Short Answer Type Questions- Circles Class 9 Notes | EduRev

Mathematics (Maths) Class 9

Class 9 : Short Answer Type Questions- Circles Class 9 Notes | EduRev

The document Short Answer Type Questions- Circles Class 9 Notes | EduRev is a part of the Class 9 Course Mathematics (Maths) Class 9.
All you need of Class 9 at this link: Class 9

Question 1. The diameter of circle is 3.8 cm. Find the length of its radius.
 Solution: 
Since, the diameter of circle is double its radius.
∴ Diameter = 2 x Radius
⇒ (1/2)x Diameter = Radius

⇒ Radius = (1/2) x 3.8 cm

  = 1/2 x 38/10 cm

  =  19/10 = 1.9 cm

 

Question 2. In the adjoining figure, O is the centre of the circle. The chord AB = 10 cm is such that OP ⊥ AB. Find the length of AP.
 Solution:
∵ OP ⊥ AB
∴ P is the mid-point of AB.

Short Answer Type Questions- Circles Class 9 Notes | EduRev

⇒ AP =(1/2)AB
⇒ AP = (1/2)x 10 cm = 5 cm
 

Question 3. In the figure AOC is a diameter of the circle and arc AXB = (1/2) arc BYC. Find ∠BOC. 
 Solution: 
∵ arc AXB = (1/2)arc BYC
∴ ∠AOB = (1/2) ∠BOC

Short Answer Type Questions- Circles Class 9 Notes | EduRev

Also ∠AOB + ∠BOC = 180°
⇒  (1/2)∠BOC + ∠BOC = 180°
⇒  (3/2) ∠BOC = 180°
⇒ ∠BOC =  (1/2) × 180° = 120°


Question 4. In the figure ∠ABC = 45°. Prove that OA ⊥ OC.
 Solution: 
Since the angle subtended at the centre by an arc is double the angle subtended by it at any other point on the remaining part of the circle.

Short Answer Type Questions- Circles Class 9 Notes | EduRev

∴ ∠ABC =(1/2)∠AOC
⇒ ∠AOC = 2 ∠ABC = 2 × 45° = 90°                  [∵ ∠ABC = 45°]
Thus, OA ⊥ OC.


Question 5. Look at the adjoining figure, in which O is the centre of the circle. If AB = 8 cm and OP = 3 cm, then find the radius of the circle.
 Solution: 
∵ OP ⊥ AB
∴ P is the mid-point of AB.

Short Answer Type Questions- Circles Class 9 Notes | EduRev

⇒ AP = (1/2)AB = (1/2)x 8 cm = 4 cm
Now, in right ΔOPA, Radius, OA = Short Answer Type Questions- Circles Class 9 Notes | EduRev = 5 cm


Question 6. In the adjoining figure, O is the centre of the circle. Find the length of AB.
 Solution:
Since chord AB and chord CD subtend equal angles at the centre,
i.e. ∠ AOB = ∠ COD                  [Each = 60°]

Short Answer Type Questions- Circles Class 9 Notes | EduRev

∴ Chord AB = Chord CD
⇒ Chord AB = 5 cm                  [∵ Chord CD = 5 cm]
Thus, the required length of chord AB is 5 cm.

 

Question 7. In the adjoining figure, O is the centre of the circle and OP = OQ. If AP = 4 cm, then find the length of CD.
 Solution:
∵ OP = OQ
∴ Chord AB and chord CD are equidistant from the centre.

Short Answer Type Questions- Circles Class 9 Notes | EduRev

Short Answer Type Questions- Circles Class 9 Notes | EduRev

Short Answer Type Questions- Circles Class 9 Notes | EduRev

Thus, the required length of CD is 8 cm.


Question 8. AB and CD are two parallel chords of a circle that are on opposite sides of the centre such that AB = 24 cm and CD = 10 cm and the distance between AB and CD is 17 cm. Find the radius of the circle.
 Solution: 
∵ Perpendicular from the centre to a chord bisects the chord.

Short Answer Type Questions- Circles Class 9 Notes | EduRev

∴ AP = (1/2)AB = (1/2)x 24 cm
= 12 cm
Similarly, CQ = (1/2)CD =(1/2)x 10 cm = 5 cm
Let OP = x cm
⇒ OQ = (17 - x) cm
Now, in right ΔAPO, x+ 122 = OA2                  ...(1)
Again, in right ΔCOQ, OC= CQ2 + (17 - x)2
= 5+ 172 + x2 - 34x
= 25 + 289 + x2 - 34x = 314 + x2 - 34x                  ...(2)
From (1) and (2),
we have x2 + 314 - 34x = x2 + 122
⇒ x2 - x2 - 34x = 144 - 314
⇒ - 34x = -170
⇒ x= (-170/34)  = 5

Now from (1), we have OA= 52 + 122 = 25 + 144
⇒ OA= 169
⇒ OA = 13 cm
Thus, the required radius of the circle = 13 cm.


Question 9. If O is the centre of the circle, then find the value of x.
 Solution:
∵ AOB is a diameter.
∴ ∠ AOC + ∠ COB = 180°                  [Linear pairs]
⇒ 130° + ∠ COB = 180°
⇒ ∠ COB = 180° - 130° = 50°
Now, the arc CB is subtending ∠COB at the centre and ∠CDB at the remaining part.

Short Answer Type Questions- Circles Class 9 Notes | EduRev

∴ ∠ CDB = (1/2)∠COB
⇒ ∠ CDB = (1/2)x 50° = 25°
Thus, the measure of x = 25°


Question 10. The radius of a circle is 17 cm. A chord of length 30 cm is drawn. Find the distance of the chord from the centre.
 Solution:
Length of chord AB = 30 cm.
Since, OP ⊥ AB
∴ P is the mid-point of AB.

Short Answer Type Questions- Circles Class 9 Notes | EduRev

AP = (1/2)AB =(1/2) x 30 cm = 15 cm
Now, in right ΔAPO, AO2 = AP2 + OP2
⇒ 172 = 15+ OP2
∴ OP2 = 172 - 152 = (17 - 15)(17 + 15)
= 2 x 32 = 64
⇒ OP = 64 = 8 cm
∴ The distance of the chord AB from the centre O is 8 cm.


Question 11. Find the length of a chord which is at a distance of 4 cm from the centre of a circle of radius 5 cm.
 Solution:
∵ The perpendicular distance, OP = 4 cm

Short Answer Type Questions- Circles Class 9 Notes | EduRev

∴ In right ΔAPO, AO2 = AP2 + OP2
⇒ 52 = AP+ 42
⇒ AP2 = 5- 4= (5 - 4)(5 + 4)
= 1 x 9 = 9
⇒ AP = √9= 3 cm
Since the perpendicular from the centre to a chord of a circle divides the chord into two equal parts.
∴ AP =  (1/2) AB
⇒ AB = 2AP
⇒ AB = 2 x 3 cm = 6 cm
Thus, the required length of chord AB is 6 cm.

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