MHB Area of a region in Rectangle: very tough question

[Johncena1]

I have been stuck on this problem for 8 hours. Can anyone please help me? it would be great if full solution is provided, but even a general overview would help.
PQRS is a rectangle with PQ=20 and QR=5. Point X lies on the line PQ and Y on RS such that angle PYQ is 90 degrees. Angle SXR is also 90. Triangle PYQ is congruent to triangle SXR.

Let the point of intersection of PY and SX be D and intersection of QY and XR be E.

Find the area of DYEX.

I hope I am clear!

 

[Prove It]

I have been stuck on this problem for 8 hours. Can anyone please help me? it would be great if full solution is provided, but even a general overview would help.
PQRS is a rectangle with PQ=20 and QR=5. Point X lies on the line PQ and Y on RS such that angle PYQ is 90 degrees. Angle SXR is also 90. Triangle PYQ is congruent to triangle SXR.

Let the point of intersection of PY and SX be D and intersection of QY and XR be E.

Find the area of DYEX.

I hope I am clear!

This took me three double sided sheets of paper to do.

Start by writing PX = YR = x and XQ = SY = 20 - x.

Then notice that the triangle PYQ you can evaluate the area of directly, as its base is 20 and its height is 5. So its area is 50 square units. But since this is a right-angle triangle, if you knew lengths PY and YQ, you could use those lengths to evaluate the area as well.

So use Pythagoras to get expressions for PY and YQ in terms of x, then evaluate the area of triangle PYQ in terms of x. Set this area equal to 50 and solve for x.

Then once you have x, you could put your rectangle on a set of Cartesian axes. Find the equation of the lines PY and SX to find where they intersect. Once you know that, evaluate the lengths of your inner rectangle and find your area.

It will take AGES, but it is doable.

 

[Opalg]

I have been stuck on this problem for 8 hours. Can anyone please help me? it would be great if full solution is provided, but even a general overview would help.
PQRS is a rectangle with PQ=20 and QR=5. Point X lies on the line PQ and Y on RS such that angle PYQ is 90 degrees. Angle SXR is also 90. Triangle PYQ is congruent to triangle SXR.

Let the point of intersection of PY and SX be D and intersection of QY and XR be E.

Find the area of DYEX.

Hi Johncena! Welcome to MHB, and thanks for posing this interesting problem.

View attachment 4261​

Let $N$ be the midpoint of $PQ$, and $O$ the midpoint of $RS$. A circle with $RS$ as diameter (so centred at $O$, and with radius $10$) will cut $PQ$ at the point $X$ (because the right angle $SXR$ is an angle in a semicircle). As far as I can see, $X$ can be either of the points where the circle crosses $PQ$, so it could equally well be at the point vertically above $Y$. In that case, the region $DYEX$ would be kite-shaped. Maybe that is what you wanted? But I am assuming, as did Prove It, that $X$ is chosen so that $PX=YR$. In that case, $DYEX$ will be a long, very narrow, rectangle, as in the above diagram.

By Pythagoras in the triangle $ONX$, $NX^2 = OX^2 - ON^2 = 100 - 25 = 75$. Therefore $OY = NX = \sqrt{75} = 5\sqrt3.$ Hence $SY = SO - OY = 5(2-\sqrt3).$

By Pythagoras again, in the triangle $PSY$, $PY^2 = PS^2 + SY^2 = 25 + 25\bigl(2-\sqrt3\bigr)^2 = 100\bigl(2-\sqrt3\bigr) = 20\cdot SY.$ Therefore $\Bigl(\dfrac{SY}{PY}\Bigr)^2 = \dfrac{SY}{20}.$

The triangles $PSY$ and $SDY$ are similar, having equal angles. The ratio of their areas is the square of the ratio of their sides. Therefore the area of the little triangle $SDY$ is given by $$\operatorname{area}(SDY) = \Bigl(\dfrac{SY}{PY}\Bigr)^2 \!\operatorname{area}(PSY) = \dfrac{SY}{20}\cdot \frac12 PS\cdot SY = \dfrac{SY}{20}\cdot\dfrac{5SY}2 = \dfrac{SY^2}8 = \dfrac{25}8\bigl(2-\sqrt3\bigr)^2 = \dfrac{25}8\bigl(7 - 4\sqrt3\bigr).$$

Next, the area of the parallelogram $PXRY$ is (base)$\times$(vertical height) $= (RO+OY)\times5 = 5\bigl(10 + 5\sqrt3\bigr) = 25\bigl(2 + \sqrt3\bigr).$

The two triangles $PYQ$ and $SXR$ together cover the whole of the parallelogram $PXRY$, but they overlap over the rectangle $DYEX$ and they also include the two (equal) little triangles $SDY$ and $QEX.$ Therefore $$\operatorname{area}(PYQ) + \operatorname{area}(SXR) = \operatorname{area}(PXRY) + \operatorname{area}(DYEX) + 2\operatorname{area}(SDY),$$ and so $$\begin{aligned} \operatorname{area}(DYEX) &= \operatorname{area}(PYQ) + \operatorname{area}(SXR) - \operatorname{area}(PXRY) - 2\operatorname{area}(SDY) \\ &= 50 + 50 - 25\bigl(2 + \sqrt3\bigr) - \dfrac{25}4\bigl(7 - 4\sqrt3\bigr) = \dfrac{25}4. \end{aligned}$$