Question 1181241
<pre>

By the secant-secant product theorem:

(whole_1)(outerPortion_1) = (whole_2)(outerPortion_2)

AC*YC = BC*XC

Let d = XC to make the work more clear:

Therefore, (32+3)(3) = (16+d)(d)

{{{ d^2 + 16d - 105 = 0 }}}

{{{ (d-5)(d+16) = 0}}}
d = 5  -->  BC = 21  -->  AB = {{{ sqrt(35^2 - 21^2) = sqrt(784) = 28 }}}

Area = {{{(1/2)(BC)(AB) }}} = {{{ (1/2)(21)(28) = highlight(294) }}} sq units