Two circle touch externally. The sum of their areas is 130 Pi sq.cm. and the distance between their centres is 14 cm. Find the radii of two circles.

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10 Apr 2014 Quadratic equation Two circle touch externally. The sum of their areas is 130 Pi sq.cm. and the distance between their centres is 14 cm. Find the radii of two circles.: OPtion
1) 9, 5
2) 11, 5
3) 3, 3
4) 9, 3
5) 11, 7
6) 13, 3
7) 11, 3
8) 12, 4
9) 7, 4
10)None of these; Correct Option: 7 Best Solution (49)

Ann Theressa 10 years AGO

Let the radius of circles be x and y

Area of 2 circles = 130*pi
pi*x^2 + pi*y^2 = 130*pi
x^2 + y^2 = 130 ----(1)

Sum of their radius = 14
x + y = 14 ----(2)

(x+y)^2 = x^2 + y^2 + 2xy
14^2 = 130 + 2xy
2xy = 66
xy = 33 ----(3)

From (2) and (3)

x*(14-x) = 33
x^2 -14x + 33 = 0
(x-11)(x-3) = 0
x = 11,3

Radii of 2 circles are 11cm and 3cm

Like? Yes (70) | No (26)

AMIT KUMAR 10 years AGO

let 1st radius = r cm
2nd will be = 14-r cm
a/c question,
(pi*r^2) + (pi*(14-r)^2) = 130 pi
=> r^2 + (14-r)^2 = 130
=> 2*r^2 - (28*r) + 66 = 0
=> r^2 - 14*r + 33 = 0
=> r^2 - 3*r - 11*r + 33 = 0
=> r(r - 3) -11(r-3) = 0
r = 3 or r = 11
if r = 3 then 2nd will be = 11
r = 11 then 2nd will be = 3 ans

Like? Yes (34) | No (7)

Gitanjali K 10 years AGO

Let r, represent the measure of the radius of one of the circles,

then 14-r, must represent the radius of the other circle.

Therefore, the area of the first circle is πr^2,
and the area of the second circle is π(14-r)^2, and given that the sum of these areas is 130π cm^2
Therefore,
πr^2+ π(14-r)^2=130π
r^2+ (14-r)^2=130
2r^2-28r+66=0
r^2-14r+33=0
therefore r=3 or r=11

Like? Yes (16) | No (4)

Clair Christ D 10 years AGO

pi(a^2+b^2)=130pi
a^2+b^2=130
a+b=14
(a+b)^2=a^2+b^2+2*a*b
196=130+2ab
ab=33
a+b=33
11,3

Like? Yes (10) | No (2)

Dr.Dipin Singh 10 years AGO

(x^2+y^2)*pi=130 *pi
(x^2+y^2)=130
x+y=14
solving
x=11 cms,y=3 cms

Like? Yes (15) | No (10)

Surya Narayana K 10 years AGO

Answer : 11 cm and 3 cm (Option 7)
Let r1 and r2 be the radius of two circles and given that they are touch externally.

So, distance between their centers is r1+r2=14cm --(1)
and sum of their areas is Pi(r1^2+r2^2)=130Pi sq.cm --(2)

Substitute r2=14-r1 (from (1)) in (2),then
r1^2+(14-r1)^2=130
=>r1^2+r1^2+196-28r1=130
=>2r1^2-28r1+66=0
=>r1^2-14r1+33=0
=> r1= 11 or 3
then r2= 3 or 11.

The radii of two circles : 11 cm and 3 cm (Option 7).

Like? Yes (15) | No (10)

vishesh saxena 10 years AGO

let areas of two circle be a1 & a2,
given: a1+a2= 130 Pi,
radius be ri & r2;
given: r1+r2=14;
area of circle= pi(r)^2;
=> a1+a2= pi(r1^2+r2^2);
calculating r1r2 from here. solving further gives
r1= 11,3; r2= 3,11;

Like? Yes (13) | No (8)

vishal gupta 10 years AGO

sum of areas=130pi
pi(r1)^2+pi(r2)^2=130pi
(r1)^2+(r2)^2=130 (first equation)
distance between centres (touch externally)=14
r1+r2=14 (second equation)
solving equqtion 1 & 2
r1=11.
r2=3.

Like? Yes (3) | No (1)

shree 10 years AGO

(3.14)*121+(3.14)*9=130(3.14)

Like? Yes (1) | No

arnold 10 years AGO

simply add two values
we have now two option
option 1 and option 7
take sum of the square of both number
option 7 gets us the answer

Like? Yes (1) | No

Satish Zod 10 years AGO

only sr no. 7 fits the situation.

Like? Yes (1) | No

subramani 10 years AGO

In given Answers only 1 and 7 is equal to 14 cms. Which is Distance b/w. their centres.
If apply both in Area of Circle ,7th Answer is equal to 130 Pi Sq.cm.

Like? Yes (1) | No

Govind 10 years AGO

x^2+y^2=130
x+y= 14
xy = (14^2-130)/2= 33
x=11, y=3

Like? Yes (2) | No (1)

karthik hariharan 10 years AGO

3.14x^2+3.14y^2=130*3.14
so x^2+y^2=130;
radius distance 14
so x+y=14
x=y-14
substitute the eqn and get radius 11,3
that is 7th option

Like? Yes (1) | No

C.Aarthi 10 years AGO

let x, y be the two radii..
And since it touch externally
x+y =14
Since sum of the area=130
x^2 pi + Y^2 pi =130 pi
x^2 + Y^2=130
x^2 +(14-X)^2 =130
x^2 +196+X^2 -28x=130
2X^2 -28x +66 =0
x^2 - 14 x + 33=0
x=11, 3
y=11,3
thus radii is 11,3

Like? Yes (3) | No (2)

Kaustubh karekar 10 years AGO

by solving the equation we get the answer as 11 and 3

Like? Yes (2) | No (1)

Roopali 10 years AGO

let r1=x,then r2=14-x
pi(r1)square+pi(r2)square=13pi
on solving we get 11,3 Ans.

Like? Yes (2) | No (1)

Anupam Banerjee 10 years AGO

Let the radii of the two circles be r1 & r2;
so, pi r1^2+ pi r2^2 = 130 pi sq.cm.
or, r1^2 + r2^2=130;

and we have, r1+r2=14cm.
or, r1^2 + 2.r1.r2 + r2^2 = 196 sq. cm;
or, 130 + 2.r1.r2 = 196;
or, 2.r1.r2 = 66;

r1^2 + r2^2 = 130;
or, (r1-r2)^2+2.r1.r2 = 130;
or, (r1-r2)^2 = 130-66;
or, r1-r2 = sqrt 64;
or, r1-r2 = 8;

and r1+r2=14;

so,2.r1 = 22;
or, r1 = 11cm;

and 2.r2 = 6;
or, r2 = 3cm.

Like? Yes (2) | No (1)

Md Shahid Akhter 10 years AGO

r1+r2=14
pi(r1^2+r2^2)=130*pi
r1^2+r2^2=130
we know
r1^2+r2^2=(r1+r2)^2-2*r1*r2
putting the values
130=14^2-2*r1*r2
r1*r2=33
r1=11,r2=3 or r1=3,r2=11
Option 7 is correct one

Like? Yes (2) | No (1)

Santosh Vishwakarma 10 years AGO

suppose radius are x and (14-x)
sum of areas is=>Pi*x*x+Pi*(14-x)*(14-x)=130Pi
=>x*x+(14-x)*(14-x)=130
=>x=11,3

Like? Yes (2) | No (1)

Arindrajit Chakraborty 10 years AGO

Easy as pie..can be done by option testing (11)^2 + (3)^2=130

Like? Yes (3) | No (2)

Patel Yogesh Rameshbhai 10 years AGO

distance between two centers of circles 14 cm so there are two possibilities from given option
the radius of two cicle are either 9,5 or 11,3
sum of area is 130pi sq.cm so if we take radius 11,3 then we get exact sum of area
so right answers is 11,3

Like? Yes (6) | No (5)

Pragnesh patel 10 years AGO

there are two possibilities from given option
the radius of two cicle are either 9,5 or 11,3
sum of area is 130pi sq.cm so if we take radius 11,3 then we get exact sum of area
so right answers is 11,3

Like? Yes (3) | No (2)

Vinotha 10 years AGO

given sum of their area=130(pi)
r^2+R^2=130;
r+R=14
(r+R)^2=196
solving we get r,R=11,3

Like? Yes (2) | No (1)

neerav 10 years AGO

π(14-x)^2 + π(x^2) = 130 π
π(196+x^2 -28x) + πx^2 = 130 π
π(196+2πx^2-28x)=130π
196+2πx^2-28x=130
66+2x^2-28x=0
33+x^2-14x=0
33-11x+x^2-3x=0
-11(x-3)+x(x-3)=0
(x-3)(x-11)=0
x=3 & x=11

Like? Yes (2) | No (1)

Raja 10 years AGO

11+3=14 ; 11^2+3^2=130

Like? Yes (2) | No (1)

MP 10 years AGO

Two circle touch externally. The sum of their areas is 130 Pi sq.cm. and the distance between their centres is 14 cm.
pi*(R^2+r^2)=130 *pi
(R^2+r^2)=130
R+r=14
solving these eqns, we get
R=11 cms,r=3 cms

Like? Yes (2) | No (2)

Aman 10 years AGO

pi*(x^2+y^2)=130 *pi
(x^2+y^2)=130...(1)
x+y=14...........(2)
solving (1) & (2)
x=11 cms,y=3 cms

Like? Yes (5) | No (5)

llamar 10 years AGO

take circle1 radius=r,circle2 radius=R
given pir^2+piR^2=130pi
=>r^2+R^2=130---(1)
given r+R=14---(2)
by solving (1)&(2)
we get r-R=8---(3)
by solving (2)&(3)
we get r=11,R=3

Like? Yes (7) | No (7)

Rahul Kesharwani 10 years AGO

As circles touch each other externally, it means their area are seprated,
So Area of Circle = pi * r^2

i. e. 11^2 * pi + 3^2 * pi = 130* pi

Like? Yes (4) | No (4)

ASHISH MITTAL 10 years AGO

By solving we get r1*r2=33
And answer which give this value is 11 and 3

Like? Yes (3) | No (3)

K.Bhagyasri 10 years AGO

circle area formula= pi*(r^2)
sum of two circle areas :
pi(r1^2+r2^2)=130*pi sq.cm--------(1)

form (1)
r1^2+r2^2=130--------------(2)

r1+r2=14 cm------------(3)
square on both sides

(r1+r2)^2 =14^2
r1^2+r2^2+2r1r2 = 196
130+2r1r2 = 196
r1r2=33
r1=33/r2 substitute in (3)

r1+r2=14
(33/r1)+r1=14

r1^2-14r1^2+33=0

sol : 11,3

Like? Yes (5) | No (5)

Mukesh Nayal 10 years AGO

area of two circles is 130 pi
so,pi*(r1)*(r1)+pi*(r2)*(r2)=130*pi
r1^2+r2^2=130
adding and subtracting 2*r1*r2 we have,
(r1+r2)^2-2*r1*r2=130
since r1+r2=14(given)
so,14^2-130=2*r1*r2
66=2*r1*r2
r1*r2=33
so only one possibility of r1 and r2 which is 11 and 3 which equates 33.

Like? Yes (4) | No (4)

Ghanshyam Gulati 10 years AGO

r1^2+r2^2=130
r1+r2=14

on solving r1=11
r2=3

Like? Yes (2) | No (2)

Saraswathy 10 years AGO

Let "r" be the radius of one circle
Then radius of another circle is (14-r)
area of the circle is πr^2

Given sum of 2 areas is 130π
πr^2 + π(14-r)^2 = 130π
π[r^2 + 196 + r^2 - 28r] = 130π
2r^2 - 28r = -66
r^2 - 14r + 33 = 0

Factorize this...
You ll get r = 11, 3

Radius of 1st circle is 11
Radius of 2nd circle is (14 - 11) = 3

So option (7)

Like? Yes (2) | No (2)

Naveen Ganesh 10 years AGO

Given r1 + r2 = 14
r1^2 + r2^2 = 130
looking at answer options, only 1 and 7 sum up to 14, among them only 7 satisfies the second equation

Like? Yes (1) | No (1)

RAKESH 10 years AGO

pi(r1^2+r2^2)= 130pi

=> r1^2 + r2^2 = 130

& r1 + r2 = 14 => r1,r2=11,3

Like? Yes (2) | No (2)

akash kumar 10 years AGO

for 2nd cndition to be fulfilled we have only option 1&7.put values of option 1&7 .we find option 7 is correct

Like? Yes (2) | No (2)

karthika kumar 10 years AGO

Let A1 & A2 be area of two circles.
Given A1+A2=130pi sq.cm
=> piR1^2+piR2^2=130pi => R1^2+R2^2=130
Also given distance between their centres is 14 cm and the two circle touch externally ,therefore R1+R2=14
(R1+R2)^2=R1^2+R2^2+2R1R2
=> 14^2=130+2R1R2
=> R1R2=33
thus we have R1+R2=14 and R1R2=33
therefore R1=11 and R2=3

Like? Yes (2) | No (2)

hanamant k 10 years AGO

A=A1+A2 =>given
130*pi= pi*r^2 + pi*(14-r)^2

2r^2-28r+66=0
so,
r=3 or r=11

Like? Yes (1) | No (1)

puja 10 years AGO

radius=r,o
r^2+o^2=130
and
r+o=14
so
from, equating both eqn. in below
(r+o)^2=14^2
r^2+o^2+2ro=196,
r=11 and o=3

Like? Yes (1) | No (1)

Rajitha 10 years AGO

Distance between the centers is 14 cm.
assume r be the radius of one circle.
the radius of another circle will be (14-r).
Area of the circle is pi*r^2.
Sum of the two circles is= (pi*r^2)+(pi*(14-r)^2)=130 pi
pi(r^2+(14-r)^2)=130 pi
r^2+(14-r)^2=130
2 r^2-28 r+66=0
r^2-14 r+33=0
(r-3)(r-11)=0
r=3 or r=11
the radius of one circle is 3 and the radius of another circle is 11.

Like? Yes (1) | No (1)

TUSHAR 10 years AGO

r1+r2=14,r1^2+r2^2=130pi. solve equation, we get r1=11 r2=3

Like? Yes (1) | No (1)

vishnuvardhan 10 years AGO

C1C2=R1+R2
14=R1+R2
R1^2+R2^2=130
(R1+R2)^2=196
R1R2=33
R1=11,R2=3

Like? Yes (1) | No (1)

NIRUPOM GHOSH 10 years AGO

pi(r1^2+r2^2)=130pi and r1+r2=14 solving r1 and r2 are 11,3.

Like? Yes (1) | No (1)

nalini singh 10 years AGO

pi*r1^2+pi*r2^2=130*pi
r1+r2=14
solve this two equation
r1=11,r2=3

Like? Yes (1) | No (1)

ravi prakash 10 years AGO

area of circle=pir^2.
sum of radius=distance between theire centre.
11+3=14.
area would be=pi(R^2+r^2)
=pi(11^2+3^2)
=130pi.

Like? Yes (2) | No (3)

Mahalakshmi 10 years AGO

let radius of one circle be r and another (14-r)
area of circle is =3.14*r^2
for area of two circles touching eachother will be
3.14r^2+3.14(14-r^2)=130*3.14
solving this equation we will get
(r-11)(r-3)=0
r=11,3
so ans is 7)11,3

Like? Yes (1) | No (3)

Saidinesh 10 years AGO

11+3=14 distance of two radius,
121pi+9Pi=130Pi;
it satisfies option 6

Like? Yes (1) | No (4)

Dimple

India

Punjab

Time: 00:01:33

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Two circle touch externally. The sum of their areas is 130 Pi sq.cm. and the distance between their

M4maths Previous Puzzles

10 Apr 2014 Quadratic equation