State which pairs of triangles in Figure, are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form:

i) Given, in \( \triangle \) ABC and \( \triangle \) PQR,

\( \angle \)A = \( \angle \)P = 60°
\( \angle \)B = \( \angle \)Q = 80°
\( \angle \)C = \( \angle \)R = 40°

Therefore by AAA similarity criterion,

\( \triangle \) ABC is similar to \( \triangle \) PQR


(ii) Given, in \( \triangle \) ABC and \( \triangle \) PQR,
\(\frac{AB}{QR} = \frac{BC}{RP} = \frac{CA}{PQ} \)
By SSS similarity criterion,

\( \triangle \) ABC is similar to \( \triangle \) QRP


(iii) Given, in \( \triangle \) LMP and \( \triangle \) DEF,

LM = 2.7, MP = 2, LP = 3, EF = 5, DE = 4, DF = 6
\(\frac{MP}{DE} = \frac{2}{4} = \frac{1}{2} \)
\(\frac{PL}{DF} = \frac{3}{6} = \frac{1}{2} \)
\(\frac{LM}{EF} = \frac{2.7}{5} = \frac{27}{50} \)
Here , \(\frac{MP}{DE} = \frac{PL}{DF} \) \(\ne\) \(\frac{LM}{EF} \)

Therefore, \( \triangle \) LMP and \( \triangle \) DEF are not similar.


(iv) In \( \triangle \) MNL and \( \triangle \) QPR, it is given,
\(\frac{MN}{QP} = \frac{ML}{QR} = \frac{1}{2} \)
\( \angle \)M = \( \angle \)Q = 70°
Therefore, by SAS similarity criterion

\( \triangle \) MNL is similar to \( \triangle \) QPR


(v) In \( \triangle \) ABC and \( \triangle \) DEF, given that,
AB = 2.5, BC = 3, \( \angle \)A = 80°, EF = 6, DF = 5, \( \angle \)F = 80°
Here , \(\frac{AB}{DF} = \frac{2.5}{5} = \frac{1}{2} \)
And, \(\frac{BC}{EF} = \frac{3}{6} = \frac{1}{2} \)

\( \angle \)B \(\ne\) \( \angle \)F

Hence, \( \triangle \) ABC and \( \triangle \) DEF are not similar.


(vi) In \( \triangle \) DEF, by sum of angles of triangles, we know that,
\( \angle \)D + \( \angle \)E + \( \angle \)F = 180°
70° + 80° + \( \angle \)F = 180°
\( \angle \)F = 180° – 70° – 80°
\( \angle \)F = 30°

Similarly, In \( \triangle \) PQR,
\( \angle \)P + \( \angle \)Q + \( \angle \)R = 180 (Sum of angles of \( \triangle \) )
\( \angle \)P + 80° + 30° = 180°
\( \angle \)P = 180° – 80° -30°
\( \angle \)P = 70°

Now, comparing both the triangles, \( \triangle \) DEF and \( \triangle \) PQR, we have
\( \angle \)D = \( \angle \)P = 70°
\( \angle \)E = \( \angle \)Q = 80°
\( \angle \)F = \( \angle \)R = 30°

Therefore, by AAA similarity criterion,

Hence, \( \triangle \) DEF is similar to \( \triangle \) PQR