \angle \)SCQ = m\( \angle \)PCR = m\( \angle \)RCQ = 90° The measure of the angle subtended at the centre by an arc is the measure of the arc.
\( \therefore \) m(arc PS) = m\( \angle \)PCS = 90° ...(i) m (arc SQ) = m\( \angle \)SCQ = 90° ...(ii)
\( \therefore \) m(arc PS) = m(arc SQ) ... [From (i) and (ii)]
\( \therefore \) arc PS \( \cong \) arc SQ ... [If the measures of two arcs of a circle are same, then the two arcs are congruent] m(arc PR) = m\( \angle \)PCR = 90° ...(iii) m (arc RQ) = m\( \angle \)RCQ = 90° ... (iv)
\( \therefore \) m(arc PS) = m(arc PR) = m(arc RQ) ... [From (i), (iii) and (iv)]
\( \therefore \) arc PS \( \cong \) arc PR \( \cong \) arc RQ [If the measures of two arcs of a circle are same, then the two arcs are congruent]
\( \therefore \) arc PR and arc RQ are congruent to arc PS.
In simple words: When two diameters of a circle are perpendicular, they divide the circle into four arcs, each measuring 90 degrees. Arcs that subtend equal angles at the center are congruent. Therefore, arc PS and arc SQ are congruent because they both measure 90 degrees, and similarly, arc PR and arc RQ are also congruent to arc PS.

🎯 Exam Tip: Remember that the measure of a central angle is equal to the measure of its intercepted arc. Congruent arcs have equal measures and vice-versa, which is crucial for proving arc congruency.

 

Question 2. In the given figure, O is the centre of the circle whose diameter is MN. Measures of some central angles are given in the figure.
(i) m\( \angle \)AOB and m\( \angle \)COD
(ii) Show that arc AB \( \cong \) arc CD
(iii) Show that chord AB \( \cong \) chord CD
ℹ️ चित्र व्याख्या (Diagram Explanation): एक वृत्त केंद्र O के साथ दिखाया गया है, जिसका व्यास MN है। बिंदु M, A, B, C, D, N वृत्त की परिधि पर हैं। कुछ केंद्रीय कोणों के माप दिए गए हैं: \( \angle \)AOM = 100°, \( \angle \)BON = 35°, \( \angle \)DOM = 100°, \( \angle \)CON = 35°।
Answer: i. Seg MN is the diameter of the circle. ... [Given]
\( \therefore \) m\( \angle \)AOM + m\( \angle \)AON = 180° ... [Angles in a linear pair]
\( \therefore \) m\( \angle \)AOM + (m\( \angle \)AOB + m\( \angle \)BON) = 180° ... [Angle addition property]
\( \therefore \) 100° + m\( \angle \)AOB + 35° = 180°
...[: m\( \angle \)AOM = 100°, m\( \angle \)BON = 35°]
\( \therefore \) m\( \angle \)AOB + 135° = 180°
\( \therefore \) m\( \angle \)AOB = 180°- 135° m\( \angle \)AOB = 45° ...(i) Also, m\( \angle \)DOM + m\( \angle \)DON = 180° ... [Angles in a linear pair]
\( \therefore \) m\( \angle \)DOM + (m\( \angle \)COD + m\( \angle \)CON) = 180° ... [Angle addition property]
\( \therefore \) 100° +m\( \angle \)COD + 35°= 180°
...[: m\( \angle \)DOM = 100°, m\( \angle \)CON = 35° ]
\( \therefore \) m\( \angle \)COD + 135° = 180°
\( \therefore \) m\( \angle \)COD = 180°- 135°
\( \therefore \) m\( \angle \)COD = 45° ...(ii)
ii. m(arc AB) = m\( \angle \)AOB = 45° ... [From (i)] m(arc DC) = m\( \angle \)DOC = 45° ...[From (ii)]
\( \therefore \) m(arc AB) = m(arc DC) ...[From (i) and (ii)]
\( \therefore \) arc AB \( \cong \) arc CD [If the measures of two arcs of a circle are same, then the two arcs are congruent]
iii. arc AB \( \cong \) arc CD
\( \therefore \) chord AB \( \cong \) chord CD ....[The chords corresponding to congruent arcs are congruent]
In simple words: We find the unknown central angles \( \angle \)AOB and \( \angle \)COD by using the property of angles in a linear pair and angle addition. Once we establish that these central angles are equal, it implies that the corresponding arcs AB and CD have equal measures and are therefore congruent. Finally, chords corresponding to congruent arcs are also congruent.

🎯 Exam Tip: Pay close attention to linear pairs of angles and the angle addition property when calculating central angles. The direct relationship between central angles, their intercepted arcs, and the corresponding chords is fundamental to solving such geometry problems.

 

Maharashtra Board Class 8 Maths Chapter 17 Circle: Chord And Arc Practice Set 17.2 Intext Questions And Activities

 

Question 1. If the measures of two arcs of a circle are same, then two arcs are congruent. Verify this property using tracing paper. (Textbook pg. no. 117)
Answer: [Students should attempt the above activities on their own.]
In simple words: This activity demonstrates a key property in circle geometry: if two arcs in a circle have the same measure, they can be perfectly superimposed on each other, proving their congruence. Tracing paper allows for a direct visual verification of this principle.

🎯 Exam Tip: Practical verification using tools like tracing paper helps solidify conceptual understanding. In exams, you'll need to apply this property logically, stating that "arcs with equal measures are congruent."

 

Question 2. With the help of following activity find out the properties of the chord and the corresponding arc.
(i) a. Draw a circle with centre O.
b. Draw \( \angle \)COD and \( \angle \)AOB of same measure. You will find that the arc AXB and arc CYD are congruent.
c. Draw chords AB and CD.
d. Using compass experience that the length of chord AB and chord CD is also same. (Textbook pg. no. 117)
ℹ️ चित्र व्याख्या (Diagram Explanation): एक वृत्त केंद्र O के साथ दिखाया गया है। इसमें दो केंद्रीय कोण \( \angle \)AOB और \( \angle \)COD हैं जो समान माप के हैं। AXB और CYD चाप इन कोणों द्वारा अंतरित हैं, और संगत जीवाएँ AB और CD खींची गई हैं।
Answer: [Students should attempt the above activities on their own.]
(ii) a. Draw a circle with centre C.
b. Draw the congruent chords AB and DE of the circle. Draw the radii CA, CB, CD and CE.
c. Check that \( \angle \)ACB and \( \angle \)DCE are congruent.
d. Hence show that measure of arc AB and arc DE is equal. Hence these arcs are congruent. (Textbook pg. no. 117)
ℹ️ चित्र व्याख्या (Diagram Explanation): एक वृत्त केंद्र C के साथ दिखाया गया है। इसमें दो सर्वांगसम जीवाएँ AB और DE हैं। CA, CB, CD और CE त्रिज्याएँ खींची गई हैं, जो केंद्रीय कोण \( \angle \)ACB और \( \angle \)DCE बनाती हैं।
Answer: [Students should attempt the above activities on their own.]
In simple words: This activity explores the relationship between central angles, arcs, and chords. Part (i) shows that if central angles are equal, their corresponding arcs and chords are congruent. Part (ii) reverses this by showing that if chords are congruent, their corresponding central angles and arcs are also congruent, reinforcing the interconnected properties within a circle.

🎯 Exam Tip: These activities illustrate fundamental theorems: "Congruent central angles subtend congruent arcs and congruent chords," and "Congruent chords subtend congruent central angles and congruent arcs." Understanding these inverse relationships is key to solving complex problems involving circles.

MSBSHSE Solutions Class 8 Maths Chapter 17 Circle Chord and Arc Set 17.2

Students can now access the MSBSHSE Solutions for Chapter 17 Circle Chord and Arc Set 17.2 prepared by teachers on our website. These solutions cover all questions in exercise in your Class 8 Maths textbook. Each answer is updated based on the current academic session as per the latest MSBSHSE syllabus.

Detailed Explanations for Chapter 17 Circle Chord and Arc Set 17.2

Our expert teachers have provided step-by-step explanations for all the difficult questions in the Class 8 Maths chapter. Along with the final answers, we have also explained the concept behind it to help you build stronger understanding of each topic. This will be really helpful for Class 8 students who want to understand both theoretical and practical questions. By studying these MSBSHSE Questions and Answers your basic concepts will improve a lot.

Benefits of using Maths Class 8 Solved Papers

Using our Maths solutions regularly students will be able to improve their logical thinking and problem-solving speed. These Class 8 solutions are a guide for self-study and homework assistance. Along with the chapter-wise solutions, you should also refer to our Revision Notes and Sample Papers for Chapter 17 Circle Chord and Arc Set 17.2 to get a complete preparation experience.

FAQs

Where can I find the latest Maharashtra Board Class 8 Maths Chapter 17 Circle Chord and Arc Set 17.2 Solutions for the 2026-27 session?

The complete and updated Maharashtra Board Class 8 Maths Chapter 17 Circle Chord and Arc Set 17.2 Solutions is available for free on StudiesToday.com. These solutions for Class 8 Maths are as per latest MSBSHSE curriculum.

Are the Maths MSBSHSE solutions for Class 8 updated for the new 50% competency-based exam pattern?

Yes, our experts have revised the Maharashtra Board Class 8 Maths Chapter 17 Circle Chord and Arc Set 17.2 Solutions as per 2026 exam pattern. All textbook exercises have been solved and have added explanation about how the Maths concepts are applied in case-study and assertion-reasoning questions.

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