geometry
Given a rectangular sheet of paper ABCD such that the lengths of AB and AD are respectively 7 and 3 cms.Suppose B' and D' are two points on AB and AD respectively such that if the paper is folded along B'D' then A falls on A' on the side DC. Determine the maximum possible area of the triangle AB'D'.
I discovered quite a few basic facts that everybody can, but cannot actually make any progress. Please help.
Let us look at $\triangle AED$ and $\triangle CED$:
So, by $SAS$ congruence, those two triangles are congruent, and hence $EC = AE$. Now let $BE$ be $x$. So, by Pythagoras Theorem, $(EC)^2=x^2+100$. But, $(EC)^2= (AE)^2 = (24-x)^2$. Therefore, we get an equation, $(24-x)^2= 576 + x^2 - 48x = x^2+100$. Which upon solving, we get, $x = \frac{119}{12}$ .
Back to $\triangle AED$, we see that $(ED)^2 = (24-x)^2 - 169$. So, we know $x$ and we can get $ED$ as well. The final touch, $$Area(BEDC) = Area(ABC) - Area(EDA) = \frac{24 \cdot 10}{2} - \frac{ED \cdot 13}{2}$$
Edit: Details of calculation: $(ED)^2 = (24 - x)^2 - 169 = (24 - \frac{119}{12})^2 - 169 = (\frac{169}{12})^2 - 169 = 169(\frac{169}{144}- 1)= \frac{169 \cdot 25 }{144}$ Therefore, $ED = \sqrt{\frac{169 \cdot 25 }{144}} = \frac{13 \cdot 5 }{12}$.
So, $$Area(BEDC) = \frac{24 \cdot 10}{2} - \frac{ED \cdot 13}{2} = 120 - \frac{13 \cdot 13 \cdot 5}{24} = 120- \frac{845}{24} = \frac{2035}{24} = 84.791666...$$
I decided to upgrade my comment into an answer.
Angle $C$ is free to vary. That would change the length of $BD$ while preserving the given lengths, which is were you went astray (the length of $BD$ is not necessarily $4$).
Now, in general, if we know two side lengths of a triangle (call them $x$ and $y$) and the angle measure between them (call it $\theta$), then the area of the triangle is given by $A = \frac{1}{2}xy\sin(\theta)$.
Applying this to our scenario, since the angle between the two given sides can vary, then so too can the area of the triangle. This lack of uniqueness is evident in the figure shown below: Every triangle shown has the prescribed side lengths, but all have different areas. In short, we cannot determine the area with the given information.
Best Answer
My suggestion is to compute AD' and AB' as a function of AD and $\alpha$. $\Delta AOD' \sim \Delta ADA' \to AD' \times AD = AO\times AA'$
and note that $AA' = \frac{AD}{\cos \alpha}$ and $AO = \frac{1}{2}AA'$
$\therefore AD' = \frac{AA'^2}{2AD}=\frac{AD}{2\cos^2 \alpha}$
$AB' = \frac{AD'}{\tan \alpha} = \frac{AD}{2\cos^2 \alpha} \times \frac{\cos \alpha}{\sin \alpha}=\frac{AD}{\sin 2\alpha}$
$S_{AB'D'} = \frac{1}{2}AB' \times AD' = \frac{AD^2}{2\cos \alpha \sin 2\alpha}$
Since AD is constant, maximizing $S_{AB'D'}$ is to maximize the expression of $\alpha$. You can find the bound of $\alpha$ by moving (1) D' to D and (2) B' to B combining with the assumption 3:7 given. Then, the min or max could be easily obtained.