euclidean-geometrygeometry
Let $ABCD$ be a trapezoid such that $\left|AC\right|$=$8$,$\left|BD\right|=6$ and $AD ||BC $. Let $P$ and $S$ be the midpoints of $AD$ and $BC$ , respectively . If $|PS|$=$5$, find the area of the trapezoid $ABCD$.
I don't chase the angle. At first I drop perpendiculars from $A,P,D$ to $BC$ and named then $X,Y,Z$ respectively and using Pythagoras theorem repeatedly but I can't write after that . Somebody please help me.
The problem is ill-posed; the perimeter isn't determined by the given information. If we move $D$ and $C$ outward, as they come to lie vertically above $A$ and $B$, respectively, $P$ and $Q$ converge at the centre. Thus the trapezoid has to be scaled up without bounds for $PQ$ to retain the value $4$, and thus it can have arbitrarily large perimeters.
Option 1. Consider the isosceles trapezoid with $BE=BD$ as diagonal:
$\hspace{3cm}$
We find: $$BF=\frac{2S_{\Delta BCD}}{CD}=\frac{240}{26}=\frac{120}{13};\\ CF=\sqrt{BC^2-BF^2}=\sqrt{100-\frac{120^2}{13^2}}=\frac{50}{13};\\ S_{ABCD}=\frac{AB+CD}{2}\cdot BF=\frac{(26-2\cdot CF)+26}{2}\cdot \frac{120}{13}=\\ =\frac{34560}{169}\approx \color{red}{204.5}.\\ $$
Option 2. Consider the point $E$ as a midpoint:
$\hspace{3cm}$
We find: $$EH=\sqrt{BE^2+BH^2}=\sqrt{24^2+5^2}=\sqrt{601};\\ BI=\frac{2S_{\Delta BEH}}{EH}=\frac{2\cdot 60}{\sqrt{601}};\\ S_{ABCD}=EH\cdot BF=\sqrt{601}\cdot \frac{4\cdot 60}{\sqrt{601}}=\color{red}{240}.$$
Conclusion: The trapezoid is not unique.
Best Answer
The required area is equal to that of the triangle whose sides are $(2 \times 5)$, 6 and 8. Just apply the heron formula to it.
Since [⊿blue] = [⊿green], it is clear that:-
[trap ABCD] = 0.5[//gm WXYZ] = [⊿XYZ] = as mentioned.