A geometric puzzle

Since the triangle is right angled its area is:
$(ABC)=\frac{1}{2}\cdot AC\cdot BC$
Also:
$sinBAC\angle=\frac{BC}{AB}\Rightarrow BC=sin15°\cdot 100$
$cosBAC\angle=\frac{AC}{AB}\Rightarrow AC=cos15°\cdot 100$
Therefor:
$(ABC)=\frac{1}{2}100^{2}sin15°cos15°$
By using the trigonometric identity $sin2a=2sina\cdot cosa\Rightarrow sin15°\cdot cos15°=\frac{sin30°}{2}=\frac{1}{4}$
Hence:
$(ABC)=\frac{1}{8}100^{2}\Rightarrow$
$\boxed{(ABC)=1250}$

Let $F$ be the midpoint of $AB$ , and $CT$ is an altitude of the triangle. From the Thales theorem $CF=AF=BF$ . From that $BCF\angle =75°$ and $BFC\angle =180°-75°-75°=30°$ . So the $CTF$ is a half-equilateral triangle. Then $CT=\frac{CF}{2}$ . Now we can see that $CT=\frac{CF}{2}=\frac{AF}{2}\frac{\frac{AB}{2}}{2}$ , so $T_{ABC}=AB*CT/2=100*(100/4)/2=100*25/2=\boxed{1250}$ .

We are going to find the area of the triangle by finding the lengths of the triangle: Let $x$ denote the length of BC and $y$ denote the length of AC.

$Sin 15=\frac{x}{100}$ .................So.................. $100\ Sin 15 = x = 25\sqrt{6}\ -\ 25\sqrt{2}$

$Sin 75=\frac{y}{100}$ .................So.................. $100\ Sin 75 = y = 25\sqrt{6}\ +\ 25\sqrt{2}$

Now we have found the lengths as:

$x = 25\sqrt{6}\ -\ 25\sqrt{2}$ ..................and................... $y = 25\sqrt{6}\ +\ 25\sqrt{2}$

Here we can go for an interesting approach using the fact that:

$\boxed{(a + b) (a - b) = a^2 - b^2}$

$(25\sqrt{6})^2 - (25\sqrt{2})^2$ = $2500$

and then we can simply divide it by 2 using the fact that $2500$ is equal to the length of the height into the length of the base

$\large \frac{2500}{2}=1250$