Question 1207163
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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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<pre>
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 = {{{sqrt(s*(s-10)*(s-16)*(s-22))}}},

where s = {{{(10+16+22)/2)}}} = 24 is the semi-perimeter.


Substituting this number into the formula, you will get

    area = {{{sqrt(24*(24-10)*(24-16)*(24-22))}}} = {{{sqrt(24*14*8*2)}}} = {{{sqrt(3*8^2*7*4)}}} = 

         = {{{8*2*sqrt(21)}}} = {{{16*sqrt(21)}}} = 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 = {{{(1/2)*22*h}}} = 11*h  square inches.


You will get an equation

    11*h = {{{16*sqrt(21)}}}.

Hence,  

    h = {{{(16/11)*sqrt(21))}}}.


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

    CD = {{{(32/11)*sqrt(21)}}} = 13.331 inches  (rounded).


At this point, the problem is just solved.


<U>ANSWER</U>.  The length CD is  {{{(32/11)*sqrt(21)}}} = 13.331 inches  (rounded).
</pre>

Solved.