Circle

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 
= 22/7 × 14
= 22 × 2
= 44 cm

(ii) Diameter, d = 28 cm
Radius, r = d2$\frac{d}{2}$ = 282$\frac{28}{2}$ = 14 cm
∴ Circumference, c = 2πr
= 2 × 22/7 × 14
= 88 cm

(iii) Circumference, c = 616 cm
Now, c = 2πr     (where 'r' is the radius)
⇒616 = 2 × 22/7× r
⇒r = 616 x 7/44
⇒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 × 22/7 × r
⇒r = 72.6 × 7/44
⇒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 × 22/7 × r
⇒r = 176 / 2 x 3.14
⇒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 × 22/7 × 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
= 22/7 × 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$\frac{\mathrm{total} \mathrm{distance}}{\mathrm{circumference}}$ 
= 11004.4$\frac{1100}{4.4}$
= 1100044$\frac{11000}{44}$
= 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$\boxed{12 eryyt4}$
 

(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^∘.