Question 1099501
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<pre>
1.  From the right-angled triangle AFE  sin( < FAE) = {{{12/13}}}.

    Hence, sin( < BAD) = sin( < FAE) = {{{12/13}}}.


2.  From the right-angled triangle BHG  sin( < GBH) = {{{8/17}}}.

    Hence,  sin( < CBD) = sin( < GBH) = {{{8/17}}}.


3.  Angles CBD and ADB are, obviously, congruent.

    Therefore, sin( < ADB) = sin( < CBD) = {{{8/17}}}.


4.  Thus in triangle ADB we know

    AD = 22,  sin( < BAD) = {{{12/13}}}  and  sin( < ADB) = {{{8/17}}}.


    Having sines of two angles at the base of the triangle ADB, we can find sin( < ABD):

    sin( < ABD) = sin(180 - < BAD - < ADB) = sin( < BAD + < ADB) = sin*cos + cos*sin = {{{(12/13)*sqrt(1-(8/17)^2) + sqrt(1-(12/13)^2)*(8/17)}}} = 

    = {{{(12/13)*(15/17) + (5/13)*(8/17)}}} = {{{220/221}}}.


5.  Now apply the sine law theorem to find the side AB:

    {{{abs(AB)/sin(ADB)}}} = {{{abs(AD)/sin(ABD)}}}  ====>  |AB| = 22*{{{((8/17))/((220/221))}}} = {{{(8*221)/(10*17)}}} = {{{(8*13)/10}}} = {{{104/10}}} = 10.4.



6.  Now calculate the area of the parallelogram ABCD by applying the formula

    {{{S[ABCD]}}} = |AD|*|AB|*sin( < A) = 22*10.4*{{{(12/13)}}} = 22*0.8*12 = 211.2.
</pre>

<U>Answer</U>.  The area of the parallelogram ABCD is 211.2 square units.