Any purely geometric solution to this problem? 
What is the largest possible area of a rectangle(in square units) inscribed in the triangle shown in the picture above?
 A: Hint:
You can prove that any such rectangle has the area at most half the triangle area via a few simple cases (of course, there always exists such a rectangle):


*

*Two sides parallel. For any right triangle, and any rectangle with sides parallel to the sides other than hypotenuse: to maximize the area the rectangle has to have proportions of the triangle.


*One side parallel. Any triangle and any rectangle with at least one side parallel to some side of the triangle: put a line perpendicular to that side which goes through one of the triangle vertices (it doesn't need to be a height) and use the previous point.

*No sides parallel. Any triangle and any rectangle: take a line that goes through one of the triangle vertices and is parallel to one of the sides of the rectangle, and use previous point.
I hope this helps $\ddot\smile$
A: Let us generalize the case to obtain the largest rectangle inscribed in any triangle.

Let $AB=b$ and $CF=h$. Let $x$ and $y$ be the length and wide of rectangle, respectively. It is easy to notice that $\Delta ABC\sim \Delta CDE$. Hence
\begin{align}
\frac{x}{b}&=\frac{h-y}{h}\\
x&=\frac{b(h-y)}{h}.\tag1
\end{align}
The area of rectangle is $xy$ and by substituting equation $(1)$ to $xy$ we will obtain
\begin{align}
xy&=\frac{b(h-y)y}{h}\\
&=\frac{-b(y^2-hy)}{h}\\
&=\frac{-b\left(y-\dfrac12h\right)^2+\dfrac14bh^2}{h}.\tag2
\end{align}
It can be seen from  the equation $(2)$ that the area of rectangle will be maximum when $y=\dfrac12h$, therefore the largest area of rectangle inscribed in any triangle is $\dfrac14bh$ or half of the area of the triangle. Substituting $y=\dfrac12h$ to equation $(1)$ yields
$$
x=\frac{b\left(h-\dfrac12h\right)}{h}=\dfrac12b.
$$
Thus, the maximum area of rectangle occurs when the midpoints of two sides of the triangle were joined to make a side of the rectangle.

Now, let $AB=21$, $BC=17$, and $AC=10$. Using cosine formula, we will obtain
$$
\cos A=\frac{10^2+21^2-17^2}{2\cdot10\cdot21}=\frac35\quad\Rightarrow\quad\sin A=\frac45.
$$
The area of $\Delta ABC$ is
$$
[\Delta ABC]=\frac12\cdot AB\cdot BC\cdot\sin A=84\text{ square units.}
$$
Thus, the maximum area of rectangle is
$$
\frac12[\Delta ABC]=\large\color{blue}{42\text{ square units}}.
$$
A: Let $\triangle ABC$ be the given triangle with $AB = 10$, $AC = 17$, and $BC = 21$. Choose an arbitrary point $M$ on the side $AB$, and let $x = AM$. Let $N$ be the point on $AC$ such that $MN \parallel  BC$, and points $P$, and $Q$ on $BC$ such that $NP \perp BC$, and $MQ \perp BC$. Thus $MNPQ$ is a rectangle. Let $h_a$ be the length of the altitude from $A$. Note that $h_a$ is constant. The we have the followings proportions:
$\dfrac{BM}{BA} = \dfrac{MQ}{h_a} \to \dfrac{10-x}{10} = \dfrac{MQ}{h_a} \to MQ = \dfrac{h_a}{10}\cdot (10-x)$. Also:
$\dfrac{AM}{AB} = \dfrac{MN}{BC} \to \dfrac{x}{10} = \dfrac{MN}{21} \to MN = \dfrac{21}{10}\cdot x$. 
Let $S(x)$ be the area of the rectangle $MNPQ$, then:
$S(x)  = MN\cdot MQ = \dfrac{21h_a}{100}\cdot x(10 - x)$. Since:
$x + (10 - x) = 10$, a constant, $S$ achieves a maximum value when $x = 10 - x$, or $x = 5$.
From this we can solve for $MN =\dfrac{21}{2}$, and $MQ = \dfrac{h_a}{2}$. Thus:
$S_{max} = \dfrac{21h_a}{4} = \dfrac{S_{\triangle ABC}}{2}$
Note: $h_a$ can be calculated using Heron formula.
A: Sharing an elegant approach provided by Dr.Shailesh A Shirali.


*

*BC = a, CA = b, AB = c

*PQRS: rectangle with PS ∥ BCand PQ ⊥ BC

*Triangle APS is similar to Triangle ABC

*Let AP/AB = t; then AS/AC = t and PS/BC = t 

*Dimensions of Triangle APS: ta,tb, tc 

*AD is an altitude of Triangle ABC 

*AD = h 

*Altitude of Triangle APS: th    (by similarity)  

*PQ = h−th = (1−t)h


Following the train of reasoning as shown below figure, we see that:
Area of triangle ABC = (1/2)ah,
Area of rectangle PQRS = t(1−t)ah,
Here 0 <= t <= 1. The maximum value attained by t(1−t) for 0 <= t <= 1 is 1/4 ,
attained when t = 1/2 . 
Hence the maximum possible area of PQRS is (1/4)ah, which is half the area of the triangle. 
