Two circle touch externally. The sum of their areas is 149*Pi sq.cm and the distance between their c

Rats   10 years AGO

let radius of one circle= xcm
So, radius of other circle = (17-x) cm
So, acc. to problem,
(pi*x^2)+(pi*(17-x)^2)=149*pi
On solving, we get x=7cm, 10cm
So, radius of two circles are 7cm, 10cm.
ratio of areas = (pi*7^2):(pi*10^2)
= 49:100

Like? Yes (7) | No (2)  

ABHISHEK KUMAR MISHRA   10 years AGO

given that
r1+r2=17....(1)
r1^2+r2^2=149
or,(r1+r2)^2-2r1*r2=149
put (1) in above equation then r1*r2=70.....(2)
solve (1) and (2)then r1=7,r2=10 or vice versa
now r1^2/r2^2=(7)^2/(10)^2=49/100 answer

Like? Yes (3) | No   

Md Intshamuddin   10 years AGO

pi*(r1^2+r2^2)=149*pi
=> (r1^2+r2^2)=149...(i)
c1-c2=r1+r2=17
=> r1+r2=17......(ii)
solving (i) and (ii)
r1=7,r2=10
hence ratio of areas=49:100

Like? Yes (2) | No   

Hasina Abdulhakkim   10 years AGO

pi*(r1^2+r2^2)=149*pi
=> (r1^2+r2^2)=149...(i)
c1-c2=17
hence r1+r2=17......(ii)
solving (i) and (ii)
r1=7,r2=10
hence ratio of areas=49:100

Like? Yes (2) | No   

Ashok singh   10 years AGO

Given that :
Sum of areas 149 * pi sq. cm
and sum of radius =17 cm
assume two radius are r1,r2
r1^2+r2^2 = 149 -----(1)
r1+r2 = 17 -----(2)
we know that
(r1+r2)^2 = r1^2 + r2^2 + 2*r1*r2
289 = 149+2*r1*r2
r1*r2 = 70 -------(3)
using eqn (2) and (3)
r1 +70/r1 = 17
take lcm and make eqn
we got:
r1^2-17*r1+70 =0
by using shri dharacharya
we get value of r1 10 or 7
then value of r2 is 7 or 10
Ratio of areas = 49/100
Thats all

Like? Yes (2) | No   

Ezhilarasi   10 years AGO

pi(r)^2+pi(17-r)^2=149pi
pi[r^2+289+r^2-34r]=149pi
2r^2-34r+140=0
r^2-17r+70=0
by solving we get
r=7,10
area1=pi(7^2)=49pi
area2=pi(10^2)=100pi
area1:area2=49pi:100pi
49:100

Like? Yes (2) | No   

Shashi Kumar Gupta   10 years AGO

given r1+r2=17----a
pi*r1^2+pi*r2^2=149*pi
r1^2+r2^2=149-----b

by using a & b

r1r2=70;
so r1:r2=7:10
so answer is 49:100

Like? Yes (1) | No   

ShAg   10 years AGO

radius of circle 1 + radius of circle 2 = 17
so, if radius of circle 1 = x
radius of circle 2 = 17-x
area of circle 1 = pi*x*x
area of circle 2 = pi*(17-x)*(17-x)
pi*x*x + pi*(17-x)*(17-x) = 149*pi
=> x = 7,10
so ratio of areas = 49:100

Like? Yes (1) | No   

Rohit Chandra Lal   10 years AGO

7^2+10^2=149 and 10+7 is 17.so the radii are 7 and 10. And hence ratio of areas is 49:100.

Like? Yes (1) | No   

Yallapu Uday tarun kumar   10 years AGO

r1^2+r2^2=149
r1+r2=17
r1-r2=-3
r1=7 r2=10
ratio is r1^2:r2^2=49:100

Like? Yes (1) | No   

Bingumalla Abinay   10 years AGO

r1+r2=17

given pi(r1^2+r2^2)=149pi

verifying with answers r1,r2 are 7,10
so area's ration 49:100

Like? Yes (1) | No   

Mayuresh Srivastava   10 years AGO

r1+r2=17
r1^2+r2^2=149
r1r2=70
r1-r2=3
r1=10,r2=7
A2/A1=49/100.

Like? Yes (1) | No   

Avinash Aryan   10 years AGO

let radius r and 17-r,
then pi*(r*r) + pi*(17-r)*(17-r)=149*pi
on solving
r=7 and r=10,
so area1/area2=(7*7)/(10*10)=49:100.

Like? Yes (1) | No   

sujeet kumar keshri   10 years AGO

pi*r1^2+pi*r2^2=149*pi
=> r1^2+r2^2=149
and r1+r2=17
thus r1*r2=70
and r1-r2=3
therefore r1=10 and r2= 7
ratio of area = 49:100

Like? Yes (1) | No   

DINKAR BHARTI   10 years AGO

pi*r^2+PI*(17-r)^2=149*pi on solving this
r=10cm & r=7cm so ratio=49:100

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KALIDASS   10 years AGO

Let be radius of 1st circle=r1,the second is=r2
3.14*(r1)^2+3.14*(r2)^2=149*3.14
(r1)^2+(r2)^2=149...1
given (r1+r2)^2=(17)^2
(r1)^2+(r2)^2+2r1r2=289
149+2r1r2=289
r1r2=70......2
solve 1 & 2 we get
r1=7 r2=10
Ratio=(3.14*49)/(3.14*100)=49/100

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Shuvankar Ghosh   10 years AGO

r1=7,r2=10
so ratio of area is 7^2:10^2

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Devendra Marghade   10 years AGO

If two radius are R & r, then
(R^2+r^2)Pi=149*Pi or R^2+r^2=149 ---(i)
Also R+r=17 ---(ii)
so (R+r)^2=17^2
R^2+2R*r+r^2=289 ---(iii)
Substituting value of R^2+r^2 from (i) in (iii)
we get R*r=70
From (ii) r=17-R, so R*(17-R)=70
R^2-17R+70=0, Solving R=10 or 7
Ratio of their areas=(Pi*r^2)/(Pi*R^2)
=7^2/10^2=49/100 or 49:100

2R

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nagashree   10 years AGO

given pi*r1^2+pi*r2^2=149*pi
r1^2+r2^2=149
nd distance between center\'s given 17 implies r1+r2=17
on solving v get r1*r2=70
which gives numbers as 10 and 7
so area\'s can be in the ratio 100:49
or 49 :100
which gives 6 option

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Rupalika senapati   10 years AGO

pi*r2+pi*(17-r)2=149*pi
by solving this eq.
we get r=10/7
so ratio= 49/100 or 100/49
so ans-(6)

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SHIVANI PRAKASH   10 years AGO

let r1,r2 be the radius of resp. circles
r1+r2=17.....eq1
r1^2+r2^2=149....eq2
thus solving 1 and 2,r1r2=70
now we have to find r1^2/r2^2,checking from options:
49:100 satisfies the conditions thus it is the answer.

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Abhishek Kumar   10 years AGO

(3.14*r*r)+(3.14*{17-r)(17-r)=149*3.14
when we calculate this we get the value of
r=7 or 10
so,the ratio of their areas is 49:100 so option (6)is correct.ans

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sugyani dalai   10 years AGO

Let radius of two circles are r1 and r2.
given, (pi*r1*r1)+(pi*r2*r2)=149*pi
=>(r1*r1)+(r2*r2)=149 -------(eq1)
Also given , Distance between thier radius is 17
=>r1+r2=17
=>r1=17-r2
Now putting the value of r1 in to the Eq.(1)
((17-r2)(17-r2))+(r2*r2)=149
=>289+2(r2*r2)-34r2=149
=>(r2*r2)-17r2+70=0
=>(r2-7)(r2-10)=0
=>r2=7 or 10

=>r1=7 or 10

Now ratio of thier areas=49/100

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sanjay kumar saha   10 years AGO

A1+A2=149*pi
r1+r2=17
r1^2+r2^2=149
(r1+r2)^2=r1^2+r2^2+2r1*r2=289
2r1*r2=140
(r1-r2)^2=r1^2+r2^2-2r1*r2=149-140=9
r1-r2=3
r1+r2=17
r1=10
r2=7
A1:A2=49:100

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rohit kumar   10 years AGO

supppose two circles r A & B,
area(A)+area(B)=149*pi sq.cm
radius(A)+radius(B)=17

pi*radius(A)^2+pi*radius(B)^2=149*pi sq.cm
after solving the above eqn we get,
radius(A) or radius(B)=7 or 10

now,
area(A):area(B)=pi*radius(A)^2 : pi*radius(B)^2
i.e 49:100

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NEHA    10 years AGO

x+y=17
x^2+y^2=149
=>xy=70
x=70/y
=>y=7,x=10
x^2/y^2=49/100.

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prasanna babu   10 years AGO

let r,R are radius ,
given pi(r^2+R^2)=149
R+r = 17
on solving we get Rr =70
also R+r =17
we get R:r = 7:10
areas ratio is 49:100

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praveen kumar   10 years AGO

let sum of areas pi r^2+pi(17-r)^2=149pi
r^2+(289+r^2-34r)=149
r^2-17r+70=0
r=7,10
then ratio of areas =pi r^2:pi (17-7)^2
ans 49:100

Like? Yes (1) | No   

Amit kumar   10 years AGO

let the radius of circles are R and r
Then,
Pi(R^2+r^2)= 149*Pi
=>R^2+r^2=149..............(1)

R-r=17............(2)
Solve both Equations
We get R=10 and r = 7.
Ratio of their Areas= r^2 : R^2 = 7^2 : 10^2
===> r^2 : R^2 = 49 :100

Like? Yes (1) | No   

Sanampreet Singh   10 years AGO

let their radius is x & y.
So we have, x+y=17. and Pi(x^2+y^2)=149*Pi.
It gives x^2+y^2=149.
By solving both eqations we get x=7,y=10 or x-10,y=7

So ratio of their areas are 7^2:10^2 = 49:100

Like? Yes (1) | No   

KARIMUL SANFUI   10 years AGO

say radius of one r and radius of another (17-r).
pi*r*r+pi*(17-r)*(17-r)=149*pi.
solving above equation we get r=20,7.but r cantbe 20.so it is 7.finaly ratio of two circle=(pi*7*7)/(pi*10*10)=49/100.

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Nitesh Panjwani   10 years AGO

here r1+r2=17
Pi*r1^2+Pi*r2^2=149*Pi
Applying (r1+r2)^2=17^2
solving equation we get r1r2=70
now r2=17-r1
hence substituting r2 in r1r2=70 we get r1=7 r2=10
hence ratio=Pi*49/Pi*100=49/100

Like? Yes (1) | No   

HardikDungrani   10 years AGO

Here,
=>pi*r1^2+pi*r2^=149*pi
=>r1^2+r2^2=149.
now,ri+r2=17 so
=>(r1+r2)^2=(17)^2
=>r1^2+r2^2+2r1*r2=289
=>149+2r1*r2=289
=>r1*r2=70
nor r1=(17-r2) so
=>(17-r2)r2=70
=>r2^2-17r2+70=0
so r2=7 or r2=10 and ri=10 or r1=7
ration of their areas
=>(r1)^2:(r2)^2=(49):(100)

Like? Yes (1) | No   

Suman Saurabh   10 years AGO

Let radius of first circles is x cm & radius of second circle is (17-x) cm.
from question,
3.14[x^2 + (17-x)^2]=149*3.14
(x-7) (x-10)=0
x= 7 or 10.
so ratio of circles = 49 : 100

Like? Yes (1) | No   

KUNAL GAURAV   10 years AGO

We have been given Pi*r1^2 + Pi*r2^2 = 149
and r1+r2 =17.
From equation 2 we get r2=17-r1
Put this value of r2 in equation 1.
We get the value of r1 and r2 = 10 and 7
So the ratio will be = 49:100

Like? Yes (1) | No   

partha   10 years AGO

149=100+49;
=10^2 +7^2;

10+7=17

therfore areas raatios=7^2/10^2

Like? Yes (1) | No   

Balakrishnan M   10 years AGO

The sum of their areas = 149*Pi sq.cm
pi*R^2+pi*r^2=149*pi
R^2+r^2=149
The distance between their centres=17 cm.
R+r=17==>R=17-r
(17-r)^2+r^2=149
r^2+289-34r+r^2=149
r^2-17r+140=0
r=10 or 7 cm
r=10 ==> R=7
r=7 ===> R=10
ratio of their areas,
pi*r^2:pi*R^2=pi*49:pi*100
=49:100

Like? Yes (1) | No   

ankit rawat   10 years AGO

r1+r2=17......(1)
pi(((r1)^2)+((r2)^2)) = pi(149)......(2)
putting r1 = 17-r2 in equation(2)
we can solve the quadratic
r2 comes out to be 10 and 7
say r2 = 10
then r1 = 7
so ratio becomes pi(49):pi(100)
i.e. 49:100

Like? Yes (1) | No   

Anil Kumar Bharti   10 years AGO

Exp:
let radius of 1st circle: x
and 2nd circle is : 17-x
by question..
pi*x^2+pi*(17-x)^2=149*pi
after solve it
x=7,10
then to find the area of circle is:
pi*7^2:pi*10^2=49:100 Ans

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Deepak   10 years AGO

A1+A2=149Pi,
R1+R2=17,
solving , Pi R1^2+ Pi R2^2=149 Pi.
R1=07,R2=10
A1/A2=49/100.

Like? Yes (1) | No   

LOGAN WOLVERINE   10 years AGO

pi(r)^2+pi(17-r)^2=149pi;
pi(r^2+r^2-34r+289)=149pi;
r^2-17r+70=0;
solving this we get,
r=7 or r=10;
then
49pi+100pi=149pi;
so ratio is 49:100;
option 1

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NITESH DAS   10 years AGO

Pi*r^2+pi*(17-r)^2=149*pi
r=7,10(i got the two value)
then ratio of area pi*7^2:pi*10^2

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richa ahuja   10 years AGO

let the radii of two circles be r1 and r2
thus
r1^2+r2^2=149
and ri+r2=17
(r1+r2)^2=r1^2+r2^2-2r1r2
solving this we get ri=10,7
therefore r2=7,10
ratio of ares=pi*r1^2/pi*r2^2
=49/100

Like? Yes (1) | No   

Harjit kaur   10 years AGO

By making a quadratic eq we get two radii 7 & 10 of 1st & 2nd circle respectively so opt is 6

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Ajit Kumar Singh   10 years AGO

calculate r1 and r2 by adding areas of both circle=given area
r1=7,10
so r2=10,7
cal r1/r2 and square it

Like? Yes (1) | No   

HEENA RATHORE   10 years AGO

sum of areas =Pi*((r1)^2+(r2)^2)=149*pi sq.cm
now, r1^2+r2^2=149, also r1+r2=17,clearly r1=7 and r2=10,
so ratio of areas=pi*r1^2/pi*r2^2
=r1^2/r2^2
=7^2/10^2
=49/100

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Dinesh Kumaar R   10 years AGO

sum of areas=(100+49)*pi sq.cm
hence radius are 7,10
ratio of area =49:100

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PRATHAM MEHROTRA   10 years AGO

becoz radii of both the circles will be 7 and 10 cm

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Raunak MAnchanda   10 years AGO

Let the radius be r and 17-r.
sum of areas pi*r*r ➕ pi*(17-r)*(17-r) equals 149*pi.
solving we get value of r as 7,10.
Ratio is r*r:(17-r)*(17-r),putting the value of r as 7,we get 49:100

Like? Yes (1) | No   

Manoji Kumar   10 years AGO

Sum of area is 149*pi
A1+A2=149*pi
Pi*r1^(2)+pi*r2^(2)=149*pi
R1^2+R2^2=149…..(1)
But it was given that distance between there center’s is 17
R1+R2=17…….(2)
By solving (1) and (2) equation we will get R1=7,R2=10
Ratio of their areas:-
A1:A2=pi*r1^(2): pi*r2^(2)
= r1^(2):r2^(2)
=49:100

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vignesh   10 years AGO

given that r1 + r2 is 17
sum of area is 149pi

thus 149pi = r1^2pi + r2^2pi

the only logic solve the above two conditions r1+r2=17 and r1^2+r2^2=49 is r1=10,r2=7

thus the ratio of area is
pi*r1^2 :: pi*r2^2
49 : 100
thus the option is 6

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Nirmalram   10 years AGO

sum of their areas=149*pi
=>pi*(r^2)+pi*(R^2)=149*pi
=>(r^2)+(R^2)=149

distance bet centres=17
=>r+R=17
=>r=17-R
(substitute in 1st)
we get,
((17-R)^2)+(R^2)=149
=>R^2-17*R+70=0
=>R=10,7

ratio of areas=r^2:R^2
answer 49:100

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sudipta sadhukhan   10 years AGO

149=r^2+(17-r)^2
r=10 or 7
r1:r2=7:10
area1:area2=49:100

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IFFAT IRAM   10 years AGO

let the radii of the circles be r1 and r2
r1+r2=17.......(1)
(r1^2+r2^2)*Pi=149*Pi
r1^2+r2^2=149..........(2)
=> (r1+r2)^2-2r1*r2=149
=> 17^2-2r1*r2=149
=> 289-149=2r1*r2
=> r1*r2= 70........(3)
(r1-r2)^2=(r1+r2)^2-4r1*r2
(r1-r2)^2=9
r1-r2=3........(4)
r1=10
r2=7
ratio of the areas=(r1^2):(r2^2)
= 100:49

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ravi singh   10 years AGO

answer is 49:100

let radius of the circles is A and B

then, pi*A^2 + pi*B^2 =149*pi

=>A^2 + B^2 =149................eq.-1

and also given that

A + B =17
squaring both side

A^2 + B^2 + 2A*B=289

=>A*B = (289-149)/2..................eq-2

by solving eq-1 and eq-2

we get A=10, and B=7

hence,
(pi*A^2)/(pi*B^2) =49/100

hence option 6

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divya arya   10 years AGO

r1+r2=17
r1^2+r2^2=149
by solving these equations we get r1=7 and r2=10
so ratio=49:100
so ans is 6

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BHARGAV ARYASOMAYAJULA   10 years AGO

(r1^2)+(r2^2)=149
r1+r2=17
r1-r2=3
r1=10
r2=7

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Sudharsan   10 years AGO

sum of their areas pi*(r1^2+r2^2)=pi*149
r1^2+r2^2=149
also they touch externally r1+r2=17
2*r1*r2=140 => r1*r2=70
therefore radius is 7 and 10 respectively
but ratio of areas pi*7^2:pi*10^2= 49:100

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Sravan   10 years AGO

since sum of areas = 149
let radiues be r1 and r2
let first circle radii be r1 = x and other r2 = 17 -x
pi(x^2+(17-x)^2)=149*pi
solving this equation we get (x-7)(x-10)=0
==> r1 = 7,r2 = 10

now ration of area pir^2/pir,^2 ==> (7/10)^2 = 49/100

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Thaiyalnayaki   10 years AGO

area of circle=pi*r^2
so it can be split as 49*pi and 100*pi which 149*pi
if area's are 49 and 100 then radius id 7 and 10 respectively. then only the distance can be 17 when externally touched

therefore ratio of areas are 49:100

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Kumar Ankit   10 years AGO

Let r1 and r2 be the radius of two circle
Given pi*r1^2+pi*r2^2=149*pi----------i
r1+r2=17-------------------------ii
r1=17-r2
putting the value of r1 in i
By solving we get r2=10,7
r1=7,10
Hence ratio of their area is 49:100

Like? Yes (1) | No (1)  

Ravi Kant Sharma   10 years AGO

r1^2+r2^2=149
2r1r2=140
r1r2=70
r1*(r1-17)=70
r1

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