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Class 10 Mathematics Unit 8 Chords and Arcs of a Circle – Unit Wise Class Test 3 | Punjab Board 2027 Model Papers
Class 10 Math Chapter 8 Test 3
Class 10 Mathematics Unit 8: Chords and Arcs of a Circle – Unit Wise Class Test 3 Exercise No. 8.2 and Review Exercise is a useful practice paper for students preparing for the Punjab Board 2027 Model Papers. This Class 10 Mathematics test covers important concepts related to chords, arcs of a circle, radius, diameter, secant lines, and perpendicular distance from the centre to a chord. Students can use this Unit Wise Class Test 3 for board-pattern practice, revision, and preparation according to the Punjab Board 2027 paper pattern.
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Exercise No.8.2 & Review Exercise
T-3
Unit 8: Chords and Arcs of a Circle
Time: 5 minutes
(Objective Type)
Total Marks: 5
1. Tick the correct answer.
(1 × 5 = 5)
(i) Set of all the points in a plane that are at the same distance
from a fixed point is called:
SOLUTION
Step 1: All points of a plane at a fixed distance (the radius) from a fixed point (the centre) form a circle. This is the definition of a circle.
Answer: (c) Circle
Step 1: All points of a plane at a fixed distance (the radius) from a fixed point (the centre) form a circle. This is the definition of a circle.
Answer: (c) Circle
(ii) If C is the circumference of a circle then its diameter can be
calculated by the formula
d =
SOLUTION
Step 1: Circumference C = 2πr, and the diameter d = 2r.
Step 2: So C = π × (2r) = πd.
Step 3: Divide both sides by π: d = C/π.
Answer: (c) C/π (C/2π is the radius)
Step 1: Circumference C = 2πr, and the diameter d = 2r.
Step 2: So C = π × (2r) = πd.
Step 3: Divide both sides by π: d = C/π.
Answer: (c) C/π (C/2π is the radius)
(iii) 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 the distance to any of the three points is the radius, so only one circle exists.
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 the distance to any of the three points is the radius, so only one circle exists.
Answer: (b) Exactly one
(iv) Relation between radius r and diameter d of a circle is:
SOLUTION
Step 1: The diameter passes through the centre and is made of two radii placed end to end.
Step 2: Hence d = r + r = 2r (equivalently r = d/2).
Answer: (b) d = 2r
Step 1: The diameter passes through the centre and is made of two radii placed end to end.
Step 2: Hence d = r + r = 2r (equivalently r = d/2).
Answer: (b) d = 2r
(v) The line shown in the figure is called:
SOLUTION
Step 1: The line extends beyond the circle on both sides and cuts it at two points, A and B.
Step 2: A line that meets a circle at two points is a secant. (A tangent touches at one point; a diameter is a chord through the centre.)
Answer: (c) Secant line
Step 1: The line extends beyond the circle on both sides and cuts it at two points, A and B.
Step 2: A line that meets a circle at two points is a secant. (A tangent touches at one point; a diameter is a chord through the centre.)
Answer: (c) Secant line
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Exercise 8.2 & Review Exercise
T-3
Unit 8: Chords and Arcs of a Circle
Time: 35 minutes
(Subjective Type)
Total Marks: 20
2. Write short answers to the following questions:
(6 × 2 = 12)
(i) Two congruent arcs in a circular track subtend angles of
60° each at the centre. If the length of one chord of the circle
is 10 metres, what is the length of other chord?
SOLUTION
Step 1: Congruent arcs of a circle subtend equal angles at the centre (here 60° each).
Step 2: Equal central angles in the same circle cut off equal chords (ΔOAB ≅ ΔOCD by SAS: OA = OC, OB = OD, ∠AOB = ∠COD).
Step 3: So the chord of the other arc is equal to the first chord.
Length of the other chord = 10 m
Step 1: Congruent arcs of a circle subtend equal angles at the centre (here 60° each).
Step 2: Equal central angles in the same circle cut off equal chords (ΔOAB ≅ ΔOCD by SAS: OA = OC, OB = OD, ∠AOB = ∠COD).
Step 3: So the chord of the other arc is equal to the first chord.
Length of the other chord = 10 m
(ii) In two congruent circular clocks, the minute hand points
from the centre to the 3 on both. A decorative string connects
3 to 9 on both clocks. Are the arcs from 3 to 9 on both clocks
congruent?
SOLUTION
Step 1: On a clock, 3 and 9 are directly opposite each other, so the string joining them passes through the centre. It is a diameter.
Step 2: A diameter divides the circle into two semicircles, so the arc from 3 to 9 is a semicircle (central angle 180°) on both clocks.
Step 3: The two clocks are congruent circles (equal radii), so their semicircles have equal length and equal central angle.
Yes, the arcs are congruent.
Step 1: On a clock, 3 and 9 are directly opposite each other, so the string joining them passes through the centre. It is a diameter.
Step 2: A diameter divides the circle into two semicircles, so the arc from 3 to 9 is a semicircle (central angle 180°) on both clocks.
Step 3: The two clocks are congruent circles (equal radii), so their semicircles have equal length and equal central angle.
Yes, the arcs are congruent.
(iii) In a circular park, two walkways
AB and
CD are both straight paths
(chords) of length 14 metres. What can be said about the
minor arcs subtended by these walkways?
SOLUTION
Step 1: AB = CD = 14 m are chords of the same circle, so they are equal (congruent) chords.
Step 2: In a circle, equal chords subtend equal central angles (SSS: OA = OB = OC = OD and AB = CD give ΔOAB ≅ ΔOCD).
Step 3: Equal central angles cut off arcs of equal measure.
The minor arcs AB and CD are congruent (equal in length).
Step 1: AB = CD = 14 m are chords of the same circle, so they are equal (congruent) chords.
Step 2: In a circle, equal chords subtend equal central angles (SSS: OA = OB = OC = OD and AB = CD give ΔOAB ≅ ΔOCD).
Step 3: Equal central angles cut off arcs of equal measure.
The minor arcs AB and CD are congruent (equal in length).
(iv) On a circular clock with radius of 6cm, the points from
2 to 10 form a chord that is 10cm, long. Find the perpendicular
distance from the centre of clock to the chord.
SOLUTION
Step 1: Let O be the centre and the chord be AB = 10 cm with OA = 6 cm. Draw OM ⊥ AB. M is the midpoint of AB.
Step 2: AM = ½ × 10 = 5 cm.
Step 3: In right ΔOMA: OM² = OA² − AM² = 6² − 5² = 36 − 25 = 11.
Step 4: OM = √11 = 3.32 cm.
Distance from centre = √11 cm ≈ 3.32 cm
Step 1: Let O be the centre and the chord be AB = 10 cm with OA = 6 cm. Draw OM ⊥ AB. M is the midpoint of AB.
Step 2: AM = ½ × 10 = 5 cm.
Step 3: In right ΔOMA: OM² = OA² − AM² = 6² − 5² = 36 − 25 = 11.
Step 4: OM = √11 = 3.32 cm.
Distance from centre = √11 cm ≈ 3.32 cm
(v) A chord 6cm long is at a distance of 4cm from the centre.
Calculate the radius of the circle.
SOLUTION
Step 1: Let chord AB = 6 cm and OM ⊥ AB with OM = 4 cm. M is the midpoint, so AM = 3 cm.
Step 2: In right ΔOMA: OA² = OM² + AM² = 4² + 3² = 16 + 9 = 25.
Step 3: OA = √25 = 5 cm.
Radius = 5 cm
Step 1: Let chord AB = 6 cm and OM ⊥ AB with OM = 4 cm. M is the midpoint, so AM = 3 cm.
Step 2: In right ΔOMA: OA² = OM² + AM² = 4² + 3² = 16 + 9 = 25.
Step 3: OA = √25 = 5 cm.
Radius = 5 cm
(vi) Jannat is designing a triangular garden with corners at
points A(2, 1), B(4, 4) and C(1, 5). Can she install a circular
fountain that touches all three corners?
SOLUTION
A circle passes through three points only if they are non-collinear. We check this using slopes.
Step 1: Slope of AB = (4 − 1)/(4 − 2) = 3/2.
Step 2: Slope of BC = (5 − 4)/(1 − 4) = −1/3.
Step 3: 3/2 ≠ −1/3, so A, B, C are not on one line (they form a triangle).
Step 4: Exactly one circle passes through three non-collinear points (the circumcircle of the triangle).
Verification: The centre is (45/22, 69/22) ≈ (2.05, 3.14), the radius is √2210 / 22 ≈ 2.14. Distance² to each corner = 2210/484 (the same for A, B and C) ✓
Yes, she can install the circular fountain.
A circle passes through three points only if they are non-collinear. We check this using slopes.
Step 1: Slope of AB = (4 − 1)/(4 − 2) = 3/2.
Step 2: Slope of BC = (5 − 4)/(1 − 4) = −1/3.
Step 3: 3/2 ≠ −1/3, so A, B, C are not on one line (they form a triangle).
Step 4: Exactly one circle passes through three non-collinear points (the circumcircle of the triangle).
Verification: The centre is (45/22, 69/22) ≈ (2.05, 3.14), the radius is √2210 / 22 ≈ 2.14. Distance² to each corner = 2210/484 (the same for A, B and C) ✓
Yes, she can install the circular fountain.
Note: Answer the following Questions.
(4 + 4 = 8)
3. A steel bar 6m long lies inside a circular
structure with ends on the circle. It is bisected by a
perpendicular rod from the centre. Find the radius of the
circular structure if the perpendicular distance from the
centre to the bar is 4m.
SOLUTION
Step 1: Let the bar be chord AB = 6 m and the perpendicular rod be OM, with OM ⊥ AB and OM = 4 m. M is the midpoint of AB.
Step 2: AM = ½ × 6 = 3 m.
Step 3: In right ΔOMA: OA² = OM² + AM² = 4² + 3² = 16 + 9 = 25.
Step 4: OA = √25 = 5 m.
Radius of the structure = 5 m
Step 1: Let the bar be chord AB = 6 m and the perpendicular rod be OM, with OM ⊥ AB and OM = 4 m. M is the midpoint of AB.
Step 2: AM = ½ × 6 = 3 m.
Step 3: In right ΔOMA: OA² = OM² + AM² = 4² + 3² = 16 + 9 = 25.
Step 4: OA = √25 = 5 m.
Radius of the structure = 5 m
4. A tree branch lies across a circular pond,
forming a 20m chord. A measuring rod having length 10m is
perpendicular to chord from the centre. What is the radius of
the pond?
SOLUTION
Step 1: Let the branch be chord AB = 20 m and the rod be OM ⊥ AB from the centre O, so OM = 10 m. M is the midpoint of AB.
Step 2: AM = ½ × 20 = 10 m.
Step 3: In right ΔOMA: OA² = OM² + AM² = 10² + 10² = 100 + 100 = 200.
Step 4: OA = √200 = 10√2 = 14.14 m.
Radius of the pond = 10√2 m ≈ 14.14 m
Step 1: Let the branch be chord AB = 20 m and the rod be OM ⊥ AB from the centre O, so OM = 10 m. M is the midpoint of AB.
Step 2: AM = ½ × 20 = 10 m.
Step 3: In right ΔOMA: OA² = OM² + AM² = 10² + 10² = 100 + 100 = 200.
Step 4: OA = √200 = 10√2 = 14.14 m.
Radius of the pond = 10√2 m ≈ 14.14 m
