In the given figure, PT = 4\(\sqrt {2}\) cm, TR = 6\(\sqrt {2}\) cm, and PQ = 20 cm. If ST is parallel to QR, then what is the value of SQ (in cm)?

12

ΔPST and ΔPQR are similar

\(\frac{PS}{PQ}\) = \(\frac{PT}{PR}\)

\(\frac{PS}{20}\) = \(\frac{4\sqrt {2}}{10\sqrt {2}}\)

PS = 20 x \(\frac{2}{5}\) = 8 cm

PQ = PS + SQ

SQ = PQ - PS = 20 - 8 = 12 cm