GSEB Solutions for Class 9 Mathematics – Circle

GSEB Solutions for Class 9 Mathematics – Circle (English Medium)

GSEB SolutionsMathsScience
Exercise – 12

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Solution 2:

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Solution 5:

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Solution 10:

Solution 11:

Solution 12(1):

a. in the interior of the circle

Solution 12(2):

a. in the interior of the circle

Solution 12(3):

d. a chord joining the end-points of the semicircle arc.

Solution 12(4):

d. a radius

Solution 12(5):

c. an equilateral triangle
For ⨀(O, x), m∠ AOB = 60°
In ∆ OAB,
Radius = OA = OB = x
∴ ∠AOB + ∠OBA + ∠OAB = 180°
∴ 60° + x + x = 180°
∴ x = 60°
∴ ∆ OAB is a triangle.

Solution 12(6):

d. isosceles triangle
In ∆ OAB,
OA = OB = Radius
Thus, Δ OAB is isosceles triangle.

Solution 12(7):

Solution 12(8):

c. 60°

Solution 12(9):

b. a minor sector

Solution 12(10):

b. the midpoint of the diameter

Solution 12(11):

Solution 12(12):

Solution 12(13):

b. one

Solution 12(14):

d. infinite

Solution 12(15):

c. the centre of the circle passing through all A, B, C.

Solution 12(16):

d. a line passing through one of the end points of the common chord.

Solution 12(17):

a. 130° and 50°
The measure of the remaining angles = 180° – 50° = 130° and 180° – 100° = 80°.

Solution 12(18):

Solution 12(19):

d. 35°
m∠ ADC = 180 – m∠ ABC = 180° – 100° = 80°
m∠ADC = m∠ADB + m∠BDC
∴ 80° = m∠ADB + 45°
∴m∠ADB = 35°

Solution 12(20):

c. 50°
∠ APB is inscribed in a semicircle.

In ∆ PAB,
m∠ APB + m∠ PAB + m∠ PBA = 180°
∴ 90° + 40° + m∠ PBA = 180°
∴ m∠ PBA = 180° – 90° – 40° = 50°

Solution 12(21):

d. 120°
In equilateral triangle ABC,
m∠ ACB = 60°
P is the circumcentre of equilateral ΔABC.
m∠ APB = 2 × m∠ ACB = 2 × 60° = 120°

Exercise – 12.1

Solution 1:

1. If two circles having centres P and Q are concentric, then points P and Q represent the same point. ∴P = Q
2. If two circles having centres P and Q are congruent, then their radii are equal.
3. In a circle with centre O, if P is in the interior and Q is in the exterior, then OQ is larger than OP.

Solution 2:

1. FalseA line-sgement joining the centre to any point of the circle is the radius of the circle.
2. TrueAn arc with the diameter as its endpoints is always a semi-circle.
3. FalseThe set of points equidistant from a fixed point in a plane form a circle whereas the set of points equidistant from a fixed point in space form a sphere.
4. FalseThe union of any two radii of a circle is not a diameter of the circle. The union of two radii of a circle is a diameter of the circle only if the radii lie on the same line.

Exercise – 12.2

Solution 1:

Exercise – 12.3

Solution 1:

The possible number of points of intersection of two distinct circles in a plane depends on the distance between their centres and the sum and positive difference between their radii.
Consider ⨀ (A, R₁) and ⨀ (B, R₂) lying in a plane. Then the distance between their centres is AB and the sum of their radii is R₁ + R₂.
1. If AB > R₁ + R₂, then the circles do not intersect each other. The number of points of intersection of the two circles is zero.

2. If AB = R₁ + R₂, the circles intersect each other at only one point.

3. If |R₁ – R₂| < AB < R₁ + R₂, the circles intersect each other at two distinct points.

4. If AB = |R₁ – R₂|, the circles intersect each other at only one point.

5. If AB < | R₁ – R₂|, the circles do not intersect each other.

Thus, the possible number of points of intersection of two distinct circles in a plane is zero, one or two.

Solution 2:

Exercise – 12.4

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Solution 2:

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Solution 5:

Exercise – 12.5

Solution 1:

The measure of the angle subtended by a minor arc of a circle at the centre is twice the measure of an angle subtended by the arc at any point on the remaining part of the circle.
m∠AOB = 2 × m∠ADB = 2 × 45° = 90°

Solution 2:

In ∆ABC,
m∠BAC + m∠ABC + m∠ACB = 180°
∴m∠BAC + 49° + 51° = 180°
∴m∠BAC + 100° = 180°
∴m∠BAC = 180° – 100°
∴m∠BAC = 80°
Also, ∠BAC and ∠BDC are angles in the same segment.
∴m∠BDC = m∠BAC
∴m∠BDC = 80°

Solution 3:

Solution 4:

Solution 5:

∠SQR and ∠SPR are angles in the same segment.
∴m∠SPR = m∠SQR = 70°
∴m∠SPQ = m∠SPR + m∠QPR (∵ R lies in the interior of ∠SPQ.)
∴m∠SPQ = 70° + 30° = 100°
In cyclic □ PQRS,
m∠QRS + m∠SPQ = 180°
∴ m∠QRS + 100° = 180°
∴ m∠QRS = 180° – 100° = 80°
In ∆PQR, PQ = PR
∴ m∠PRQ = m∠PQR
Now,
m∠PRQ + m∠PQR + m∠QPR = 180°
∴ m∠PQR + m∠PQR + 30° = 180° …[ m∠QPR = 30° ]
∴ 2m∠PQR = 150°
∴ m∠PQR = 75°
m∠PQR = m∠PQS + m∠SQR
∴ 75° = m∠PQS + 70° …[m∠PQR = 75° and m∠SQR = 70°]
∴ m∠PQS = 5°
∠PQS and ∠PRS are angles in the same segment.
∴m∠PRS = m∠PQS
∴m∠PRS = 5°
∴m∠ERS = 5° (∵ P – E – R)

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Solution 10: