TITLE: Find the area of the trapezoid where diagonals are 8 and 6.
QUESTION [1 upvotes]: 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.

REPLY [0 votes]: Take $M$ to be the midpoint of $AB$. Then $MP$ is a midsegment in triangle $ABD$ and thus $MP$ is parallel to $BD$ and $MP = \frac{1}{2} \, BD = 3$. Analogously, $MS$ is a midsegment in triangle $ABC$ and thus $MS$ is parallel to $AC$ and $MS = \frac{1}{2} \, AC = 4$. Therefore, the edge-lengths of triangle $MPS$ are $$MP = 3,\,\, MS = 4,\,\, PS = 5$$ Since $$MP^2 + MS^2 = 3^2 + 4^2 = 5^2 = PS^2$$ by the inverse direction of Pythagoras' theorem the triangle $MPS$ is right and $\measuredangle \, PMS = 90^{\circ}\,$ :D Therefore the diagonals $AC$ and $BD$ are orthogonal. Thus the area of the trapezoid is
$$\text{Area}({ABCD}) = \frac{1}{2} \, AC \cdot BD = \frac{1}{2} \, \cdot 8 \cdot 6 = 24$$
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