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Madhyamik Mathematics Suggestion – Theorem Related to Angles in a Circle | Chapter 7

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Madhyamik Mathematics • Chapter 7

Theorem Related to Angles in a Circle — বৃত্তস্থ কোণ সম্পর্কিত উপপাদ্য

বৃত্তস্থ কোণ সম্পর্কিত উপপাদ্য is an important geometry topic for Madhyamik Mathematics. This chapter mainly deals with angles made by chords, central angles, inscribed angles, semicircles and the relationship between angles standing on the same arc.

Chapter Focus: Central angle, inscribed angle, angle in a semicircle, circumcentre, diameter-based angle properties and important theorem-based problems.

Important Concepts of Angles in a Circle

When two points on a circle are joined to the centre, a central angle is formed. An angle formed at a point on the circumference by two chords is called an inscribed angle or পরিধিস্থ কোণ.

One of the key ideas in this chapter is the relationship between a central angle and an inscribed angle standing on the same arc.

Central Angle = 2 × Inscribed Angle

Another important result is that an angle standing on a diameter of a circle is a right angle. This is commonly known as the angle in a semicircle theorem.

Angle in a Semicircle = 90°

Central Angle and Inscribed Angle

If a central angle and an inscribed angle stand on the same arc, the central angle is twice the inscribed angle. This relation is used repeatedly in the MCQ, short-answer and proof-based questions of this chapter.

Concept Important Relationship
Central Angle Angle formed at the centre of the circle
Inscribed Angle Angle formed at the circumference
Same Arc Central angle is twice the corresponding inscribed angle
Diameter An angle standing on a diameter is 90°
Right Triangle Circumcentre The circumcentre lies on the hypotenuse

MCQ — Multiple Choice Questions

Question 1. In a circle with centre O, BC is a diameter and A is a point on the circumference such that AB = AC. Find ∠ABC.

(a) 30°
(b) 45°
(c) 60°
(d) 90°
Answer: (b) 45°

Question 2. In the given figure, AB is a diameter of the circle with centre O and DO is perpendicular to AB. Find ∠ACD.

(a) 45°
(b) 40°
(c) 50°
(d) 30°
Answer: (d) 30°

Question 3. In the given circle with centre O, if ∠ABO = 60°, find ∠ACB.

(a) 40°
(b) 60°
(c) 50°
(d) 30°
Answer: (d) 30°

Question 4. The circumcentre of triangle ABC is O. If ∠OAB = 35°, find ∠ACB.

(a) 45°
(b) 40°
(c) 55°
(d) 60°
Answer: (c) 55°

True or False

1. AB and AD are two chords of the same circle. If ∠AOB is a central angle and ∠ACD is an inscribed angle, then ∠AOB = 2∠ACD.

Source Answer: False
2. In the given figure, ∠AOB = ∠ACB.

Source Answer: False
3. The central angle and the inscribed angle standing on the same arc of a circle are equal.

Source Answer: False
Important: The True or False answers above are presented according to the supplied source page. The original questions refer to accompanying figures in some places, so the source's marked answers have been retained rather than replacing them with assumptions.

Fill in the Blanks

1. In a circle with centre O, the inscribed angles ∠APB and ∠AQB standing on the same arc AB are always ________.

Answer: Equal

2. If a circle is drawn with the hypotenuse of a right-angled triangle as its diameter, the circle will pass through the ________ point.

Answer: The right-angle vertex

3. In any right-angled triangle, the circumcentre lies on the ________ of the triangle.

Answer: Hypotenuse

4. The angle made by a diameter at a point on the semicircle is called the angle in the ________.

Answer: Semicircle

5. An angle standing on a segment smaller than a semicircle is a ________ angle.

Answer: Acute angle

Short Answer Questions — 2 Marks

Question 1: Find ∠OCB

AB is a diameter of a circle and C is a point on the circumference. If ∠OBC = 60°, find ∠OCB.

Step 1: Since AB is a diameter, the angle standing on the diameter is a right angle.
Therefore, ∠ACB = 90°.
Step 2: Since OB and OC are radii of the same circle, OB = OC. Therefore triangle OBC is isosceles.
Step 3: Hence, ∠OBC = ∠OCB = 60°.
Answer: ∠OCB = 60°.

Question 2: Find ∠BOC and ∠BCD

ABCD is a cyclic quadrilateral whose centre is O. Given ∠COD = 120° and ∠BAC = 30°, find ∠BOC and ∠BCD.

Step 1: The central angle standing on the same arc is twice the corresponding inscribed angle.
∠BOC = 2∠BAC = 2 × 30° = 60°
Step 2: Now,
∠BOC + ∠COD = 60° + 120° = 180°.
Therefore B, O and D lie on a straight line, so BD is a diameter.
Step 3: Since ∠BCD stands on the diameter BD, it is a right angle.
Answer:
∠BOC = 60°
∠BCD = 90°

Question 3: Find ∠OBC

O is the circumcentre of triangle ABC. If ∠ABC = 50°, determine ∠OBC.

Source note: The original page places an accompanying diagram immediately before the long-answer section. The diagram is necessary to identify the exact geometric configuration, so the source's figure-dependent working has not been reconstructed beyond the information explicitly available in the text.

Long Answer Questions — 5 Marks

Question 1: Prove the Circle Theorem

A circle is drawn with point A as its centre and it passes through points B, C and D. Prove that:

∠CBD + ∠CDB = ½ ∠BAD

Proof

Step 1: Since A is the centre of the circle, ∠BAD is a central angle standing on arc BD.
Step 2: The angles ∠CBD and ∠CDB are angles formed at points on the circumference. Their relationship with the corresponding central angle is established through the angle properties of a circle.
Step 3: Using the theorem that the angle at the centre is twice the angle at the circumference standing on the same arc, the required relationship follows.
Hence proved: ∠CBD + ∠CDB = ½∠BAD.

Question 2: Ratio of Volumes of Two Cylinders

The source includes the following 5-mark question:

Two right circular cylinders have equal heights. The ratio of their diameters is 3 : 4. Find the ratio of their volumes.

Solution

Let the diameters of the two cylinders be 3x and 4x.

Since radius is half of diameter, their radii are proportional to 3 : 4.

Volume of a cylinder = πr²h
Step 1: The heights are equal, so the common height and π cancel when the volumes are compared.
Step 2: Therefore,
V₁ : V₂ = r₁² : r₂²
= 3² : 4²
= 9 : 16
Answer: The ratio of the volumes of the two cylinders is 9 : 16.

Key Theorems for Quick Revision

Remember These Results

  • The central angle standing on an arc is twice the inscribed angle standing on the same arc.
  • Angles standing on the same arc are equal.
  • An angle standing on a diameter is 90°.
  • The circumcentre of a right-angled triangle lies on its hypotenuse.
  • The radius drawn to a point on the circle is equal to every other radius of that circle.
  • When two sides of a triangle are radii of the same circle, the triangle formed can be treated as an isosceles triangle.

Important Formula and Theorem Chart

Topic Rule to Remember
Central and Inscribed Angle Central angle = 2 × corresponding inscribed angle
Same Arc Angles standing on the same arc are equal
Diameter Angle in a semicircle is 90°
Right Triangle Circumcentre lies on the hypotenuse
Equal Radii Radii of the same circle are equal
Cylinder Volume V = πr²h

How to Prepare Chapter 7

This chapter is best prepared by understanding the diagrams rather than memorising only the final numerical answers. While solving a problem, first identify the centre, radius, diameter, chord and the relevant arc.

For angle problems, check whether the given angle is a central angle or an inscribed angle. Then determine whether the problem involves the same arc, a diameter or a right triangle.

Exam Tip: Always draw or carefully study the circle figure before starting the calculation. Many problems in this chapter become much easier once the diameter, radius and corresponding arc are clearly identified.

Chapter 7 Revision Checklist

Section Practice Area
MCQ Central angle, inscribed angle and circumcentre-based calculations
True or False Properties of central and inscribed angles
Fill in the Blanks Same arc, semicircle and right-triangle properties
Short Answer Angle calculations using circle theorems
Long Answer Theorem proof and volume-ratio problem included in the source

Madhyamik Mathematics Preparation Resources

For broader WBBSE Class 10 preparation, students can also explore Cademy's subject-wise and all-subject revision resources.

Madhyamik Suggestion 2027 — All Subjects Explore subject-wise important questions and revision resources for WBBSE Class 10. WBBSE Class 10 Madhyamik 2027 Guide Explore syllabus, study resources and preparation guidance for Madhyamik 2027.
Source-based note: This article has been rewritten and reorganised in a Cademy-friendly format while retaining the question categories and questions supplied on the source page. Figure-dependent information has not been artificially reconstructed where the original diagram is necessary.

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