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Class 10 Mathematics Unit 8 Chords and Arcs of a Circle | Unit Wise Class Test 1 | Punjab Board 2027 Model Papers
Class10 Math Chapter 8 Test 1 New Board Paper 2027
Class 10 Mathematics Unit 8: Chords and Arcs of a Circle Exercise No. 8.1 – Download and practice the Unit Wise Class Test 1 prepared according to the Punjab Board 2027 Model Papers pattern. This Class 10 Mathematics test covers important concepts of Chords and Arcs of a Circle, including chords, arcs, circles, secant lines, congruent arcs, and related mathematical concepts. It is useful for students preparing for the Punjab Board Class 10 Mathematics 2027 examinations and regular classroom assessment.
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Exercise No. 8.1
T-1
Unit 8: Chords and Arcs of a Circle
1. Tick the correct answer.
(1 × 5 = 5)
(i) Number of different circles can pass through three non-collinear points is:
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, and its distance to any of the three points is the radius, so only one circle is possible.
Answer: (b) Exactly 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, and its distance to any of the three points is the radius, so only one circle is possible.
Answer: (b) Exactly one
(ii) Given that a diameter bisects a given chord, then both the diameter and chord:
SOLUTION
Step 1: A diameter passes through the centre O. If it bisects the chord at P, then OP joins the centre to the midpoint of the chord.
Step 2: The line from the centre to the midpoint of a chord is perpendicular to the chord (theorem).
Answer: (c) Intersect each other at right angle
Step 1: A diameter passes through the centre O. If it bisects the chord at P, then OP joins the centre to the midpoint of the chord.
Step 2: The line from the centre to the midpoint of a chord is perpendicular to the chord (theorem).
Answer: (c) Intersect each other at right angle
(iii) Set of all the points in a plane that are at the same distance from a fixed point is called:
SOLUTION
Step 1: This is the definition of a circle: all points in a plane at a fixed distance (radius) from a fixed point (centre).
Answer: (c) Circle
Step 1: This is the definition of a circle: all points in a plane at a fixed distance (radius) from a fixed point (centre).
Answer: (c) Circle
(iv) The line shown in the figure is called:
SOLUTION
Step 1: The line extends beyond the circle in both directions and cuts the circle at two points, A and B.
Step 2: A line meeting a circle at two points is a secant (a tangent touches at only one point; a diameter is a chord through the centre).
Answer: (c) Secant line
Step 1: The line extends beyond the circle in both directions and cuts the circle at two points, A and B.
Step 2: A line meeting a circle at two points is a secant (a tangent touches at only one point; a diameter is a chord through the centre).
Answer: (c) Secant line
(v) Two circles can intersect each other at:
SOLUTION
Step 1: Three non-collinear points determine exactly one circle.
Step 2: So two different circles cannot share three points. They can meet at 0, 1 or 2 points only.
Answer: (b) At most two points
Step 1: Three non-collinear points determine exactly one circle.
Step 2: So two different circles cannot share three points. They can meet at 0, 1 or 2 points only.
Answer: (b) At most two points
Exercise No. 8.1
T-1
Unit 8: Chords and Arcs of a Circle
2. Write short answers to the following questions:
(6 × 2 = 12)
(i) What is the difference between major arcs and minor arcs?
ANSWER
Minor arc: the smaller arc of a circle between two points, which is less than a semicircle (its central angle is less than 180°).
Major arc: the larger arc between the same two points, which is greater than a semicircle (its central angle is greater than 180°).
The two arcs together make the complete circle (360°).
Minor arc: the smaller arc of a circle between two points, which is less than a semicircle (its central angle is less than 180°).
Major arc: the larger arc between the same two points, which is greater than a semicircle (its central angle is greater than 180°).
The two arcs together make the complete circle (360°).
(ii) Prove that a unique circle can pass through the points
A(1, 1), B(4, 2) and C(3, 5).
SOLUTION
A unique circle passes through three points if and only if they are non-collinear. We show A, B, C are not on one line.
Step 1: Slope of AB = (2 − 1)/(4 − 1) = 1/3.
Step 2: Slope of BC = (5 − 2)/(3 − 4) = 3/(−1) = −3.
Step 3: 1/3 ≠ −3, so A, B, C are not collinear.
Step 4: Three non-collinear points lie on exactly one circle. Hence a unique circle passes through A, B and C. Proved.
Verification (optional):
AB² = 3² + 1² = 10, BC² = 1² + 3² = 10, AC² = 2² + 4² = 20.
Since AB² + BC² = AC², ∠B = 90°, so AC is a diameter. The centre is the midpoint of AC = (2, 3), and the radius = ½√20 = √5.
Check: distance² from (2, 3) to A = 1 + 4 = 5, to B = 4 + 1 = 5, to C = 1 + 4 = 5 ✓
Circle: (x − 2)² + (y − 3)² = 5
A unique circle passes through three points if and only if they are non-collinear. We show A, B, C are not on one line.
Step 1: Slope of AB = (2 − 1)/(4 − 1) = 1/3.
Step 2: Slope of BC = (5 − 2)/(3 − 4) = 3/(−1) = −3.
Step 3: 1/3 ≠ −3, so A, B, C are not collinear.
Step 4: Three non-collinear points lie on exactly one circle. Hence a unique circle passes through A, B and C. Proved.
Verification (optional):
AB² = 3² + 1² = 10, BC² = 1² + 3² = 10, AC² = 2² + 4² = 20.
Since AB² + BC² = AC², ∠B = 90°, so AC is a diameter. The centre is the midpoint of AC = (2, 3), and the radius = ½√20 = √5.
Check: distance² from (2, 3) to A = 1 + 4 = 5, to B = 4 + 1 = 5, to C = 1 + 4 = 5 ✓
Circle: (x − 2)² + (y − 3)² = 5
(iii) In a circle, mAB = 8 cm is bisected at point P by a line segment from the centre O. Find mOP if radius is 5 cm.
SOLUTION
Step 1: OP joins the centre to the midpoint P of chord AB, so OP ⊥ AB. ΔOPA is right-angled at P.
Step 2: AP = ½ × AB = ½ × 8 = 4 cm, and OA = radius = 5 cm.
Step 3: By Pythagoras: OP² = OA² − AP² = 5² − 4² = 25 − 16 = 9.
Step 4: OP = √9 = 3.
m OP = 3 cm
Step 1: OP joins the centre to the midpoint P of chord AB, so OP ⊥ AB. ΔOPA is right-angled at P.
Step 2: AP = ½ × AB = ½ × 8 = 4 cm, and OA = radius = 5 cm.
Step 3: By Pythagoras: OP² = OA² − AP² = 5² − 4² = 25 − 16 = 9.
Step 4: OP = √9 = 3.
m OP = 3 cm
(iv) Calculate the length of a chord which stands at a distance of 5 cm from the centre of a circle whose radius is 13 cm.
SOLUTION
Step 1: Let the chord be AB and the perpendicular from the centre O meet it at M. Then OM = 5 cm, OA = 13 cm, and M is the midpoint of AB.
Step 2: In right ΔOMA: AM² = OA² − OM² = 13² − 5² = 169 − 25 = 144.
Step 3: AM = √144 = 12 cm.
Step 4: AB = 2 × AM = 2 × 12 = 24 cm.
Length of chord = 24 cm
Step 1: Let the chord be AB and the perpendicular from the centre O meet it at M. Then OM = 5 cm, OA = 13 cm, and M is the midpoint of AB.
Step 2: In right ΔOMA: AM² = OA² − OM² = 13² − 5² = 169 − 25 = 144.
Step 3: AM = √144 = 12 cm.
Step 4: AB = 2 × AM = 2 × 12 = 24 cm.
Length of chord = 24 cm
(v) In construction, three steel rods are fixed at points A, B, and C (not in a straight line). A circular hoop needs to pass through all three. How many hoops can be used?
SOLUTION
Step 1: The rods are at three points that are not in a straight line, so they are non-collinear.
Step 2: Exactly one circle passes through three non-collinear points.
Only one hoop can be used.
Step 1: The rods are at three points that are not in a straight line, so they are non-collinear.
Step 2: Exactly one circle passes through three non-collinear points.
Only one hoop can be used.
(vi) Two holes A and B are drilled 12 cm apart on a circular tabletop of radius 10 cm. Find the perpendicular distance from the centre to AB.
SOLUTION
Step 1: AB is a chord of length 12 cm. The perpendicular from the centre O to AB meets it at its midpoint M.
Step 2: AM = ½ × 12 = 6 cm, and OA = 10 cm.
Step 3: In right ΔOMA: OM² = OA² − AM² = 10² − 6² = 100 − 36 = 64.
Step 4: OM = √64 = 8 cm.
Perpendicular distance = 8 cm
Step 1: AB is a chord of length 12 cm. The perpendicular from the centre O to AB meets it at its midpoint M.
Step 2: AM = ½ × 12 = 6 cm, and OA = 10 cm.
Step 3: In right ΔOMA: OM² = OA² − AM² = 10² − 6² = 100 − 36 = 64.
Step 4: OM = √64 = 8 cm.
Perpendicular distance = 8 cm
Note: Answer the following Questions:
(4 + 4 = 8)
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.
Step 2: AM = ½ × 12 = 6 m.
Step 3: In right ΔOMA: OM² = OA² − AM² = 8² − 6² = 64 − 36 = 28.
Step 4: OM = √28 = 2√7 = 5.29 m.
Distance from O to 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.
Step 2: AM = ½ × 12 = 6 m.
Step 3: In right ΔOMA: OM² = OA² − AM² = 8² − 6² = 64 − 36 = 28.
Step 4: OM = √28 = 2√7 = 5.29 m.
Distance from O to AB = 2√7 m ≈ 5.29 m
4. Chord AB of 24 m is across a circular arch of radius 13 m. Find the height of the chord from the centre.
SOLUTION
Step 1: Given: chord AB = 24 m, radius OA = 13 m. Draw OM ⊥ AB. M is the midpoint of AB, and OM is the required height of the chord from the centre.
Step 2: AM = ½ × 24 = 12 m.
Step 3: In right ΔOMA: OM² = OA² − AM² = 13² − 12² = 169 − 144 = 25.
Step 4: OM = √25 = 5 m.
Height of the chord from the centre = 5 m
Step 1: Given: chord AB = 24 m, radius OA = 13 m. Draw OM ⊥ AB. M is the midpoint of AB, and OM is the required height of the chord from the centre.
Step 2: AM = ½ × 24 = 12 m.
Step 3: In right ΔOMA: OM² = OA² − AM² = 13² − 12² = 169 − 144 = 25.
Step 4: OM = √25 = 5 m.
Height of the chord from the centre = 5 m
