A circular road runs round a circular ground. If the difference between the circumferences of the ou

RAKESH   9 years AGO

2*22/7 *(r1-r2) = 66
=> r1-r2 = 10.5

Like? Yes (7) | No (1)  

KSuresh   9 years AGO

let the circumference of the inner circle be 2*pi*r1,
and that of the outer be 2*pi*r2.
given difference i.e., 2*pi*r2 - 2*pi*r1=66
therefore r2-r1=66/2*3.14=10.509

Like? Yes (6) | No (1)  

Kajal Manohar Roshni Rangwani   9 years AGO

2(pi) [(R-r)]=66
therefore difference between the two radii will be the width of the road

Like? Yes (5) | No   

RAHUL DEB BARMAN   9 years AGO

let radius of road and ground be r1 metre, r metre respectively
2*pi*r1-2*pi*r=66
there for , r1-r=(66*7)/(2*22)=42/4=10.5 metre

Like? Yes (4) | No   

Gaurav Kumar   9 years AGO

let the outer radius be R and inner radius is r.
A/q
2?R-2?r=66
(R-r)=21/2=10.5

Like? Yes (4) | No   

mithilesh kumar   9 years AGO

Ans:2*pi*R-2*pi*r=66
so R-r=66*7/2*22 where pi=22/7
=21/2
=10.5m which is width of the road

Like? Yes (4) | No   

suraj bhatt   9 years AGO

2*pi(R1-R2)=66
R1-R2 is the width of the road

Like? Yes (4) | No   

Balkrishna Meena   9 years AGO

R= radius of circular road
r= radius of ground

width of road = R-r = (66/2(pi))= 10.5 m

Like? Yes (4) | No   

Jyotsna    9 years AGO

Let the outer radius be R and inner radius be r, now 2*pi*R-2*pi*r=66
therefore, R-r=66/(2*pi)=66/6.28=10.5

Like? Yes (4) | No   

neha gupta   9 years AGO

let the radius of circular ground be r and final radius after a circular road is constructed around it be R
so , 2*pi*R - 2*pi*r = 66
R - r = 10.5

Like? Yes (3) | No   

ABHINAV PUSHPAM   9 years AGO

Ans: 2
2π(R-r)=60 => 2*22/7 * (R-r)=60.
Therefore, (R-r)= (66*7/44)= 10.5m

Like? Yes (3) | No   

A M ANKIT KALLURAYA   9 years AGO

comparing the difference of the circumferences,we will get the width of the road

Like? Yes (3) | No   

sanjeev kumar   9 years AGO

Suppose r1=radius of ground and r2=redius of road
(2pi)r1-(2pi)r2=66
width of road =r1-r2=10.5

Like? Yes (3) | No   

Megha Goel   9 years AGO

difference between circumference is 66
i.e. 2*pi*R-2*pi*r=66
(R=outer and r=inner)
R can be written as (r+w)
where w= width of road
hence , 2*pi*w= 66
pi*w=33m
w=33*7/22m
w=21/2m
w=10.5m

Like? Yes (4) | No (1)  

ADITYA SARIPALLI   9 years AGO

Let the outer circle radius be "r2"
Let the inner circle radius be "r1"

difference in the circumferences of the 2 circles = 66 m
Hence we have ...

2(pi)(r2-r1) = 66 {where pi = 22/7}
=> (r2-r1) = 66 / (2 * pi)
= (66 * 7) / (2 * 22)
= 21/2
= 10.5 m

Thus width of the road is 10.5 m

Like? Yes (3) | No   

Surit Datta   9 years AGO

2*pi*R - 2*pi*r = 66
(R - r ) = 21/2 = 10.5 m
so option 2
that's it.

Like? Yes (2) | No   

shivjee yadav   9 years AGO

2πR-2πr=66
2π(R-r)=66
R-r=66/2π=66*7/44=21/2=10.5

Like? Yes (2) | No   

Debabrata Dey   9 years AGO

2*3.14*r1-2*3.14*r2=66
=>r1-r2=10.5 m
so width is 10.5m

Like? Yes (2) | No   

Testing Account   9 years AGO

2*pi*outer-2*pi*inner=66
diff.=width of road=66/6.28
=10.5m

Like? Yes (2) | No   

Piyush Telang   9 years AGO

Let radius of outer circle be R and that of inner circle be r meters.
Then according to given condition
2 x pi x R - 2 x pi x r = 66
=> 2 x 22 x ( R - r ) / 7 = 66
=> ( R - r ) = 10.5 m Width of the road is determined by the difference between the Radii i.e. R - r.
Hence 10.5 m is the width of the road.

Like? Yes (2) | No   

sudhir kumar maurya   9 years AGO

we need to find r2-r1
2 * pi *r2 - 2*pi*r1 =66 ( pi=22/7)
r2-r1=66/2*pi
r2-r1=10.5

Like? Yes (2) | No   

Shweta Sharma   9 years AGO

let r of inner circle be x so radius of outer circle will be x+width
diffrence in circumfrence=66=2X22/7(r+width-r)
width=(66X7)/(2X22)
width=10.5m

Like? Yes (2) | No   

RAVAL   9 years AGO

2*pi*outer-2*pi*inner=66
so.diff= width of road = 66/6.28
=10.5m

Like? Yes (2) | No   

Pihu Mascott   9 years AGO

let radius of road and ground be r1 metre, r metre respectively
2*pi*r1-2*pi*r=66
there for , r1-r=(66*7)/(2*22)=42/4=10.5 metre

Like? Yes (2) | No   

santosh kumar patel   9 years AGO

2*22/7*(R-r)=60
therefore, R-r = 66*7/44 = 10.5

Like? Yes (2) | No   

shreya saklecha   9 years AGO

diffrence in circumference=2*(22/7)[R-r]=66 (given)
where R=radius of outer circle and r=radius of inner circle
[R-r]=66/[2*(22/7)]=10.5m

Like? Yes (3) | No (1)  

pratik avinash   9 years AGO

As given the difference between their circumference is 66.
As all know circumference of a circle is 2(pi)r.
So, if we take radius of outer circle is r1 and radius of inner circle is r2.
A/q
2(pi)r1-2(pi)r2=66
so 2(pi)[r1-r2]=66
after solving this we get r1-r2=10.5
and that is thw width of the road.

Like? Yes (3) | No (1)  

velmala gangadhar   9 years AGO

R is the radius of outer circle
r is the radius of inner circle
2πR-2πr=66
width of the road is R-r=(66/2π)=10.5

Like? Yes (2) | No   

SOMA NARESH   9 years AGO

(2*PI*R)-(2*PI*r)=44

2*(22/7)(R-r)=66

R-r=(66*7)/44=10.5

Road width=(R-r)=10.5 meter

Like? Yes (2) | No   

Soma Naresh   9 years AGO

(2*PI*R)-(2*PI*r)=44

2*(22/7)(R-r)=66

R-r=(66*7)/44=10.5

Road width=(R-r)=10.5 meter

Like? Yes (2) | No   

Kuheli   9 years AGO

let R is the radius of the big circle and r is the radius of the small circle
so given that
2*pi*R - 2*pi*r = 66
=> R - r = 33/pi = 10.5

Like? Yes (3) | No (1)  

jayaram   9 years AGO

Because he asked for road width that is the difference between the radius of both circles..
2*3.14(R-r)=64
(R-r)=64/6.28;
=10.5

Like? Yes (2) | No   

Ravi Chauhan   9 years AGO

if outer radius is 2*3.14*R and inner radius is 2*3.14*r and difference is 66m,then R-r=10.5m

Like? Yes (2) | No   

Rishikesh Ranjan   9 years AGO

let r1 is the radius of outer circle, and r2 be the radius of inner circle
2*pi*r1- 2*pi*r2 =66
so r1-r2= 10.5m

Like? Yes (2) | No   

Angeline   9 years AGO

2*pi*R - 2*pi*r = 66
2*pi(R-r) = 66
R-r = 66/(2*pi)
R-r = (66*7)/(2*22)
R-r = 21/2
R-r = 10.5m

Like? Yes (2) | No   

anand vijay   9 years AGO

r=radius of outer circle & r1=radius of inner circle
2pi*r=66+2pi*r1
r-r1=66/2pi=10.5 m (pi=22/7)
option 2)

Like? Yes (2) | No   

laxmikant chhipa   9 years AGO

circumfrences of cirle =2Pir
diff of cir = (2pir1=2pir2) =66
2pi(r1-r2)=66
r1-r2=66/(2x3.14)
r1-r2 =10.5

Like? Yes (2) | No   

rahul shaw   9 years AGO

let inner and outer radius be x and r resp.
thn acc to prblm,
2?(x+r)-2?x=66
r=10.5mt
option 2

Like? Yes (2) | No   

pooja   9 years AGO

(2*r1*pi)-(2*r2*pi)=66
2pi(r2-r1)=66
hence width =10.5m

Like? Yes (2) | No   

Sonu Khandelwal   9 years AGO

outer circle's radius- r1
inner circle's radius- r2
then according question- (2*pi*r1)-(2*pi*r2)= 66
so r1-r2 = 66/2pi= 10.5

Like? Yes (2) | No   

Dipanjan Saha   9 years AGO

2πR-2πr=66;
2π(R-r)=66;
R-r=66/2π
R-r=10.5
Ans. 2)10.5

Like? Yes (2) | No   

ramya   9 years AGO

Inner circle radius =r1
Outer circle radius=r2
Width of the road = r2-r1
So, 2π(r2-r1)=66
r2-r1=10.5

Like? Yes (2) | No   

Rashmi Verma   9 years AGO

2?r2-2?r1=66
2?(r2-r1)=66
r2-r1=66/2?
r2-r1=21/2
r2-r1=10.5

Like? Yes (2) | No   

Anuj Rai   9 years AGO

let the outer radius be r1 and inner radius be r2
as per the question
2*22/7*(r1-r2)=66
where r1-r2= width of the road
r1-r2=10.5

Like? Yes (2) | No   

Ankesh   9 years AGO

given:
difference between the circumferences
of outer and inner circlr is = 66 m
now, we know that circumference = 22/7
to find :
As we have to find width of road between
circle 1 and circle 2 so;
2*(22/7)(R-r) = 66
44/7(R-r) = 66
44(R-r) = 66 * 7 = 462/44 = 10.5
therefore, (R - r) = 10.5

Like? Yes (2) | No   

shubham kumar soni    9 years AGO

Circumference of outer circle =2*pi*r1
Circumference of inner circle =2*pi*r2
Then
2pir1- 2pir2=66
2pi(r1-r2)=66
r1-r2=33/pi
Pi =22/7
So
r1-r2 =10.5

Like? Yes (2) | No   

jaganathan   9 years AGO

It is two circles one around the other.
Let the radius of the inner circle be "r1" and
He radius of the outer circle be " r2" .
We know circumference of a circle is 2*pi*r ( pi=22/7 and r is radius).
Given that:
Difference between circumference of outer and inner circle is "66"
That is.( 2*pi*r2)-(2*pi*r1)=66
r2-r1=66/(2*pi). pi=22/7
( the difference between the radius gives the width of the road )
=(66*7)/(2*22)
=10.5m. Which is the width of the road

Like? Yes (2) | No   

shubhika katarey   9 years AGO

as diffrence b/w circumference of outer and inner circle is 66mts,hence calculating difference 2pie(r+x)-2pie(r)=66, taking r=radius of inner circle and x=width of road,we get x=10.5.

Like? Yes (2) | No   

Shiva Sharma   9 years AGO

2*pi*(R-r) = 66
pi = 22/7
width(R-r) = 21/2
width=10.5

Like? Yes (2) | No   

preksha   9 years AGO

2*pi*(r1-r2)=66
r1-r2=66*7/(2*22)
r1-r2=21/2
10.5

Like? Yes (2) | No   

Narendra Gopi   9 years AGO

Circumference of a circle is 2*pi*r
so difference b/w innner and upper is 2*pi*R-2*pi*r=66
2*pi(R-r)=66
2*22/7(R-r)=66
R-r=66*7/44=10.5

Like? Yes (2) | No   

Ashu Kumar   9 years AGO

LET GROUND & ROAD HAVE RADIUS r &R RESPECTIVELY.
SO 2*22(R-r)/7=66
R-r=10.5
SO CORRECT OPTION IS 2

Like? Yes (2) | No   

AKANSHA SUMAN   9 years AGO

Let the radius of inner circle be r and that of outer circle be R.Now the difference between the radii of these ciercles will give the width of the road i.e.(2*pi*R-2*pi*r)=66.After solving this,we get R-r=66/2*pi = 10.5

Like? Yes (2) | No   

Sushmita Ghanty   9 years AGO

let the radius of the outer circle and inner circle are r1 and r2
so, 2pi(r1-r2)=66
ri-r2=66/2pi=10.5
therefore the width of the road is 10.5m

Like? Yes (2) | No   

Rupa Kumari   9 years AGO

circumference of the circle= 2 Pi * r.
Now, A/c Q, 2 Pi*(R-r) = 66 m
then, width of the road, i.e. (R-r) = 10.5 m

Like? Yes (2) | No   

SOUVIK   9 years AGO

2*22/7(R-r)=66
=>R-r=66*7/44
=>R-r=21/2=10.5 m
[here R is the radius of the circular outer road
r, is the radius of the circular ground
here pi=22/7]

Like? Yes (2) | No   

abhishek kumar   9 years AGO

2*227*R-2*227*r=66 where R is radius of circular road and r bis radius of circular ground. so (R-r)=10.5 width of road

Like? Yes (2) | No   

sai madhuri   9 years AGO

let the radii of ground and road be r,R.
given that 2*pi*(R-r)=66m
from this we can get R-r=10.5m
thus,10.5m is width of road.

Like? Yes (2) | No   

Kshitiz Khandelwal   9 years AGO

let r be inner radius
let x be width of road

then
2*pi(x+r) - 2*pi(r) = 66
on solving
x=10.5

Like? Yes (2) | No   

jaya   9 years AGO

2*22/7(R-r)=66
width of road=R-r=(66*7)/(22*2)
=10.5m
where r and R are radius of outer and inner circle resp.

Like? Yes (2) | No   

Zainab Naseem   9 years AGO

using the formula of circle circumference the value of their differencee comes 10.5m which is not among the given options.

Like? Yes (2) | No   

Subham Mahanta   9 years AGO

Let, the radius of the outer circle is R.
So, the perimeter is 2*pi*R.

Let, thee radius of inner circle is r.
So, the perimeter is 2*pi*r.

Mathematically,
2*pi*R - 2*pi*r = 66
which gives,
(R - r)= 66/ (2*pi)
= 10.5 m
So, option 2 is correct.

Like? Yes (2) | No   

ARJYA HALDER   9 years AGO

Let, the radius of the outer circle is R.
So, the perimeter is 2*pi*R.

Let, thee radius of inner circle is r.
So, the perimeter is 2*pi*r.

Mathematically,
2*pi*R - 2*pi*r = 66
which gives,
(R - r)= 66/ (2*pi)
= 10.5 m
So, option 2 is correct.

Like? Yes (2) | No   

shahida   9 years AGO

let the inner and outer radii be r and R metres..
then 2*22/7(R-r)=66
R-r=66*7/22*2
R-r=10.5

Like? Yes (2) | No   

vimal tiwari   9 years AGO

2* .3.14* r1 - 2*3.14* r2 = 66

r1 -r2 = 10.5m

Like? Yes (2) | No   

Mugesh sivagnanam Krishnapillai   9 years AGO

Let the radius of outer round be R and inner round be r,
2πR-2πr=66
2π(R-r)=66
R-r=10.504.
Thus the width of road is 10.5m.

Like? Yes (2) | No   

Khan   9 years AGO

Circumference of outer circle be 2*pi*R.
Circumference of inner circle be 2*pi*r.
Given difference is 66m.
Hence, 2*pi*R-2*pi*r=66
R-r= 66 / (2 * 3.14)
R-r= 10.5

Like? Yes (2) | No   

Kethineni Vinodh   9 years AGO

2piR-2pir=66
R-r=66/2*pi
Width=R-r=10.5095metres

Like? Yes (2) | No   

Sangeetha G   9 years AGO

diff of circum=2*22/7(r2-r1)=66
so,r2-r1=10.5
ans:10.5

Like? Yes (2) | No   

Nishith HM   9 years AGO

let R be outer radius n r be inner radius given 2piR-2pir=66
solving we will get R-r=10.5m

Like? Yes (2) | No   

Anupam Mukherjee   9 years AGO

2*pi*(outer rad.-inner rad.)= 66 ; therfore, (outer rad.-inner rad.)= 10.5= width of the road

Like? Yes (2) | No   

Archie   9 years AGO

(2*pi*R)-(2*pi*r)=66
R-r=66/(2*pi)
=66/(2*(22/7))
= 10.5 m
where R is radius og road and r i s radius of ground :)

Like? Yes (3) | No (1)  

Prabakaran   9 years AGO

W - width
2*pi(R+W)-2*pi*R=66
2*pi*W=66
W=(66*7)/(2*22)
=10.5

Like? Yes (2) | No   

Devendra Marghade   9 years AGO

If radius of outer circle=R & inner circle=r, then
Width of the road=R-r
Difference between circumference=(2*Pi*R)- (2*Pi*r)=66 , 2*Pi*(R-r)=66, R-r=66/(2*22/7)=10.5
Option 2)

Like? Yes (2) | No (1)  

Kundan Gope   9 years AGO

Let r1 is outer radius and r2 is inner radius
2*3.14*r1-2*3.14*r2=66
r1-r2=10.5 m

Like? Yes (2) | No (1)