+0

0
178
2

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

[asy]
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);
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]

Jul 12, 2021

#1
+26222
+3

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

$$\text{Let \angle QPR = 180^\circ - 2A} \\ \text{Let \angle RQT = 180^\circ - 2A} \\ \text{Let \angle TRS = 180^\circ - 2A} \\ \text{Let \angle PTQ = 180^\circ - A}\\ \text{Let PQ=x } \\ \text{Let QR = QT = 8 } \\ \text{Let TR = SR = y }$$

$$\begin{array}{|rcll|} \hline ST&=&QT-SQ \\ ST&=&8-2\\ ST& =& 6 \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline \text{In \triangle [QPT]:} \\ \hline \dfrac{\sin(180^\circ-2A)}{8} &=& \dfrac{\sin(180^\circ-A)}{x} \\\\ \dfrac{\sin(2A)}{8} &=& \dfrac{\sin(A)}{x} \\\\ \dfrac{\sin(2A)}{\sin(A)} &=& \dfrac{8}{x} \qquad (1)\\ \hline \end{array}\\ \begin{array}{|rcll|} \hline \text{In \triangle [QTR]:} \\ \hline \dfrac{\sin(180^\circ-2A)}{y} &=& \dfrac{\sin(A)}{8} \\\\ \dfrac{\sin(2A)}{y} &=& \dfrac{\sin(A)}{8} \\\\ \dfrac{\sin(2A)}{\sin(A)} &=& \dfrac{y}{8} \qquad (2)\\ \hline \end{array}\\ \begin{array}{|rcll|} \hline \text{In \triangle [QSR]:} \\ \hline \dfrac{\sin(180^\circ-2A)}{6} &=& \dfrac{\sin(A)}{y} \\\\ \dfrac{\sin(2A)}{6} &=& \dfrac{\sin(A)}{y} \\\\ \dfrac{\sin(2A)}{\sin(A)} &=& \dfrac{6}{y} \qquad (3)\\ \hline \end{array}$$

$$\begin{array}{|lrcll|} \hline (2)=(3):& \dfrac{y}{8} &=& \dfrac{6}{y} \\\\ & y^2 &=& 48 \\ & y^2 &=& 16*3 \\ & \mathbf{y} &=& \mathbf{4\sqrt{3}} \\ \hline \end{array} \begin{array}{|lrcll|} \hline (2)=(1):& \dfrac{y}{8} &=& \dfrac{8}{x} \\\\ & xy &=& 64 \\\\ & x &=& \dfrac{64}{y} \\\\ & x &=& \dfrac{64}{4\sqrt{3}} \\\\ & x &=& \dfrac{16}{\sqrt{3}} \\\\ & x &=& \dfrac{16}{\sqrt{3}} * \dfrac{\sqrt{3}}{\sqrt{3}} \\\\ & \mathbf{x} &=& \mathbf{\dfrac{16}{3}\sqrt{3}} \\ \hline \end{array}$$

$$PQ = \mathbf{\dfrac{16}{3}\sqrt{3}}$$

Jul 12, 2021
#2
+1607
+2

Nice work, heureka, as always!!!

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

QS = TU = 2

QT = QR = 8

QW = 5

RV = 4

TW = sqrt(QT2 - QW2)

RT = sqrt(TW2 + RW2)

RW / RT = RV / PR

PQ = PR = (RT * RV) / RW

civonamzuk  Jul 14, 2021