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by goyalsau » Wed Jan 05, 2011 9:03 pm
Two circles have their centres 21 cm apart. The radii of the circles are 10 cm and 17 cm. Find the length (in cm) of the common chord of the two circles.



8

12

16

18

24
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by Rahul@gurome » Wed Jan 05, 2011 9:47 pm
Solution:
The two radii and the line joining the centres form a triangle with sides as 10, 17 and 21.
So semi perimeter s = 1/2(10+17+21) = 24.
Or area of triangle is sqrt(s(s-10)(s-17)(s-21)) = sqrt(24*14*7*3) = sqrt(3^2 * 4^2 * 7^2) = 84.
Also area of triangle is 1/2*(height over side which is 21)*21 = 84.
So height of triangle over side which is 21 is 8.
So total length of chord is 2*8 = 16.
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by goyalsau » Wed Jan 05, 2011 10:53 pm
Rahul@gurome wrote:Solution:
The two radii and the line joining the centres form a triangle with sides as 10, 17 and 21.
So semi perimeter s = 1/2(10+17+21) = 24.
Or area of triangle is sqrt(s(s-10)(s-17)(s-21)) = sqrt(24*14*7*3) = sqrt(3^2 * 4^2 * 7^2) = 84.
Also area of triangle is 1/2*(height over side which is 21)*21 = 84.
So height of triangle over side which is 21 is 8.
So total length of chord is 2*8 = 16.
Great Work Rahul, just for the knowledge i want to know whether is there any other method to solve this problem.
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by Anurag@Gurome » Thu Jan 06, 2011 12:19 pm
Image

Refer to the above image.

AB = 21, AC = 10, BC = 17. We have to find 2CM.
Say, CM = x, AM = a and BM = b
Thus, (a + b) = 21 ............................................................................ (1)

In triangle ACM, (AM)² + (CM)² = (AC)² => a² = (100 - x²) .............. (2)
In triangle BCM, (BM)² + (CM)² = (BC)² => b² = (289 - x²) .............. (3)

Now we have three equations in three unknown. We can solve for x. This apparent cumbersome task can be easily solved if we apply the basic formula of algebra: (a² - b²) = (a + b)(a - b)

From (2) and (3), (b² - a²) = (289 - x²) - (100 - x²) = 189
=> (b + a)(b - a) = 189
=> 21*(b - a) = 189
=> (b - a) = 9 ................................................................................... (4)

From (1) and (4), a = 6 and b = 15

Hence from (2), x² = (100 - a²) = (100 - 36) = 64 => x = 8 => 2x = 16

Hence length of the common chord is 16 cm.

The correct answer is C.
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by thebigkats » Mon Jan 17, 2011 2:46 pm
(disclaimer - didn;t know what the cord was.... had to look up the meaning of common chord. so cheated :-))

Once I drew the picture, I came up with following:

0.5 * height * base = sqrt (s * (s-base) * (s-side1) * (s-side2))
s = (base + side1 + side2) / 2 = 21 + 10 + 17 = 48 / 2 = 24

0.5 * h * 21 = sqrt ( 24 * (24-21) * (24-10) * (24-17))
10.5 * h = sqrt (24 * 3 * 14 * 7)
h = sqrt (2*2*2*3*3*7*2*7) /10.5
h = 84/10.5 = 8
2h = 16
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by GMATj » Mon Jan 24, 2011 9:55 am
Hi!
For some reason I don't get the same answer. Where's my error? I will refer to Anurag/Guromes figure for clarity:

First, figuring out the overlap:
(1) max length of overlap = combined distance of radii - distance between center = 17+10-21=6
So the proportions are:
small circle to overlap = 10-6 = 4
overlap 6
large circle to overlap = 17-6 = 11

(2) to find the straight distance from each circle's origin to the chord (AM and MB)
length without overlap + ½ of overlap = 4+3=7 (AM) and 11+3=14 (MB)

(3) now I have divided the line between the circle's origins into two parts (point M ), and if one goes perpendicularly up from their meeting point, one finds the chords top point (point C). Thus one can create 2 triangles, using the bases (which are known), the hypotenuse (which = radius) and solving for CM.
So AM^2 + AC^2 = CM^2 --> 7^2 + 10^2 = 149 = CM^2 --> CM = sqrt(149)

Chords length = 2CM = sqrt(2x149) = sqrt (298) which =! 16

Somethings wrong, but i can't find it : )
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by Oneva » Mon Jan 24, 2011 8:00 pm
This problem took me 25 minutes to solve. Would something this challenging really be on the GMAT? It seems to be on a whole new level compared to the CAT tests.
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by santoshs » Mon Jan 24, 2011 8:58 pm
Use can also make use of simple plugin method. As we know that gmat use common side ratio for right triangle such as
3:4:5, 5:12:13, 7:24:25, 8:15:17 and 9:40:41

Now lets plugin code length as 16.
Then we have one side of triangle as 10:8:6 following the ratio 3:4:5 and another side of triangle as 8:15:17 following the ratio 8:15:17

Now by adding 6 and 15 we get 21.
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by monge1980 » Sun May 01, 2011 2:20 am
For this problem I used the method suggested by Santoshs. I think it is not just a simple method, but the right method. This problem has only one solution, so if you find a solution it must be the right one.

When you look at triangles, one has a hypotenuse of 10 ... suggesting 3:4:5 ratio, the other has hypotenuse of 17 ... suggesting 8:15:17 triangle. Thus, we get one possible solution ... 6:8:10 and 8:15:17, where the sum of the two sides 6 and 15 get 21. Since this problem has only one solution, this must be the solution. So the chord is 2*8=16.
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by prashant misra » Wed Nov 09, 2011 10:01 am
can anyone please tell me that the way in which i did this question is right though i know that coming to the right answer is not the only thing.i added the two diameters 20+34 and the distance 17 that gives us 84 then subtracted it from 10+21+17 which gives us 16.please anyone do surely give me a reply because i know its wrong.
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by nskandan » Wed Feb 01, 2012 10:08 am
Anurag@Gurome wrote:Image

Refer to the above image.

AB = 21, AC = 10, BC = 17. We have to find 2CM.
Say, CM = x, AM = a and BM = b
Thus, (a + b) = 21 ............................................................................ (1)

In triangle ACM, (AM)² + (CM)² = (AC)² => a² = (100 - x²) .............. (2)
In triangle BCM, (BM)² + (CM)² = (BC)² => b² = (289 - x²) .............. (3)

Now we have three equations in three unknown. We can solve for x. This apparent cumbersome task can be easily solved if we apply the basic formula of algebra: (a² - b²) = (a + b)(a - b)

From (2) and (3), (b² - a²) = (289 - x²) - (100 - x²) = 189
=> (b + a)(b - a) = 189
=> 21*(b - a) = 189
=> (b - a) = 9 ................................................................................... (4)

From (1) and (4), a = 6 and b = 15

Hence from (2), x² = (100 - a²) = (100 - 36) = 64 => x = 8 => 2x = 16

Hence length of the common chord is 16 cm.

The correct answer is C.
Are the triangles ACM, BCM congruent . If that's the case I was thinking of applying the principle of congruency.
please advise.
NSK
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by somsubhra86 » Mon Feb 06, 2012 4:49 am
Thanks Rahul.
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by mourinhogmat1 » Tue Feb 07, 2012 11:55 pm
Common chord is not a GMAT concept per OG12, right?
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by ronnie1985 » Thu Mar 29, 2012 9:46 am
Drawing the figure makes it easy to formulate the chord length. 16 cm is answer
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by rajeshsinghgmat » Sat Mar 23, 2013 11:21 pm
(B) 12
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