$AB,CD$ and $EF$ are three parallel lines, in that order. Let $d_{1}$ and $d_{2}$ be the distance from $CD$ to $AB$ and $EF$ respectively. $d_{1}$ and $d_{2}$ are integers where $d_{2}:d_{2} = 2:1$, $P$ is a point on $AB$, $Q$ and $S$ are points on $CD$ and $R$ is a point on $EF$. If the area of the quadrilateral $PQRS$ is $30$ square units, what is the value of $QR$ when value of $SR$ is the least?
In my solution approach I had originally taken $P$ and $R$ in between of $Q$ and $S$ and assumed $d_{2} = x$ and $d_{2} =2x$. Now summing up the area of the two triangles formed i.e. $RQS$ and $PQS$ :-
$area(RQS) + area(PQS) = 30$
We get the relation that $QS.x = 20$ and $area(RQS) = 10$ units and $area(PQS)=20$ units.
Now when $RS$ will be minimum i.e. when $RS$ will be perpendicular to the line $CD$ then at that time I fixed all the points i.e. $R,S$ and $P$ and moved $Q$ to $Q'$ in order to keep the area of the quadrilateral same and $Q'S = QS + y$ and again using the sum of area equation I came to conclusion that $y = 0$ and $Q=Q'$. Then this means that $P$ will have to be moved on the line $AB$ somewhere in order to keep the total area of the quadrilateral the same but at this step only I am getting stuck as to how to determine the distance by which $P$ would have moved and also how to arrive at the length of $QR$.
Please help !
Thanks in advance !
Triangles $PQS$ and $RQS$ stand on same base $QS$ and their heights are in ratio $2:1$. Clearly $[RQS]=30/(2+1)=10$.
Let $QS=z$. Since $SR \perp QS$,
$$[RQS]=\frac{1}{2}xz=10\Rightarrow xz=20$$
Again since $SR=d_1$ is least integer, $SR=1$. Can you finish?