SOLUTION: Two circles intersect and have a common chord. THe radii of the circles are 13 cm and 15 cm. The distance between their centers is 14 cm. Find the length of their common chord. I a

Algebra ->  Circles -> SOLUTION: Two circles intersect and have a common chord. THe radii of the circles are 13 cm and 15 cm. The distance between their centers is 14 cm. Find the length of their common chord. I a      Log On


   

Question 194449: Two circles intersect and have a common chord. THe radii of the circles are 13 cm and 15 cm. The distance between their centers is 14 cm. Find the length of their common chord. I already drew the picture. I drew two circles intersecting each other like a venn diagram. Then, i connected the two centers and that equals 14 and drew a line from the center of each circle to an edge of the circle equaling the radius of each circle. Then, I drew a dashed line down the intersection of the two circles in the middle and a perpendicular line shows up with the line connected by the center. What step am I supposed to take from there?
Found 2 solutions by Edwin McCravy, vleith:
Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!

Let the circles' centers be A and B.
Draw the common chord CD. Draw radii AC and BC.

We know that AC=13, BC=15, AB=14 



We are given all three sides of triangle ABC, so
we can use the law of cosines to find cos(A)

BC%5E2=AB%5E2%2BAC%5E2-2%2AAB%2AAC%2Acos%28A%29

2%2AAB%2AAC%2Acos%28A%29=AB%5E2%2BAC%5E2-BC%5E2

cos%28A%29=%28AB%5E2%2BAC%5E2-BC%5E2%29%2F%282%2AAB%2AAC%29

cos%28A%29=%2814%5E2%2B13%5E2-15%5E2%29%2F%282%2A14%2A13%29

cos%28A%29=%28196%2B169-225%29%2F%282%2A14%2A13%29

cos%28A%29=140%2F364

cos%28A%29=5%2F13

We know that the line through their centers
is the perpendicular bisector of the common 
chord CD.  

This tells us two things. 

1. If we find CE we can just
double it to get the common chord CD.  

2. Triangle AEC is a right triangle and therefore

%28CE%29%2F%28AC%29=sin%28A%29

CE=AC%2Asin%28A%29

CE=13%2Asin%28A%29

Now we use the identity 

Sin%5E2%28A%29+%2B+Cos%5E2%28A%29+=+1

Sin%5E2%28A%29+=+1+-+Cos%5E2%28A%29

Substitute 5%2F13 for Cos%28A%29

Sin%5E2%28A%29+=+1+-+%285%2F13%29%5E2

Sin%5E2%28A%29+=+1+-+25%2F169

Sin%5E2%28A%29+=+169%2F169-25%2F169

Sin%5E2%28A%29+=+144%2F169

sin%28A%29+=+sqrt%28144%2F169%29

sin%28A%29+=+12%2F13

Now we substitute that in

CE=13%2Asin%28A%29

CE=13%2A%2812%2F13%29

CE=cross%2813%29%2A%2812%2Fcross%2813%29%29

CE=12

CE is half the common chord CD, so

the common chord CD is twice CE or 

2x12 or 24.

Edwin

Answer by vleith(2983) About Me  (Show Source):
You can put this solution on YOUR website!
So far so good.
The line that connects the two centers will cut the chord into 2 equal lengths and be perpendicular to it. Let the length of the chord be 2L. Then the line connecting the centers cuts that chord into 2 segments, each L long.
So Now you have 2 sets of right triangles. Both sets have one leg of length L. One set has a hypotenuse of 15; the other of 13. The sum of the two third legs is 14. Let the length of the thrid leg on the larger triangle be x. Then the lenght of the third leg on the shorter one is (14-x).
Use Pythagorean theorem twice
L%5E2+=+sqrt%2815%5E2+-+x%5E2%29
L%5E2+=+sqrt%2813%5E2+-+%2814-x%29%5E2%29+
I bet you can take it in from here