Let $c$ be the length of the hypotenuse of a right triangle. Show that the area of the triangle is at most $1/4c^2$. I approached it by drawing the right triangle, and then drawing $c^2$ as a square. I then split the square into fourths, and showed that the right triangle was a fourth of $c^2$. Is this how I approach the question?  
 A: A pure Euclidean Geometry approach:

Name the triangle $ABC$ with $AB$ the hypotenuse. Let $D$ be the midpoint of $AB$ and $E$ the foot of altitude $CE$.  
Now it's well known that $CD=\frac{1}{2}AB.$ Also $CE\le CD$ since $CE$ is the shortest distance from $C$ to $AB.$  So $$\text{Area of }\triangle ABC=\frac12\cdot AB\cdot CE\le \frac12\cdot AB\cdot CD=\frac12\cdot AB\frac12 AB=\frac14\cdot AB^2=\frac14c^2.$$
That's what we had to prove. $\blacksquare$
A: 
Geometrically, four replica of the triangle are less than the square. This suggest the following proof:
Denoting $a$ and $b$ (with $a\le b$) the sides of the triangle. The side of the inner square is $b-a$, then the total area of the large square is
$$(b-a)^2+4\times\frac{ab}{2}=a^2+b^2$$
which is an obvious identity. Then, with $a^2+b^2=c^2$,
$$\frac{ab}{2}=\frac{1}{4}\left[c^2-(b-a)^2\right]\le \frac{c^2}{4}$$
A: Using Sine Law, $$\dfrac a{\sin A}=\dfrac b{\sin B}=\dfrac c{\sin90^\circ}$$
$\implies a=c\sin A$ and $b=c\sin B=c\sin(90^\circ-A)=c\cos A$
So, the area $$\dfrac12(c\sin A)(c\cos A)=\dfrac{c^2\sin2A}4$$
A: $$c^2=a^2+b^2$$
Now area $=\dfrac12ab\sin90^\circ$
Again $(a-b)^2\ge0\iff2ab\le a^2+b^2=c^2$
