Perpendiculars are drawn from the vertex of the obtuse angles of a rhombus to its sides. Perpendiculars are drawn from the vertex of the obtuse angles of a rhombus to its sides.the length of each perpendicular is equal to a units.The distance between their feet being equal to b units.Find the area of the rhombus
 A: Set the $A,B,C,D \,\,$ vertices of the rhombus in $(m,0) $, $(0,n)\,\,   $,    $(-m,0)\,\,   $, and $(0, -n) \,\,  $, respectively. The slope of the side $AB\,\,  $ is $-n/m \,\,  $. So the slope of the perpendicular to $AB\,\,   $, drawn from its opposite vertex $C $, is $m/n\,\,   $. Since this perpendicular line passes through $(-n,0)  \,\,  $, its equation is $y=m/n x +m^2n    \,\, \,\,   $. The $(x_0,y_0 ) \,\,  $ coordinates of the crossing point between this perpendicular line and $AB $ (let us call this foot point $E $) can be found by solving 
$$ m/n x +m^2/n=-n/m x +n$$
which gives 
$$x_0=\frac {m (n^2-m^2)}{n^2+m^2}$$
and then 
$$y_0=\frac {2n m^2}{n^2+m^2}$$
Since the distance between $E $ and $C $ is $a $, we have
$$\left ( \frac {m (n^2-m^2)}{(n^2+m^2)}+m \right)^2+\frac {4n^2 m^4}{(n^2+m^2)^2}=a^2$$
which reduces to
$$\frac {4n^2 m^2}{n^2+m^2}=a^2$$
Also note that, as a result of the symmetry of the problem, the distance between $E $ and its symmetric point on side $CD\,\,   $ (i.e. the foot of the perpendicular drawn from vertex $A $ to  $CD\,\,    $) is  twice  the distance between $E $ and the origin. Since this distance is given as equal to $b $, we have
$$\frac {m^2 (n^2-m^2)^2}{(n^2+m^2)^2}+\frac {4n^2 m^4}{(n^2+m^2)^2}=\frac {b^2}{4}$$
which reduces to
$$m^2=\frac {b^2}{4}$$
and then $m=b/2  \,\,  $. Substituting in the equation above, this also leads to 
$$\frac{4n^2 b^2/4}{n^2+b^2/4}=a^2$$
$$n=\frac {ab}{2\sqrt {b^2-a^2}} $$
Since the area $A $  of the rhombus is $2mn \,\,  $, it follows
$$A=  2 \frac {b}{2}  \cdot \frac{ab}{2\sqrt {b^2-a^2}}$$
$$=\frac {ab^2}{2 \sqrt {b^2-a^2}} $$
Note that, when $n=m  \,\,  $, that is to say the rhombus is a square, $a $ corresponds to the side of the square and  $b $ to its diagonal, so in this case $a=\sqrt {2}n  \,\,  $ and $b=2n   \,\, $. Accordingly, the formula reduces to $A=2n^2\,\,    $, which is the area of a square with vertices in $(n,0)\,\,    $, $(0,n) \,\,   $,    $(-n,0) \,\,  $, $(0, -n)  \,\, $.
