Mathematics Solutions Solutions for Class 7 Maths Chapter 12 Circle are provided here with simple step-by-step explanations. These solutions for Circle are extremely popular among class 7 students for Maths Circle Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Mathematics Solutions Book of class 7 Maths Chapter 12 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Mathematics Solutions Solutions. All Mathematics Solutions Solutions for class 7 Maths are prepared by experts and are 100% accurate.

Page No 77:

Question 1:

Complete the table below

   Sr. No.   Radius (r)  Diameter (d)     Circumference (c)
    (i)     7 cm   ....................  .................
    (ii)  ............    28 cm    ..............
    (iii)  ..............   ...............    616 cm
    (iv)   ...........   .............    72.6 cm

Answer:


(i) Radius, r = 7 cm
 Diameter, d = 2r = 2 × 7 = 14 cm
∴ Circumference, c = πd 
227 × 14
= 22 × 2
= 44 cm

(ii) Diameter, d = 28 cm
Radius, rd2 = 282 = 14 cm
∴ Circumference, c = 2πr
= 2 × 227 × 14
= 88 cm

(iii) Circumference, c = 616 cm
Now, c = 2πr     (where 'r' is the radius)
⇒616 = 2 × 227 × r
r = 616 × 12 × 722
r = 98
So, radius = 98 cm
Diameter, d = 2r = 2 × 98 = 196 cm

(iv) Circumference, c = 72.6 cm
Now, c = 2πr     (where 'r' is the radius)
⇒72.6 = 2 × 227 × r
r = 72.6 × 12 × 722
r = 11.55
So, radius = 11.55 cm
Diameter, d = 2r = 2 × 11.55 = 23.1 cm

The complete table is shown below.
 

   Sr. No.   Radius (r)  Diameter (d)     Circumference (c)
    (i)     7 cm   14 cm    44 cm
    (ii) 14 cm    28 cm    88 cm
    (iii)  98 cm   196 cm    616 cm
    (iv)   11.55 cm  23.1 cm    72.6 cm

Page No 77:

Question 2:

If the circumference of a circle is 176 cm, find its radius.

Answer:


Circumference, c = 176 cm
Now, c = 2πr    (where 'r' is the radius of circle)
⇒176 = 2 × 227 × r
r = 176 × 12 × 722
r = 28
∴ Radius of the circle = 28 cm

Page No 77:

Question 3:

The radius of a circular garden is 56 m. What would it cost to put a 4-round fence around this garden at a rate of 40 rupees per metre ?

Answer:


Radius of the circular garden, r = 56 m
Circumference of the circular garden, c = 2πr
= 2 × 227 × 56
= 352 m
∴ Length of the wire needed for one round of fencing = c = 352 m
Cost of one round of fencing = length of wire × cost per metre
= 352 × 40
= 14080 rupees
Cost of four round of fencing = 4 × 14080 = 56320 rupees

Page No 77:

Question 4:

The wheel of a bullock cart has a diameter of 1.4 m. How many rotations will the wheel complete as the cart travels 1.1 km ?

Answer:


Diameter of the wheel, d = 1.4 m
Circumference, c = πd
227 × 1.4
= 4.4 m
When the wheel completes 1 rotation, it covers a distance that is equal to its circumference.
So, number of rotations taken by the wheel to cover 4.4 m = 1
Now, the wheel covered a total distance of 1.1 km.
We know that, 1 km = 1000 m
∴ 1.1 km = 1.1 × 1000 m = 1100 m
∴ Total number of rotations taken by wheel = total distancecircumference 
11004.4
1100044
= 250
Hence, the wheel completes 250 rotations to cover a distance of 1.1 km.



Page No 79:

Question 1:

Choose the correct option.
If arc AXB and arc AYB are corresponding arcs and m(arc AXB) = 120° then m(arc AYB) = 12 eryyt4
 
(i) 140°   (ii) 60°   (iii) 240°  (iv) 160°
 

Answer:


Consider that arc AXB is the minor arc and arc AYB is the corresponding major arc.
It is known that, measure of major arc = 360∘ − measure of the corresponding minor arc.
We have, m(arc AXB) = 120°.
So, m(arc AYB) = 360∘ − m(arc AXB) = 360∘ − 120∘ = 240∘
Hence, the correct answer is option (iii).

Page No 79:

Question 2:

Some arcs are shown in the circle with centre ‘O’. Write the names of the minor arcs, major arcs and semicircular arcs from among them.

Answer:

Minor arc : An arc of a circle having measure less than 180∘.
Major arc : An arc of a circle having measure greater than 180∘.
Semicircular arc : An arc of a circle having measure equal to 180∘.


Names of minor arcs : 
(i) arc PXQ
(ii) arc PR
(iii) arc RY
(iv) arc XP
(v) arc XQ
(vi) arc QY

Names of major arcs :
(i) arc PYQ
(ii) arc PQR
(iii) arc RQY
(iv) arc XQP
(v) arc QRX

Names of semicircular arcs :
(i) arc QPR
(ii) arc QYR

Page No 79:

Question 3:

In a circle with centre O, the measure of a minor arc is 110°. What is the measure of the major arc PYQ?

Answer:



Suppose PQ is the minor arc and then m(arc PQ) = 110∘.
We know that, measure of major arc = 360∘ − measure of corresponding minor arc.
∴ m(arc PYQ) = 360∘ − m(arc PQ) 
=  360∘ − 110∘
= 250∘
Hence, the measure of major arc PYQ is 250∘.



View NCERT Solutions for all chapters of Class 7