Education
class 10th maths chapter 9 exercise 9.1 solutions Tangents and Angles of a Circle | Unit Wise Class Test 1| Punjab Board 2027
class 10th math chapter 9 exercise 9.1
Class 10 Mathematics Unit 9: Tangents and Angles of a Circle — Prepare for Punjab Board 2027 Model Papers with this Unit Wise Class Test 1 based on Exercise No. 9.1. This test includes objective-type MCQs, short-answer questions, and numerical problems covering important concepts such as tangents, angles, external and internal touching of circles, and tangent lengths. Students can use this Class 10 Mathematics Unit 9 test for exam preparation, revision, self-assessment, and practice according to the Punjab Board syllabus.
786Times
786Times
786Times
786Times
786Times
786Times
786Times
786Times
786Times
786Times
786Times
786Times
786Times
786Times
786Times
Exercise No. 9.1
T-1
Unit 9: Tangents and Angles of a circle
Time: 5 minutes
(Objective Type) Ex 9.1
Total Marks: 5
1. Tick the correct answer.
(1 × 5 = 5)
(i) Angle between tangent line and radial line segment joining
point of contact to the origin is:
SOLUTION
Step 1: The radius drawn to the point of contact is perpendicular to the tangent at that point (theorem).
Step 2: Hence the angle between them is a right angle.
Answer: (d) 90°
Step 1: The radius drawn to the point of contact is perpendicular to the tangent at that point (theorem).
Step 2: Hence the angle between them is a right angle.
Answer: (d) 90°
(ii) Number of tangents drawn to a circle from a point
outside the circle is:
SOLUTION
Step 1: From a point outside a circle we can draw one tangent on each side of the circle.
Step 2: There are exactly two tangents, and they are equal in length. (From a point on the circle there is one tangent; from a point inside there is none.)
Answer: (c) Exactly two
Step 1: From a point outside a circle we can draw one tangent on each side of the circle.
Step 2: There are exactly two tangents, and they are equal in length. (From a point on the circle there is one tangent; from a point inside there is none.)
Answer: (c) Exactly two
(iii) Two circles of radius 3 and 4 touches each other
externally, then distance between their centers is:
SOLUTION
Step 1: When two circles touch externally, the point of contact lies on the line joining the centres, between them.
Step 2: Distance between centres = r₁ + r₂ = 3 + 4 = 7.
Answer: (b) 7
Step 1: When two circles touch externally, the point of contact lies on the line joining the centres, between them.
Step 2: Distance between centres = r₁ + r₂ = 3 + 4 = 7.
Answer: (b) 7
(iv) Two circles of radii 3 and 4 touches each other
internally then distance between their centers is:
SOLUTION
Step 1: When two circles touch internally, the distance between the centres is the difference of the radii.
Step 2: Distance = r₂ − r₁ = 4 − 3 = 1.
Answer: (a) 1
Step 1: When two circles touch internally, the distance between the centres is the difference of the radii.
Step 2: Distance = r₂ − r₁ = 4 − 3 = 1.
Answer: (a) 1
(v) A round table has a decorative quadrilateral pattern
inscribed on its surface. If one angle of the quadrilateral
is 105°, what is the measure of the opposite angle.
SOLUTION
Step 1: A quadrilateral inscribed in a circle is a cyclic quadrilateral.
Step 2: Opposite angles of a cyclic quadrilateral are supplementary (sum = 180°).
Step 3: Opposite angle = 180° − 105° = 75°.
Answer: (a) 75°
Step 1: A quadrilateral inscribed in a circle is a cyclic quadrilateral.
Step 2: Opposite angles of a cyclic quadrilateral are supplementary (sum = 180°).
Step 3: Opposite angle = 180° − 105° = 75°.
Answer: (a) 75°
Exercise No. 9.1
T-1
Unit 9: Tangents and Angles of a circle
Time: 35 minutes
(Subjective Type)
Total Marks: 20
2. Write short answers to the following questions:
(6 × 2 = 12)
(i) The Radii of the two circles are 4cm and 6cm.
What is the distance between their centres if they
touch externally?
SOLUTION
Step 1: For circles touching externally, distance between centres = r₁ + r₂.
Step 2: = 4 + 6 = 10.
Distance = 10 cm
Step 1: For circles touching externally, distance between centres = r₁ + r₂.
Step 2: = 4 + 6 = 10.
Distance = 10 cm
(ii) A radar antenna scans an arc subtending 90° at the
centre. What is the angle subtended by the arc at the
edge of its circular path?
SOLUTION
Step 1: The central angle of the arc is 90°.
Step 2: The angle subtended by an arc at a point on the circumference (in the major arc) is half the central angle.
Step 3: Angle at the edge = ½ × 90° = 45°.
Angle = 45°
Step 1: The central angle of the arc is 90°.
Step 2: The angle subtended by an arc at a point on the circumference (in the major arc) is half the central angle.
Step 3: Angle at the edge = ½ × 90° = 45°.
Angle = 45°
(iii) Two circular gears touch each other externally for
proper rotation in a machine. The radii of the two
circular gears are 5cm and 7cm. What is the distance
between their centres if they touch externally?
SOLUTION
Step 1: External contact: distance between centres = r₁ + r₂.
Step 2: = 5 + 7 = 12.
Distance = 12 cm
Step 1: External contact: distance between centres = r₁ + r₂.
Step 2: = 5 + 7 = 12.
Distance = 12 cm
(iv) A small sensor lies inside a satellite dish and
touches its wall internally. If the dish has radius
15cm and the sensor has radius 2.5cm, find the distance
between their centres.
SOLUTION
Step 1: Internal contact: distance between centres = difference of the radii = R − r.
Step 2: = 15 − 2.5 = 12.5.
Distance = 12.5 cm
Step 1: Internal contact: distance between centres = difference of the radii = R − r.
Step 2: = 15 − 2.5 = 12.5.
Distance = 12.5 cm
(v) An inner holder touches the outer cylindrical
container internally. If the outer container has radius
8cm and the inner holder has radius 6cm, find the
distance between their centres.
SOLUTION
Step 1: Distance between centres = R − r (internal contact).
Step 2: = 8 − 6 = 2.
Distance = 2 cm
Step 1: Distance between centres = R − r (internal contact).
Step 2: = 8 − 6 = 2.
Distance = 2 cm
(vi) An inner decorative circle touches the outer clock
face internally. If the outer circle has radius 12cm
and the inner has radius 7.5cm, find the distance
between their centres.
SOLUTION
Step 1: Distance between centres = R − r (internal contact).
Step 2: = 12 − 7.5 = 4.5.
Distance = 4.5 cm
Step 1: Distance between centres = R − r (internal contact).
Step 2: = 12 − 7.5 = 4.5.
Distance = 4.5 cm
Note: Answer the following Questions.
(4 + 4 = 8)
3. An architect designs a circular dome on a building.
A lighting rod is placed 12m away from the centre of the
dome which has a radius of 5m. Support cables are attached
from the rod to the dome, touching the dome at two points.
What is the length of each cable?
SOLUTION
Step 1: Let O be the centre of the dome, P the rod and T the point where a cable touches the dome. Then PT is a tangent. OP = 12 m and OT = 5 m (radius).
Step 2: The radius to the point of contact is perpendicular to the tangent, so ∠OTP = 90° and ΔOTP is right-angled at T.
Step 3: By Pythagoras: PT² = OP² − OT² = 12² − 5² = 144 − 25 = 119.
Step 4: PT = √119 = 10.91.
Step 5: The two tangents from P are equal, so both cables have the same length.
Length of each cable = √119 m ≈ 10.91 m
Step 1: Let O be the centre of the dome, P the rod and T the point where a cable touches the dome. Then PT is a tangent. OP = 12 m and OT = 5 m (radius).
Step 2: The radius to the point of contact is perpendicular to the tangent, so ∠OTP = 90° and ΔOTP is right-angled at T.
Step 3: By Pythagoras: PT² = OP² − OT² = 12² − 5² = 144 − 25 = 119.
Step 4: PT = √119 = 10.91.
Step 5: The two tangents from P are equal, so both cables have the same length.
Length of each cable = √119 m ≈ 10.91 m
4. A pyramid-shaped sculpture is placed in the center of
a circular plaza with a radius of 10m. A decorative pole
stands 26m from the center of the circle. Two guide wires
are attached from the pole to the plaza edge, just touching
the circle. Find the length of each wire.
SOLUTION
Step 1: Let O be the centre of the plaza, P the pole and T a point where a wire touches the circle. PT is a tangent, with OP = 26 m and OT = 10 m.
Step 2: ∠OTP = 90° (radius ⟂ tangent).
Step 3: PT² = OP² − OT² = 26² − 10² = 676 − 100 = 576.
Step 4: PT = √576 = 24.
Step 5: Both tangents from P are equal.
Length of each wire = 24 m
Step 1: Let O be the centre of the plaza, P the pole and T a point where a wire touches the circle. PT is a tangent, with OP = 26 m and OT = 10 m.
Step 2: ∠OTP = 90° (radius ⟂ tangent).
Step 3: PT² = OP² − OT² = 26² − 10² = 676 − 100 = 576.
Step 4: PT = √576 = 24.
Step 5: Both tangents from P are equal.
Length of each wire = 24 m
