Is there any better way to solve this geometric area problem?

Question

Problem:

I took quite a long time to solve (8 mins?). My work is in the photo.

Short explanation: basically I set point C as (0,0) in a Cartesian plane, and progressively calculated the slopes of line $\overline{BP}$(denoted as $\alpha$ in my diagram), $\overline{QA}$($b$), $\overline{RD}$($c$), $\overline{CS}$($d$), and then the coordinate points of Q, R, S, T, which are denoted as φ, $\pi$, $\sigma$, $\phi$ respectively. Afterwards I use the shoelace formula and calculated the area of the central quadrilateral.

Anyone have a faster way?

Answer 1

Although using coordinates is a sure (yet brute-force) way to get the result, it may be more natural, or quicker, to solve it geometrically. First, conclude QP || RD and T is the midpoint of CS as well, from the trapezoid AQPX and the midpoints R, D and P.

Denote areas as [.] and I = [ABCD]. Then, the midpoint P implies [BCP] = $\frac14$I and the midpoint Q implies [ABQ] = $\frac14$I. Moreover, the distance ratio of Q to BC and AD is 1:3, implying [AQD] = $\frac38$I; and the distance ratio of R to BA and CD is 1:3, implying [RCD] = $\frac38$I. To summarize,

$$[BCP] =[ABQ]= \frac14I,\>\>\>\>\>\>\>[AQD] =[RCD] =\frac38I\tag 1$$

Then, the midpoints R, S and T lead to, respectively,

$$[ARD] =\frac12[AQD] =\frac3{16}I,\>\>\>\>\> [CSD] =\frac12[RCD] =\frac3{16}I,\>\>\>\>\> [TCP] =\frac14[CSD] =\frac3{64}I\tag 2$$

Recognize that the shaded area can be constructed with

$$[RQTS] = I - [BCP] - [ABQ] - [ARD] - [CSD] + [TCP]$$

Plug in the results (1) and (2), as well as the given I=1024, to arrive at,

$$[RQTS] = \left(1- \frac14-\frac14 -\frac3{16}-\frac3{16} + \frac3{64}\right)I=\frac{11}{64}\cdot 1024 = 176$$

Answer 2

Note that $\text{slope} \ \overleftrightarrow{QT} =\text{slope} \ \overleftrightarrow{RS} = -2$

Hence $\square RSTQ$ is a trapezoid.

$\left(\text{Note: The slope of the line $Ax + By = C$ is $-\dfrac AB$} \right)$.

Since the line $\overleftrightarrow{RS}$ passes through the point $R=(20,24)$ its equation is

$$2x+y = 64 \quad (= 2(20) + 1(24))$$

The altitude of the trapezoid $\square RSTQ$ is the distance from the line $\overleftrightarrow{RS}$ to the point $T=(13,6)$ is

$$h = \dfrac{|2(13)+1(6)-64|}{\sqrt{2^2+1^2}} = \dfrac{32}{\sqrt 5}$$

Since $QT = 5\sqrt 5$ and $RS = 6\sqrt 5$ Then

$$\text{area} \square QRST = h \cdot \dfrac{QT + RS}{2} = 176$$

Answer 3

Hint: Let $RD$ intersect $AB$ at $U$. Since $PT\parallel DS$, $BPDU$ is a parallelogram that has the same area as the rectangle $BCPU$. It's now easy to figure out $\frac{QT}{BP}$ and $\frac{RS}{UD}$ and finish up.

Answer 4

Let $[\cdot]$ denote the area. Starting from the square $[ABCD]=a^2$, we can cut $[QTSR]$ from it by removing triangular pieces.

\begin{align} [BCP]&=\tfrac14\,[ABCD]=\tfrac14\,a^2 ,\\ [ABQ]&=\tfrac12\,[ABP]=\tfrac14\,a^2 ,\\ [AQD]&=[ABCD]-2\,[ABQ]-\tfrac12\,[BCP] =a^2-\tfrac12\,a^2-\tfrac18\,a^2 =\tfrac38\,a^2 ,\\ [ARD]&=\tfrac12\,[AQD] =\tfrac 3{16}\,a^2 ,\\ [DRC]&= \tfrac 34\cdot \tfrac12\,a^2=\tfrac 38\,a^2 ,\\ [DSC]&=\tfrac12\,[DRC]=\tfrac 3{16}\,a^2 ,\\ [DSP]&=\tfrac12\,[DSC]=\tfrac 3{32}\,a^2 ,\\ [PST]&=\tfrac12\,[DSP]=tfrac 3{64}\,a^2 ,\\ [QTSR]&= [ABCD]-[BCP]-[ABQ]-[ARD]-[DSP]-[PST] \\ &=a^2\cdot(1-\tfrac14-\tfrac14-\tfrac 3{16} -\tfrac 3{32}-\tfrac 3{64}) \\ &=\tfrac{11}{64}\,a^2=176 . \end{align}