# What is the area of $PXQY$ in the rectangle $ABCD$ in the following diagram?

In $$ABCD$$ rectangle $$AB=6$$, $$AD=8$$, $$AE=ED$$, $$BF=FC$$, $$EP=PQ=QF$$. Find the area of $$PXQY$$.

I cant prove that which type of quadrilateral it is. How can I get the diagonal or the sides of $$PQXY$$? Please help me with a hint.

$$PF=FC$$ and $$PF\perp FC,$$ which says $$\measuredangle QPY=\measuredangle FPC=45^{\circ},$$ which gives that $$PXQY$$ is a square and $$S_{PXQY}=\frac{XY\cdot PQ}{2}=\frac{2^2}{2}=2.$$

with Q being in the middle of PF, and the symmetry of the graph making XY cut the middle of PQ, you can actually visualise triangle PBC as the following diagram:

Can you continue?

• how can you prove that $PY$ and $XQ$ have are same? – Shromi Feb 19 at 6:41
• do you know how to prove PYQX is parallelogram? – qsmy Feb 19 at 7:00

As $$\triangle XAB\sim \triangle XQP$$ and $$AB= 3PQ$$, the distance from $$X$$ to $$AB$$ is equal to 3 times the distance from $$X$$ to $$PQ$$.

Therefore, the distance from $$X$$ to $$PQ$$ is $$\displaystyle 8\times\frac{1}{2}\times\frac{1}{4}=1$$.

$$PQ=6\div3=2$$.

The area of $$PXQY$$ is $$\displaystyle 2\times\frac{1}{2}\times2\times 1=2$$.