In the adjoining figure (not drawn to scale), PS = 4 cm, SR = 2 cm, PT = 3 cm and QT = 5 cm.

(i) Show that ΔPQR ∼ ΔPST.

(ii) Calculate ST, if QR = 5.8 cm.

(i) Given,

PS = 4 cm, SR = 2 cm, PT = 3 cm and QT = 5 cm.

PR = PS + SR = 4 + 2 = 6 cm

PQ = PT + TQ = 3 + 5 = 8 cm

In ΔPQR and ΔPST,

$\Rightarrow \dfrac{PQ}{PS} = \dfrac{8}{4} = 2 \\[1em] \Rightarrow \dfrac{PR}{PT} = \dfrac{6}{3} = 2 \\[1em] \therefore \dfrac{PQ}{PS} = \dfrac{PR}{PT}$

∠QPR = ∠SPT [Common angle]

∴ ΔPQR ∼ ΔPST [By S.A.S. axiom]

Hence, ΔPQR ∼ ΔPST.

(ii) We know that,

Corresponding sides of similar triangles are proportional.

$\Rightarrow \dfrac{QR}{ST} = \dfrac{PQ}{PS} \\[1em] \Rightarrow \dfrac{5.8}{ST} = \dfrac{8}{4} \\[1em] \Rightarrow ST = \dfrac{5.8}{2} \\[1em] \Rightarrow ST = 2.9 \text{ cm.}$

Hence, ST = 2.9 cm.