CBSE Class 10 Mathematics Circles Worksheet Set H

Read and download free pdf of CBSE Class 10 Mathematics Circles Worksheet Set H. Students and teachers of Class 10 Mathematics can get free printable Worksheets for Class 10 Mathematics Chapter 10 Circles in PDF format prepared as per the latest syllabus and examination pattern in your schools. Class 10 students should practice questions and answers given here for Mathematics in Class 10 which will help them to improve your knowledge of all important chapters and its topics. Students should also download free pdf of Class 10 Mathematics Worksheets prepared by school teachers as per the latest NCERT, CBSE, KVS books and syllabus issued this academic year and solve important problems with solutions on daily basis to get more score in school exams and tests

Worksheet for Class 10 Mathematics Chapter 10 Circles

Class 10 Mathematics students should refer to the following printable worksheet in Pdf for Chapter 10 Circles in Class 10. This test paper with questions and answers for Class 10 will be very useful for exams and help you to score good marks

Class 10 Mathematics Worksheet for Chapter 10 Circles

 

Circles

Q.-A chord PQ of a circle is parallel to the tangent drawn at a point R of the circle. Prove that R bisects the arc PRQ. 
 
Ans- Given: In a circle a chord PQ and a tangent MRN at R such that QP || MRN
WT_circles test 29
To prove: R bisects the arc PRQ.
Construction: Join RP and RQ.
Proof: Chord RP subtends ∠1 with tangent MN and ∠2 in alternates segment of circle
so 1 = 2.
MRN || PQ
∴∠1 =∠3 [Alternate interior angles]
∠2 = ∠3
PR = RQ [Sides opp. to equal s in Δ RPQ]
Equal chords subtend equal arcs in a circle so
arc PR = arc RQ
or R bisect the arc PRQ. Hence, proved.
 
Q.-In figure, O is the centre of the circle and TP is the tangent to the circle from an external point T. If ∠PBT = 30°, prove that BA : AT = 2:1. 
WT_circles test 31
Ans- According to the question,
 
WT_circles test 30
 
AB is the chord passing through the centre
So, AB is the diameter
Since, angle in a semi circle is a right angle
∠APB= 90°
By using alternate segment theorem
We have ∠APB = ∠PAT = 30°
Now, in APB
∠BAP + ∠APB + ∠BAP = (Angle sum property of triangle)
BAP = 180° – 90° – 30° = 60°
Now, ∠BAP =∠ APT +∠ PTA (Exterior angle property)
60° = 30° +∠ PTA
∠PTA = 60° – 30° = 30°
We know that sides opposite to equal angles are equal
AP = AT
In right triangle ABP:
sin ∠ ABP= AP/BBA
sin30°= AP/BBA
BA : AT = 2 : 1
 
Q.-In the adjacent figure, if TP and TQ are two tangents to a circle with centre O, so that, ∠POQ =1000 then ∠PTQ is equal to 
WT_circles test 32
a. 60°
b. 40°
c. 80°
d. 90°
 
Ans- c. 80°
Explanation: Since the angle between the two tangents drawn from an external point to a circle in supplementary of the angle between the radii of the circle through the points of contact.
∴∠PTQ = 180° - 100° = 80° 
 
Q.- In the given figure, the measure of OQP is 
WT_circles test 33
a. 90°
b. 40°
c. 60°
d. 35°
 
Ans-  b. 40°
Explanation: Here OPB = 90° [Angle between tangent and radius through the point of contact]
∠OPQ + QPB = 90°
∠OPQ + 50° = 90°
∠OPQ = 40° But OPQ = OQP
[Angle opposite to equal radii]
∴∠OQP = 40°
 
Q.- In figure, AB is a chord of a circle and AT is a tangent at A such that ,∠BAT=60measure ∠ACB of is : 
WT_circles test 34
a. 120°
b. 150°
c. 90°
d. 110°
 
Ans- a. 120°
Explanation: Since OA is perpendicular to AT, then OAT = 90°
∠OAB +∠BAT = 90°
∠OAB + 60° = 90°
∠OAB = 30°
∠OAB= ∠OBA = [Angles opposite to radii]
∠AOB = 180° - (30° + 30°) = 120° [Angle sum property of a triangle]
Reflex ∠AOB = 360° - 120° = 240°
Now, since the arc AB of a circle makes an angle which is equal to twice the
angle ACB subtended by it at the circumference.
Reflex ∠AOB = 2 ∠ACB
240° = 2 ∠ACB
∠ACB = 120°
 
Q.- Quadrilateral PQRS circumscribes a circle as shown in the figure. The side of the quadrilateral which is equal to PD + QB is 
WT_circles test 35
a. PS
b. PR
c. QR
d. PQ
 
Ans- d. PQ
Explanation: PD + QB = PA + QA [Tangents from an external point to a circle are equal]
 
Q.- In the figure, QR is a common tangent to given circle which meet at T. Tangent at T meets QR at P. If QP = 3.8 cm, then find length of QR. 
WT_circles test 36
 
Ans- QP = 3.8
QP = PT (Length of tangents from the same external point are equal)
Therefore, PT = 3.8 cm
Also, PR = PT = 3.8 cm
Now, QR = QP + PR
QR = 3.8 + 3.8 = 7.6 cm

 

AREAS RELATED TO CIRCLES SECTION A: (1 MARK)

1.If area of a circle is numerically double its perimeter, then find the radius of the circle. (CBSE 2011)(4 units)

2.If the area of a sector is 5/18 of the area of the circle, find the angle subtended by the sector at the centre.(100:)

3.If the diameter of a semicircular protractor is 14 cm, then find its perimeter. (CBSE 2009)(36 cm)

SECTION B: (2 MARKS)

4.A square is inscribed in a circle. What is the ratio of the areas of the circle and the square.(π : 2)

5.The area enclosed between the two concentric circles is 346.5 cm2. The circumference of the inner circle is 88 cm. Find the radius of the outer circle. (CBSE 2011)(17.5 cm)

6.Find the number of revolutions made by a circular wheel of area 1.54 cm2 in rolling a distance of 176 cm. (CBSE 2013)(40)

7.If r and R are the respective radii of the smaller and bigger semi-circles, find the area of the shaded portion. ( CBSE 2010)(πR2)

SECTION C: (3 MARKS)

8.The long and short hands of a clock are 6 cm and 4 cm long respectively. Find the sum of the distances travelled by their tips in 24 hours. (Use π = 3.14)(954.56cm)

9.OABC is a rhombus whose three vertices A, B and C lie on a circle with centre O. Find the area of the rhombus, if the area of the circle is 1256 cm2. (Use π = 3.14)(20000√3cm2)

10.ABCD is a field in the shape of a trapezium with AD ‖ BC, ABC = 90: and ADC = 60:. Four sectors are formed with centres A, B, C and D. The radius at each sector is 14 m. Find the total area of four sectors. (CBSE 2012)

11.A steel wire when bent in the form of a square encloses an area of 121 sq. cm. If the same wire is bent into the form of a circle, find the area of the circle.(154 cm2)

SECTION D: (4 MARKS)

12.The diameters of front and rear wheels of a tractor are 80 cm and 2 m respectively. Find the number of revolutions that rear wheel will make in covering a distance in which the front wheel makes 1400 revolutions. (CBSE 2013)(280)

13.Find the area of the shaded region for figure 1.{(180-8π) cm2)}

14.Find the area of shaded region in fig.2, where a circle of radius 6 cm has been drawn with vertex O of an equilateral triangle OAB of side 12 cm.(Use π = 3.14 and 3=1.73) (EXAMPLAR QUESTION)(317.64cm2)

Fig.1

Fig.2

15.In fig.3, ΔABC is right angled at A. Semicircles are drawn on AB, AC and BC as diameters. Find the area of shaded region. (CBSE 2014)

 

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CBSE Class 10 Mathematics Chapter 10 Circles Worksheet

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Worksheet for Mathematics CBSE Class 10 Chapter 10 Circles

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