15. Chapter 15. Circles NIOS Class 10 Mathematics Textbook Solutions
Chapter 15. Circles NIOS Class 10 Mathematics Textbook Solutions for Exam Preparation
Chapter 15. CIRCLES
CHECK YOUR PROGRESS 15.1
In questions 1 to 5, fill in the blanks to make each of the statements true.
1. In Fig. 15.25,
(i) AB is a ... of the circle.
(ii) Minor arc corresponding to AB is... .
Answer: (i) AB is a chord of the circle.
(ii) Minor arc corresponding to AB is APB .
2. A ... is the longest chord of a circle.
Answer: A diameter is the longest chord of a circle.
3. The ratio of the circumference to the diameter of a circle is always ... .
Answer: The ratio of the circumference to the diameter of a circle is always constant .
4. The value of π as 3.1416 was given by great Indian Mathematician... .
Answer: The value of π as 3.1416 was given by great Indian Mathematician aryabhata-I
5. Circles having the same centre are called ... circles.
Answer: Circles having the same centre are called concentric circles.
6. Diameter of a circle is 30 cm. If the length of a chord is 20 cm, find the distance of the chord from the centre.
Solution: Here, AB = 30 cm , OA=OB =OC=AB2=302=15 cm 
, CD=20 cm 
and CM=DM=12CD=202=10 cm 
Photo
In ∆OCM
, we have
OM2+CM2=OC2
⇒OM2=OC2-CM2
⇒OM=152-102=225-100=125=5²×5
⇒OM=55cm 
Therefore, the distance of the chord from the centre is 55cm 
.
7. Find the circumference of a circle whose radius is
(i) 7 cm (ii) 11 cm (Take π=227 
)
Solution: (i) Here, r=7 cm
Therefore, the circumference of a circle =2πr 
=2×227×7=2×22=44 cm
(ii) Here, r=11 cm
Therefore, the circumference of a circle =2πr
=2×227×11=447×11=4847=69.14 cm 
8. In the Fig. 15.26, RS is a diameter which bisects the chords PQ and AB at the points M and N respectively. Is PQ || AB ? Given reasons.
Solution: We know that, the perpendicular drawn from the centre of a circle to a chord bisects the chord.
Since , RS is a diameter which bisects the chords PQ and AB at the points M and N respectively.
∠RMP=90° 
and ∠RNB=90° 
So, ∠RMP=∠RNB=90° 
[Alternative interior angles, If RS is transversal .]
Therefore, PQ||AB .
9. In Fig. 15.27, a line l intersects the two concentric circles with centre O at points A,B, C and D. Is AB = CD ? Give reasons.
Solution: Given, a line l intersects the two concentric circles with centre O at points A,B, C and D.
Since, O is the centre of a circle whose diameter is BC.
∴ OB=OC 
………………..(i)
Again, O is the centre of a circle whose diameter is AD.
OA=OD 
…………………(ii)
ii-i⇒OA-OB=OD-OC 
⇒AB+OB-OB=OC+CD-OC 
⇒AB=CD 
Yes , AB = CD .
TERMINAL EXERCISE
1. If the length of a chord of a circle is 16 cm and the distance of the chord from the centre is 6 cm, find the radius of the circle.
Solution: Here , PQ=16 cm and OM=6 cm
Since , OM⊥PQ 
So, PM=MQ=12×16=8 cm 
In ∆OPM 
, we have
OP2=OM2+PM2 
⇒OP=OM2+PM2 
⇒OP=62+82=36+64=100 
⇒OP=10 cm 
Therefore, the radius of the circle is 10 cm .
2. Two circles with centres O and O′ (See Fig. 15.28) are congurent. Find the length of the arc CD.
Photo
Solution:
3. A regular pentagon is inscribed in a circle. Find the angle which each side of the pentagon subtends at the centre.
4. In Fig. 15.29, AB = 8 cm and CD = 6 cm are two parallel chords of a circle with centre O. Find the distance between the chords.
Photo
Solution: Here, AB = 8 cm , CD = 6 cm and OA = OC = 5 cm .
Since , OM⊥AB and ON⊥CD 
, we have
AM=BM=12AB=12×8=4 cm 
and CN=DN=12CD=12×6 cm=3 cm 
In ∆OMA 
, we have
OM2+AM2=OA2
⇒OM2=OA2-AM2
⇒OM=OA2-AM2 
⇒OM=52-42=25-16=9
⇒OM=3 cm 
In ∆ONC 
, we have
ON2+CN2=OC2
⇒ON2=OC2-CN2
⇒ON=OC2-CN2=52-32=25-9=16
⇒ON=4 cm 
∴MN=ON-OM=4-3=1 cm 
Therefore, the distance between the chords is 1 cm .
5. In Fig.15.30 arc PQ = arc QR, ∠ POQ = 15° and ∠ SOR = 110° . Find ∠ SOP.
Solution:
6. In Fig. 15.31, AB and CD are two equal chords of a circle with centre O. Is chord BD = chord CA? Give reasons.
Solution:
7. If AB and CD are two equal chords of a circle with centre O (Fig. 15.32) and OM ⊥ AB, ON ⊥ CD. Is OM = ON ? Give reasons.
Solution:
8. In Fig. 15.33, AB = 14 cm and CD = 6 cm are two parallel chords of a circle with centre O. Find the distance between the chords AB and CD.
Solution: Here, AB = 14 cm , CD = 6 cm and OA = OC = 9 cm
Since , OM⊥AB and ON⊥CD , we have
AM=BM=12AB=12×14=7 cm 
and CN=DN=12CD=12×6 cm=3 cm
In ∆OMA , we have
OM2+AM2=OA2
⇒OM2=OA2-AM2
⇒OM=92-72 
⇒OM=81-49=32=2×42
⇒OM=42 cm 
In ∆ONC , we have
ON2+CN2=OC2
⇒ON2=OC2-CN2
⇒ON=92-32=81-9=72=2×36
⇒ON=62 cm 
∴MN=OM+ON=42+62 cm=102 cm 
Therefore, the distance between the chords AB and CD is 102 cm 
.
9. In Fig. 15.34, AB and CD are two chords of a circle with centre O, intersecting at a point P inside the circle.
OM ⊥ CD, ON ⊥ AB and ∠ OPM = ∠ OPN. Now answer:
Is (i) OM = ON, (ii) AB = CD ? Give reasons.
Solution:
10. C1
and C2
are concentric circles with centre O (See Fig. 15.35), l is a line intersecting C1 at points P and Q and C2 at points A and B respectively, ON ⊥ l, is PA = BQ? Give reasons.
Solution:
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