SOLUTION: two circles intersect (a nd b) and share a common chord (cd). the radius of circle a is 10in, radius circle b = 16 in. distance between centers, 22 in. find cd.

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Question 1207163: two circles intersect (a nd b) and share a common chord (cd). the radius of circle a is 10in, radius circle b = 16 in. distance between centers, 22 in. find cd.
Found 2 solutions by MathLover1, ikleyn:
Answer by MathLover1(20850)   (Show Source): You can put this solution on YOUR website!
circle :

circle :

distance between centers=
if we place the center of the circle a on -ais, at point (,), than the center of the circle b will be at (,) and the distance between them is

now we have equations:
circle



circle



find intersection points by solving this system:


-----------------------------subtract










plug it in





or

solutions:
,
or
,

intersection points are
(, ) and (, )

distance between is: two times value


The length of is .



Answer by ikleyn(52932)   (Show Source): You can put this solution on YOUR website!
.
two circles intersect (A nd B) and share a common chord (CD).
the radius of circle a is 10in, radius circle b = 16 in. distance between centers, 22 in.
find cd.
~~~~~~~~~~~~~~~~~~~ 

Draw the radii of the circles from their centers to the intersection point C.
Connect the centers by the straight line AB.


You will get a triangle ABC with the sides 10 in, 16 in and 22 in.


Find its area using the Heron's formula

    area = ,

where s =  = 24 is the semi-perimeter.


Substituting this number into the formula, you will get

    area =  =  =  = 

         =  =  = 73.32121112... square inches.


Write the formula for the area of triangle ABC in other way, using the base AB = 22 in and the height h
from point C to the base

    area =  = 11*h  square inches.


You will get an equation

    11*h = .

Hence,  

    h = .


This value of h is half of the length CD;  so

    CD =  = 13.331 inches  (rounded).


At this point, the problem is just solved.


ANSWER.  The length CD is   = 13.331 inches  (rounded).

Solved.



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