Geometry Problem:
$PQRS$ is a rectangle. A point $X$ is halfway along $PQ$, a point $Y$ is a third of the way along $QR$ ($Y$ is closer to $Q$) and a point $Z$ is a quarter of the way along $RS$ ($Z$ is closer to $R$). What fraction of the area $PQRS$ is represented by the quadrilateral $XYZS$?
I am struggling with this problem and do not know how to solve it. To make this better for me to explain and for you to understand, please let me know how to construct a picture/diagram of what I am referring to. Since as of this moment, I do not have a picture attached to this post, I advise that you grab a pen and paper and draw what has been mentioned in the yellow sandbox above. Then, continue drawing to make sure my working out is correct.
My Attempt:
Draw a line from $X$ to the halfway point of $RS$. Call that halfway point $W$, then we have formed a triangle $XWS$ with a right angle at point $W$. Since our line $XW$ divides $PQRS$ into halves then the area of $\Delta XWS$ is exactly $1/4$ of the area of $PQRS$.
So the entire area of $XYZS$ is greater than $1/4$ of the area of $PQRS$.
Now, draw a line from $Y$ to a point we can call $V$ a third of the way along $XW$ such that $YV$ is perpendicular to $XW$. We have now formed another triangle, namely, $XVY$ with a right angle at point $V$. Since the point $Y$ is located a third of the way along $QR$ then the newly rectangle constructed, namely, $XQYV$ is $1/3$ the area of $1/2$ the area of $PQRS$ (since as mentioned before, $XW$ splits $PQRS$ into two sections). Therefore, the area of $XQYV$ is $1/6$ the area of $PQRS$. And, since the area of $\Delta XVY$ is half the area of $XQYV$, then the area of $\Delta XVY$ is $1/12$ the area of $PQRS$.
So the entire area of $XYZS$ is greater than $\frac 14 +\frac {1}{12} = \frac 13$ of the area of $PQRS$.
Lastly, draw a line from $Z$ to a point $U$ halfway along $YV$ such that $ZU$ is perpendicular to $YV$. We have now formed one more triangle, namely, $ZUY$ with a right angle at $U$. We have also formed a rectangle $ZUYR$. Now, since $Z$ is located $1/4$ along $RS$ then $ZUYR$ is $1/4$ the area of $2/3$ the area of $PQRS$. Therefore, the area of $\Delta ZUY$ is $1/2$ the area of $1/4$ the area of $2/3$ the area of $PQRS$, which is also $1/12$ the area of $PQRS$.
So the entire area of $XYZS$ is greater than $\frac 13 + \frac{1}{12} = \frac{5}{12}$ of the area of $PQRS$.
Now, if you were drawing, there should be another area within rectangle $XYZS$ that is the last remainding area... but that, I do not know how to calculate. Please help me.
Apologies for the long question.
Thank you in advance.
Let |ABC| denote the area of triangle ABC. $$|XYZS|=|PQRS|-|XQY|-|YRZ|-|PXS|$$ $$=1-\frac{1}{12}-\frac{1}{12}-\frac{1}{4}$$ $$=\frac{7}{12}$$