Answer: The correct option is
![(B)~\triangle TRS\sim \triangle TPQ,~SAS\sim.](https://tex.z-dn.net/?f=%28B%29~%5Ctriangle%20TRS%5Csim%20%5Ctriangle%20TPQ%2C~SAS%5Csim.)
Step-by-step explanation: We are given to check whether the triangles in the figure are similar or not. If so, we are to state the similarity statement.
From the figure, we note that
in the triangles TPQ and TRS, we have
![TP=42,~TQ=28,TR=42+6=48,~~TS=28+4=32.](https://tex.z-dn.net/?f=TP%3D42%2C~TQ%3D28%2CTR%3D42%2B6%3D48%2C~~TS%3D28%2B4%3D32.)
Therefore, the ratios of the corresponding sides of two triangles are
![\dfrac{TP}{TR}=\dfrac{42}{48}=\dfrac{7}{8},\\\\\\\dfrac{TQ}{TS}=\dfrac{28}{32}=\dfrac{7}{8}.](https://tex.z-dn.net/?f=%5Cdfrac%7BTP%7D%7BTR%7D%3D%5Cdfrac%7B42%7D%7B48%7D%3D%5Cdfrac%7B7%7D%7B8%7D%2C%5C%5C%5C%5C%5C%5C%5Cdfrac%7BTQ%7D%7BTS%7D%3D%5Cdfrac%7B28%7D%7B32%7D%3D%5Cdfrac%7B7%7D%7B8%7D.)
Now, in ΔTPQ and ΔTRS, we have
![\dfrac{TP}{TR}=\dfrac{TQ}{TS},\\\\\\m\angle TPQ=m\angle TRS~~~\textup{[common angle]}](https://tex.z-dn.net/?f=%5Cdfrac%7BTP%7D%7BTR%7D%3D%5Cdfrac%7BTQ%7D%7BTS%7D%2C%5C%5C%5C%5C%5C%5Cm%5Cangle%20TPQ%3Dm%5Cangle%20TRS~~~%5Ctextup%7B%5Bcommon%20angle%5D%7D)
So, triangles TPQ and TRS are similar by SAS proportionality postulate.
Thus, the correct option is
![(B)~\triangle TRS\sim \triangle TPQ,~SAS\sim.](https://tex.z-dn.net/?f=%28B%29~%5Ctriangle%20TRS%5Csim%20%5Ctriangle%20TPQ%2C~SAS%5Csim.)