Class 9 Exam  >  Class 9 Notes  >  Mathematics (Maths) Class 9  >  Practice Questions: Cirlces

Class 10 Maths Chapter 10 Practice Question Answers - Circles

Question 1. In the figure, OD is perpendicular to chord AB of a circle whose centre is O. If BC is a diameter, prove that CA = 2OD.
 Solution:
We have a circle whose centre is O. BC is a diameter and AB a chord such that OD ⊥ AB. Let us join AC.
∵ The perpendicular from centre of a circle to a chord bisects the chord

Class 10 Maths Chapter 10 Practice Question Answers - Circles

∴ D is the mid-point of AB
∵ O is the mid-point of the diameter BC
∴ OD is the line segment joining the mid-points of two sides of ΔABC.
∴ OD is half of the third side of ΔABC.

Class 10 Maths Chapter 10 Practice Question Answers - Circles

i.e. OD =(1/2)AC
or 2OD = AC


 Question 2. l is a line intersecting two concentric circles having common centre O, at A, B, C and D. Prove that AB = CD.
 Solution: 
We have two concentric circles with common centre O.
Line l intersects these circles at A, B, C and D.
Let us draw OP ⊥ ℓ For the inner circle

Class 10 Maths Chapter 10 Practice Question Answers - Circles

∵ OP ⊥ BC                 [construction]
∴ P is the mid-point of BC                 [∵ Perpendicular from the centre to a chord bisects the chord]
∴ PB = PC                 …(1)
For the outer circle.

Class 10 Maths Chapter 10 Practice Question Answers - Circles

∵ OP ⊥ AD
∴ PA = PD                ..(2)
Subtracting (1) from (2), we have PA - PB = PD - PC
⇒ AB = CD


Question 3. AB and CD are equal chords of a circle whose centre is O. When produced, these chords meet at E. Prove that EB = ED.

Class 10 Maths Chapter 10 Practice Question Answers - Circles

Solution: We have a circle with centre O. Chord AB = chord CD and on production, AB and CD meet at E.

Class 10 Maths Chapter 10 Practice Question Answers - Circles

Let us join OE and draw OP ⊥ AB and OQ ⊥ CD.
∵ In a circle, equal chords are equidistant from the centre.
∴ OP = OQ.                [∵ AB = CD]
Now, in right ΔOPE and right ΔOQE, we have OP = OQ                 [Proved],
OE = OE                [Common]
∵ Using RHS criterion, ΔOPE ≌ ΔOQE
⇒ P E = QE                 [c.p.c.t.]           …(1)
But AB = CD                [Given]
⇒ (1/2) AB = (1/2)CD or PB = QD                 [∵ OP ⊥ AB and OQ ⊥ CD] …(2)
Subtracting (2) from (1), we have PE ∠ PB = QE ∠ QD
⇒ EB = ED


Question 4. If O be the centre of the circle, find the value of ‘x’ in each of the following figures.

Class 10 Maths Chapter 10 Practice Question Answers - Circles

Class 10 Maths Chapter 10 Practice Question Answers - Circles

Solution: (i) ∵ OA = OB                 [Radii of the same circle]
∴ ∠A = ∠B                 [Angles opposite to equal side in a triangle are equal]
In ΔABC, ∠A + ∠B + ∠O = 180º
∴ x + x + 70º = 180º                 [∵ ∠O = 70º (given) and ∠A = ∠B]
⇒ 2x + 70º = 180º
⇒ 2x = 180º - 70º = 110º
⇒ x= (1100/2)= 55º
Thus, x = 55°

(ii) In ΔAOC, ∠A + ∠ACO + ∠AOC = 180º
⇒ 40º + ∠ACO + 90º = 180º
⇒ ∠ACO = 180º - 40º - 90º = 50º
∵ AB is a diameter.
∴ ∠ACB = 90º                 [Angle in a semicircle]
∴ 50º + x = 90º
⇒ xº = 90º - 50º = 40º
Thus, x = 40º

(iii) ∵ ∠AOC + ∠COB = 180º                 [Linear pairs]
∴ 120º + ∠COB = 180º
⇒ ∠COB = 180º - 120º = 60º
∵ The arc CB is subtending ∠COB at the centre and ∠CDB at the remaining part.
∴ ∠CDB = (1/2)∠COB
⇒ x= (1/2)(60º) = 30º
⇒ x = 30º

(iv) In ΔAOC,
∵ AO = OC                 [Radii of the same circle]
∴ ∠OAC = ∠OCA                [Angles opposite to equal sides are equal]
⇒ ∠OAC = 50º
∴ Exterior ∠COB = 50º + 50º = 100º.
Now, the arc BC is subtending ∠BOC at the centre and ∠BDC at the remaining part of the circle.
∴ ∠BDC = (1/2)∠BOC
⇒ x= (1/2)(100º) = 50º
Thus, x = 50º

(v) In ΔOAC, OA = OC                 [Radii of the same circle]
∴ ∠AOC = ∠ACO                [∵ Angles opposite to equal sides are equal]
Now, ∠AOC + ∠ACO + ∠OAC = 180º
⇒ ∠AOC + ∠ACO + 50º = 180º
⇒ ∠AOC + ∠ACO = 180º - 50º = 130º
⇒ ∠AOC = ∠ACO = (1300/2) = 65º
Now, ∠AOB + ∠AOC = 180º                [Linear pairs]
∴ ∠AOB + 65º = 180º
⇒ ∠AOB = 180º - 65º = 125º
∵ The arc AB subtends ∠AOB at the centre and ∠ADB at the remaining part of the circle.  
∴ ∠ADB = (1/2) ∠AOB
⇒ x= (1/2)(125º) = 62 (1/2)
∴ x= 62(1/2)

(vi) ∵ ∠BDC = ∠BAC                [Angles in the same segment]
∴ ∠BDC = 40º
Now, in ΔBDC, we have ∠BDC + ∠CBD + ∠BCD = 180º
∴ 40º + 80º + x = 180º
⇒ 120º + x = 180º
⇒ x = 180º - 120º = 60º
Thus, x = 60º


Question 5. In the adjoining figure, O is the centre of the circle. Prove that ∠ XOZ = 2(∠ XZY + ∠ YXZ).
 Solution:
Let us join OY.
∵ The arc XY subtends ∠XOY at the centre and ∠XZY at a point Z on the remaining part of the circle.

Class 10 Maths Chapter 10 Practice Question Answers - Circles

∴ ∠XOY = 2∠XZY                …(1)
Similarly, ∠YOZ = 2∠YXZ                …(2)
Adding (1) and (2), we have ∠XOY + ∠YOZ = 2∠XZY + 2∠YXZ
⇒ ∠XOZ = 2[∠XZY + ∠YXZ]


Question 6. Show that the sum of the opposite angles of a cyclic quadrilateral is 180º.
 Solution:
We have a cyclic quadrilateral. Let us join AC and BD. Since, angles in the same segment are equal.
∴ ∠ACB = ∠ADB                …(1)
and ∠BAC = ∠BDC                …(2)
Adding (1) and (2), we have

Class 10 Maths Chapter 10 Practice Question Answers - Circles

∠ACB + ∠BAC = ∠ADB + ∠BDC
⇒ ∠ACB + ∠BAC = ∠ADC
Adding ∠ABC to both sides, we have ∠ACB + ∠BAC + ∠ABC = ∠ADC + ∠ABC
But, ∠ACB + ∠BAC + ∠ABC = 180º                 [Sum of the angles of ΔABC = 180º]
∴ ∠ADC + ∠ABC = 180º
⇒ ∠B + ∠D = 180º
Since, ∠A + ∠B + ∠C + ∠D = 360º
⇒ ∠A + ∠C = 360º ∠ 180º = 180º

Class 10 Maths Chapter 10 Practice Question Answers - Circles


Question 7. Prove that the quadrilateral formed by angle bisectors of a cyclic quadrilateral is also cyclic.
 Solution: 
We have a cyclic quadrilateral ABCD in which the bisectors ∠A, ∠B, ∠C and ∠D for a quadrilateral PQRS.
From ΔABP, we have ∠PAB + ∠PBA + ∠P = 180º                [Sum of the three angles of ΔABP]

Class 10 Maths Chapter 10 Practice Question Answers - Circles⇒ (1/2) ∠A + (1/2)∠B + ∠P = 180º                …(1)
From ΔCDR, we have
∠RCD + ∠RDC + ∠R = 180º                [Sum of the three angles of ΔCDR.]
⇒ (1/2)∠C +(1/2)∠D + ∠R = 180°                 …(2)
Adding (1) and (2), we have (1/2)∠A + (1/2)∠B + (1/2)∠C +(1/2)∠D + ∠P + ∠R = 360º
⇒ (1/2) (∠A + ∠B + ∠C + ∠D) + ∠P + ∠R = 360º
⇒ (1/2)(360º) + ∠P + ∠R = 360º                 [∵ ∠A + ∠B + ∠C + ∠D = 360°]
⇒ ∠P + ∠R = 360º -(1/2) (360º) = 180º
Similarly, ∠Q + ∠S = 180º
Thus, the pairs of opposite angles of quadrilateral PQRS are supplementary.
Hence, PQRS is cyclic.

The document Class 10 Maths Chapter 10 Practice Question Answers - Circles is a part of the Class 9 Course Mathematics (Maths) Class 9.
All you need of Class 9 at this link: Class 9
48 videos|378 docs|65 tests

FAQs on Class 10 Maths Chapter 10 Practice Question Answers - Circles

1. What is a circle?
Ans. A circle is a closed curve in which all the points are equidistant from a fixed point called the center of the circle.
2. What is the formula to calculate the circumference of a circle?
Ans. The formula to calculate the circumference of a circle is C = 2πr, where C represents the circumference and r represents the radius of the circle.
3. How is the area of a circle calculated?
Ans. The area of a circle is calculated using the formula A = πr^2, where A represents the area and r represents the radius of the circle.
4. What is the relationship between the diameter and radius of a circle?
Ans. The diameter of a circle is twice the length of its radius. In other words, the diameter is equal to 2 times the radius.
5. How can the radius of a circle be determined if only the circumference is known?
Ans. If only the circumference of a circle is known, the radius can be determined by using the formula r = C / (2π), where r represents the radius and C represents the circumference of the circle.
48 videos|378 docs|65 tests
Download as PDF
Explore Courses for Class 9 exam
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
10M+ students study on EduRev
Download the FREE EduRev App
Track your progress, build streaks, highlight & save important lessons and more!
Related Searches

Sample Paper

,

ppt

,

Extra Questions

,

Semester Notes

,

Objective type Questions

,

MCQs

,

Previous Year Questions with Solutions

,

Class 10 Maths Chapter 10 Practice Question Answers - Circles

,

video lectures

,

Free

,

Class 10 Maths Chapter 10 Practice Question Answers - Circles

,

shortcuts and tricks

,

Viva Questions

,

mock tests for examination

,

Class 10 Maths Chapter 10 Practice Question Answers - Circles

,

Important questions

,

Exam

,

Summary

,

past year papers

,

study material

,

practice quizzes

,

pdf

;