Check the picture below.
now, we have three triangles, a purple one, a red one and a green one.
we know that TU || PR, we also know that PQR is a triangle, meaning that if P, S, R are collinear and Q, U, R are also collinear, then P, T, Q must also be collinear in order to have a triangle, if that's so, the angle made at point P, is the same angle QTU, and the angle at TQU is shared by both the red triangle and the purple triangle, that means those two are similar by AA.
the angle made at point P is also shared by the green triangle, and since TS || QR, then the angle PST is equal to the angle PRQ, thus we can say that the green triangle is simiilar with the purple triangle by AA, and likewise with the red triangle, since the red triangle by association, since we know that the red one is similar to the purple one.
let's find "x", by using proportions on the green and purple triangles.
[tex]\cfrac{x}{30}=\cfrac{x+10}{45}\implies 45x=30x+300\implies 15x=300\implies x=\cfrac{300}{15}\implies x=20[/tex]
now let's use the red and purple triangles to get "y".
[tex]\cfrac{10}{y}=\cfrac{x+10}{y+20}\implies \cfrac{10}{y}=\cfrac{\stackrel{x}{20}+10}{y+20}\implies \cfrac{10}{y}=\cfrac{30}{y+20} \\\\\\ 10y+200=30y\implies 200=20y\implies \cfrac{200}{20}=y\implies 10=y[/tex]
now, for the areas hmmm we know that triangles TQU and PQR are similar, so let's pick a side of each and their corresponding areas using proportions, hmmm say let's use the side TQ which we know is 10 and PQ which we know is 30.
[tex]~\hspace{5em} \textit{ratio relations of two similar shapes} \\\\ \begin{array}{ccccllll} &\stackrel{\stackrel{ratio}{of~the}}{Sides}&\stackrel{\stackrel{ratio}{of~the}}{Areas}&\stackrel{\stackrel{ratio}{of~the}}{Volumes}\\ \cline{2-4}&\\ \cfrac{\stackrel{similar}{shape}}{\stackrel{similar}{shape}}&\cfrac{s}{s}&\cfrac{s^2}{s^2}&\cfrac{s^3}{s^3} \end{array}~\hspace{6em} \cfrac{s}{s}=\cfrac{\sqrt{Area}}{\sqrt{Area}}=\cfrac{\sqrt[3]{Volume}}{\sqrt[3]{Volume}} \\\\[-0.35em] \rule{34em}{0.25pt}[/tex]
[tex]\cfrac{TQ}{PQ}=\cfrac{\sqrt{\stackrel{tqu}{Area}}}{\sqrt{\stackrel{pqr}{Area}}}\implies \cfrac{10}{30}=\cfrac{\sqrt{15}}{\sqrt{A}}\implies \cfrac{1}{3}=\sqrt{\cfrac{15}{A}}\implies \left( \cfrac{1}{3} \right)^2=\cfrac{15}{A} \\\\\\ \cfrac{1}{9}=\cfrac{15}{A}\implies A=135[/tex]
