+0

0
81
6

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]

Jan 10, 2021

#1
0

By similar triangles, PQ = 5*sqrt(3).

Jan 10, 2021
#2
0

That is still wrong can someonelse please try thank You. But still thanks for the effort!

Guest Jan 10, 2021
#3
0

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

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

Jan 10, 2021
#4
0

Can someone please do a step by step explanation on how to find the answer.

Jan 11, 2021
#5
+1164
+1

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

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

QR = QT = 8          XT = QS = 2           QL = 5            QZ = 4

XY = TL = sqrt( QT² - QL² ) = √39

∠PQR = tan-1(√39 / 3)

PZ = tan∠PQR * QZ = 8.326663998

PQ = sqrt(QZ2 + PZ2) = 9.237604326

jugoslav  Jan 11, 2021
#6
0

Can you please tell me that in square root form. THX

Jan 12, 2021