Education
Class 10 Mathematics Unit 8 Chords and Arcs of a Circle – Unit Wise Class Test 2 | Punjab Board 2027 Model Papers
Class 10 Mathematics Chapter 8 Test 2 New syllabus 2027
Class 10 Mathematics Unit 8: Chords and Arcs of a Circle – Unit Wise Class Test 2 is a useful practice paper for students preparing for the Punjab Board 2027 Model Papers. This Mathematics class test covers important concepts related to chords, arcs, circumference, diameter, central angles, major arcs, minor arcs, segments and circles. Students can use this Class 10 Mathematics Unit 8 Class Test 2 for exam preparation and practice according to the Punjab Board paper pattern.
786Times
786Times
786Times
786Times
786Times
786Times
786Times
786Times
786Times
786Times
786Times
786Times
786Times
786Times
786Times
Unit 08:
Total Time: 35 min
CHAPTER WISE CLASS TEST
(Test # 2)
Mathematics
Total Marks: 25
| A | B | C | D | A | B | C | D | A | B | C | D |
|---|---|---|---|---|---|---|---|---|---|---|---|
| 1 A |
B | C | D | 4 A |
B | C | D | 7 A |
B | C | D |
| 2 A |
B | C | D | 5 A |
B | C | D | 8 A |
B | C | D |
| 3 A |
B | C | D | 6 A |
B | C | D | ||||
NOTE: Four possible answers A, B, C and D to each question are given.
The choice which you think is correct, fill that circle in front of that question with
Marker or Pen ink. Cutting or filling two or more circles will result in zero mark in
that question.
Q.1:
(8)
1.
The boundary of the circle is called:
SOLUTION
Step 1: The complete curved line that forms a circle is its boundary, and its length is called the circumference.
Step 2: A chord is a line segment joining two points of the circle, a segment is a region, and a diameter is a chord through the centre.
Answer: (C) circumference
Step 1: The complete curved line that forms a circle is its boundary, and its length is called the circumference.
Step 2: A chord is a line segment joining two points of the circle, a segment is a region, and a diameter is a chord through the centre.
Answer: (C) circumference
2.
A chord passing through the centre of a circle is called:
SOLUTION
Step 1: A chord joins two points on the circle.
Step 2: When that chord also passes through the centre, it is the longest chord, called the diameter (d = 2r).
Answer: (B) diameter
Step 1: A chord joins two points on the circle.
Step 2: When that chord also passes through the centre, it is the longest chord, called the diameter (d = 2r).
Answer: (B) diameter
3.
______ circle(s) can pass through the three non-collinear points?
SOLUTION
Step 1: The perpendicular bisectors of the two chords joining three non-collinear points meet at exactly one point.
Step 2: That point is the centre, so only one circle passes through the three points.
Answer: (A) one
Step 1: The perpendicular bisectors of the two chords joining three non-collinear points meet at exactly one point.
Step 2: That point is the centre, so only one circle passes through the three points.
Answer: (A) one
4.
Perpendicular bisector of a chord always passes through the ______ of circle.
SOLUTION
Step 1: Every point on the perpendicular bisector of a chord is equidistant from the chord’s two end points.
Step 2: The centre of the circle is equidistant from both end points (each distance is the radius), so it lies on that bisector.
Answer: (C) centre
Step 1: Every point on the perpendicular bisector of a chord is equidistant from the chord’s two end points.
Step 2: The centre of the circle is equidistant from both end points (each distance is the radius), so it lies on that bisector.
Answer: (C) centre
5.
An arc whose central angle measures less than 180° is called:
SOLUTION
Step 1: Central angle less than 180° → minor arc. Exactly 180° → semi-circle. More than 180° → major arc.
Answer: (B) minor arc
Step 1: Central angle less than 180° → minor arc. Exactly 180° → semi-circle. More than 180° → major arc.
Answer: (B) minor arc
6.
A circle whose radius is zero is called:
SOLUTION
Step 1: If r = 0, every point of the circle is at distance 0 from the centre, so the circle shrinks to a single point.
Answer: (B) point circle
Step 1: If r = 0, every point of the circle is at distance 0 from the centre, so the circle shrinks to a single point.
Answer: (B) point circle
7.
What is the maximum number of points at which two distinct circles can intersect?
SOLUTION
Step 1: Exactly one circle passes through three non-collinear points.
Step 2: So two distinct circles cannot share three points. They can meet at 0, 1 (touching) or at most 2 points.
Answer: (B) two
Step 1: Exactly one circle passes through three non-collinear points.
Step 2: So two distinct circles cannot share three points. They can meet at 0, 1 (touching) or at most 2 points.
Answer: (B) two
8.
Perpendicular from the centre of a circle on a chord ______ it.
SOLUTION
Step 1: By the theorem, a perpendicular drawn from the centre to a chord cuts the chord into two equal parts.
Answer: (A) bisect
Step 1: By the theorem, a perpendicular drawn from the centre to a chord cuts the chord into two equal parts.
Answer: (A) bisect
(PART – I)
Q.2. Write short answers to following questions.
(6×2=12)
(i) What is the chord of a circle?
ANSWER
A chord of a circle is a line segment whose two end points lie on the circle. The chord that passes through the centre is the diameter, the longest chord.
A chord of a circle is a line segment whose two end points lie on the circle. The chord that passes through the centre is the diameter, the longest chord.
(ii) Define the term ‘Arc’.
ANSWER
An arc is a continuous part (portion) of the circumference of a circle. Any two points on a circle divide it into two arcs: a minor arc and a major arc.
An arc is a continuous part (portion) of the circumference of a circle. Any two points on a circle divide it into two arcs: a minor arc and a major arc.
(iii) Define a major arc.
ANSWER
An arc of a circle that is greater than a semi-circle (its central angle is greater than 180°) is called a major arc.
An arc of a circle that is greater than a semi-circle (its central angle is greater than 180°) is called a major arc.
(iv) What is an angle in the segment?
ANSWER
If A and B are the end points of a chord and C is any point on the arc of a segment, then ∠ACB, formed by joining C to A and B, is called an angle in that segment.
If A and B are the end points of a chord and C is any point on the arc of a segment, then ∠ACB, formed by joining C to A and B, is called an angle in that segment.
(v) Define the central angle of a circle.
ANSWER
The angle formed at the centre of a circle by two radii drawn to the end points of an arc is called a central angle. It is the angle the arc subtends at the centre.
The angle formed at the centre of a circle by two radii drawn to the end points of an arc is called a central angle. It is the angle the arc subtends at the centre.
(vi) What is the condition for three points to determine a unique circle?
ANSWER
The three points must be non-collinear (not on the same straight line). Exactly one circle passes through any three non-collinear points; if they are collinear, no circle can pass through them.
The three points must be non-collinear (not on the same straight line). Exactly one circle passes through any three non-collinear points; if they are collinear, no circle can pass through them.
(PART – II)
Note: Attempt the following question.
(5)
Q.3.
In a circular fountain of radius 8 m, a pipe AB spans across the fountain.
If mAB = 12 m and O is centre of the circle.
Find the distance from O to the chord AB.
SOLUTION
Step 1: Given: radius OA = 8 m, chord AB = 12 m. Draw OM ⊥ AB. The perpendicular from the centre bisects the chord, so M is the midpoint of AB. OM is the required distance.
Step 2: AM = ½ × AB = ½ × 12 = 6 m.
Step 3: ΔOMA is right-angled at M. By Pythagoras: OM² = OA² − AM² = 8² − 6² = 64 − 36 = 28.
Step 4: OM = √28 = 2√7 = 5.29 m.
Distance from O to chord AB = 2√7 m ≈ 5.29 m
Step 1: Given: radius OA = 8 m, chord AB = 12 m. Draw OM ⊥ AB. The perpendicular from the centre bisects the chord, so M is the midpoint of AB. OM is the required distance.
Step 2: AM = ½ × AB = ½ × 12 = 6 m.
Step 3: ΔOMA is right-angled at M. By Pythagoras: OM² = OA² − AM² = 8² − 6² = 64 − 36 = 28.
Step 4: OM = √28 = 2√7 = 5.29 m.
Distance from O to chord AB = 2√7 m ≈ 5.29 m
