Question 1149735
<pre>
{{{drawing(400,700/3,-3,33,-3,18,

locate(2.4,8.7,17),locate(11,8.7,17), locate(19,8.7,27),
locate(7,0,x), locate(21,0,x),locate(7.4,16.6,A),
locate(0,0,B), locate(14.7,0,C), locate(29.7,0,D), 
triangle(0,0,7.416198487,15.29705854,2sqrt(55),0),
triangle(0,0,7.416198487,15.29705854,4sqrt(55),0) )}}}

Use the law of cosines on triangles ABC and ABD

{{{cos(B)=(a^2+c^2-b^2)/(2ac)}}}

Using it on ABC

{{{cos(B)=(17^2+x^2-17^2)/(2*17*x)}}}

which simplifies to

{{{cos(B)=x/34}}}

Using it on ABD

{{{cos(B)=(17^2+(2x)^2-27^2)/(2*17*2x)}}}

which simplifies to

{{{(x^2-110)/17x}}}

Equating the two expressions for cos(B),


{{{x/34=(x^2-110)/17x}}}

Solve that and get

{{{x=2sqrt(55)}}}

The height of both triangles is the green line h = AE:

{{{drawing(400,700/3,-3,33,-3,18,
green(line(7.416198487,0,7.416198487,15.29705854),locate(7.47,7.6,h),

locate(7.7,1.3,E)), 
locate(2.4,8.7,17),locate(11,8.7,17), locate(19,8.7,27),
locate(7,0,x), locate(21,0,x),locate(7.4,16.6,A),
locate(0,0,B), locate(14.7,0,C), locate(29.7,0,D), 
triangle(0,0,7.416198487,15.29705854,2sqrt(55),0),
triangle(0,0,7.416198487,15.29705854,4sqrt(55),0) )}}}

{{{AE^2=AB^2-BE^2}}}
{{{AE^2=17^2-(sqrt(55)/4)^2}}}
{{{AE^2=234}}}
{{{AE=3sqrt(26)}}}

{{{Area=expr(1/2)base*height}}}

{{{Area=expr(1/2)4sqrt(55)*3sqrt(26)}}}

{{{Area=6sqrt(1430)}}}

approx. 226.8920448

Edwin</pre>