Question

In the diagram below, $\angle PQR = \angle PRQ = \angle STR = \angle TSR$, $RQ = 8$, and $SQ = 2$. Find $PQ$.

Answer 1

Answer:

$\boxed{\frac{16\sqrt{3}}{3}}.\]$

Step-by-step explanation:

From the given information, we know that $\triangle PQR \sim\triangle QTR \sim \triangle RST$. That also means that all these triangles are isosceles.

[asy]

import markers;

pair A,B,C,D,E;

A = (0, 0.9);

B = (-0.4, 0);

C = (0.4, 0);

D = (-0.275, 0.16);

E = (0.11, 0.65);

markangle(1, C, B, A, radius = 5mm, red);

markangle(1, A, C, B, radius = 5mm, red);

markangle(1, D, E, C, radius = 5mm, red);

markangle(1, C, D, E, radius = 5mm, red);

draw(A--B);

draw(A--C);

draw(B--C);

draw(B--E);

draw(C--D);

label("$P$",A,N);

label("$Q$", B, S);

label("$R$", C, S);

label("$S$", D, S);

label("$T$", E, W);

[/asy]

Since triangle $RQT$ is isosceles, we have

\[ST = QT - QS = QR - QS = 8 - 2 = 6.\]

From $\triangle QTR \sim\triangle RST$, we know that $\frac{RT}{8}=\frac{6}{RT}$, and we can prove that $RT=4\sqrt{3}$.

From $\triangle PQR \sim \triangle QRT$, we have $\frac{PQ}{QR}=\frac{QR}{RT}$, so that would mean that

\[PQ=\frac{QR^2}{RT}=\frac{8^2}{4\sqrt{3}}=\boxed{\frac{16\sqrt{3}}{3}}.\]

You can use this LaTex converter to read the answer:

https://w2.syronex.com/jmr/latex-symbols-converter

Sorry, this is my first time using this website, so I might mess up a lot :D