**(i)** Given, in ΔABC and ΔPQR,

∠A = ∠P = 60°

∠B = ∠Q = 80°

∠C = ∠R = 40°

Therefore by AAA similarity criterion,

∴ ΔABC ~ ΔPQR

**(ii) **Given, in ΔABC and ΔPQR,

\( \dfrac{AB}{QR}=\dfrac{2}{4}=\dfrac{1}{2}\),

\( \dfrac{BC}{RP}=\dfrac{2.5}{5}=\dfrac{1}{2}\),

\( \dfrac{CA}{PA}=\dfrac{3}{6}=\dfrac{1}{2}\)

By SSS similarity criterion,

ΔABC ~ ΔQRP

**(iii)** Given, in ΔLMP and ΔDEF,

LM = 2.7, MP = 2, LP = 3, EF = 5, DE = 4, DF = 6

\( \dfrac{MP}{DE}=\dfrac{2}{4}=\dfrac{1}{2}\)

\( \dfrac{PL}{DF}=\dfrac{3}{6}=\dfrac{1}{2}\)\(\)

\( \dfrac{LM}{EF}=\dfrac{2.7}{5}=\dfrac{27}{50}\)

Here , \( \dfrac{ MP}{DE}=\dfrac{PL}{DF} ≠\dfrac{LM}{EF}\)

Therefore, ΔLMP and ΔDEF are not similar.

**(iv)** In ΔMNL and ΔQPR, it is given,

\( \dfrac{MN}{QP}=\dfrac{ML}{QR}=\dfrac{1}{2}\)

∠M = ∠Q = 70°

Therefore, by SAS similarity criterion

∴ ΔMNL ~ ΔQPR

**(v)** In ΔABC and ΔDEF, given that,

AB = 2.5, BC = 3, ∠A = 80°, EF = 6, DF = 5, ∠F = 80°

Here , \( \dfrac{AB}{DF}=\dfrac{2.5}{5}=\dfrac{1}{2}\)

And, \( \dfrac{BC}{EF}=\dfrac{3}{6}=\dfrac{1}{2}\)

⇒ ∠B ≠ ∠F

Hence, ΔABC and ΔDEF are not similar.

**(vi)** In ΔDEF, by sum of angles of triangles, we know that,

∠D + ∠E + ∠F = 180°

⇒ 70° + 80° + ∠F = 180°

⇒ ∠F = 180° – 70° – 80°

⇒ ∠F = 30°

Similarly, In ΔPQR,

∠P + ∠Q + ∠R = 180 (Sum of angles of Δ)

⇒ ∠P + 80° + 30° = 180°

⇒ ∠P = 180° – 80° -30°

⇒ ∠P = 70°

Now, comparing both the triangles, ΔDEF and ΔPQR, we have

∠D = ∠P = 70°

∠F = ∠Q = 80°

∠F = ∠R = 30°

Therefore, by AAA similarity criterion,

Hence, ΔDEF ~ ΔPQR

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