Question

bcdef ~ tsrqp. find qr and pt.
f
e
3
s
1
r
2
b
6
t
6
d
r
2
c
3
p
q
2
qr =
pt =

Explanation:

Step1: Determine the scale factor

Since \( BCDEF\sim TSRQP\), the scale factor \(k\) is \(\frac{SR}{FE}=\frac{1}{3}\)

Step2: Find \(QR\)

Corresponding sides of similar polygons are proportional. \(CD\) corresponds to \(QR\). Given \(CD = 3\), using the proportion \(\frac{QR}{CD}=k\). Substitute \(k=\frac{1}{3}\) and \(CD = 3\), we get \(QR=\frac{1}{3}\times3 = 1\)

Step3: Find \(PT\)

\(DE\) corresponds to \(PT\). Given \(DE = 6\), using the proportion \(\frac{PT}{DE}=k\). Substitute \(k = \frac{1}{3}\) and \(DE=6\), we get \(PT=\frac{1}{3}\times6=2\)

Answer:

\(QR = 1\), \(PT=2\)