1
The diameter equals —
💡 Hint:
Both ways from the centre.
Explanation:
2 × the radius.
Class 9 · Maths · Chapter 10 · बिहार बोर्ड (BSEB)CBSE · NCERT 2026-27
Circles
Circles
How to use this page:
1. Read — Terms · Exercise 10.1 · angle at the centre · Exercise 10.2 · perpendicular · three points · Exercise 10.3 · distance · Exercise 10.4 · arc · cyclic · Exercise 10.5 · Exercise 10.6, diagram, worked example, board tip
2. Check — each lesson has its own questions; the number follows the lesson
3. Mastery ★ — all of that lesson correct. Redo the wrong ones
4. The memory figure shows the angle at the centre and the angle at the circumference on the same arc, with the centre angle double.
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- 1 Terms of a circle
- 2 Exercise 10.1 — the missing terms
- 3 Equal angles at the centre — Exercise 10.2
- 4 The perpendicular, three points, and Exercise 10.3
- 5 Distance from the centre — Exercise 10.4
- 6 The angle of an arc and a semicircle
- 7 Cyclic quadrilaterals — Exercise 10.5
- 8 Optional Exercise 10.6
- Chapter winner — every lesson at mastery ★
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📄 PDF download — notes + answer key
Terms of a circle
वृत्त के पद · Textbook 10.2 · radius · chord · diameter
Points at a fixed distance from the centreNotes
A circle is the path of points at a fixed distance from the centre. That distance is the radius. A segment joining two points is a chord. A diameter is the chord through the centre, and diameter = 2 × radius.
An arc is a part of the circle. The shorter arc is minor and the longer is major. The part between an arc and a chord is a segment. The part between an arc and two radii is a sector. The centre is inside the circle. A point outside is farther than the radius.
Worked exampleExample
Question: The radius is 7 cm. Find the diameter.
Formula: diameter = 2 × radius.
Substitution: 2 × 7 cm = 14 cm. The unit is the centimetre.
10-second revision
- The radius runs from the centre to the circle
- Diameter = 2 × radius
- A segment and a sector are different
Board tip · BSEBBoard tip
Calling the diameter the radius is a common slip. If the radius is 7 cm, the diameter is 14 cm.
Board tip · CBSEBoard tip
A chord and an arc are not the same. The chord is straight. The arc bends along the circle.
Check your understandingall correct = mastery ★
2
The longest chord is the diameter.
Explanation:
True — it passes through the centre.
3
If the radius is 7 cm, the diameter is ______ cm.
💡 Hint:
2 × 7.
Explanation:
14 cm.
4
The part between an arc and a chord is called —
💡 Hint:
The chord cuts it.
Explanation:
A segment. A sector lies between two radii.
5
Write the radius, a chord and the diameter in one line each.
Model answer (board format)
The radius goes from the centre to the circle. A chord joins two points. The diameter is the chord through the centre and is twice the radius.
Check your answer:
Exercise 10.1 — the missing terms
अभ्यास 10.1 — रिक्त पद · Textbook 10.2 · Exercise 10.1
The name is fixed by distance and placeNotes
Exercise 10.1 fills blanks for centre, radius, chord, diameter, arc, sector and segment. If the distance from the centre equals the radius, the point is on the circle. If it is less, the point is inside. If it is more, the point is outside.
A circle together with its inside is a disk, but if the question asks for the circle do not leave only a loose phrase. A diameter is two radii joined. Keep the unit cm.
Exercise 10.1 — one reason for each blankActivity
Write a short reason beside each blank. For a diameter, “the chord through the centre”. For a sector, “an arc and two radii”. For a segment, “an arc and a chord”. If there is a number, show 2 × radius. This exercise comes before the theorems.
Worked exampleExample
Question: The distance of a point from the centre is 5 cm and the radius is 5 cm. Where is the point?
Rule: if the distance equals the radius, the point is on the circle.
Substitution: 5 cm = 5 cm, so the point is on the circle.
10-second revision
- Distance = radius, the point is on the circle
- Sector and segment are different names
- Exercise 10.1 comes before the theorems
Board tip · BSEBBoard tip
Swapping inside and outside is common. If the distance is smaller, the point is inside.
Board tip · CBSEBoard tip
Do not write only the word in the blank. On a diameter also add “2 × radius” when a number is given.
Check your understandingall correct = mastery ★
1
If the distance from the centre is less than the radius, the point is —
💡 Hint:
Compare with the radius.
Explanation:
Inside.
2
A sector is the part between an arc and a chord.
Explanation:
False — that is a segment. A sector is between an arc and two radii.
3
If the radius is 9 cm, the diameter is ______ cm.
💡 Hint:
2 × 9.
Explanation:
18 cm.
4
If the distance equals the radius, the point is —
💡 Hint:
The definition.
Explanation:
On the circle.
5
If the radius is 4 cm, write the diameter and say where a point 6 cm away lies.
Model answer (board format)
The diameter = 8 cm. 6 cm is greater than 4 cm, so the point is outside the circle.
Check your answer:
6
Exercise 10.1 comes before Theorem 10.1.
Explanation:
True — terms first, then the angle at the centre.
Equal angles at the centre — Exercise 10.2
केंद्र पर बराबर कोण — अभ्यास 10.2 · Textbook 10.3 · Theorem 10.1 · Theorem 10.2 · Exercise 10.2
Equal chords, equal angles at the centreNotes
Theorem 10.1: equal chords of a circle make equal angles at the centre. Theorem 10.2 is the converse: equal angles at the centre make the chords equal. The proof uses congruence of the triangles formed by two radii.
The same fact runs on congruent circles, because the radii are equal. The angle of a chord at the circumference is not yet this result. That arrives in Theorem 10.8.
Exercise 10.2 — the chord or the angleActivity
Exercise 10.2 asks for the angle at the centre from equal chords, or the chord from equal angles. First draw the two radii. Then prove the triangles congruent by SAS or SSS. Finish with the number of Theorem 10.1 or 10.2. If the circles are congruent, add one sentence that the radii are equal.
Worked exampleExample
Question: AB = CD and both are chords of the same circle. At the centre O, ∠AOB = 80°. Find ∠COD.
Theorem: 10.1, equal chords make equal angles.
Substitution: ∠COD = 80°. The unit is the degree.
10-second revision
- Theorem 10.1: equal chords, equal angles
- Theorem 10.2 is the converse
- The same fact runs on congruent circles
Board tip · BSEBBoard tip
Do not halve 80° and write 40°. This angle is at the centre, not at the circumference.
Board tip · CBSEBoard tip
In the converse write the chords equal, not the angles. See whether the question gives the angle or the chord.
Check your understandingall correct = mastery ★
1
Equal chords at the centre make —
💡 Hint:
Theorem 10.1.
Explanation:
Equal angles.
2
Theorem 10.2 is the converse of Theorem 10.1.
Explanation:
True — equal angles at the centre make the chords equal.
3
If ∠AOB = 80° and the chords are equal, ∠COD is ______ degrees.
💡 Hint:
The same measure.
Explanation:
80°.
4
Two angles at the centre are 50° and 50°. What will you say about the chords? Give the theorem number.
Model answer (board format)
The chords are equal. Theorem 10.2.
Check your answer:
The perpendicular, three points, and Exercise 10.3
लंब, तीन बिंदु और अभ्यास 10.3 · Textbook 10.4–10.5 · Theorems 10.3, 10.4, 10.5 · Exercise 10.3
The perpendicular halves, three points give one circleNotes
Theorem 10.3: the perpendicular from the centre to a chord bisects the chord. Theorem 10.4 is the converse: the line from the centre to the mid-point is perpendicular to the chord. The two right triangles are congruent.
Theorem 10.5: one and only one circle passes through three non-collinear points. The perpendicular bisectors of two sides meet at the centre. If the three points lie on one line, no such circle is formed.
The folding check, then Exercise 10.3Activity
The book draws a circle and a chord on tracing paper and folds through the centre so the pieces of the chord meet. This is headed Activity, not number 10.1. The fold does not replace Theorem 10.3. Exercise 10.3 asks about common points of two circles and the circle through three points. For three non-collinear points write one circle.
Worked exampleExample
Question: The perpendicular from the centre meets the chord at M. The chord is 10 cm. Find AM.
Theorem: 10.3, the perpendicular bisects it. AM = chord / 2.
Substitution: AM = 10 cm / 2 = 5 cm.
10-second revision
- Theorem 10.3: the perpendicular halves the chord
- Theorem 10.4 is the converse
- Theorem 10.5: three non-collinear points, one circle
Board tip · BSEBBoard tip
10 cm is the whole chord. After the perpendicular write half, 5 cm, not 10 cm.
Board tip · CBSEBoard tip
Trying to draw a circle through three collinear points is wrong. First check that they are not on one line.
Check your understandingall correct = mastery ★
1
The perpendicular from the centre — the chord
💡 Hint:
Theorem 10.3.
Explanation:
Bisects it.
2
Infinitely many circles pass through three non-collinear points.
Explanation:
False — Theorem 10.5 says one and only one circle.
3
If the chord is 10 cm and the perpendicular meets it in the middle, the half is ______ cm.
💡 Hint:
10 / 2.
Explanation:
5 cm.
4
The line from the centre to the mid-point is, to the chord, —
💡 Hint:
Theorem 10.4.
Explanation:
Perpendicular.
5
Write Theorems 10.3 and 10.5 in one line each.
Model answer (board format)
The perpendicular from the centre bisects the chord. One circle passes through three non-collinear points.
Check your answer:
Distance from the centre — Exercise 10.4
केंद्र से दूरी — अभ्यास 10.4 · Textbook 10.6 · Theorem 10.6 · Theorem 10.7 · Exercise 10.4
The distance is the length of the perpendicularNotes
Theorem 10.6: equal chords are at equal distances from the centre. Theorem 10.7 is the converse. The distance is the length of the perpendicular from the centre to the chord. Half the chord, this perpendicular and the radius make a right triangle.
The formula: (distance)² = (radius)² − (half the chord)². If the radius is 5 cm and the chord is 8 cm, half the chord is 4 cm. The book’s second check is also headed Activity: measure the perpendiculars on equal chords. Do not number it 10.4. Exercise 10.4 comes later.
The measuring check, then Exercise 10.4Activity
The book draws perpendiculars from the centre to equal chords and matches the distances. This Activity has no number. Exercise 10.4 asks for the common chord of two circles of radii 5 cm and 3 cm whose centres are 4 cm apart. Find half the chord from the right triangle, then double it.
Worked exampleExample
Question: Circles of radii 5 cm and 3 cm meet at two points. The centres are 4 cm apart. Find the common chord.
Formula: (half the chord)² = (radius)² − (perpendicular from the centre)². M is the foot. OM = x, and the other piece is 4 − x.
Substitution: 25 − x² = 9 − (4 − x)². 25 − x² = 9 − 16 + 8x − x². 25 = −7 + 8x. x = 4 cm. Half the chord = √(25 − 16) = √9 = 3 cm. The whole chord = 2 × 3 cm = 6 cm.
10-second revision
- Theorem 10.6: equal chords, equal distance
- distance² = radius² − (half chord)²
- 5 cm, 3 cm, centres 4 cm: chord 6 cm
Board tip · BSEBBoard tip
√9 = 3 cm is half the chord. The whole chord is 6 cm. Stopping at 3 cm is half an answer.
Board tip · CBSEBoard tip
x = 4 cm means the smaller centre sits at the mid-point of the chord. These are the 3-4-5 numbers. Write the unit cm.
Check your understandingall correct = mastery ★
1
Equal chords are, from the centre, —
💡 Hint:
Theorem 10.6.
Explanation:
At equal distances.
2
Theorem 10.7 says chords at equal distances are equal.
Explanation:
True — it is the converse of 10.6.
3
With radius 5 cm and half-chord 4 cm, the distance is ______ cm.
💡 Hint:
√(25 − 16).
Explanation:
3 cm.
4
For 5 cm, 3 cm and centres 4 cm apart, the common chord is —
💡 Hint:
Half is 3, whole is 6.
Explanation:
6 cm.
5
Radius 13 cm and chord 10 cm. Find the distance from the centre. Show the formula.
Model answer (board format)
Half the chord = 5 cm. Distance = √(13² − 5²) = √(169 − 25) = √144 = 12 cm.
Check your answer:
6
The common chord is 3 cm because √9 = 3.
Explanation:
False — 3 cm is half the chord. The whole chord is 6 cm.
The angle of an arc and a semicircle
चाप का कोण और अर्धवृत्त · Textbook 10.7 · Theorem 10.8 · Theorem 10.9
Double at the centre, a right angle in a semicircleNotes
Theorem 10.8: the angle an arc makes at the centre is double the angle it makes on the remaining part of the circumference. So the angle at the circumference = the angle at the centre / 2.
In a semicircle the angle at the centre is 180°. The angle on the circumference = 180° / 2 = 90°. Angles in the same segment are equal. That is Theorem 10.9. For a major arc the reflex angle at the centre may be needed.
Worked exampleExample
Question: The same arc makes 70° at the centre. Find the angle on the remaining circumference. Also write one line for a semicircle.
Formula: angle at the circumference = angle at the centre / 2. In a semicircle, 180° / 2.
Substitution: 70° / 2 = 35°. In a semicircle, 180° / 2 = 90°. The unit is the degree.
10-second revision
- Theorem 10.8: centre = 2 × circumference
- 90° in a semicircle
- Theorem 10.9: angles in one segment are equal
Board tip · BSEBBoard tip
Do not double 70° into 140° when the question asks for the angle at the circumference. Watch the direction.
Board tip · CBSEBoard tip
The 90° of a semicircle is a special case of Theorem 10.8. Remember it apart, but the reason is the same double rule.
Check your understandingall correct = mastery ★
1
If the angle at the centre is 70°, the angle at the circumference is —
💡 Hint:
Half.
Explanation:
35°.
2
The angle on the circumference in a semicircle is 90°.
Explanation:
True — 180° / 2 = 90°.
3
Half of 70° at the centre is ______ degrees.
💡 Hint:
70 / 2.
Explanation:
35°.
4
One angle in a segment is 42°. Write the other angle and the theorem.
Model answer (board format)
The other is also 42°. Theorem 10.9.
Check your answer:
Cyclic quadrilaterals — Exercise 10.5
चक्रीय चतुर्भुज — अभ्यास 10.5 · Textbook 10.8 · Theorem 10.11 · Theorem 10.12 · Exercise 10.5
Opposite angles add to 180°Notes
If four points lie on a circle, the quadrilateral is cyclic. Theorem 10.10 says that if a segment makes equal angles at two points on the same side, the four points lie on a circle. Theorem 10.11: opposite angles of a cyclic quadrilateral add to 180°.
Theorem 10.12 is the converse. One pair of 180° is enough. An exterior angle equals the opposite interior angle, because a linear pair replaces the other opposite angle. Exercise 10.5 finds these angles.
Exercise 10.5 — subtract from 180°Activity
Exercise 10.5 brings the angle at the centre, an angle in the same segment, and the angles of a cyclic quadrilateral together. First see whether the angle is at the centre or on the circumference. If it is cyclic, the opposite angle = 180° − the given angle. If there is a diameter, check 90° apart. Optional 10.6 is not yet.
Worked exampleExample
Question: In cyclic ABCD, ∠A = 70°. Find ∠C.
Formula: opposite angles add to 180°. Theorem 10.11. ∠C = 180° − ∠A.
Substitution: ∠C = 180° − 70° = 110°. The unit is the degree.
10-second revision
- Theorem 10.11: opposite sum 180°
- Theorem 10.12 is the converse
- The opposite of 70° is 110°
Board tip · BSEBBoard tip
The neighbour of 70° is not the opposite angle. The facing angle is 110°.
Board tip · CBSEBoard tip
If one pair is 180°, write cyclic by 10.12. Measuring all four angles is not required.
Check your understandingall correct = mastery ★
1
In a cyclic quadrilateral if ∠A = 70°, ∠C is —
💡 Hint:
180 − 70.
Explanation:
110°.
2
If one pair of opposite angles is 180°, the quadrilateral is cyclic.
Explanation:
True — Theorem 10.12.
3
The opposite angle of 70° is ______ degrees.
💡 Hint:
180 − 70.
Explanation:
110°.
4
Angles in the same segment are —
💡 Hint:
Theorem 10.9.
Explanation:
Equal.
5
∠A = 85° in a cyclic quadrilateral. Find ∠C and name the theorem. Also give one line for the exterior angle if the side at A is extended.
Model answer (board format)
∠C = 180° − 85° = 95°, Theorem 10.11. The exterior angle at A equals the opposite interior angle ∠C, which is 95°.
Check your answer:
Optional Exercise 10.6
वैकल्पिक अभ्यास 10.6 · Textbook · Exercise 10.6 optional
The number 10.6 staysNotes
The book calls Exercise 10.6 optional. The number does not change, and it comes only after Exercise 10.5. The questions are about the line of centres, intersecting circles, and equal angles.
The line joining the centres bisects the common chord at right angles. That is the language of 10.3 and 10.4. If two circles meet, the line of centres makes equal angles at the two intersection points. Do not invent a new theorem number. The old 10.1 to 10.12 are enough.
Exercise 10.6 — from the old theoremsActivity
In each question first say which old theorem applies. If there is a common chord, use the perpendicular bisector. If the angles are equal, use 10.1 or 10.9. If the angle is cyclic, use 10.11. Optional does not mean skip. Write the number 10.6.
Worked exampleExample
Question: Two circles meet at A and B. The centres are O and O′. The common chord AB = 6 cm. What angle does OM make with the chord, where M is the mid-point?
Rule: the line of centres cuts the common chord at right angles. This is the language of Theorem 10.4.
Substitution: the angle = 90°. Half the chord = 6 cm / 2 = 3 cm. The units are the degree and cm.
10-second revision
- Exercise 10.6 is optional, and the number stays
- 90° on the common chord
- The old theorems are enough
Board tip · BSEBBoard tip
Do not write 10.6 as 10.5. The order is 10.6 after 10.5.
Board tip · CBSEBoard tip
Half of 6 cm is 3 cm and the angle is 90°. Writing only one number leaves out the other part.
Check your understandingall correct = mastery ★
1
Exercise 10.6 in the book is —
💡 Hint:
At the end.
Explanation:
Optional 10.6.
2
The line of centres cuts the common chord at right angles.
Explanation:
True — the perpendicular bisector.
3
If the common chord is 6 cm, half of it is ______ cm.
💡 Hint:
6 / 2.
Explanation:
3 cm.
4
This exercise needs a new theorem number —
💡 Hint:
The old theorems.
Explanation:
No.
5
Write the order of Exercises 10.1 to 10.6 in one line. Which one is optional?
Model answer (board format)
10.1, 10.2, 10.3, 10.4, 10.5, then 10.6. The optional one is 10.6.
Check your answer:
6
Exercise 10.6 is placed before Exercise 10.4.
Explanation:
False — 10.4 is about distance. 10.6 is the optional set at the end.
❓ Full question bank — with answers and explanations — 66 questions
Pick a type. The 41 lesson checks are separate — each lesson has as many as its topic needs. All correct earns mastery ★.
BSEB modelpractice · not annual60 tried · 0 correctMultiple choiceMCQ130 tried · 0 correctTrue / falseTrue / false80 tried · 0 correctFill in the blanksFill80 tried · 0 correctMatchMatch20 tried · 0 correctAssertion–reasonAssertion–reason50 tried · 0 correctBalancingBalancing40 tried · 0 correctIdentifyIdentify20 tried · 0 correctShort and longVSA, SA, LA120 tried · 0 correctCBSE-stylenot a PYQ60 tried · 0 correct
No question is marked as a verified past paper. The BSEB set is a model for practice. CBSE items are CBSE-style, not a copy of any year’s paper.
Multiple choice
0/13Pick one option. A wrong try brings a hint.
1
The diameter equals —
💡 Hint:
Both ways from the centre.
Explanation:
2 × the radius.
2
The longest chord is —
💡 Hint:
Through the centre.
Explanation:
The diameter.
3
Equal chords at the centre make —
💡 Hint:
Theorem 10.1.
Explanation:
Equal angles.
4
The perpendicular from the centre — the chord
💡 Hint:
Theorem 10.3.
Explanation:
Bisects it.
5
Through three non-collinear points there pass —
💡 Hint:
Theorem 10.5.
Explanation:
One.
6
With radius 5 cm and half-chord 4 cm the distance is —
💡 Hint:
√(25 − 16).
Explanation:
3 cm.
7
The common chord for 5 cm, 3 cm and centres 4 cm is —
💡 Hint:
The whole chord.
Explanation:
6 cm.
8
If the angle at the centre is 80°, the angle at the circumference is —
💡 Hint:
Half.
Explanation:
40°.
9
The angle in a semicircle is —
💡 Hint:
180 / 2.
Explanation:
90°.
10
In a cyclic quadrilateral the opposite of 70° is —
💡 Hint:
180 − 70.
Explanation:
110°.
11
Angles in the same segment are —
💡 Hint:
Theorem 10.9.
Explanation:
Equal.
12
Exercise 10.6 is —
💡 Hint:
The end.
Explanation:
Optional.
13
For radius 7 cm the diameter is —
💡 Hint:
2 × 7.
Explanation:
14 cm.
True or false
0/81
The diameter is half the radius.
Explanation:
False — the diameter is double.
2
Theorem 10.1 says equal chords make equal angles at the centre.
Explanation:
True.
3
The perpendicular from the centre does not bisect the chord.
Explanation:
False — Theorem 10.3 says that it does bisect the chord.
4
A circle must still pass through three collinear points.
Explanation:
False — the points must be non-collinear.
5
Equal chords are at equal distances from the centre.
Explanation:
True — Theorem 10.6.
6
The angle in a semicircle is 60°.
Explanation:
False — it is 90°.
7
Opposite angles of a cyclic quadrilateral are equal.
Explanation:
False — their sum is 180°.
8
Exercise 10.6 is optional and the number stays 10.6.
Explanation:
True.
Fill in the blanks
0/81
For radius 7 cm the diameter is ______ cm.
💡 Hint:
2 × 7.
Explanation:
14.
2
After the perpendicular, half of a 10 cm chord is ______ cm.
💡 Hint:
10 / 2.
Explanation:
5.
3
√(25 − 16) = ______ .
💡 Hint:
√9.
Explanation:
3.
4
The whole common chord is ______ cm when the half is 3 cm.
💡 Hint:
Double.
Explanation:
6.
5
The circumference angle of 70° at the centre is ______ degrees.
💡 Hint:
Half.
Explanation:
35.
6
The angle in a semicircle is ______ degrees.
💡 Hint:
180 / 2.
Explanation:
90.
7
In a cyclic quadrilateral the opposite of 70° is ______ degrees.
💡 Hint:
180 − 70.
Explanation:
110.
8
The converse number of the equal-angle theorem at the centre is ______.
💡 Hint:
10.1 is the direct rule.
Explanation:
10.2.
Match
0/21
Match the theorem with the statement.
Column B: A. Opposite angles 180° · B. Equal chords, equal centre angles · C. Perpendicular halves the chord · D. Centre angle is double
1. 10.1
2. 10.3
3. 10.8
4. 10.11
Explanation:
10.1 is the chord, 10.3 the perpendicular, 10.8 the double angle, and 10.11 cyclic.
2
Match the number with the result.
Column B: A. 6 cm · B. 2 × radius · C. 90° · D. 3 cm
1. Diameter
2. Semicircle
3. Distance from 5 and 4
4. Common chord
Explanation:
The diameter is double, the semicircle is 90°, the distance is 3 cm, and the whole chord is 6 cm.
Assertion–reason
0/5Check both statements, then see whether the reason explains the assertion.
1
Assertion (A): The diameter is twice the radius.
Reason (R): The diameter joins one radius on each side of the centre.
Explanation:
Both are true and R is the reason.
2
Assertion (A): Equal chords make equal angles at the centre.
Reason (R): Opposite angles of a cyclic quadrilateral add to 180°.
Explanation:
Both are true, but R does not explain the chord statement.
3
Assertion (A): The angle in a semicircle is 90°.
Reason (R): The angle in a semicircle is 60°.
Explanation:
A is true. R is false.
4
Assertion (A): Opposite angles of a cyclic quadrilateral are equal.
Reason (R): Their sum is 180°.
Explanation:
A is false. R is true, Theorem 10.11.
5
Assertion (A): The angle at the centre is double the angle at the circumference.
Reason (R): The same arc makes the larger angle at the centre and half of it on the remaining circumference.
Explanation:
Both are true and R explains Theorem 10.8.
Coefficient practice
0/4These are not results of this maths chapter. They are only coefficient practice for H2 + O2 → H2O.
1
This is coefficient practice, not a result of this maths chapter. Fill coefficients for H2 + O2 making H2O (blank = 1). This is not a circle result.
H2 + O2 → H2O
Explanation: 2H2 + O2 → 2H2O
2
This is coefficient practice, not a result of this maths chapter. Fill coefficients for H2 + O2 making H2O (blank = 1). Match the hydrogen count.
H2 + O2 → H2O
Explanation: 2H2 + O2 → 2H2O
3
This is coefficient practice, not a result of this maths chapter. Fill coefficients for H2 + O2 making H2O (blank = 1). Match the oxygen count.
H2 + O2 → H2O
Explanation: 2H2 + O2 → 2H2O
4
This is coefficient practice, not a result of this maths chapter. Fill coefficients for H2 + O2 making H2O (blank = 1). A blank coefficient is 1.
H2 + O2 → H2O
Explanation: 2H2 + O2 → 2H2O
Classify
0/21
Place each measure in the right slot.
Diameter and radius
Centre and circumference angles
Cyclic opposite angles
Semicircle angle 90°
Explanation:
The diameter, the centre angle and the semicircle belong to the double rule. Cyclic angles give 180°.
2
Place each statement on the theorem.
Equal chords, equal angles
Centre angle is double
Opposite angles 180°
Equal chords of congruent circles
Explanation:
The chord family is 10.1. The double angle is 10.8. Cyclic is 10.11.
Very short answer
0/51
Write the link between diameter and radius.
Model answer (board format)
Diameter = 2 × radius.
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2
Write Theorem 10.3.
Model answer (board format)
The perpendicular from the centre to a chord bisects the chord.
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3
Find the distance from the centre for radius 5 cm and chord 8 cm.
Model answer (board format)
Half the chord is 4 cm. Distance = √(25 − 16) = 3 cm.
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4
What is the angle in a semicircle?
Model answer (board format)
90°.
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5
Write Theorem 10.11 in one line.
Model answer (board format)
Opposite angles of a cyclic quadrilateral add to 180°.
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Short answer
0/41
The angle at the centre is 100°. Find the angle at the circumference on the same arc. Write one line on how a semicircle differs.
Model answer (board format)
100° / 2 = 50°. In a semicircle the centre is 180°, so the circumference angle is 90°.
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2
In a cyclic quadrilateral ∠A = 75°. Find ∠C.
Model answer (board format)
∠C = 180° − 75° = 105°. Theorem 10.11.
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3
Radii 5 cm and 3 cm, centres 4 cm apart. Find the common chord.
Model answer (board format)
The foot of the perpendicular falls at the smaller centre. Half the chord is 3 cm. The whole chord is 6 cm.
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4
∠AOB = ∠COD = 64° in one circle. Write about the chords with the theorem.
Model answer (board format)
AB = CD by Theorem 10.2, because the angles at the centre are equal.
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Long answer
0/31
Write Theorems 10.1, 10.3 and 10.8. Find the distance for radius 5 cm and chord 6 cm.
Model answer (board format)
Equal chords, equal centre angles. The perpendicular halves the chord. The centre angle is double. Half the chord is 3 cm. Distance = √(25 − 9) = √16 = 4 cm.
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2
Write Theorem 10.11 and the 90° of a semicircle. Find the opposite of 100° in a cyclic quadrilateral. Also find the circumference angle of 120° at the centre.
Model answer (board format)
The opposite sum is 180°. In a semicircle, 90°. The opposite angle = 80°. The circumference angle = 120° / 2 = 60°.
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3
Write the order of Exercises 10.1 to 10.6. Why is 10.6 optional, and why are the half and the whole chord different in the 6 cm example?
Model answer (board format)
The order is 10.1 to 10.6. The book calls 10.6 optional, and the number stays. √9 = 3 cm is the half. The whole chord is 2 × 3 = 6 cm.
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BSEB model paper · practice
0/6This model set is for practice. It is not a question from any year’s annual examination. Annual questions will be added only when a source page is available.
1
One angle of a cyclic quadrilateral is 120°. The opposite angle is —
💡 Hint:
180 − 120.
Explanation:
60°.
2
The angle at the centre is 150°. The angle at the circumference is —
💡 Hint:
Half.
Explanation:
75°.
3
For radius 11 cm the diameter is ______ cm.
💡 Hint:
2 × 11.
Explanation:
22.
4
Theorem 10.8 calls the angle at the centre double.
Explanation:
True — double the angle at the circumference.
5
The chord is 24 cm and the radius is 13 cm. Find the distance from the centre.
Model answer (board format)
Half the chord = 12 cm. Distance = √(13² − 12²) = √(169 − 144) = √25 = 5 cm.
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6
This is coefficient practice, not a result of the circles chapter. Balance H2 + O2 making H2O by filling coefficients (blank = 1).
H2 + O2 → H2O
Explanation: 2H2 + O2 → 2H2O
CBSE-style questions
0/6These are competency-based practice questions. They are not copies of a CBSE paper.
1
The mouth of a well is circular. If the radius is 70 cm, the diameter is —
💡 Hint:
2 × 70.
Explanation:
140 cm.
2
Two hands of a clock make 60° at the centre. The angle on the circumference on that arc is —
💡 Hint:
Theorem 10.8.
Explanation:
30°, because the angle at the circumference is half.
3
Assertion (A): The angle in a semicircle is a right angle.
Reason (R): All angles of a cyclic quadrilateral are 90°.
Explanation:
A is true. R is false. Only the opposite angles add to 180°.
4
Measuring a chord on a graph is the full proof of Theorem 10.3.
Explanation:
False — a measurement is a check. The proof uses congruent triangles.
5
A park has a circular path. A chord is 30 m and someone claims the distance from the centre is 8 m. The radius is 17 m. Check whether 8 m is right.
Model answer (board format)
Half the chord = 15 m. Distance = √(17² − 15²) = √(289 − 225) = √64 = 8 m. The claim is right. The unit is the metre.
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6
A window frame is a cyclic quadrilateral. One corner is 95°. Write the opposite corner, and say why this is not an annual-exam question.
Model answer (board format)
The opposite corner = 180° − 95° = 85°. This is model practice, not a question from any year’s annual examination.
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🏛️ Board exam corner — Bihar Board (BSEB)— CBSE
Switch board with BSEB | CBSE above. The lessons follow the same NCERT chapter.
BSEB model · practiceBoard tip
This page has no verified annual-exam question, because no source page has been added. The model set below is practice in the board pattern.
🏛️ Model questions on one page →
The verified label will be used only when a source page for the question is available.
CBSE-style · competency-based
These are case and assertion-reason practice items. Do not treat them as past CBSE questions.
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🧠 What you learned + equation sheet
What you learned
- The diameter is the longest chord and twice the radius.
- Equal chords make equal angles at the centre, and the converse is also true.
- The perpendicular from the centre halves the chord. One circle passes through three non-collinear points.
- Equal chords are at equal distances from the centre.
- The angle at the centre is double the angle at the circumference on the same arc. In a semicircle the angle is 90°.
- Opposite angles of a cyclic quadrilateral add to 180°. Exercises 10.1 to 10.6 stay in that order.
| What | Keep this |
|---|---|
| Diameter | 2 × radius |
| Chord | equal chords, equal angles |
| Perpendicular | bisects the chord |
| Distance | equal chords, equal distance |
| Angle at centre | double the circumference |
| Cyclic | opposite angles 180° |
The notes are original writing. The textbook was used only for activity order and numbers. “Verified” will be used only on a question that has a source page.